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‘ Questions Advanced.txt 2014 03 Copyright VE2AAY.

‘ If you reproduce this without asking permission and giving due credit, may your CQ calls go forever unanswered.

‘ Sequence of questions is immaterial. They have been regrouped here under arbitrary “Lessons” used by the author.

‘ 2014 06 14 Tweak 3-2-4.

‘ Header format: ^ Qualification ^ Number of Questions ^ Pass Mark ^Designation of first Answer record^

^ Advanced Qualification ^ 50 ^ 70 ^A ^

{L02} Inductors and Capacitors; resonant circuits

Upon the application or removal of voltage, the RC or RL time constant `tau` describes the changing rate at which current or voltage rises or decays.

One time constant is the time to reach about 63% of the maximum value (or 36% from the minimum value).

After two time constants, the value reaches 86% of the maximum value (or 13% from the minimum).

A-1-1-1 (D) What is the meaning of the term “time constant” in an RL circuit?
A The time required for the current in the circuit to build up to 36.8% of the maximum value
B The time required for the voltage in the circuit to build up to 63.2% of the maximum value
C The time required for the voltage in the circuit to build up to 36.8% of the maximum value
D The time required for the current in the circuit to build up to 63.2% of the maximum value

Inductance is a property of circuits that oppose changes in current. The time constant is the time current WOULD need to reach final value IF the initial rate of change COULD be maintained. The time constant in seconds equals L in henrys divided by R in ohms: the lower the resistance, the greater the rate of change resulting in greater opposition. The current after 1, 2 and 5 time constants is respectively 63%, 87% and 100% of the final value. With capacitors, the ratios are the same but they relate to voltage; the time constant then becomes R times C.

A-1-1-2 (D) What is the term for the time required for the capacitor in an RC circuit to be charged to 63.2% of the supply voltage?
A An exponential rate of one
B A time factor of one
C One exponential period
D One time constant

Capacitance is a property of circuits that oppose changes in voltage. Under charging conditions, the time constant is the time voltage WOULD need to reach the final value IF the initial rate of change COULD be maintained. The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. The voltage after 1, 2 and 5 times constants is respectively 63%, 87% and 100% of the final value. With inductors, the ratios are the same but they relate to current; the time constant then becomes L divided by R.

A-1-1-3 (A) What is the term for the time required for the current in an RL circuit to build up to 63.2% of the maximum value?
A One time constant
B An exponential period of one
C A time factor of one
D One exponential rate

Inductance is a property of circuits that oppose changes in current. The time constant is the time current WOULD need to reach final value IF the initial rate of change COULD be maintained. The time constant in seconds equals L in henrys divided by R in ohms: the lower the resistance, the greater the rate of change resulting in greater opposition. The current after 1, 2 and 5 time constants is respectively 63%, 87% and 100% of the final value. With capacitors, the ratios are the same but they relate to voltage; the time constant then becomes R times C.

A-1-1-4 (A) What is the term for the time it takes for a charged capacitor in an RC circuit to discharge to 36.8% of its initial value of stored charge?
A One time constant
B A discharge factor of one
C An exponential discharge of one
D One discharge period

Key word: DISCHARGE. The time constant is the time voltage WOULD need to reach the final value IF the initial rate of change COULD be maintained. The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. The voltage after 1, 2 and 5 times constants is respectively 63%, 87% and 100% of the final value. Heading towards ZERO, we are left with 37% (100 minus 63) and 13% (100 minus 87) respectively after 1 and 2 time constants.

A-1-1-6 (A) After two time constants, the capacitor in an RC circuit is charged to what percentage of the supply voltage?
A 86.5%
B 63.2%
C 95%
D 36.8%

Capacitance is a property of circuits that oppose changes in voltage. Under charging conditions, the time constant is the time voltage WOULD need to reach the final value IF the initial rate of change COULD be maintained. The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. The voltage after 1, 2 and 5 times constants is respectively 63%, 87% and 100% of the final value. With inductors, the ratios are the same but they relate to current; the time constant then becomes L divided by R.

A-1-1-7 (A) After two time constants, the capacitor in an RC circuit is discharged to what percentage of the starting voltage?
A 13.5%
B 36.8%
C 86.5%
D 63.2%

Key word: DISCHARGE. The time constant is the time voltage WOULD need to reach the final value IF the initial rate of change COULD be maintained. The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. The voltage after 1, 2 and 5 times constants is respectively 63%, 87% and 100% of the final value. Heading towards ZERO, we are left with 37% (100 minus 63) and 13% (100 minus 87) respectively after 1 and 2 time constants.

The formula is
`tau = R L` or `tau = R C`

A-1-1-8 (C) What is the time constant of a circuit having a 100 microfarad capacitor in series with a 470 kilohm resistor?
A 470 seconds
B 0.47 seconds
C 47 seconds
D 4700 seconds

The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. In multiplying microfarads and megohms, the prefixes cancel one another. 100 microfarads times 0.470 megohm = 100 times 0.47 = 47 seconds.

A-1-1-9 (C) What is the time constant of a circuit having a 470 microfarad capacitor in series with a 470 kilohm resistor?
A 47 000 seconds
B 470 seconds
C 221 seconds
D 221 000 seconds

The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. In multiplying microfarads and megohms, the prefixes cancel one another. 470 microfarads times 0.470 megohm = 470 times 0.47 = 221 seconds.

A-1-1-10 (C) What is the time constant of a circuit having a 220 microfarad capacitor in series with a 470 kilohm resistor?
A 470 seconds
B 220 seconds
C 103 seconds
D 470 000 seconds

The time constant in seconds equals R in ohms times C in farads: the higher the resistance, the longer the time. In multiplying microfarads and megohms, the prefixes cancel one another. 220 microfarads times 0.470 megohm = 220 times 0.47 = 103 seconds.

The skin effect causes RF current to flow in a layer at the surface of a conductor, which becomes thinner at higher frequencies and results in greater resistance.

The skin effect causes most RF current to flow in a thin layer along the surface of a conductor.

A-1-2-2 (C) What effect causes most of an RF current to flow along the surface of a conductor?
A Resonance effect
B Layer effect
C Skin effect
D Piezoelectric effect

Skin Effect is the tendency of AC to flow in an increasingly thinner layer at the surface of a conductor as frequency increases.

A-1-2-3 (D) Where does almost all RF current flow in a conductor?
A In a magnetic field in the centre of the conductor
B In a magnetic field around the conductor
C In the centre of the conductor
D Along the surface of the conductor

Skin Effect is the tendency of AC to flow in an increasingly thinner layer at the surface of a conductor as frequency increases.

A-1-2-4 (B) Why does most of an RF current flow within a very thin layer under the conductor’s surface?
A Because of heating of the conductor’s interior
B Because of skin effect
C Because the RF resistance of a conductor is much less than the DC resistance
D Because a conductor has AC resistance due to self-inductance

Skin Effect is the tendency of AC to flow in an increasingly thinner layer at the surface of a conductor as frequency increases.

As frequency increases, skin effect increases and the conduction layer is thinner.

A-1-2-1 (C) What is the result of skin effect?
A Thermal effects on the surface of the conductor increase impedance
B Thermal effects on the surface of the conductor decrease impedance
C As frequency increases, RF current flows in a thinner layer of the conductor, closer to the surface
D As frequency decreases, RF current flows in a thinner layer of the conductor, closer to the surface

Skin Effect is the tendency of AC to flow in an increasingly thinner layer at the surface of a conductor as frequency increases.

The practical consequence of this is that for the same size wire at higher and higher frequencies, less and less of the cross-section conducts current. So, it starts to look like a thinner wire, with a consequently higher resistance. (Therefore, you start to need larger or different types of conductors for long runs at UHF and microwave frequencies, for example.)

A-1-2-5 (D) Why is the resistance of a conductor different for RF currents than for direct currents?
A Because of the Hertzberg effect
B Because conductors are non-linear devices
C Because the insulation conducts current at high frequencies
D Because of skin effect

Skin Effect is the tendency of AC to flow in an increasingly thinner layer at the surface of a conductor as frequency increases.

Capacitors and inductors establish electrostatic and electromagnetic fields respectively, both of which store potential energy.

Capacitors set up an electrostatic field between their charged plates.

A-1-2-10 (A) Between the charged plates of a capacitor there is:
A an electrostatic field
B a magnetic field
C a cloud of electrons
D an electric current

Voltage across a capacitor creates an electrostatic field between the plates. An electrostatic field is the electric field present between objects with different static electrical charges. An electric field is a space where an electrical charge exerts a force (attraction or repulsion) on other charges.

The readiness of a capacitor to store an electric charge is called capacitance, measured in Farads.

A-1-2-6 (D) What unit measures the ability of a capacitor to store electrical charge?
A Coulomb
B Watt
C Volt

Capacitors store energy in an electrostatic field. The capacitance in farads is one factor influencing how much energy can be stored in a capacitor. The coulomb is a quantity of electrons ( 6 times 10 exponent 18 ). One farad accepts a charge of one coulomb when subjected to one volt. The watt is a rate of doing work (one joule per second). One volt, a force, moves one coulomb with one joule of energy.

Wires carrying current (such as inductors) are surrounded by an electromagnetic field.

A-1-2-7 (A) A wire has a current passing through it. Surrounding this wire there is:
A an electromagnetic field
B an electrostatic field
C a cloud of electrons
D a skin effect that diminishes with distance

An electromagnetic field is the magnetic field created around a conductor carrying current. A magnetic field is a space around a magnet or a conductor where a magnetic force is present. A magnetic field is composed of magnetic lines of force. An electrostatic field is the electric field present between objects with different static electrical charges. An electric field is a space where an electrical charge exerts a force (attraction or repulsion) on other charges.

The unit of inductance is the Henry.

A-1-2-11 (B) Energy is stored within an inductor that is carrying a current. The amount of energy depends on this current, but it also depends on a property of the inductor. This property has the following unit:
A watt
B henry
C coulomb

Inductors store energy in an electromagnetic field. The inductance in henrys is one factor influencing how much energy can be stored in an inductor. One henry produces one volt of counter EMF with current changing at a rate of one ampere per second. The coulomb is a quantity of electrons ( 6 times 10 exponent 18 ). One farad accepts a charge of one coulomb when subjected to one volt. The watt is a rate of doing work (one joule per second).

The left hand rule describes the orientation of the magnetic field relative to electron flow.

A-1-2-8 (C) In what direction is the magnetic field oriented about a conductor in relation to the direction of electron flow?
A In the same direction as the current
B In the direct opposite to the current
C In the direction determined by the left-hand rule
D In all directions

The ‘Left-hand Rule’: position the left hand with your thumb pointing in the direction of electron flow; encircle the conductor with the remaining fingers, the fingers point in the direction of the magnetic lines of force. [ Using conventional current flow, this would become the Right-hand rule. ]

Both these electromagnetic and electrostatic fields store potential energy.

A-1-2-9 (A) What is the term for energy that is stored in an electromagnetic or electrostatic field?
A Potential energy
B Kinetic energy
C Ampere-joules
D Joule-coulombs

Key word: STORED. Potential: “capable of coming into being or action (Canadian Oxford)”. Kinetic: “of or due to motion (Canadian Oxford)”.

In an inductor, a Back-EMF is a voltage that opposes an applied voltage.

A-1-1-5 (A) What is meant by “back EMF”?
A A voltage that opposes the applied EMF
B A current that opposes the applied EMF
C An opposing EMF equal to R times C percent of the applied EMF
D A current equal to the applied EMF

‘Back EMF’ or ‘counter electromotive force’ is the voltage induced by changing current in an inductor. It is the force opposing changes in current through inductors.

The resonant frequency of an RLC circuit depends only on the value of L and C.

``````                    1
f_r,series = ----------------
2 pi sqrt( L C )``````

Series-resonant circuits admit much current at the resonant frequency (impedance is minimum).

At resonance, the reactances are equal and opposite, resulting in zero net reactance. The two components appear as a short. Net current flow is at a maximum, determined by the value of R.

Parallel LC circuits admit little current at the resonant frequency (impedance is maximum).

Again, at resonance, the reactances are equal and opposite.

Therefore, current in the capacitor and inductor is equal and opposite (totally out of phase).

The net current flow through them is zero, and the two components together appear as an open.

The only current that flows is through the resistor, and again, the current flow at resonance is determined solely by R.

This time, however, the net current is at minimum. Off-resonance, the inductor and capacitor admit more current.

Off resonance, current flows through all three components, net current is greater, impedance is lesser, and admittance is greater.

The resonant frequency of a series L C circuit is given by:

``````                    1
f_r,series = ----------------
2 pi sqrt( L C )``````

A-1-3-2 (D) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 40 microhenrys and C is 200 picofarads?
A 1.99 kHz
B 1.99 MHz
C 1.78 kHz
D 1.78 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 40 times 200 equals 8000 ; The square root of 8000 is 89.4 ; 89.4 times 2 times 3.14 is 561.4 ; 1000 divided by 561.4 is 1.78 MHz.

A-1-3-3 (C) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 50 microhenrys and C is 10 picofarads?
A 3.18 MHz
B 3.18 kHz
C 7.12 MHz
D 7.12 kHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 50 times 10 equals 500 ; The square root of 500 is 22.4 ; 22.4 times 2 times 3.14 is 140.7 ; 1000 divided by 140.7 is 7.11 MHz.

A-1-3-4 (D) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 25 microhenrys and C is 10 picofarads?
A 63.7 MHz
B 10.1 kHz
C 63.7 kHz
D 10.1 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 25 times 10 equals 250 ; The square root of 250 is 15.8 ; 15.8 times 2 times 3.14 is 99.2 ; 1000 divided by 99.2 is 10.08 MHz.

A-1-3-5 (D) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 3 microhenrys and C is 40 picofarads?
A 13.1 MHz
B 13.1 kHz
C 14.5 kHz
D 14.5 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 3 times 40 equals 120 ; The square root of 120 is 11 ; 11 times 2 times 3.14 is 69.1 ; 1000 divided by 69.1 is 14.47 MHz.

A-1-3-6 (D) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 4 microhenrys and C is 20 picofarads?
A 19.9 MHz
B 19.9 kHz
C 17.8 kHz
D 17.8 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 4 times 20 equals 80 ; The square root of 80 is 8.9 ; 8.9 times 2 times 3.14 is 55.9 ; 1000 divided by 55.9 is 17.89 MHz.

A-1-3-7 (B) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 8 microhenrys and C is 7 picofarads?
A 2.13 MHz
B 21.3 MHz
C 28.4 MHz
D 2.84 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 8 times 7 equals 56 ; The square root of 56 is 7.5 ; 7.5 times 2 times 3.14 is 47.1 ; 1000 divided by 47.1 is 21.23 MHz.

A-1-3-8 (B) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 3 microhenrys and C is 15 picofarads?
A 23.7 kHz
B 23.7 MHz
C 35.4 MHz
D 35.4 kHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 3 times 15 equals 45 ; The square root of 45 is 6.7 ; 6.7 times 2 times 3.14 is 42.1 ; 1000 divided by 42.1 is 23.75 MHz.

A-1-3-9 (A) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 4 microhenrys and C is 8 picofarads?
A 28.1 MHz
B 49.7 MHz
C 49.7 kHz
D 28.1 kHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 4 times 8 equals 32 ; The square root of 32 is 5.7 ; 5.7 times 2 times 3.14 is 35.8 ; 1000 divided by 35.8 is 27.93 MHz.

A-1-3-10 (A) What is the resonant frequency of a series RLC circuit, if R is 47 ohms, L is 1 microhenry and C is 9 picofarads?
A 53.1 MHz
B 5.31 MHz
C 17.7 MHz
D 1.77 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 1 times 9 equals 9 ; The square root of 9 is 3 ; 3 times 2 times 3.14 is 18.8 ; 1000 divided by 18.8 is 53.19 MHz.

A-1-4-1 (A) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 1 microhenry and C is 10 picofarads?
A 50.3 MHz
B 15.9 kHz
C 50.3 kHz
D 15.9 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 1 times 10 equals 10 ; The square root of 9 is 3.2 ; 3.2 times 2 times 3.14 is 20.1 ; 1000 divided by 20.1 is 49.75 MHz.

A-1-4-2 (D) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 2 microhenrys and C is 15 picofarads?
A 29.1 kHz
B 5.31 MHz
C 5.31 kHz
D 29.1 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 2 times 15 equals 30 ; The square root of 30 is 5.5 ; 5.5 times 2 times 3.14 is 34.5 ; 1000 divided by 34.5 is 28.99 MHz.

A-1-3-1 (D) What is the resonant frequency of a series RLC circuit if R is 47 ohms, L is 50 microhenrys and C is 40 picofarads?
A 1.78 MHz
B 7.96 MHz
C 79.6 MHz
D 3.56 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 50 times 40 equals 2000 ; The square root of 2000 is 44.7 ; 44.7 times 2 times 3.14 is 280.7 ; 1000 divided by 280.7 is 3.56 MHz.

The resonant frequency of a parallel L C circuit is given by:

``````                      1
f_r,parallel = ----------------
2 pi sqrt( L C )``````

A-1-4-3 (B) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 5 microhenrys and C is 9 picofarads?
A 3.54 kHz
B 23.7 MHz
C 23.7 kHz
D 3.54 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 5 times 9 equals 45 ; The square root of 45 is 6.7 ; 6.7 times 2 times 3.14 is 42.1 ; 1000 divided by 42.1 is 23.75 MHz.

A-1-4-4 (A) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 2 microhenrys and C is 30 picofarads?
A 20.5 MHz
B 2.65 MHz
C 2.65 kHz
D 20.5 kHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 2 times 30 equals 60 ; The square root of 60 is 7.7 ; 7.7 times 2 times 3.14 is 48.4 ; 1000 divided by 48.4 is 20.66 MHz.

A-1-4-8 (D) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 10 microhenrys and C is 50 picofarads?
A 7.12 kHz
B 3.18 MHz
C 3.18 kHz
D 7.12 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 10 times 50 equals 500 ; The square root of 500 is 22.4 ; 22.4 times 2 times 3.14 is 140.7 ; 1000 divided by 140.7 is 7.11 MHz.

A-1-4-5 (B) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 15 microhenrys and C is 5 picofarads?
A 18.4 kHz
B 18.4 MHz
C 2.12 kHz
D 2.12 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 15 times 5 equals 75 ; The square root of 75 is 8.7 ; 8.7 times 2 times 3.14 is 54.6 ; 1000 divided by 54.6 is 18.32 MHz.

A-1-4-9 (B) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 200 microhenrys and C is 10 picofarads?
A 7.96 kHz
B 3.56 MHz
C 3.56 kHz
D 7.96 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 200 times 10 equals 2000 ; The square root of 2000 is 44.7 ; 44.7 times 2 times 3.14 is 280.7 ; 1000 divided by 280.7 is 3.56 MHz.

A-1-4-10 (A) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 90 microhenrys and C is 100 picofarads?
A 1.68 MHz
B 1.77 kHz
C 1.77 MHz
D 1.68 kHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 90 times 100 equals 9000 ; The square root of 9000 is 94.9 ; 94.9 times 2 times 3.14 is 596 ; 1000 divided by 596 is 1.68 MHz.

A-1-4-6 (D) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 3 microhenrys and C is 40 picofarads?
A 1.33 kHz
B 1.33 MHz
C 14.5 kHz
D 14.5 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 3 times 40 equals 120 ; The square root of 120 is 11 ; 11 times 2 times 3.14 is 69.1 ; 1000 divided by 69.1 is 14.47 MHz.

A-1-4-7 (D) What is the resonant frequency of a parallel RLC circuit if R is 4.7 kilohms, L is 40 microhenrys and C is 6 picofarads?
A 6.63 MHz
B 6.63 kHz
C 10.3 kHz
D 10.3 MHz

Resonant frequency equals 1 over ( 2 pi times the square root of L times C ). Restating for frequency in megahertz becomes 1000 over ( 2 pi times the square root of microhenrys times picofarads ). 40 times 6 equals 240 ; The square root of 240 is 15.5 ; 15.5 times 2 times 3.14 is 97.3 ; 1000 divided by 97.3 is 10.28 MHz.

Note that in both cases, the formulas are the same, and resistance does not affect the resonant frequency.

Going backward is just a little tricky because of the square root.

A-1-3-11 (C) What is the value of capacitance (C) in a series R-L-C circuit, if the circuit resonant frequency is 14.25 MHz and L is 2.84 microhenrys?

Method A: Reactances are equal at resonance. XL = 2 times 3.14 times 14.25 times 2.84 = 254.2 ohms. XC = 1 over ( 2 pi f C ). Restating for f in megahertz and C in picofarads, XC = one million over 2 pi megahertz times picofarads. Thus, C = one million over ( 2 pi f XC ) ; 2 times 3.14 times 14.25 times 254.2 = 22 748 ; one million divided by 22 748 = 43.96 picofarads. Method B: at 14 MHz, C has to be in picofarads; test the two answers in picofarads with “resonant frequency in megahertz equals 1000 over ( 2 pi times the square root of microhenrys times picofarads )”.

A-1-4-11 (B) What is the value of inductance (L) in a parallel RLC circuit, if the resonant frequency is 14.25 MHz and C is 44 picofarads?
A 0.353 microhenry
B 2.8 microhenrys
C 253.8 millihenrys
D 3.9 millihenrys

Method A: Reactances are equal at resonance. XC = 1 over ( 2 pi f C ). Restating for f in megahertz and C in picofarads, XC = one million over (2 pi times megahertz times picofarads). XC = one million divided by ( 2 times 3.14 times 14.25 times 44 ) = 254 ohms. With XL = 2 pi f L, L is XL divided by 2 pi f: 254 divided by ( 2 times 3.14 times 14.25 ) = 2.8 microhenrys. Method B: at 14 MHz, L has to be in microhenrys; test the two answers in microhenrys with “resonant frequency in megahertz equals 1000 over ( 2 pi times the square root of microhenrys times picofarads )”.

The Q-factor of an RLC circuit depends on the resistance of R relative to the reactance X of L and C.

``````      R
Q_p = - ; X_C = 1/(2 pi f C)
X   X_L = 2 pi f L`````` Crystal filters have high Q-factor.

A-2-11-5 (C) A quartz crystal filter is superior to an LC filter for narrow bandpass applications because of the:
A LC circuit’s high Q
B crystal’s simplicity
C crystal’s high Q
D crystal’s low Q

Piezoelectric crystals behave like tuned circuits with an extremely high “Q” (“Quality Factor”, in excess of 25 000). Their accuracy and stability are outstanding.

In a parallel tuned circuit, the so called damping resistor increases the filter bandwidth at the expense of Q

A-1-5-11 (B) Why is a resistor often included in a parallel resonant circuit?
A To increase the Q and decrease bandwidth
B To decrease the Q and increase the bandwidth
C To increase the Q and decrease the skin effect
D To decrease the Q and increase the resonant frequency

A Damping Resistor can be placed across a parallel resonant circuit, or in series with a series resonant circuit, to lower the Q. Reducing the Quality factor increases bandwidth.

‘ — — — —

The formula for Q-factor is different for series and parallel circuits, but the exam only asks about parallel circuits.

``````      R
Q_p = - ; X_C = 1/(2 pi f C)
X   X_L = 2 pi f L``````

A-1-5-1 (A) What is the Q of a parallel RLC circuit, if it is resonant at 14.128 MHz, L is 2.7 microhenrys and R is 18 kilohms?
A 75.1
B 7.51
C 0.013
D 71.5

Reactance = 2 pi f L = 2 times 3.14 times 14.128 times 2.7 = 240 ( the mega in megahertz cancels the micro in microhenrys). Q = 18 000 divided by 240 = 75 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-10 (D) What is the Q of a parallel RLC circuit, if it is resonant at 3.625 MHz, L is 43 microhenrys and R is 1.8 kilohms?
A 0.543
B 54.3
C 23
D 1.84

Reactance = 2 pi f L = 2 times 3.14 times 3.625 times 43 = 979 ( the mega in megahertz cancels the micro in microhenrys). Q = 1800 divided by 979 = 1.84 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-9 (B) What is the Q of a parallel RLC circuit, if it is resonant at 3.625 MHz, L is 42 microhenrys and R is 220 ohms?
A 0.00435
B 0.23
C 2.3
D 4.35

Reactance = 2 pi f L = 2 times 3.14 times 3.625 times 42 = 956 ( the mega in megahertz cancels the micro in microhenrys). Q = 220 divided by 956 = 0.23 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-8 (D) What is the Q of a parallel RLC circuit, if it is resonant at 3.625 MHz, L is 3 microhenrys and R is 2.2 kilohms?
A 25.6
B 31.1
C 0.031
D 32.2

Reactance = 2 pi f L = 2 times 3.14 times 3.625 times 3 = 68 ( the mega in megahertz cancels the micro in microhenrys). Q = 2200 divided by 68 = 32.3 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-7 (D) What is the Q of a parallel RLC circuit, if it is resonant at 7.125 MHz, L is 12.6 microhenrys and R is 22 kilohms?
A 22.1
B 0.0256
C 25.6
D 39

Reactance = 2 pi f L = 2 times 3.14 times 7.125 times 12.6 = 564 ( the mega in megahertz cancels the micro in microhenrys). Q = 22 000 divided by 564 = 39 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-6 (C) What is the Q of a parallel RLC circuit, if it is resonant at 7.125 MHz, L is 10.1 microhenrys and R is 100 ohms?
A 0.00452
B 4.52
C 0.221
D 22.1

Reactance = 2 pi f L = 2 times 3.14 times 7.125 times 10.1 = 452 ( the mega in megahertz cancels the micro in microhenrys). Q = 100 divided by 452 = 0.22 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-5 (A) What is the Q of a parallel RLC circuit, if it is resonant at 7.125 MHz, L is 8.2 microhenrys and R is 1 kilohm?
A 2.73
B 36.8
C 0.368
D 0.273

Reactance = 2 pi f L = 2 times 3.14 times 7.125 times 8.2 = 367 ( the mega in megahertz cancels the micro in microhenrys). Q = 1000 divided by 367 = 2.7 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-3 (A) What is the Q of a parallel RLC circuit, if it is resonant at 4.468 MHz, L is 47 microhenrys and R is 180 ohms?
A 0.136
B 7.35
C 0.00735
D 13.3

Reactance = 2 pi f L = 2 times 3.14 times 4.468 times 47 = 1319 ( the mega in megahertz cancels the micro in microhenrys). Q = 180 divided by 1319 = 0.136 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-2 (A) What is the Q of a parallel RLC circuit, if it is resonant at 14.128 MHz, L is 4.7 microhenrys and R is 18 kilohms?
A 43.1
B 13.3
C 0.023
D 4.31

Reactance = 2 pi f L = 2 times 3.14 times 14.128 times 4.7 = 417 ( the mega in megahertz cancels the micro in microhenrys). Q = 18 000 divided by 417 = 43 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

A-1-5-4 (B) What is the Q of a parallel RLC circuit, if it is resonant at 14.225 MHz, L is 3.5 microhenrys and R is 10 kilohms?
A 71.5
B 31.9
C 7.35
D 0.0319

Reactance = 2 pi f L = 2 times 3.14 times 14.225 times 3.5 = 313 ( the mega in megahertz cancels the micro in microhenrys). Q = 10 000 divided by 313 = 31.9 . In a PARALLEL circuit loaded by a resistor, Q = Resistance divided by Reactance: the higher the parallel resistance, the lesser the effect on the response curve. Parallel resistance lowers the Q of a parallel tuned circuit. A parallel Damping Resistor is used to increase bandwidth.

Filters are resonant circuits characterized by their frequency response.

The three general groupings of filters are high-pass, low-pass, and band-pass.

A-2-12-1 (D) What are the three general groupings of filters?
A Hartley, Colpitts and Pierce
B Audio, radio and capacitive
C Inductive, capacitive and resistive
D High-pass, low-pass and band-pass

There are 4 categories of filters: high-pass, low-pass, band-pass and band-stop. Hartley, Colpitts and Pierce are oscillator configurations. “Capacitive” is not a range of frequencies like audio or radio. Resistors do not discriminate frequency.

Resonant cavities have a narrow bandpass and are used at VHF and above. Their length is roughly a quarter wavelength long. Consequently, their size becomes unwieldy below roughly 50 MHz.

A-2-12-5 (B) Resonant cavities are used by amateurs as a:
A high-pass filter above 30 MHz
B narrow bandpass filter at VHF and higher frequencies
C power line filter
D low-pass filter below 30 MHz

The quarter wavelength Resonant Cavity behaves like a very high “Q” filter. Due to their physical size, they become practical only at VHF frequencies: at 50 MHz (6 m), the length of the cavity is 1.5 m (one quarter wavelength).

A-2-12-6 (D) On VHF and above, 1/4 wavelength coaxial cavities are used to give protection from high-level signals. For a frequency of approximately 50 MHz, the diameter of such a device would be about 10 cm (4 in). What would be its approximate length?
A 0.6 metres (2 ft)
B 2.4 metres (8 ft)
C 3.7 metres (12 ft)
D 1.5 metres (5 ft)

The quarter wavelength Resonant Cavity behaves like a very high “Q” filter (around 3000). Due to their physical size, they become practical only at VHF frequencies: at 50 MHz (6 m), the length of the cavity is 1.5 m (one quarter wavelength).

A-2-12-11 (D) Which of the following filter types is not suitable for use at audio and low radio frequencies?
A Elliptical
B Chebyshev
C Butterworth
D Cavity

The quarter wavelength Resonant Cavity behaves like a very high “Q” filter. Due to their physical size, they become practical only at VHF frequencies: at 50 MHz (6 m), the length of the cavity is 1.5 m (one quarter wavelength).

Butterworth filters have a flat response. (Mnemonic: “Smooth as butter.”)

A-2-12-2 (C) What are the distinguishing features of a Butterworth filter?
A It only requires conductors
B It only requires capacitors
C It has a maximally flat response over its pass-band
D The product of its series and shunt-element impedances is a constant for all frequencies

The Butterworth class of filters exhibit “maximally flat response”: smooth response, no passband ripple. Their frequency response is as flat as mathematically possible in the passband, no bumps or variations (ripple) [first described by British engineer Stephen Butterworth]. The Chebyshev class of filters [in honour of Pafnuty Chebyshev, a Russian mathematician] have steeper cutoff slopes and more ripple than Butterworth filters. Elliptic filters are sharper than the previous two. Here is a mnemonic trick: “The Butterworth’s response is smooth as butter”.

A-2-12-9 (B) What is the primary advantage of the Butterworth filter over the Chebyshev filter?
A It requires only capacitors
B It has maximally flat response over its passband
C It allows ripple in the passband in return for steeper skirts
D It requires only inductors

The Butterworth class of filters exhibit “maximally flat response”: smooth response, no passband ripple. Their frequency response is as flat as mathematically possible in the passband, no bumps or variations (ripple) [first described by British engineer Stephen Butterworth]. The Chebyshev class of filters [in honour of Pafnuty Chebyshev, a Russian mathematician] have steeper cutoff slopes and more ripple than Butterworth filters. Elliptic filters are sharper than the previous two. Here is a mnemonic trick: “The Butterworth’s response is smooth as butter”.

Chebyshev filters have a sharper cuttoff, but a rippled bandpass.

A-2-12-3 (B) Which filter type is described as having ripple in the passband and a sharp cutoff?
A A Butterworth filter
B A Chebyshev filter
C An active LC filter
D A passive op-amp filter

The Butterworth class of filters exhibit “maximally flat response”: smooth response, no passband ripple. Their frequency response is as flat as mathematically possible in the passband, no bumps or variations (ripple) [first described by British engineer Stephen Butterworth]. The Chebyshev class of filters [in honour of Pafnuty Chebyshev, a Russian mathematician] have steeper cutoff slopes and more ripple than Butterworth filters. Elliptic filters are sharper than the previous two. Here is a mnemonic trick: “The Butterworth’s response is smooth as butter”.

A-2-12-4 (B) What are the distinguishing features of a Chebyshev filter?
A It has a maximally flat response in the passband
B It allows ripple in the passband in return for steeper skirts
C It requires only inductors
D It requires only capacitors

The Butterworth class of filters exhibit “maximally flat response”: smooth response, no passband ripple. Their frequency response is as flat as mathematically possible in the passband, no bumps or variations (ripple) [first described by British engineer Stephen Butterworth]. The Chebyshev class of filters [in honour of Pafnuty Chebyshev, a Russian mathematician] have steeper cutoff slopes and more ripple than Butterworth filters. Elliptic filters are sharper than the previous two. Here is a mnemonic trick: “The Butterworth’s response is smooth as butter”.

A-2-12-10 (C) What is the primary advantage of the Chebyshev filter over the Butterworth filter?
A It requires only inductors
B It has maximally flat response over the passband
C It allows ripple in the passband in return for steeper skirts
D It requires only capacitors

The Butterworth class of filters exhibit “maximally flat response”: smooth response, no passband ripple. Their frequency response is as flat as mathematically possible in the passband, no bumps or variations (ripple) [first described by British engineer Stephen Butterworth]. The Chebyshev class of filters [in honour of Pafnuty Chebyshev, a Russian mathematician] have steeper cutoff slopes and more ripple than Butterworth filters. Elliptic filters are sharper than the previous two. Here is a mnemonic trick: “The Butterworth’s response is smooth as butter”.

A helical resonator is a filter formed from coiled feedline, approriate for VHF & UHF applications They are used to sanitize receiver inputs.

A-2-12-7 (C) A device which helps with receiver overload and spurious responses at VHF, UHF and above may be installed in the receiver front end. It is called a:
A directional coupler
B duplexer
C helical resonator
D diplexer

The Helical Resonator, based on the concept of a resonant helically-wound section of transmission line within a shielded enclosure, achieves selectivity comparable to the quarter-wave resonant cavity but with a substantial size reduction.

None of the filters described here have a bandpass that is 6 MHz wide.

A-2-12-8 (C) Where you require bandwidth at VHF and higher frequencies about equal to a television channel, a good choice of filter is the:
A Butterworth
B Chebyshev
C none of the other answers
D resonant cavity

The bandwidth of a fast-scan TV channel is 6 MHz; that is much too wide for any of the filters listed.

(Which makes me wonder how CATV filters work.)

Crystal elements use the piezoelectric effect, in which electrical energy and mechanical strain are transformed.

A-2-11-10 (D) Crystal oscillators, filters and microphones depend upon which principle?
A Hertzberg effect
B Ferro-resonance
C Overtone effect
D Piezoelectric effect

The piezoelectric property of quartz (generating electricity under mechanical stress, bending when subjected to electric field) is used in crystal-based oscillators, radio-frequency crystal filters, such as the lattice filter, and crystal microphones. The Active Filter is based on an active device, generally an operational amplifier, and a network of resistors and capacitors.

A-2-11-6 (D) Piezoelectricity is generated by:
A touching crystals with magnets
B adding impurities to a crystal
C moving a magnet near a crystal
D deforming certain crystals

The piezoelectric property of quartz is two-fold: apply mechanical stress to a crystal and it produces a small electrical field; subject quartz to an electrical field and the crystal changes dimensions slightly. Crystals are capable of resonance either at a fundamental frequency depending on their physical dimensions or at overtone frequencies near odd-integer multiples (3rd, 5th, 7th, etc.). Piezoelectric crystals can serve as filters because of their extremely high “Q” (> 25 000) or as stable, noise-free and accurate frequency references.

Crystal elements are not appropriate for active filters, but are used in lattice filters, oscillators, and microphones.

A-2-11-11 (C) Crystals are not applicable to which of the following?
A Lattice filters
B Oscillators
C Active filters
D Microphones

The piezoelectric property of quartz (generating electricity under mechanical stress, bending when subjected to electric field) is used in crystal-based oscillators, radio-frequency crystal filters, such as the lattice filter, and crystal microphones. The Active Filter is based on an active device, generally an operational amplifier, and a network of resistors and capacitors.

‘ — — — —

Crystal lattice filters are made up of multiple crystals with different frequencies. The choice of frequencies determine the filter characteristics.

A-2-11-2 (C) What factor determines the bandwidth and response shape of a crystal lattice filter?
A The gain of the RF stage following the filter
B The amplitude of the signals passing through the filter
C The relative frequencies of the individual crystals
D The centre frequency chosen for the filter

A filter with narrow bandwidth and steep skirts made with quartz crystals. “Lattice: a structure of crossed laths with spaces between, used as a screen or fence.” The frequency separation between the crystals sets the bandwidth and the response shape. Crystal lattice filter: uses two matched pairs of series crystals and a higher-frequency matched pair of shunt crystals in a balanced configuration. Half-lattice crystal filter: uses two crystals in an unbalanced configuration. Such filters can be cascaded. A ‘Crystal Gate’ uses a single crystal.

The frequency response of a crystal lattice filter has a narrow bandwidth and steep skirts, that is, they have very high Q.

A-2-11-1 (A) What is a crystal lattice filter?
A A filter with narrow bandwidth and steep skirts made using quartz crystals
B A filter with wide bandwidth and shallow skirts made using quartz crystals
C An audio filter made with four quartz crystals that resonate at 1 kHz intervals
D A power supply filter made with interlaced quartz crystals

A filter with narrow bandwidth and steep skirts made with quartz crystals. “Lattice: a structure of crossed laths with spaces between, used as a screen or fence.” The frequency separation between the crystals sets the bandwidth and the response shape. Crystal lattice filter: uses two matched pairs of series crystals and a higher-frequency matched pair of shunt crystals in a balanced configuration. Half-lattice crystal filter: uses two crystals in an unbalanced configuration. Such filters can be cascaded. A ‘Crystal Gate’ uses a single crystal.

A-2-11-7 (B) Electrically, what does a crystal look like?
A A variable tuned circuit
B A very high Q tuned circuit
C A very low Q tuned circuit
D A variable capacitance

Piezoelectric crystals behave like tuned circuits with an extremely high “Q” (“Quality Factor”, in excess of 25 000). Their accuracy and stability are outstanding.

For example, the filter bandwidth of an SSB filter should be about 3 kHz.

A-2-11-3 (A) For single-sideband phone emissions, what would be the bandwidth of a good crystal lattice filter?
A 2.4 kHz
B 15 kHz
C 500 Hz
D 6 kHz

Speech frequencies on a communication-grade SSB voice channel range from 300 hertz to 3000 hertz and thus require a bandwidth of 2.7 kHz; 2.1 kHz is a good compromise between fidelity and selectivity. 15 kHz is the bandwidth of FM, 6 kHz is for AM, 500 Hz is a common filter width for CW.

{L03} Semiconductors.

Semiconductors in general
The exam deals with:

• Common semiconductor elements (Si, Ge, GaAs).
• Basics of doped silicon (P-type, N-type).

Semiconductors act as conductors, insulators, or in-between.

A-2-1-8 (B) An element which is sometimes an insulator and sometimes a conductor is called a:
A P-type conductor
B semiconductor
C intrinsic conductor
D N-type conductor

The most basic semiconductor materials are silicon and germanium. Atoms in metallic elements hold their peripheral electrons loosely, such materials make good conductors. Peripheral electrons in non-metallic elements are tightly bound, such materials are insulators. Germanium and silicon fall somewhere between the two categories but are mostly insulators when pure. Doping with impurities increases their conductivity.

Some common semiconducting elements are:

• Silicon,
• germanium, and
• gallium arsenide (for microwave applications).

A-2-1-9 (B) Which of the following materials is used to make a semiconductor?
A Sulphur
B Silicon
C Tantalum
D Copper

The most basic semiconductor materials are silicon and germanium. Atoms in metallic elements hold their peripheral electrons loosely, such materials make good conductors. Peripheral electrons in non-metallic elements are tightly bound, such materials are insulators. Germanium and silicon fall somewhere between the two categories but are mostly insulators when pure. Doping with impurities increases their conductivity.

A-2-1-1 (D) What two elements widely used in semiconductor devices exhibit both metallic and non-metallic characteristics?
A Galena and germanium
B Galena and bismuth
C Silicon and gold
D Silicon and germanium

The most basic semiconductor materials are silicon and germanium. Atoms in metallic elements hold their peripheral electrons loosely, such materials make good conductors. Peripheral electrons in non-metallic elements are tightly bound, such materials are insulators. Germanium and silicon fall somewhere between the two categories but are mostly insulators when pure. Doping with impurities increases their conductivity.

A-2-1-2 (B) In what application is gallium-arsenide used as a semiconductor material in preference to germanium or silicon?
A In bipolar transistors
B At microwave frequencies
C In high-power circuits
D At very low frequencies

Gallium arsenide (GaAs) devices can work at higher frequencies with less noise than their silicon counterparts.

Pure silicon is an insulator.

A-2-1-7 (A) Silicon, in its pure form, is:
A an insulator
B a superconductor
C a semiconductor
D a conductor

The most basic semiconductor materials are silicon and germanium. Atoms in metallic elements hold their peripheral electrons loosely, such materials make good conductors. Peripheral electrons in non-metallic elements are tightly bound, such materials are insulators. Germanium and silicon fall somewhere between the two categories but are mostly insulators when pure. Doping with impurities increases their conductivity.

A-2-1-10 (D) Substances such as silicon in a pure state are usually good:
A conductors
B tuned circuits
C inductors
D insulators

The most basic semiconductor materials are silicon and germanium. Atoms in metallic elements hold their peripheral electrons loosely, such materials make good conductors. Peripheral electrons in non-metallic elements are tightly bound, such materials are insulators. Germanium and silicon fall somewhere between the two categories but are mostly insulators when pure. Doping with impurities increases their conductivity.

Adding impurities is called “doping.”

A-2-1-11 (C) A semiconductor is said to be doped when it has added to it small quantities of:
A ions
B electrons
C impurities
D protons

Pure germanium and silicon are doped with impurities to produce the basic semiconductor materials. Certain doping impurities add free electrons, forming N-Type material while others accept electrons, thus creating ‘holes’ found in P-Type material.

N-type silicon contains more electrons than normal.

The charge carries are electrons.

A-2-1-4 (C) What type of semiconductor material contains more free electrons than pure germanium or silicon crystals?
A Bipolar
B Superconductor
C N-type
D P-type

Pure germanium and silicon are doped with impurities to produce the basic semiconductor materials. Certain doping impurities add free electrons, forming N-Type material while others accept electrons, thus creating ‘holes’ found in P-Type material.

A-2-1-6 (B) What are the majority charge carriers in N-type semiconductor material?
A Free neutrons
B Free electrons
C Holes
D Free protons

N-Type material comprises extra electrons which serve as the electric charge carriers. P-Type material was robbed of free electrons, positive ‘holes’ are the electric charge carriers.

P-type silicon contains fewer electrons than normal.

The charge carries are called “holes.”

A-2-1-3 (D) What type of semiconductor material contains fewer free electrons than pure germanium or silicon crystals?
A N-type
B Bipolar type
C Superconductor type
D P-type

Pure germanium and silicon are doped with impurities to produce the basic semiconductor materials. Certain doping impurities add free electrons, forming N-Type material while others accept electrons, thus creating ‘holes’ found in P-Type material.

A-2-1-5 (D) What are the majority charge carriers in P-type semiconductor material?
A Free electrons
B Free protons
C Free neutrons
D Holes

P-Type material was robbed of free electrons, positive ‘holes’ are the electric charge carriers. N-Type material comprises extra electrons which serve as the electric charge carriers.

Diodes

There are few types of diodes to know:

• Point contact detectors
• Varactors with variable capacitance (reactance)
• Schottky mixers and detectors
• PIN switches
• Zener voltage regulators

The two main categories of diode are

• Junction diodes, and
• point contact diodes.

A-2-2-6 (A) Structurally, what are the two main categories of semiconductor diodes?
A Junction and point contact
B Vacuum and point contact
C Electrolytic and point contact
D Electrolytic and junction

Point-contact diodes, where a small metal whisker touches the semiconductor material, exhibit low capacitance and serve as RF detectors or UHF mixers. Junction diodes are formed with adjacent blocks of P and N material; these are usable from DC to microwave.

Point contact diodes are used as RF detectors.

A-2-2-7 (D) What is a common use for point contact diodes?
A As a constant current source
B As a constant voltage source
C As a high voltage rectifier
D As an RF detector

Point-contact diodes, where a small metal whisker touches the semiconductor material, exhibit low capacitance and serve as RF detectors or UHF mixers. Junction diodes are formed with adjacent blocks of P and N material; these are usable from DC to microwave.

Aside from the ordinary PN junction diodes used for power rectification, there are also a few other special types of junction diodes:

• Varactor diodes vary their capacitance in relation to voltage. (Think var - variable and ractor - reactance.) (Used in e.g. tuning circuits.)

A-2-2-2 (C) What type of semiconductor diode varies its internal capacitance as the voltage applied to its terminals varies?
A Silicon-controlled rectifier
B Hot-carrier (Schottky)
C Varactor
D Zener

Zener diodes maintain a constant voltage across a range of currents. The Varactor (or Varicap) is a diode used under reverse bias as a “voltage-variable capacitor”. Hot-carrier (or Schottky-barrier) diodes have lower forward voltage and good high-frequency response: their speed make them useful in Very High Frequency mixers or detectors; in power circuits, they are excellent rectifiers in switching power supplies. PIN diodes (with a layer of undoped or lightly doped ‘intrinsic’ silicon between the P and N regions) are used as switches or attenuators. A-2-2-3 (D) What is a common use for the hot-carrier (Schottky) diode?
A As balanced mixers in FM generation
B As a variable capacitance in an automatic frequency control (AFC) circuit
C As a constant voltage reference in a power supply
D As VHF and UHF mixers and detectors

Zener diodes maintain a constant voltage across a range of currents. The Varactor (or Varicap) is a diode used under reverse bias as a “voltage-variable capacitor”. Hot-carrier (or Schottky-barrier) diodes have lower forward voltage and good high-frequency response: their speed make them useful in Very High Frequency mixers or detectors; in power circuits, they are excellent rectifiers in switching power supplies. PIN diodes (with a layer of undoped or lightly doped ‘intrinsic’ silicon between the P and N regions) are used as switches or attenuators.

• PIN diodes are used as RF switches.

(Used to mute or protect the front-end of a receiver when, e.g. in the presence of a receiver. As opposed to, say, relays.)

A-2-2-8 (D) What is one common use for PIN diodes?
A As a constant current source
B As a high voltage rectifier
C As a constant voltage source
D As an RF switch

Zener diodes maintain a constant voltage across a range of currents. The Varactor (or Varicap) is a diode used under reverse bias as a “voltage-variable capacitor”. Hot-carrier (or Schottky-barrier) diodes have lower forward voltage and good high-frequency response: their speed make them useful in Very High Frequency mixers or detectors; in power circuits, they are excellent rectifiers in switching power supplies. PIN diodes (with a layer of undoped or lightly doped ‘intrinsic’ silicon between the P and N regions) are used as switches or attenuators.

• Zener diodes have a precise and constant forward voltage that is used in voltage regulator circuits. A-2-2-9 (D) A Zener diode is a device used to:
A dissipate voltage
B decrease current
C increase current
D regulate voltage

Zener diodes maintain a constant voltage across a range of currents. The Varactor (or Varicap) is a diode used under reverse bias as a “voltage-variable capacitor”. Hot-carrier (or Schottky-barrier) diodes have lower forward voltage and good high-frequency response: their speed make them useful in Very High Frequency mixers or detectors; in power circuits, they are excellent rectifiers in switching power supplies. PIN diodes (with a layer of undoped or lightly doped ‘intrinsic’ silicon between the P and N regions) are used as switches or attenuators.

A-2-2-1 (C) What is the principal characteristic of a Zener diode?
A A negative resistance region
B An internal capacitance that varies with the applied voltage
C A constant voltage under conditions of varying current
D A constant current under conditions of varying voltage

Zener diodes maintain a constant voltage across a range of currents. The Varactor (or Varicap) is a diode used under reverse bias as a “voltage-variable capacitor”. Hot-carrier (or Schottky-barrier) diodes have lower forward voltage and good high-frequency response: their speed make them useful in Very High Frequency mixers or detectors; in power circuits, they are excellent rectifiers in switching power supplies. PIN diodes (with a layer of undoped or lightly doped ‘intrinsic’ silicon between the P and N regions) are used as switches or attenuators.

The major ratings of junction diodes are:

• Maximum forward current, and
• peak inverse voltage (PIV).

A-2-2-5 (D) What are the major ratings for junction diodes?
A Maximum reverse current and capacitance
B Maximum forward current and capacitance
C Maximum reverse current and peak inverse voltage (PIV)
D Maximum forward current and peak inverse voltage (PIV)

Diodes conduct in one direction only: under forward bias, maximum forward current is limited by acceptable junction temperature. The voltage drop across the junction (volts) multiplied by the forward current (amperes) gives rise to heat dissipation (watts). Surviving a reverse bias is determined by the Peak Inverse Voltage (PIV) rating.

A-2-2-4 (C) What limits the maximum forward current in a junction diode?
A Back EMF
B Peak inverse voltage
C Junction temperature
D Forward voltage

Diodes conduct in one direction only: under forward bias, maximum forward current is limited by acceptable junction temperature. The voltage drop across the junction (volts) multiplied by the forward current (amperes) gives rise to heat dissipation (watts). Surviving a reverse bias is determined by the Peak Inverse Voltage (PIV) rating.

At the PN junction, a thin depletion layer forms, across which an intrinsic voltage drop occurs.

Current flowing across this voltage drop dissipates power, resulting in heat.

Their forward current is limited by temperature, which is due to the power dissipated by the intrinsic voltage drop across the depletion layer. The layer can only dissipate so much heat before failing.

A-2-2-11 (C) The power-handling capability of most Zener diodes is rated at 25 degrees C or approximately room temperature. If the temperature is increased, the power handling capability is:
A much greater
B slightly greater
C less
D the same

Heat flows from hot to cold. If ambient temperature is higher, less heat can be drained from the junction, the junction will reach maximum safe operating temperature quicker.

A-2-2-10 (D) If a Zener diode rated at 10 V and 50 watts was operated at maximum dissipation rating, it would conduct __ amperes:
A 50
B 0.05
C 0.5
D 5

P = E times I. Watts = volts times amperes. Thus, I = P divided by E: 50 watts divided by 10 volts = 5 amperes.

Transistors

The two types of transistors of interest here are:

• Bipolar junction transistors (BJTs), and
• Field-effect transistors (FETs) including junction FETs (JFETS) and MOSFETS.

The distinction between BJTs and FETs has to do with input impedance; that is, how much current they draw for a given control voltage.

• BJTs have low input (gate) impedance and therefore consume a lot of current to operate at a given voltage.
• FETs have high input impedance, and consume little current for the same voltage.

A-2-4-5 (A) How does the input impedance of a field-effect transistor (FET) compare with that of a bipolar transistor?
A An FET has high input impedance; a bipolar transistor has low input impedance
B One cannot compare input impedance without knowing supply voltage
C An FET has low input impedance; a bipolar transistor has high input impedance
D The input impedance of FETs and bipolar transistors is the same

Bipolar transistors are operated with a forward-biased (conductive) Base-Emitter junction. Bipolar transistors are current amplifiers. Impedance, as a ratio of voltage to current, is necessarily low when voltage is low and current is high. The Field Effect Transistor, with a reverse biased Gate to channel junction, and the Insulated Gate Field Effect Transistor (IGFET or metal-oxide-semiconductor FET, MOSFET) with a Gate separated from the channel by a dielectric, are high impedance devices.

BJT
The middle letter indicates the polarity of the activating voltage.

The figure of merit for the common-base configuration is alpha.

The figure of merit for the common-emitter or common-collector configurations is beta.

There are two types of BJTs; PNP and NPN. The middle letter refers to the base, which responds to voltage of the same polarity:

• The NPN likes to get tickled by positive voltage, and
• the PNP likes to get tickled by negative voltage.

A-2-3-3 (D) Which component conducts electricity from a negative emitter to a positive collector when its base voltage is made positive?
A A varactor
B A triode vacuum tube
C A PNP transistor
D An NPN transistor

The terms Emitter, Collector and Base refer to bipolar transistors, of which there are two types: NPN and PNP. The Base-Emitter junction must be forward-biased for Base current to exist. A positive voltage on the Base supposes P material for conduction to take place, the ‘sandwich’ is thus NPN. Inversely, a negative Base voltage relates to a PNP.

A-2-3-8 (C) Which component conducts electricity from a positive emitter to a negative collector when its base is made negative?
A A varactor
B An NPN transistor
C A PNP transistor
D A triode vacuum tube

The terms Emitter, Collector and Base refer to bipolar transistors, of which there are two types: NPN and PNP. The Base-Emitter junction must be forward-biased for Base current to exist. A positive voltage on the Base supposes P material for conduction to take place, the ‘sandwich’ is thus NPN. Inversely, a negative Base voltage relates to a PNP.

There are two BJT configurations of interest:

For the purpose of the exam, the parameter of interest in both cases is the current gain (alpha and beta).

The forward current gain alpha is specified for the common base configuration. In this configuration, the emitter is the control terminal (?).

The quantity describes how much collector (output) current you get for how much emitter (control) current, which is mathematically:

``````        d I_c
alpha = -----
d I_e``````

Alpha is related to beta by:

``````           beta
alpha = ----------
(1 + beta)``````

A-2-3-4 (A) What is the alpha of a bipolar transistor in common base configuration?
A Forward current gain
B Forward voltage gain
C Reverse current gain
D Reverse voltage gain

The Alpha being a number smaller than 1, many authors refer to it as the “common base forward current transfer ratio” rather than a gain. In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

A-2-3-6 (D) The alpha of a bipolar transistor is specified for what configuration?
A Common collector
B Common gate
C Common emitter
D Common base

In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

A-2-3-1 (C) What is the alpha of a bipolar transistor?
A The change of base current with respect to collector current
B The change of collector current with respect to gate current
C The change of collector current with respect to emitter current
D The change of collector current with respect to base current

In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

A-2-3-9 (B) Alpha of a bipolar transistor is equal to:
A beta / (1 - beta)
B beta / (1 + beta)
C beta x (1 + beta)
D beta x (1 - beta)

Alpha (‘common base’) is always a number lesser than 1 ( the Emitter current is necessarily larger than the Collector current because the Base current also flows through the Emitter ). Beta (‘common emitter’) is normally a number greater than 10 ( the Collector current is always several times the Base current ). The Alpha is equal to Beta divided by 1 plus Beta. The Beta is equal to Alpha divided by 1 minus Alpha.

Common emitter and common collector

Common emitter:

``````emitter   o +V
|
base      N
+V o-----P
N
|
collector o -V``````

In the more typical common emitter or common collector configuration, the base is the control terminal, and the forward current gain is called beta.

(Think: base, beta)

A-2-3-7 (A) The beta of a bipolar transistor is specified for what configurations?
A Common emitter or common collector
B Common emitter or common gate
C Common base or common collector
D Common base or common emitter

In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

The quantity describes how much collector (output) current you get for how much base (control) current, which is mathematically:

``````        d I_c
beta = -----
d I_b``````

A-2-3-2 (C) What is the beta of a bipolar transistor?
A The change of collector current with respect to emitter current
B The change of base current with respect to gate current
C The change of collector current with respect to base current
D The change of base current with respect to emitter current

In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

A-2-3-5 (C) In a bipolar transistor, the change of collector current with respect to base current is called:
A delta
B alpha
C beta
D gamma

In a ‘common base’ configuration where the Emitter is the input and the Collector is the output, the Alpha factor (or common base forward current transfer ratio) is a ratio of a change in Collector current to the corresponding change in Emitter current. In a ‘common emitter’ configuration where the Base is the input and the Collector is the output, the Beta factor (or common emitter forward current gain) is a ratio of a change in Collector current to a given change in Base current. The Beta factor applies equally to a Common Collector configuration where the Base is also the input.

Beta is related to alpha by:

``````           alpha
beta = ----------
(1 - alpha)``````

A-2-3-11 (C) Beta of a bipolar transistor is equal to:
A alpha x (1 - alpha)
B alpha x (1 + alpha)
C alpha / (1 - alpha)
D alpha / (1 + alpha)

Alpha (‘common base’) is always a number lesser than 1 ( the Emitter current is necessarily larger than the Collector current because the Base current also flows through the Emitter ). Beta (‘common emitter’) is normally a number greater than 10 ( the Collector current is always several times the Base current ). The Alpha is equal to Beta divided by 1 plus Beta. The Beta is equal to Alpha divided by 1 minus Alpha.

Alpha is less than one. Beta is greater than ten.

A-2-3-10 (B) The current gain of a bipolar transistor in common emitter or common collector compared to common base configuration is:
A usually about half
B high to very high
C very low
D usually about double

Alpha (‘common base’) is always a number lesser than 1 ( the Emitter current is necessarily larger than the Collector current because the Base current also flows through the Emitter ). Beta (‘common emitter’) is normally a number greater than 10 ( the Collector current is always several times the Base current ). The Alpha is equal to Beta divided by 1 plus Beta. The Beta is equal to Alpha divided by 1 minus Alpha.

FET

The terminals are source, gate, and drain.

They act in enhancement-mode or depletion mode, akin to normally-on or normally-off.

They type determines the charge carriers: N-channel types support electron-conduction, and P-channel supports hole-conduction.

The terminals of a FET are source, gate, and drain. The gate is the control terminal.

A-2-4-6 (B) What are the three terminals of a junction field-effect transistor (JFET)?
A Gate 1, gate 2, drain
B Gate, drain, source
C Emitter, base 1, base 2
D Emitter, base, collector

Remember your Basic Qualification? The FET comprises a Source, a Gate and a Drain. They come in two types: N-Channel and P-Channel.

FETs act either in enhancement-mode or depletion-mode, depending on whether the conduction channel is enhanced or depleted by the application of a control voltage.

In an enhancement-mode FET, initially there is no channel. The channel is established (enhanced) by the application of voltage. (I.e., normally-off.)

A-2-4-1 (A) What is an enhancement-mode FET?
A An FET without a channel; no current occurs with zero gate voltage
B An FET with a channel that blocks voltage through the gate
C An FET with a channel that allows current when the gate voltage is zero
D An FET without a channel to hinder current through the gate

An Enhancement-mode Insulated Gate Field Effect Transistor (IGFET) is constructed without a channel. There is no Drain current with zero Gate voltage. A voltage applied to the gate leads to the creation of a channel. A forward bias on the gate heightens the concentration of charge carriers which, in turn, ‘enhances’ conduction. A Depletion-mode Insulated Gate Field Effect Transistor has a channel. Drain current is possible even without a Gate voltage. A reverse bias on the Gate depletes charge carriers in the channel, thus reducing Drain current. A forward bias on the Gate can make the channel even more conductive.

In a depletion mode FET, initially there is a channel that is depleted by the application of voltage. (I.e., normally-on.)

A-2-4-2 (B) What is a depletion-mode FET?
A An FET that has a channel that blocks current when the gate voltage is zero
B An FET that has a channel with no gate voltage applied; a current flows with zero gate voltage
C An FET without a channel; no current flows with zero gate voltage
D An FET without a channel to hinder current through the gate

An Enhancement-mode Insulated Gate Field Effect Transistor (IGFET) is constructed without a channel. There is no Drain current with zero Gate voltage. A voltage applied to the gate leads to the creation of a channel. A forward bias on the gate heightens the concentration of charge carriers which, in turn, ‘enhances’ conduction. A Depletion-mode Insulated Gate Field Effect Transistor has a channel. Drain current is possible even without a Gate voltage. A reverse bias on the Gate depletes charge carriers in the channel, thus reducing Drain current. A forward bias on the Gate can make the channel even more conductive.

FETs come in N-channel and P-channel, which determine the polarity of the ON or OFF states dependent on the configuration.

A-2-4-7 (A) What are the two basic types of junction field-effect transistors (JFET)?
A N-channel and P-channel
B High power and low power
C MOSFET and GaAsFET
D Silicon and germanium

Remember your Basic Qualification? The FET comprises a Source, a Gate and a Drain. They come in two types: N-Channel and P-Channel.

P-channel (MOS)FETs support hole conduction, in either-enhancement mode or depletion-mode.

A-2-4-10 (A) Hole conduction in a p-channel depletion type MOSFET is associated with:
A p-channel depletion
B n-channel enhancement
C q-channel depletion
D n-channel depletion

This seems too simple to be true, the words in the question give the answer away.

A-2-4-11 (D) Hole conduction in a p-channel enhancement type MOSFET is associated with:
A n-channel depletion
B n-channel enhancement
C q-channel depletion
D p-channel enhancement

This seems too simple to be true, the words in the question give the answer away.

N-channel (MOS)FETs support electron conduction, in either enhancement mode or depletion-mode.

A-2-4-9 (D) Electron conduction in an n-channel enhancement MOSFET is associated with:
A q-channel depletion
B p-channel enhancement
C p-channel depletion
D n-channel enhancement

This seems too simple to be true, the words in the question give the answer away.

A-2-4-8 (D) Electron conduction in an n-channel depletion type MOSFET is associated with:
A p-channel depletion
B p-channel enhancement
C q-channel enhancement
D n-channel depletion

This seems too simple to be true, the words in the question give the answer away.

MOSFET gates are sensitive to ESD, and include Zener diodes to protect them.

A-2-4-4 (D) Why are special precautions necessary in handling FET and CMOS devices?
A They are light-sensitive
B They have micro-welded semiconductor junctions that are susceptible to breakage
C They have fragile leads that may break off
D They are susceptible to damage from static charges

The Gate in an Insulated Gate Field Effect Transistor (IGFET or metal-oxide-semiconductor FET, MOSFET) is insulated from the channel by a thin oxide layer. Static electricity or excessive voltage can easily destroy the dielectric layer.

A-2-4-3 (B) Why do many MOSFET devices have built-in gate protective Zener diodes?
A The gate-protective Zener diode provides a voltage reference to provide the correct amount of reverse-bias gate voltage
B The gate-protective Zener diode prevents the gate insulation from being punctured by small static charges or excessive voltages
C The gate-protective Zener diode keeps the gate voltage within specifications to prevent the device from overheating
D The gate-protective Zener diode protects the substrate from excessive voltages

The Gate in an Insulated Gate Field Effect Transistor (IGFET or metal-oxide-semiconductor FET, MOSFET) is insulated from the channel by a thin oxide layer. Static electricity or excessive voltage can easily destroy the dielectric layer.

Transistor-Like Diodes

Silicon Controlled-Rectifiers are three-terminal, diode-like switches.

The SCR is a type of thyristor, which is made from four layers of doped silcon forming a PNPN junction. (“The big mac of diodes.”)

A-2-5-7 (D) The silicon controlled rectifier (SCR) is a member of which family?
A Phase locked loops
B Varactors
C Varistors
D Thyristors

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-5 (D) The silicon controlled rectifier (SCR) is what type of device?
A NPPN
B PNNP
C PPNN
D PNPN

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-11 (B) Which of the following is a PNPN device?
A Zener diode
B Silicon controlled rectifier (SCR)
C PIN diode
D Hot carrier (Schottky) diode

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

They are used in crowbar circuits for overvoltage protection, in conjunction with a zener diode to set the trigger voltage.

A-2-5-8 (C) In amateur radio equipment, which is the major application for the silicon controlled rectifier (SCR)?
A Microphone preamplifier circuit
B SWR detector circuit
C Power supply overvoltage “crowbar” circuit
D Class C amplifier circuit

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

The three terminals are anode, cathode, and gate. The gate is the control terminal.

A-2-5-1 (D) What are the three terminals of a silicon controlled rectifier (SCR)?
A Gate, base 1 and base 2
B Base, collector and emitter
C Gate, source and sink
D Anode, cathode and gate

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-9 (A) Which of the following devices has anode, cathode, and gate?
A The silicon controlled rectifier (SCR)
B The bipolar transistor
C The field effect transistor
D The triode vacuum tube

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-6 (B) The control element in the silicon controlled rectifier (SCR) is called the:
A emitter
B gate
C anode
D cathode

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

The two stable states are conducting and non-conducting.

A-2-5-2 (C) What are the two stable operating conditions of a silicon controlled rectifier (SCR)?
A NPN conduction and PNP conduction
B Oscillating and quiescent
C Conducting and non-conducting
D Forward conducting and reverse conducting

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

Most of the time, they do not conduct; however, if there’s an overvoltage across the diode, it will conduct.

When triggered or gated on (i.e. conducting), they appear like a forward-biased junction or rectifier diode.

A-2-5-3 (D) When a silicon controlled rectifier (SCR) is triggered, to what other semiconductor diode are its electrical characteristics similar (as measured between its cathode and anode)?
A The PIN diode
B The hot-carrier (Schottky) diode
C The varactor diode
D The junction diode

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-4 (C) Under what operating condition does a silicon controlled rectifier (SCR) exhibit electrical characteristics similar to a forward-biased silicon rectifier?
A When it is used as a detector
B During a switching transition
C When it is gated “on”
D When it is gated “off”

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

A-2-5-10 (D) When it is gated “on”, the silicon controlled rectifier (SCR) exhibits electrical characteristics similar to a:
A reverse-biased silicon rectifier
B forward-biased PIN diode
C reverse-biased hot-carrier (Schottky) diode
D forward-biased silicon rectifier

The SCR, part of the Thyristor family, is made of four layers of alternating P and N type material, namely PNPN. It comprises three electrodes: Anode, Gate and Cathode. As can be expected, the two outermost electrodes, the Anode and the Cathode are respectively type P and type N material. Without gate current, the SCR looks like a regular non-conducting junction diode. Once triggered via the Gate, the SCR resembles a forward-biased (conducting) junction diode. Conduction continues unless current falls below a critical level. One typical application is an overvoltage protection circuit in a power supply.

{L04} Power Supplies.

block diagram

A-4-4-10 (C) In a regulated power supply, the output of the electrolytic filter capacitor is connected to the:
A solid-state by-pass circuit
B matching circuit for the load
C voltage regulator
D pi filter

Remember your Basic Qualification? The Regulated Power Supply comprises: the input, the transformer, the rectifier, the filter, the regulator and the output.

Fuses are placed before the transformer, and before the output.

A-4-4-9 (B) In a regulated power supply, components that conduct alternating current at the input before the transformer and direct current before the output are:
A chokes
B fuses
C capacitors
D diodes

Only fuses can be expected to be found on either side of the transformer and pass AC or DC equally.

rectifiers

The full-wave center-tap toplogy rectifier is somewhat inferior to the full-wave bridge topology.

A-4-1-1 (D) For the same transformer secondary voltage, which rectifier has the highest average output voltage?
A Half-wave
B Quarter-wave
C Full-wave centre-tap
D Bridge

The half-wave configuration is not as efficient at rectification as the full-wave. The real choice lies between the two full-wave alternatives: bridge or full-wave with centre-tap. The bridge configuration rectifies the full secondary winding voltage. In contrast, the two-diode with centre-tapped transformer secondary only rectifies half the secondary voltage on each half-cycle.

A-4-1-4 (A) A full-wave bridge rectifier circuit makes use of both halves of the AC cycle, but unlike the full-wave centre-tap rectifier circuit it does not require:
A a centre-tapped secondary on the transformer
B any output filtering
C a centre-tapped primary on the transformer
D diodes across each leg of the transformer

A four-diode bridge rectifier makes use of the full secondary winding without the need for a centre-tap.

A-4-1-3 (B) In a full-wave centre-tap power supply, regardless of load conditions, the peak inverse voltage (PIV) will be _ times the RMS voltage:
A 1.4
B 2.8
C 0.636
D 0.707

With the RMS voltage defined in this example as the voltage from one extremity of the secondary winding to the centre-tap, each diode is subjected to 2.8 times the RMS voltage, peak reverse voltage from half the transformer winding adds to the peak DC output as a reverse bias on the diode during non-conduction ( each diode serves as a half-wave rectifier ).

A-4-4-8 (C) In a regulated power supply, four diodes connected together in a BRIDGE act as:
A matching between the secondary of the power transformer and the filter
B a tuning network
C a rectifier
D equalization across the transformer

Four diodes in a bridge configuration permit full-wave rectification with a single secondary winding.

full-wave center-tap

half-wave supply

A-4-1-2 (A) In a half-wave power supply with a capacitor input filter and a load drawing little or no current, the peak inverse voltage (PIV) across the diode can reach _ times the RMS voltage.
A 2.8
B 0.45
C 5.6
D 1.4

During conduction, the capacitor charges up to the peak value of the waveform (1.4 times the RMS value). On the opposite half-cycle, and from the diode’s standpoint, the transformer winding reaching opposite peak value (1.4 times the RMS value) adds to the capacitor charge for a total of 2.8 times the RMS value.

full-wave

A-4-1-6 (B) The ripple frequency produced by a full-wave power supply connected to a normal household circuit is:
A 30 Hz
B 120 Hz
C 60 Hz
D 90 Hz

Key word: FULL-WAVE. The two half-cycles are put to contribution. The output goes from zero to peak and back 120 times per second. Half-wave would be 60 hertz.

A-4-1-5 (A) For a given transformer the maximum output voltage available from a full-wave bridge rectifier circuit will be:
A double that of the full-wave centre-tap rectifier
B half that of the full-wave centre-tap rectifier
C the same as the full-wave centre-tap rectifier
D the same as the half-wave rectifier

The discussion relates to rectification, thus DC voltage output is the criteria. Imagine a 12 volts AC secondary with a centre tap. A bridge rectifier across the full secondary will obviously provide twice the voltage of a full-wave centre-tap rectifier where each diode draws from half the secondary. The bridge rectifier will also output slightly more DC voltage after filtering than a half-wave rectifier across the same full secondary.

half-wave supply

A-4-1-7 (A) The ripple frequency produced by a half-wave power supply connected to a normal household circuit is:
A 60 Hz
B 90 Hz
C 120 Hz
D 30 Hz

Key word: HALF-WAVE. One half-cycle only is put to contribution. The output goes from zero to peak and back 60 times per second. Full-wave would be 120 hertz.

voltage doubler

A-4-1-8 (A) Full-wave voltage doublers:
A use both halves of an AC wave
B create four times the output voltage of half-wave doublers
C use less power than half-wave doublers
D are used only in high-frequency power supplies

A voltage doubler returns a DC voltage approximately twice the supplied AC voltage. Through combinations of diodes and capacitors, both half-cycles are rectified and added together. Two ubiquitous configurations are respectively designated as “half-wave” doubler and “full-wave” doubler. The designation has more to do with the ripple frequency than how energy is transferred to the output. Ripple frequency in the “full-wave” doubler is twice the supply frequency. They can be implemented at normal line frequency or in switching power supplies.

diode specifications

A-4-1-9 (D) What are the two major ratings that must not be exceeded for silicon-diode rectifiers used in power-supply circuits?
A Average power; average voltage
B Capacitive reactance; avalanche voltage
C Peak load impedance; peak voltage
D Peak inverse voltage; average forward current

During conduction, the diode must support the average forward current. Under reverse bias, the diode must support the peak inverse voltage present across it.

output filtering

A-4-1-10 (B) In a high voltage power supply, why should a resistor and capacitor be wired in parallel with the power-supply rectifier diodes?
A To ensure that the current through each diode is about the same
B To equalize voltage drops and guard against transient voltage spikes
C To smooth the output waveform
D To decrease the output voltage

Parallel capacitors are used to bypass voltage spikes. Parallel resistors across each diode in a chain of diodes equalize reverse voltage.

full-wave rectifier

A-4-1-11 (D) What is the output waveform of an unfiltered full-wave rectifier connected to a resistive load?
A A steady DC voltage
B A sine wave at half the frequency of the AC input
C A series of pulses at the same frequency as the AC input
D A series of pulses at twice the frequency of the AC input

A full-wave rectifier puts both half-cycles to contribution: pulsating direct current with 120 zero-to-peak transitions per second is produced.

filter chokes

A-4-2-1 (C) Filter chokes are rated according to:
A power loss
B breakdown voltage
C inductance and current-handling capacity
D reactance at 1000 Hz

Filter chokes are wired in series with the rectifier output. The choke must support the current drawn by the load. Its inductance influences the reduction in ripple.

input filter

series chokes

A-4-2-7 (C) In a power supply, series chokes will:
A impede the passage of DC but will pass the AC component
B impede both DC and AC
C readily pass the DC but will impede the flow of the AC component
D readily pass the DC and the AC component

Inductors oppose changes in current. Stable DC current is not affected but AC ripple is minimized.

A bleeder resistor provides a path for a continuous trickle of current through the choke, to ameliorate inrush current and consequent back-EMF.

A-4-2-8 (C) When using a choke input filter, a minimum current should be drawn all the time when the device is switched on. This can be accomplished by:
A placing an ammeter in the output circuit
B increasing the value of the output capacitor
C including a suitable bleeder resistance
D utilizing a full-wave bridge rectifier circuit

Only the expected answer has an impact on DC current flowing through the filter whether an external load is connected or not.

input filtering

capacitor filter

A-4-2-3 (D) The advantage of the capacitor input filter over the choke input filter is:
A better filtering action or smaller ripple voltage
B improved voltage regulation
C lower peak rectifier currents
D a higher terminal voltage output

Regulation is the change in voltage from no-load to full-load. The first filter element determines the classification. Capacitor-input filters ensure high output voltage but poor regulation: voltage soars to the peak AC value under no load and drops under load. Capacitor-input leads to high peak rectifier current. Choke-input filters limit the soar in voltage through counter-EMF and by opposing capacitor charge current. Peak rectifier current is constrained but output voltage approximates the average value of the AC waveform. Half-wave circuits have the poorest regulation.

series choke filter - opposes changes in current due to back-EMF

A-4-2-4 (B) With a normal load, the choke input filter will give the:
A highest output voltage
B best regulated output
C greatest percentage of ripple
D greatest ripple frequency

Regulation is the change in voltage from no-load to full-load. The first filter element determines the classification. Capacitor-input filters ensure high output voltage but poor regulation: voltage soars to the peak AC value under no load and drops under load. Capacitor-input leads to high peak rectifier current. Choke-input filters limit the soar in voltage through counter-EMF and by opposing capacitor charge current. Peak rectifier current is constrained but output voltage approximates the average value of the AC waveform. Half-wave circuits have the poorest regulation.

A-4-2-5 (C) There are two types of filters in general use in a power supply. They are called:
A choke input and capacitor output
B choke output and capacitor input
C choke input and capacitor input
D choke output and capacitor output

Regulation is the change in voltage from no-load to full-load. The first filter element determines the classification. Capacitor-input filters ensure high output voltage but poor regulation: voltage soars to the peak AC value under no load and drops under load. Capacitor-input leads to high peak rectifier current. Choke-input filters limit the soar in voltage through counter-EMF and by opposing capacitor charge current. Peak rectifier current is constrained but output voltage approximates the average value of the AC waveform. Half-wave circuits have the poorest regulation.

regulation

A-4-2-2 (D) Which of the following circuits gives the best regulation, under similar load conditions?
A A half-wave bridge rectifier with a capacitor input filter
B A half-wave rectifier with a choke input filter
C A full-wave rectifier with a capacitor input filter
D A full-wave rectifier with a choke input filter

Regulation is the change in voltage from no-load to full-load. The first filter element determines the classification. Capacitor-input filters ensure high output voltage but poor regulation: voltage soars to the peak AC value under no load and drops under load. Capacitor-input leads to high peak rectifier current. Choke-input filters limit the soar in voltage through counter-EMF and by opposing capacitor charge current. Peak rectifier current is constrained but output voltage approximates the average value of the AC waveform. Half-wave circuits have the poorest regulation.

bleeder resistor

A-4-2-6 (B) The main function of the bleeder resistor in a power supply is to provide a discharge path for the capacitor in the power supply. But it may also be used for a secondary function, which is to:
A act as a secondary smoothing device in conjunction with the filter
B improve voltage regulation
C provide a ground return for the transformer
D inhibit the flow of current through the supply

Regulation is the change in voltage from no-load to full-load. By ensuring a certain minimum current draw on the supply, the bleeder prevents the capacitors from fully charging up to peak AC values when no external load is connected.

A-4-2-11 (D) In a properly designed choke input filter power supply, the no-load voltage across the filter capacitor will be about nine-tenths of the AC RMS voltage; yet it is advisable to use capacitors rated at the peak transformer voltage. Why is this large safety margin suggested?
A Resonance can be set up in the filter producing high voltages
B Under heavy load, high currents and voltages are produced
C Under no-load conditions, the current could reach a high level
D Under no-load conditions and a burned-out bleeder, voltages could reach the peak transformer voltage

Inductors oppose changes in current. If no current at all is drawn from a choke-input filter, the effect of the inductor vanishes: no more counter-EMF or opposition to peak capacitor charging current. Subsequent capacitors are allowed to fully charge to peak AC values. [ nine tenths the RMS: RMS is 0.707 times peak, average is 0.637 times peak, 0.637 is nine tenths of 0.707 ]

resonance

A-4-2-9 (C) In the design of a power supply, the designer must be careful of resonance effects because the ripple voltage could build up to a high value. The components that must be carefully selected are:
A first capacitor and second capacitor
B first choke and second capacitor
C first choke and first capacitor
D the bleeder resistor and the first choke

Series resonance in the first choke and first capacitor across the rectifier may cause excessive rectifier peak current and abnormally high peak reverse voltages on the diodes.

A-4-2-10 (B) Excessive rectifier peak current and abnormally high peak inverse voltages can be caused in a power supply by the filter forming a:
A tuned inductance in the filter choke
B series resonant circuit with the first choke and first capacitor
C short circuit across the bleeder
D parallel resonant circuit with the first choke and second capacitor

Series resonance in the first choke and first capacitor across the rectifier may cause excessive rectifier peak current and abnormally high peak reverse voltages on the diodes.

linear regulator

A-4-3-1 (C) What is one characteristic of a linear electronic voltage regulator?
A A pass transistor switches from its “on” state to its “off” state
B The control device is switched on or off, with the duty cycle proportional to the line or load conditions
C The conduction of a control element is varied in direct proportion to the line voltage or load current
D It has a ramp voltage at its output

In a ‘linear’ voltage regulator, a voltage higher than necessary is first produced; this voltage is brought down through a voltage dropping component. A regulator circuit (e.g., a Zener in a shunt configuration) may draw more or less current through a passive resistor to compensate for external changes. The dropping element, in a series configuration, may be a tube or transistor whose conduction may be varied. In a switching regulator, the incoming DC is switched on and off; the on time is varied so that the average DC output is maintained regardless of current draw.

A-4-3-3 (C) What device is typically used as a stable reference voltage in a linear voltage regulator?
A A varactor diode
B A junction diode
C A Zener diode
D An SCR

Remember your Basic Qualification? Zener diodes maintain a constant voltage across their terminals.

A-4-3-4 (C) What type of linear regulator is used in applications requiring efficient utilization of the primary power source?
A A constant current source
B A shunt current source
C A series regulator
D A shunt regulator

Key word: EFFICIENT. A linear regulator with an active series dropping device (tube or transistor) wastes less energy as dropping resistance is adjusted to whatever current is drawn. A linear regulator with a passive dropping resistor and a control device in a shunt configuration (Zener, tube or transistor) is wasteful because a fixed amount of current is needed to maintain a given drop in voltage regardless of load current; the device draws less when load current increases and vice-versa. The shunt configuration may be needed if the unregulated source demands a constant load.

A-4-3-5 (C) What type of linear voltage regulator is used in applications requiring a constant load on the unregulated voltage source?
A A shunt current source
B A series regulator
C A shunt regulator
D A constant current source

Key words: CONSTANT LOAD. A linear regulator with an active series dropping device (tube or transistor) wastes less energy as dropping resistance is adjusted to whatever current is drawn. A linear regulator with a passive dropping resistor and a control device in a shunt configuration (Zener, tube or transistor) is wasteful because a fixed amount of current is needed to maintain a given drop in voltage regardless of load current; the device draws less when load current increases and vice-versa. The shunt configuration may be needed if the unregulated source demands a constant load.

A-4-3-6 (C) How is remote sensing accomplished in a linear voltage regulator?
A A load connection is made outside the feedback loop
B By wireless inductive loops
C A feedback connection to an error amplifier is made directly to the load
D An error amplifier compares the input voltage to the reference voltage

Key word: REMOTE. Voltage regulation relies on comparing the output voltage to a set reference and using the error to adjust conduction in the control element of the regulator. Sensing the voltage at the load rather than at the output terminals of the power supply compensates for losses all the way out to the load.

switching regulator

duty cycle is proportional to load

A-4-3-2 (B) What is one characteristic of a switching voltage regulator?
A It gives a ramp voltage at its output
B The control device is switched on and off, with the duty cycle proportional to the line or load conditions
C The conduction of a control element is varied in direct proportion to the line voltage or load current
D It provides more than one output voltage

In a ‘linear’ voltage regulator, a voltage higher than necessary is first produced; this voltage is brought down through a voltage dropping component. A regulator circuit (e.g., a Zener in a shunt configuration) may draw more or less current through a passive resistor to compensate for external changes. The dropping element, in a series configuration, may be a tube or transistor whose conduction may be varied. In a switching regulator, the incoming DC is switched on and off; the on time is varied so that the average DC output is maintained regardless of current draw.

three terminal regulator

consists of:

• voltage reference,
• error amplifier,
• sensing resistors and transistors, and
• a pass element

A-4-3-9 (A) What type of voltage regulator contains a voltage reference, error amplifier, sensing resistors and transistors, and a pass element in one package?
A A three-terminal regulator
B An op-amp regulator
C A switching regulator
D A Zener regulator

A three-terminal regulator is a single integrated circuit comprising a voltage reference, a comparator, an error amplifier, sensing resistors and a pass transistor. Some include thermal shutdown, current foldback and over-voltage protection. The three terminals are: unregulated DC input, regulated DC output and ground. Specifications include: maximum output current, maximum output voltage, maximum input voltage and minimum input voltage (because a minimum voltage differential is needed to maintain regulation, the drop-out voltage).

A-4-3-7 (C) What is a three-terminal regulator?
A A regulator containing three error amplifiers and sensing transistors
B A regulator that supplies three voltages with variable current
C A regulator containing a voltage reference, error amplifier, sensing resistors and transistors, and a pass element
D A regulator that supplies three voltages at a constant current

A three-terminal regulator is a single integrated circuit comprising a voltage reference, a comparator, an error amplifier, sensing resistors and a pass transistor. Some include thermal shutdown, current foldback and over-voltage protection. The three terminals are: unregulated DC input, regulated DC output and ground. Specifications include: maximum output current, maximum output voltage, maximum input voltage and minimum input voltage (because a minimum voltage differential is needed to maintain regulation, the drop-out voltage).

rated by output votlage, and (maximum) output current

A-4-3-8 (B) In addition to an input voltage range what are the important characteristics of a three-terminal regulator?
A Output voltage and minimum output current
B Output voltage and maximum output current
C Maximum output voltage and minimum output current
D Minimum output voltage and maximum output current

A three-terminal regulator is a single integrated circuit comprising a voltage reference, a comparator, an error amplifier, sensing resistors and a pass transistor. Some include thermal shutdown, current foldback and over-voltage protection. The three terminals are: unregulated DC input, regulated DC output and ground. Specifications include: maximum output current, maximum output voltage, maximum input voltage and minimum input voltage (because a minimum voltage differential is needed to maintain regulation, the drop-out voltage).

A-4-3-11 (A) A modern type of regulator, which features a reference, high-gain amplifier, temperature-compensated voltage sensing resistors and transistors as well as a pass element is commonly referred to as a:
A three-terminal regulator
B nine-pin terminal regulator
C twenty-four pin terminal
D regulator six-terminal regulator

A three-terminal regulator is a single integrated circuit comprising a voltage reference, a comparator, an error amplifier, sensing resistors and a pass transistor. Some include thermal shutdown, current foldback and over-voltage protection. The three terminals are: unregulated DC input, regulated DC output and ground. Specifications include: maximum output current, maximum output voltage, maximum input voltage and minimum input voltage (because a minimum voltage differential is needed to maintain regulation, the drop-out voltage).

types of regulators

Two types of regulators:

• linear
• switching

A-4-3-10 (D) When extremely low ripple is required, or when the voltage supplied to the load must remain constant under conditions of large fluctuations of current and line voltage, a closed-loop amplifier is used to regulate the power supply. There are two main categories of electronic regulators. They are:
A non-linear and switching
B linear and non-linear
C stiff and switching
D linear and switching

In a ‘linear’ voltage regulator, a voltage higher than necessary is first produced; this voltage is brought down through a voltage dropping component. A regulator circuit (e.g., a Zener in a shunt configuration) may draw more or less current through a passive resistor to compensate for external changes. The dropping element, in a series configuration, may be a tube or transistor whose conduction may be varied. In a switching regulator, the incoming DC is switched on and off; the on time is varied so that the average DC output is maintained regardless of current draw.

A-4-4-2 (D) In any regulated power supply, the output is cleanest and the regulation is best:
A across the secondary of the pass transistor
B across the load
C at the output of the pass transistor
D at the point where the sampling network or error amplifier is connected

A voltage regulator maintains a stable output by comparing a sample of the output voltage with a reference and adjusting conduction in the pass transistor accordingly. The corrective action is only accurate for the precise point where the measurement is taken. Because of losses, the load itself may find itself at a lower voltage: this is the reason for ‘remote sensing’ in certain applications.

series-regulated

A-4-4-1 (D) In a series-regulated power supply, the power dissipation of the pass transistor is:
A the inverse of the load current and the input/output voltage differential
B dependent upon the peak inverse voltage appearing across the Zener diode
C indirectly proportional to the load voltage and the input/output voltage differential
D directly proportional to the load current and the input/output voltage differential

The pass transistor is the device acting as a variable resistor to drop the unregulated DC source down to the regulated output. Power is voltage times current: in this case, the difference in voltage from input to output times the current drawn by the load.

regulation

The exam distinguished between “static” and “dynamic” regulation; that is, the response to fast or slow changes in load

A-4-4-4 (A) The regulation of long-term changes in the load resistance of a power supply is called:
A static regulation
B active regulation
C analog regulation
D dynamic regulation

Key words: LONG-TERM. Regulation is the change in voltage from no-load to full-load. Static regulation relates to the supply’s performance in relation with long-term changes in load resistance or line variations (AC source). Dynamic regulation is required when the current draw varies as a Morse key is pressed (CW) or with each syllable (SSB) in a final amplifier. A large output capacitor, the last in the filter configuration, can improve dynamic regulation.

A-4-4-5 (A) The regulation of short-term changes in the load resistance of a power supply is called:
A dynamic regulation
B static regulation
C analog regulation
D active regulation

Key words: SHORT-TERM. Regulation is the change in voltage from no-load to full-load. Static regulation relates to the supply’s performance in relation with long-term changes in load resistance or line variations (AC source). Dynamic regulation is required when the current draw varies as a Morse key is pressed (CW) or with each syllable (SSB) in a final amplifier. A large output capacitor, the last in the filter configuration, can improve dynamic regulation.

A-4-4-6 (C) The dynamic regulation of a power supply is improved by increasing the value of:
A the input capacitor
B the bleeder resistor
C the output capacitor
D the choke

Regulation is the change in voltage from no-load to full-load. Static regulation relates to the supply’s performance in relation with long-term changes in load resistance or line variations (AC source). Dynamic regulation is required when the current draw varies as a Morse key is pressed (CW) or with each syllable (SSB) in a final amplifier. A large output capacitor, the last in the filter configuration, can improve dynamic regulation.

A-4-4-7 (D) The output capacitor, in a power supply filter used to provide power for an SSB or CW transmitter, will give better dynamic regulation if:
A the negative terminal of the electrolytic capacitor is connected to the positive and the positive terminal to ground
B a battery is placed in series with the output capacitor
C it is placed in series with other capacitors
D the output capacitance is increased

Regulation is the change in voltage from no-load to full-load. Static regulation relates to the supply’s performance in relation with long-term changes in load resistance or line variations (AC source). Dynamic regulation is required when the current draw varies as a Morse key is pressed (CW) or with each syllable (SSB) in a final amplifier. A large output capacitor, the last in the filter configuration, can improve dynamic regulation.

The voltage and current drawn by the load can be expressed as a resistance

A-4-4-3 (A) When discussing a power supply the_ resistance is equal to the output voltage divided by the total current drawn, including the current drawn by the bleeder resistor:
B ideal
C rectifier
D differential

Per Ohm’s Law, resistance is voltage divided by current. Output voltage and total current drawn describe the load placed on the power supply.

reverse voltage protection

A-4-4-11 (A) In a regulated power supply, a diode connected across the input and output terminals of a regulator is used to:
A protect the regulator from reverse voltages
B provide an RF by-pass for the voltage control
C provide additional capacity
D protect the regulator from voltage fluctuations in the primary of the transformer

A fast reverse-biased diode between the output and input terminals prevents a large capacitor following the three-terminal regulator from discharging through the regulator if the input was ever short-circuited.

{L05} Amplifiers, Mixers and Frequency Multipliers.

There are four classes of amplifer, distinguished by the conduction angle of their transistors, and efficiency. They are:

• Class A (full-cycle, linear, inefficient)
• Class B (half-cycle)
• Class AB (compromise for SSB)
• Class C (FM, RTTY, CW, mixer) The performance metrics of interest are:

• linearity,
• distortion, and
• efficiency.
• Class A amplifiers operate over a full cycle and have good linearity and distortion characteristics, but are inefficient. The transistor in a Class A amplifier operates over a full cycle.

A-2-6-1 (A) For what portion of a signal cycle does a Class A amplifier operate?
A The entire cycle
B Exactly 180 degrees
C More than 180 degrees but less than 360 degrees
D Less than 180 degrees

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

A-2-6-10 (C) Which class of amplifier operates over the full cycle?
A Class B
B Class C
C Class A
D Class AB

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

Consequently, they have low distortion.

They also have high linearity.

A-2-6-2 (A) Which class of amplifier has the highest linearity and least distortion?
A Class A
B Class AB
C Class B
D Class C

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

They are least efficient.

A-2-6-8 (B) Which class of amplifier provides the least efficiency?
A Class AB
B Class A
C Class C
D Class B

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

• Class B amplifiers operate for a half-cycle. Class B amplifer consists of two amplifying elements in a “push-pull” configuration. Each transistor conducts for a half-cycle.

A-2-6-4 (A) For what portion of a cycle does a Class B amplifier operate?
A 180 degrees
B Less than 180 degrees
C More than 180 degrees but less than 360 degrees
D The entire cycle

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

• In Between Class A and Class B are Class AB amplifiers, which operate for somewhat more than a cycle. The transistors in a Class AB amplifiers operate for more than half-a-cycle, but less than a full cycle.

A-2-6-3 (D) For what portion of a cycle does a Class AB amplifier operate?
A Exactly 180 degrees
B The entire cycle
C Less than 180 degrees
D More than 180 degrees but less than 360 degrees

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

Most-often used for ham linear power amplifiers, as the amplitude of e.g. an SSB or AM signal varies in amplitude over a wide dynamic range.

• Class C amplifiers operate for less than half-a-cycle, but are relatively efficient. Class C amplifiers also operate for less than half-a-cycle.

A-2-6-5 (C) For what portion of a signal cycle does a Class C amplifier operate?
A The entire cycle
B 180 degrees
C Less than 180 degrees
D More than 180 degrees but less than 360 degrees

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

A-2-6-11 (A) Which class of amplifier operates over less than 180 degrees of the cycle?
A Class C
B Class AB
C Class A
D Class B

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

Because the resulting waveform is very much not sinusoidal, they have high distortion (lots of harmonics).

Because they are non-linear they bad for signals with a varying amplitude, and good for signals at a constant amplitude (like CW, FM, and RTTY.)

A-2-6-9 (A) Which class of amplifier has the poorest linearity and the most distortion?
A Class C
B Class AB
C Class A
D Class B

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

However, they have high efficiency.

A-2-6-6 (B) Which of the following classes of amplifier provides the highest efficiency?
A Class B
B Class C
C Class A
D Class AB

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

A-2-6-7 (A) Which of the following classes of amplifier would provide the highest efficiency in the output stage of a CW, RTTY or FM transmitter?
A Class C
B Class AB
C Class B
D Class A

Class A: 360 degrees, best linearity, least distortion, poor efficiency [25 to 30%]. Class AB: significantly more than 180 but less than 360 degrees, very acceptable linearity, medium efficiency [50 to 60%]. Class B: 180 degrees, acceptable linearity, medium efficiency [up to 65%]. Class C: much less than 180 degrees, poor linearity, high distortion, best efficiency [up to 80%]. Usable with constant amplitude signals (CW,FM) where ‘flywheel’ effect in tank circuit maintains the waveform. Harmonic-rich output is useful in frequency multiplier.

Efficiency in this contact refers to DC power input compared to RF power output.

There are three FET amplifier configurations of interest:

• common source
• common gate
• common drain

In the common source amplifer, the source terminal is common to input and output.

The gate is the input terminal.

The drain is the output terminal. The gate biasing network (R1 and R2 in the schamtic below) determines the input impedance.

A-2-7-1 (B) What determines the input impedance of a FET common-source amplifier?
A The input impedance is essentially determined by the resistance between the drain and substrate
B The input impedance is essentially determined by the gate biasing network
C The input impedance is essentially determined by the resistance between the source and substrate
D The input impedance is essentially determined by the resistance between the source and the drain

The Junction FET is considered a high impedance device. Because the Gate in a Junction FET is always reversed-biased, its input impedance is very high; the input impedance of the whole circuit is determined by the external Gate bias resistor. The output impedance is determined primarily by the resistor acting as a load in the Drain circuit. The drain resistor (R_D) determines the output impedance.

A-2-7-2 (A) What determines the output impedance of a FET common-source amplifier?
A The output impedance is essentially determined by the drain resistor
B The output impedance is essentially determined by the drain supply voltage
C The output impedance is essentially determined by the gate supply voltage
D The output impedance is essentially determined by the input impedance of the FET

The Junction FET is considered a high impedance device. Because the Gate in a Junction FET is always reversed-biased, its input impedance is very high; the input impedance of the whole circuit is determined by the external Gate bias resistor. The output impedance is determined primarily by the resistor acting as a load in the Drain circuit.

The BJT common-emitter configuration is similar to the FET common-source configuration.

A-2-7-9 (A) The FET amplifier common source circuit is similar to which of the following bipolar transistor amplifier circuits?
A Common emitter
B Common collector
C Common base
D Common mode

Remember your Basic Qualification? Source, Gate, Drain in the FET compare to Emitter, Base, Collector in the bipolar.

The input and output are 180 degrees out of phase.

A-2-7-6 (D) In the common emitter amplifier, when the input and output signals are compared:
A the output signal leads the input signal by 90 degrees
B the output signal lags the input signal by 90 degrees
C the signals are in phase
D the signals are 180 degrees out of phase

Common Emitter: low input Z, medium output Z, 180-degrees phase shift. Common Base: very low input Z, high output Z, no phase shift. Common Collector (Common Drain, Common Plate): high input Z, low output Z, no phase shift, also known as Emitter (Source, Cathode) follower, used for isolation or impedance matching.

common-base amplifier The input and output signals are in phase.

A-2-7-4 (B) In the common base amplifier, when the input and output signals are compared:
A the signals are 180 degrees out of phase
B the signals are in phase
C the output signal lags the input signal by 90 degrees
D the output signals leads the input signal by 90 degrees

Common Emitter: low input Z, medium output Z, 180-degrees phase shift. Common Base: very low input Z, high output Z, no phase shift. Common Collector (Common Drain, Common Plate): high input Z, low output Z, no phase shift, also known as Emitter (Source, Cathode) follower, used for isolation or impedance matching.

The input impedance is very low.

The common gate or grounded gate circuit configuration is used for VHF and UHF RF amplifiers where the low input impedance allows accurate matching to the feeder impedance which is typically 50Ω or 75Ω. (Source)

A-2-7-5 (C) In the common base amplifier, the input impedance, when compared to the output impedance is:
A only slightly lower
B very high
C very low
D only slightly higher

Common Emitter: low input Z, medium output Z, 180-degrees phase shift. Common Base: very low input Z, high output Z, no phase shift. Common Collector (Common Drain, Common Plate): high input Z, low output Z, no phase shift, also known as Emitter (Source, Cathode) follower, used for isolation or impedance matching.

The FET equivalent is the common gate configuration. A-2-7-11 (B) The FET amplifier common gate circuit is similar to which of the following bipolar transistor amplifier circuits?
A Common emitter
B Common base
C Common mode
D Common collector

Remember your Basic Qualification? Source, Gate, Drain in the FET compare to Emitter, Base, Collector in the bipolar.

common-collector amplifier

The FET common drain and the BJT common collector amplifers are similar.

A-2-7-10 (B) The FET amplifier common drain circuit is similar to which of the following bipolar transistor amplifier circuits?
A Common mode
B Common collector
C Common emitter
D Common base

Remember your Basic Qualification? Source, Gate, Drain in the FET compare to Emitter, Base, Collector in the bipolar.

The common-drain is also called a source-follower.

A-2-7-8 (C) The FET amplifier source follower circuit is another name for:
A common mode circuit
B common gate circuit
C common drain circuit
D common source circuit

In a source follower stage, the Source constitutes the output; the Drain, by opposition, must be tied to a common reference (a zero reference for signals): hence the expression, common drain.

The input and output are in phase.

A-2-7-7 (A) In the common collector amplifier, when the input and output signals are compared:
A the signals are in phase
B the output signal leads the input signal by 90 degrees
C the output signal lags the input signal by 90 degrees
D the signals are 180 degrees out of phase

Common Emitter: low input Z, medium output Z, 180-degrees phase shift. Common Base: very low input Z, high output Z, no phase shift. Common Collector (Common Drain, Common Plate): high input Z, low output Z, no phase shift, also known as Emitter (Source, Cathode) follower, used for isolation or impedance matching.

A Darlington pair amplifier consists of two well-behaving cascaded transistors. Consequently, it has high gain, and good input and output impedance (high and low).

A-2-7-3 (C) What are the advantages of a Darlington pair audio amplifier?
A Mutual gain, low input impedance and low output impedance
B Low output impedance, high mutual impedance and low output current
C High gain, high input impedance and low output impedance
D Mutual gain, high stability and low mutual inductance

The Darlington pair cascades two direct-coupled emitter-follower stages; Beta parameters multiply one another. The emitter follower, just like the cathode follower or the source follower features high input impedance and low output impedance. The Darlington configuration features high gain, high input impedance and low output impedance. “High gain” is enough to identify the correct answer.

Frequency multipliers make use of the harmonics produced by Class C amplifiers.

`need work here`

A multiplier is simply a class C amplifier. The (low-frequency) input. The output signal is rich in harmonics, with images at multiples of the input frequency.

A-2-9-5 (A) A frequency multiplier circuit must be operated in:
A class C
B class AB
C class B
D class A

A Frequency-Multiplier stage relies on harmonics produced by a gain device operated in Class C. The output circuit is tuned to an exact multiple of the input frequency (harmonic, typically two to four times). If greater multiplication is required, a chain of stages will be used.

The output incorporates a tuned circuit.

This could be a tank circuit, with a high impedance at the frequency of interest, shunting unwanted signals away, leaving signals at the frequency of interest.

A-2-9-9 (B) In a circuit where the components are tuned to resonate at a higher frequency than applied, the circuit is most likely a:
A a frequency divider
B a frequency multiplier
C a VHF/UHF amplifier
D a linear amplifier

A Frequency-Multiplier stage relies on harmonics produced by a gain device operated in Class C. The output circuit is tuned to an exact multiple of the input frequency (harmonic, typically two to four times). If greater multiplication is required, a chain of stages will be used.

A-2-9-6 (A) In a frequency multiplier circuit, an inductance (L1) and a variable capacitor (C2) are connected in series between VCC+ and ground. The collector of a transistor is connected to a tap on L1. The purpose of the variable capacitor is to:
A tune L1 to the desired harmonic
B by-pass RF
C tune L1 to the frequency applied to the base
D provide positive feedback

A tuned circuit is present in the output of a frequency multiplier to select the desired harmonic and reject unwanted signals.

A-2-9-7 (C) In a frequency multiplier circuit, an inductance (L1) and a variable capacitor (C2) are connected in series between VCC+ and ground. The collector of a transistor is connected to a tap on L1. A fixed capacitor (C3) is connected between the VCC+ side of L1 and ground. The purpose of C3 is to:
A resonate with L1
B by-pass any audio components
C provide an RF ground at the VCC connection point of L1
D form a pi filter with L1 and C2

A capacitor between the supply and ground is a “bypass” capacitor, it serves two purposes: it provides a low-impedance path to complete the AC circuit and it keeps AC signals out of the supply line (through which they could affect other stages). This being a frequency multiplier, the capacitor is an RF bypass.

A-2-9-8 (B) In a frequency multiplier circuit, an inductance (L1) and a variable capacitor (C2) are connected in series between VCC+ and ground. The collector of a transistor is connected to a tap on L1. C2 in conjunction with L1 operate as a:
A voltage doubler
B frequency multiplier
C frequency divider
D voltage divider

A tuned circuit is present in the output of a frequency multiplier to select the desired harmonic and reject unwanted signals.

Input coupled to base through DC blocking capacitor.

A-2-9-4 (D) In a frequency multiplier circuit, the input signal is coupled to the base of a transistor through a capacitor. A radio frequency choke is connected between the base of the transistor and ground. The capacitor is:
A part of the input tuned circuit
B a by-pass for the circuit
C part of the output tank circuit
D a DC blocking capacitor

A capacitor used for coupling let AC signals through but blocks DC.

A-2-9-10 (D) In a frequency multiplier circuit, an inductance (L1) and a variable capacitor (C2) are connected in series between VCC+ and ground. The collector of a transistor is connected to a tap on L1. A fixed capacitor (C3) is connected between the VCC+ side of L1 and ground. C3 is a:
A DC blocking capacitor
B tuning capacitor
C coupling capacitor
D RF by-pass capacitor

A capacitor between the supply and ground is a “bypass” capacitor, it serves two purposes: it provides a low-impedance path to complete the AC circuit and it keeps AC signals out of the supply line (through which they could affect other stages). This being a frequency multiplier, the capacitor is an RF bypass.

Mixers upconvert and downconvert signals in a way that the frequencies are additive and subtractive.

Mixers combine two input signals, with the outputs consisting of both the orignal signals, and signals at the sum and difference of the input frequncies.

A-2-9-1 (C) What is the mixing process?
A The recovery of intelligence from a modulated signal
B The elimination of noise in a wideband receiver by phase comparison
C The combination of two signals to produce sum and difference frequencies
D The elimination of noise in a wideband receiver by phase differentiation

A Mixer receives two inputs. They combine within the Mixer to produce two new frequencies: the sum of the inputs and the difference between the inputs. Four frequencies are present at the output: the sum, the difference and the two original frequencies. If a Mixer is driven into non-linearity by excessively strong signals, spurious responses will be produced.

A-2-9-2 (A) What are the principal frequencies that appear at the output of a mixer circuit?
A The original frequencies and the sum and difference frequencies
B 1.414 and 0.707 times the input frequencies
C The sum, difference and square root of the input frequencies
D Two and four times the original frequency

A Mixer receives two inputs. They combine within the Mixer to produce two new frequencies: the sum of the inputs and the difference between the inputs. Four frequencies are present at the output: the sum, the difference and the two original frequencies. If a Mixer is driven into non-linearity by excessively strong signals, spurious responses will be produced.

To transform singals to frequencies that are not multiples of each other, mixers are used.

For example, 5.3 MHz input + 9 MHz IF = 14.3 MHz output

A-2-9-11 (C) What stage in a transmitter would change a 5.3-MHz input signal to 14.3 MHz?
A A frequency multiplier
B A beat frequency oscillator
C A mixer
D A linear translator

The second frequency is not a multiple of the first, this excludes the multiplier. A Frequency Multiplier stage relies on harmonics produced by a gain device operated in Class C. The output circuit is tuned to an exact multiple of the input frequency (harmonic, typically two to four times). If greater multiplication is required, a chain of stages will be used.

When mixers are overdriven, nonlinearity and results in spurious signals.

A-2-9-3 (B) What occurs when an excessive amount of signal energy reaches the mixer circuit?
A Mixer blanking occurs
B Spurious signals are generated
C Automatic limiting occurs
D A beat frequency is generated

A Mixer receives two inputs. They combine within the Mixer to produce two new frequencies: the sum of the inputs and the difference between the inputs. Four frequencies are present at the output: the sum, the difference and the two original frequencies. If a Mixer is driven into non-linearity by excessively strong signals, spurious responses will be produced.

Op amps are used to amplify and filter small signals. They do not exhibit loss.

An op-amp is a differential amplifier.

The inputs and outputs are directly coupled (as opposed to capacitive or inductive coupling, with vacuum tube amplifers).

External components determine the amplifier characteristics, such as gain.

A-2-8-1 (C) What is an operational amplifier (op-amp)?
A An amplifier used to increase the average output of frequency modulated amateur signals to the legal limit
B A program subroutine that calculates the gain of an RF amplifier
C A high-gain, direct-coupled differential amplifier whose characteristics are determined by components mounted externally
D A high-gain, direct-coupled audio amplifier whose characteristics are determined by internal components of the device

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

A-2-8-11 (B) What term is most appropriate for a high gain, direct-coupled differential amplifier whose characteristics are determined by components mounted externally?
A Summing amplifier
B Operational amplifier
C Difference amplifier
D High gain audio amplifier

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

The ideal characterisitcs are:

• infinite gain
• flat frequency response
• infinite input impedance
• zero output impedance

A-2-8-2 (D) What would be the characteristics of the ideal op-amp?
A Zero input impedance, zero output impedance, infinite gain, and flat frequency response
B Infinite input impedance, infinite output impedance, infinite gain and flat frequency response
C Zero input impedance, infinite output impedance, infinite gain, and flat frequency response
D Infinite input impedance, zero output impedance, infinite gain, and flat frequency response

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

The input impedance is ideally very high, and the output impedance very low.

Practically, this means that they sink very little current and do not significantly load the signal source.

A-2-8-6 (D) What is the output impedance of a theoretically ideal op-amp?
A Very high
B Exactly 100 ohms
C Exactly 1000 ohms
D Very low

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

A-2-8-5 (A) What is the input impedance of a theoretically ideal op-amp?
A Very high
B Very low
C Exactly 100 ohms
D Exactly 1000 ohms

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

The gain of an op-amp is determined by an external feedback network.

A-2-8-3 (C) What determines the gain of a closed-loop op-amp circuit?
A The voltage applied to the circuit
B The collector-to-base capacitance of the PNP stage
C The external feedback network
D The PNP collector load

An Operational Amplifier is a high gain, direct-coupled differential amplifier whose characteristics are determined mainly by external components. For example, circuit gain is determined by the feedback network from output to input. The “ideal” Op-Amp would have infinite gain, infinite bandwidth (i.e., constant gain at any frequency), infinite input impedance and zero output impedance.

The offset voltage is the potential between the input terminals in a closed-loop condition.

A-2-8-4 (D) What is meant by the term op-amp offset voltage?
A The difference between the output voltage of the op-amp and the input voltage required for the next stage
B The potential between the amplifier input terminals of the op-amp in an open-loop condition
C The output voltage of the op-amp minus its input voltage
D The potential between the amplifier input terminals of the op-amp in a closed-loop condition

“Offset voltage is the potential between the amplifier input terminals in the closed-loop condition. Ideally, this voltage would be zero. Offset results from imbalance between the differential input transistors (ARRL Handbook)”.

In an inverting configuration, the output is a half-cycle out of phase from the input.

A-2-8-9 (B) What is an inverting op-amp circuit?
A An operational amplifier circuit connected such that the input impedance is held to zero, while the output impedance is high
B An operational amplifier circuit connected such that the input and output signals are 180 degrees out of phase
C An operational amplifier circuit connected such that the input and output signals are in phase
D An operational amplifier circuit connected such that the input and output signals are 90 degrees out of phase

An “inverting” Op-Amp circuit introduces a 180-degrees shift: when the input goes up, the output comes down and vice-versa. With the “non-inverting” Op-Amp circuit, the output is in phase with the input.

In a non-inverting configuration, the output is in-phase with the input.

output is in-phase with the oA-2-8-10 (D) What is a non-inverting op-amp circuit?
A An operational amplifier circuit connected such that the input and output signals are 90 degrees out of phase
B An operational amplifier circuit connected such that the input impedance is held low, and the output impedance is high
C An operational amplifier circuit connected such that the input and output signals are 180 degrees out of phase
D An operational amplifier circuit connected such that the input and output signals are in phase

An “inverting” Op-Amp circuit introduces a 180-degrees shift: when the input goes up, the output comes down and vice-versa. With the “non-inverting” Op-Amp circuit, the output is in phase with the input.

Op-amp filters are used in audio circuits, because they do not introduce loss, unlike passive filters.

A-2-8-8 (D) What are the principal uses of an op-amp RC active filter in amateur circuitry?
A Op-amp circuits are used as low-pass filters at the output of transmitters
B Op-amp circuits are used as filters for smoothing power supply output
C Op-amp circuits are used as high-pass filters to block RFI at the input of receivers
D Op-amp circuits are used as audio filters for receivers

Inductors and Capacitors are passive components; they inevitably introduce loss. Op-Amps used in filter applications can provide a controlled amount of gain. Op-Amps are commonly used in active AUDIO filter circuits; all types of responses can be implemented (low-pass, high-pass, bandpass, band-stop, a.k.a., notch).

A-2-8-7 (D) What are the advantages of using an op-amp instead of LC elements in an audio filter?
A Op-amps are more rugged and can withstand more abuse than can LC elements
B Op-amps are available in more styles and types than are LC elements
C Op-amps are fixed at one frequency
D Op-amps exhibit gain rather than insertion loss

Inductors and Capacitors are passive components; they inevitably introduce loss. Op-Amps used in filter applications can provide a controlled amount of gain. Op-Amps are commonly used in active AUDIO filter circuits; all types of responses can be implemented (low-pass, high-pass, bandpass, band-stop, a.k.a., notch).

{L11} Test Equipment and measurements.

peak-to-peak & RMS

``````V_rms = 0.707 V_pk

V_pkpk = 2 V_pk``````

A-3-1-1 (B) What is the easiest amplitude dimension to measure by viewing a pure sine wave on an oscilloscope?
A Average voltage
B Peak-to-peak voltage
C Peak voltage
D RMS voltage

Key word: EASIEST. Reading the peak-to-peak amplitude of a waveform is easier than attempting to precisely center the waveform to determine the peak value. RMS and Average values can only be computed from the peak value or peak-to-peak values.

A-3-1-2 (B) What is the RMS value of a 340 volt peak-to-peak pure sine wave?
A 300 volts
B 120 volts
C 170 volts
D 240 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 340 volts divided by 2 times 0.707 = 120 volts RMS.

A-3-1-11 (D) A sine wave of 17 volts peak is equivalent to how many volts RMS?
A 24 volts
B 34 volts
C 8.5 volts
D 12 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 17 volts times 0.707 = 12 volts RMS.

A-3-1-10 (D) What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts?
A 84.8 volts
B 169.7 volts
C 204.8 volts
D 339.5 volts

Knowing that RMS is peak times 0.707, peak must be RMS divided by 0.707 . Peak-to-peak is twice the peak value. 120 volts RMS divided by 0.707 = 170 volts peak. 170 volts peak times 2 = 340 volts peak-to-peak.

A-3-1-6 (D) The effective value of a sine wave of voltage or current is:
A 50% of the maximum value
B 100% of the maximum value
C 63.6% of the maximum value
D 70.7% of the maximum value

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value.

A-3-1-9 (D) Which AC voltage value will produce the same amount of heat as a DC voltage, when applied to the same resistance?
A The average value
B The peak value
C The peak-to-peak value
D The RMS value

A kettle would bring the same quantity of water to a boil within the same time whether it runs on 120 volts DC or 120 volts RMS AC.

A-3-1-5 (D) In applying Ohm’s law to AC circuits, current and voltage values are:
A average values
B average values times 1.414
C none of the proposed answers
D peak values times 0.707

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value.

A-3-1-4 (C) If the peak value of a 100 Hz sinusoidal waveform is 20 volts, the RMS value is:
A 7.07 volts
B 16.38 volts
C 14.14 volts
D 28.28 volts

Peak is half the peak-to-peak. RMS (or effective) is 0.707 times the peak value. 20 volts times 0.707 = 14.14 volts RMS.

A-3-1-3 (B) What is the equivalent to the RMS value of an AC voltage?
A The AC voltage found by taking the square root of the average AC value
B The AC voltage causing the same heating of a given resistor as a DC voltage of the same value
C The AC voltage found by taking the square root of the peak AC voltage
D The DC voltage causing the same heating of a given resistor as the peak AC voltage

A kettle would bring the same quantity of water to a boil within the same time whether it runs on 120 volts DC or 120 volts RMS AC.

peak envelope power

Peak Envelope Power (PEP) is distinct from RMS power.

A-3-2-10 (C) What is the output PEP from a transmitter, if an oscilloscope measures 800 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
A 6400 watts
B 3200 watts
C 1600 watts
D 800 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 800 divided by 2 = 400 volts peak ; (400 times 0.707) squared = 80 000 ; 80 000 divided by 50 = 1600 watts PEP.

A-3-2-11 (A) An oscilloscope measures 500 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output during unmodulated carrier conditions. What would an average-reading power meter indicate under the same transmitter conditions?
A 625 watts
B 427.5 watts
C 884 watts
D 442 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 500 divided by 2 = 250 volts peak ; (250 times 0.707) squared = 31 240 ; 31 240 divided by 50 = 625 watts PEP.

A-3-2-9 (C) What is the output PEP from a transmitter, if an oscilloscope measures 400 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
A 600 watts
B 1000 watts
C 400 watts
D 200 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 400 divided by 2 = 200 volts peak ; (200 times 0.707) squared = 20 000 ; 20 000 divided by 50 = 400 watts PEP.

A-3-2-8 (C) What is the output PEP of an unmodulated carrier transmitter if a wattmeter connected to the transmitter output indicates an average reading of 1060 watts?
A 1500 watts
B 530 watts
C 1060 watts
D 2120 watts

Key word: UNMODULATED. This is a catch. There would be no difference between output power and PEP in the absence of modulation. [ With modulation, the answer would have been different. Some wattmeters are calibrated to read peak power. ]

A-3-2-7 (C) What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
A 2500 watts
B 500 watts
C 625 watts
D 1250 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 500 divided by 2 = 250 volts peak ; (250 times 0.707) squared = 31 240 ; 31 240 divided by 50 = 625 watts PEP.

A-3-2-6 (D) What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
A 400 watts
B 1000 watts
C 200 watts
D 100 watts

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage. In this example, 200 divided by 2 = 100 volts peak ; (100 times 0.707) squared = 5000 ; 5000 divided by 50 = 100 watts PEP.

A-3-2-5 (B) How is the output Peak-Envelope Power of a transmitter calculated if an oscilloscope is used to measure the Peak-Envelope Voltage across a dummy resistive load (where PEP = Peak-Envelope Power, PEV = Peak-Envelope Voltage, Vp = peak-voltage, RL = load resistance)?
A PEP = [(1.414 PEV)(1.414 PEV)] / RL
B PEP = [(0.707 PEV)(0.707 PEV)] / RL
C PEP = [(Vp)(Vp)] / (RL)
D PEP = (Vp)(Vp)(RL)

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Instantaneous power is given by

``````    E^2
P = ---
R``````

A-3-2-4 (D) The formula to be used to calculate the power output of a transmitter into a resistor load using a voltmeter is:
A P = EI/R
B P = EI cos 0
C P = IR
D P = (E exponent 2) / R

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

Peak envelope power is given by:

``````                    2       2
(0.707 * V_pk)   V_rms
PEP = -------------- = ------
R           2``````

A-3-2-3 (B) Peak-Envelope Power (PEP) for SSB transmission is:
A a hypothetical measurement
B Peak-Envelope Voltage (PEV) multiplied by 0.707, squared and divided by the load resistance
C peak-voltage multiplied by peak current
D equal to the RMS power

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

A-3-2-2 (B) To compute one of the following, multiply the peak-envelope voltage by 0.707 to obtain the RMS value, square the result and divide by the load resistance. Which is the correct answer?
A power factor
B PEP
C PIV
D ERP

PIV = Peak Inverse Voltage (diode rating). ERP = Effective Radiated Power (transmit power minus line losses plus antenna gain). Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

A-3-2-1 (D) The power supplied to the antenna transmission line by a transmitter during an RF cycle at the highest crest of the modulation envelope is known as:
A mean power
B carrier power
C full power
D peak-envelope power

Peak Envelope Power (PEP) in SSB is defined as the power during one cycle on a modulation peak. Knowing that P = E squared divided by R, that peak voltage is half the peak-to-peak value and that RMS voltage is peak times 0.707 . PEP can be computed as ((peak voltage times 0.707) squared ) divided by R. When given peak-to-peak, first divide by 2 to obtain peak voltage.

A (grid) dip meter is a tunable oscillator with a feedback current meter.

They work best for parallel tuned circuits. The are fixed amplutuidwe with no moduilation. Interactio between the (tu ned) oscillator and the (resonant) circuit under test. Frequency must be checked with an external receiver or coutner. Only tells you presence or absence of resonance; no other quantities.

A-3-3-5 (D) A dip meter may not be used directly to:
A align transmitter-tuned circuits
B determine the frequency of oscillations
C align receiver-tuned circuits
D measure the value of capacitance or inductance

Key word: DIRECTLY. A ‘Dip Meter’ cannot measure inductance or capacitance DIRECTLY but if an unknown component is first attached to a known inductance or capacitance, the unknown value can be computed once the resonant frequency of the combination is obtained. The oscillator in a dip meter can usually be disabled to turn the instrument into an absorption wavemeter.

A-3-3-4 (B) A dip meter supplies the radio frequency energy which enables you to check:
A the adjustment of an inductor
B the resonant frequency of a circuit
C the calibration of an absorption-type wavemeter
D the impedance mismatch in a circuit

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

A-3-3-3 (C) What two ways could a dip meter be used in an amateur station?
A To measure antenna resonance and percentage modulation
B To measure resonant frequency of antenna traps and percentage modulation
C To measure resonant frequencies of antenna traps and to measure a tuned circuit resonant frequency
D To measure antenna resonance and impedance

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

A-3-3-2 (A) What does a dip meter do?
A It gives an indication of the resonant frequency of a circuit
B It measures transmitter output power accurately
C It measures field strength accurately
D It measures frequency accurately

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

A-3-3-1 (C) What is a dip meter?
A A marker generator
B A field-strength meter
C A variable frequency oscillator with metered feedback current
D An SWR meter

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

A-3-3-8 (C) A dip meter:
A may be used only with series tuned circuits
B accurately measures frequencies
C should be loosely coupled to the circuit under test
D should be tightly coupled to the circuit under test

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

A-3-3-11 (B) Which of the following is not a factor affecting the frequency accuracy of a dip meter?
A Over coupling
B Transmitter power output
C Hand capacity
D Stray capacity

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

AC voltmeter

A-3-1-8 (B) An AC voltmeter is calibrated to read the:
A peak value
B effective value
C peak-to-peak value
D average value

Voltmeters normally read the RMS (or effective) value; i.e., 0.707 times the peak value.

A-3-1-7 (D) AC voltmeter scales are usually calibrated to read:
A peak voltage
B instantaneous voltage
C average voltage
D RMS voltage

Voltmeters normally read the RMS (or effective) value; i.e., 0.707 times the peak value.

Signal generator

A-3-3-7 (C) What is a signal generator?
A A low-stability oscillator used to inject a signal into a circuit under test
B A high-stability oscillator which generates reference signals at exact frequency intervals
C A high-stability oscillator which can produce a wide range of frequencies and amplitudes
D A low-stability oscillator which sweeps through a range of frequencies

A ‘Signal Generator’ is an instrument capable of producing any of a wide range of frequencies (RF frequencies for radio work). Signal generators sometimes include a calibrated output attenuator so a given amplitude can be produced (e.g., a certain number of microvolts). Calibration in that case is only accurate if the generator feeds a circuit with the expected impedance. An instrument which “generates reference signals at exact frequency intervals” is a marker generator or crystal calibrator.

A-3-3-6 (B) The dial calibration on the output attenuator of a signal generator:
A reads half the true output when the attenuator is properly terminated
B reads accurately only when the attenuator is properly terminated
C always reads the true output of the signal generator
D reads twice the true output when the attenuator is properly terminated

A ‘Signal Generator’ is an instrument capable of producing any of a wide range of frequencies (RF frequencies for radio work). Signal generators sometimes include a calibrated output attenuator so a given amplitude can be produced (e.g., a certain number of microvolts). Calibration in that case is only accurate if the generator feeds a circuit with the expected impedance. An instrument which “generates reference signals at exact frequency intervals” is a marker generator or crystal calibrator.

A-3-3-10 (C) The dip meter is most directly applicable to:
A digital logic circuits
B series tuned circuits
C parallel tuned circuits
D operational amplifier circuits

A ‘Dip Meter’ is an instrument incorporating a Variable Frequency Oscillator whose resonant circuit can be placed near circuits to be tested. The oscillator’s activity is monitored with a built-in meter. When the external circuit’s resonant frequency matches the oscillator’s frequency, energy is absorbed and a dip is observed on the meter. It works best on parallel-tuned circuits. It can be used to test antenna traps and tuned circuits in static conditions (i.e., not operating). Overcoupling, hand capacity and stray capacity all introduce errors.

SINAD is a measurement applicable to FM receivers.

A-3-3-9 (C) Which two instruments are needed to measure FM receiver sensitivity for a 12 dB SINAD ratio (signal + noise + distortion over noise + distortion)?
A Oscilloscope and spectrum analyzer
B Receiver noise bridge and total harmonic distortion analyser
C Calibrated RF signal generator with FM tone modulation and total harmonic distortion (THD) analyzer
D RF signal generator with FM tone modulation and a deviation meter

The SINAD (signal + noise + distortion over noise + distortion) ratio takes the SNR (signal + noise over noise ratio) one step further by including distortion. A 12 dB SINAD ratio ensures that speech remains intelligible. Sensitivity expressed in those terms is the lowest RF level that will produce a usable message. The RF signal generator must be calibrated so the number of microvolts is precisely determined. Total Harmonic Distortion compares unwanted harmonic components added to the desired fundamental frequency, an audio tone in this instance.

``````SINAD = Signal + Noise + Distortion
---------------------------
Noise + Distortion``````

Microvolts for 12 dB quieting

A-6-3-5 (C) How is receiver sensitivity often expressed for UHF FM receivers?
A Noise Figure in decibels
B Overall gain in decibels
C RF level for 12 dB SINAD
D RF level for a given Bit Error Rate (BER)

The SINAD (signal + noise + distortion over noise + distortion) ratio takes the SNR (signal + noise over noise ratio) one step further by including distortion. A 12 dB SINAD ratio ensures that speech remains intelligible. Sensitivity expressed in those terms is the lowest RF level that will produce a usable message. The RF signal generator must be calibrated so the number of microvolts is precisely determined. Total Harmonic Distortion compares unwanted harmonic components added to the desired fundamental frequency, an audio tone in this instance.

Frequency counter

A-3-4-1 (C) What does a frequency counter do?
A It generates broad-band white noise for calibration
B It produces a reference frequency
C It makes frequency measurements
D It measures frequency deviation

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

A-3-4-2 (A) What factors limit the accuracy, frequency response and stability of a frequency counter?
A Time base accuracy, speed of the logic, and time base stability
B Time base accuracy, temperature coefficient of the logic and time base stability
C Number of digits in the readout, speed of the logic, and time base stability
D Number of digits in the readout, external frequency reference and temperature coefficient of the logic

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

A-3-4-3 (C) How can the accuracy of a frequency counter be improved?
A By using faster digital logic
B By improving the accuracy of the frequency response
C By increasing the accuracy of the time base
D By using slower digital logic

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

A-3-4-4 (D) If a frequency counter with a time base accuracy of +/- 0.1 PPM (parts per million) reads 146 520 000 Hz, what is the most that the actual frequency being measured could differ from that reading?
A 0.1 MHz
B 1.4652 Hz
C 1.4652 kHz
D 14.652 Hz

The reading can be off by as much as the time base accuracy in Parts Per Million times the measured frequency in megahertz: 1 part per million is one hertz per megahertz. [ In reality, an added error of plus or minus 1 count exists as the last digit displayed may have been rounded up or down. ]

A-3-4-5 (B) If a frequency counter, with a time base accuracy of 10 PPM (parts per million) reads 146 520 000 Hz, what is the most the actual frequency being measured could differ from that reading?
A 1465.2 kHz
B 1465.2 Hz
C 146.52 Hz
D 146.52 kHz

The reading can be off by as much as the time base accuracy in Parts Per Million times the measured frequency in megahertz: 1 part per million is one hertz per megahertz. [ In reality, an added error of plus or minus 1 count exists as the last digit displayed may have been rounded up or down. ]

A-3-4-6 (A) The clock in a frequency counter normally uses a:
A crystal oscillator
B self-oscillating Hartley oscillator
C mechanical tuning fork
D free-running multivibrator

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

A-3-4-7 (D) The frequency accuracy of a frequency counter is determined by:
A the size of the frequency counter
B type of display used in the counter
C the number of digits displayed
D the characteristics of the internal time-base generator

Two methods exist for determining frequency: counting the number of cycles in a fixed time interval or counting the time interval for some integer number of input cycles. In both cases, the instrument relies on an internal time base ( a crystal reference oscillator ). The accuracy of the measurement depends on the accuracy and stability (short and long-term) of the time base. Speed limitations in the logic circuitry limit the frequency range for a given instrument.

A marker generator produces a harmonic series of signals used to calibrate receiver dials.

A-3-4-8 (D) Which device relies on a stable low-frequency oscillator, with harmonic output, to facilitate the frequency calibration of receiver dial settings?
A Signal generator
B Harmonic calibrator
C Frequency counter
D Frequency-marker generator

A frequency marker generator (or crystal calibrator) is a relatively stable and precise oscillator rich in harmonics. It is typically used to calibrate analog station receivers by providing reference signals at known intervals on the dial. For an HF receiver, harmonics may be generated every 25, 50 or 100 kHz throughout the HF spectrum. For microwave work, harmonics of 144 MHz are useful for 432 MHz (3x), 1296 (9x), 2304 (16x), 3456 (24x), 5760 (40x), 10368 (72x) and 24192 MHz (168x). The “harmonic calibrator” is a bogus answer.

Zero-beat

A-3-4-9 (D) What is the traditional way of verifying the accuracy of a crystal calibrator?
A Compare the oscillator with your transmitter
B Use a dip-meter to determine the oscillator’s fundamental frequency
C Compare the oscillator with your receiver
D Zero-beat the crystal oscillator against a standard frequency station such as WWV

One way to calibrate a crystal calibrator (frequency marker generator) is to turn it on so it is heard while receiving a standard frequency station on shortwave. “Zero beating”, which can be done audibly, is bringing, through adjustment, the difference between two frequencies down to zero. When two RF signals (calibrator and reference station) are nearly at the same frequency, they produce a heterodyne (beat) which falls in the audio range; the lower the beat note, the closer the match.

A-3-4-11 (D) You want to calibrate your station frequency reference to the WWV signal on your receiver. The resulting beat tone must be:
A a combined frequency above both
B the mathematical mean of both frequencies
C at the highest audio frequency possible
D of a frequency as low as possible and with a period as long as possible

One way to calibrate a crystal calibrator (frequency marker generator) is to turn it on so it is heard while receiving a standard frequency station on shortwave. “Zero beating”, which can be done audibly, is bringing, through adjustment, the difference between two frequencies down to zero. When two RF signals (calibrator and reference station) are nearly at the same frequency, they produce a heterodyne (beat) which falls in the audio range; the lower the beat note, the closer the match. [ Station WWV broadcasts time and frequency information from Fort Collins, Colorado; it is maintained by the Physical Measurement Laboratory (PML) of the National Institute of Standards and Technology (NIST). ]

VXCO

A-3-4-10 (C) Out of the following oscillators, one is NOT, by itself, considered a high-stability reference:
A oven-controlled crystal oscillator (OCXO)
B GPS disciplined oscillator (GPSDO)
C voltage-controlled crystal oscillator (VCXO)
D temperature compensated crystal oscillator (TCXO)

Key word: NOT. The voltage-controlled crystal oscillator (VCXO) is always part of a larger ensemble, its stability is not controlled, but can be controlled by an external circuit. The temperature compensated crystal oscillator (TCXO) relies on a temperature sensor and some circuitry (analog or digital) to correct (compensate) the frequency. The oven-controlled crystal oscillator (OCXO) maintains the crystal at a precise and constant temperature well above ambient temperature. The GPS disciplined oscillator (GPSDO) uses a GPS (Global Positioning System) receiver to synchronize an oscillator with the highly accurate reference signals from the satellites.

Lissajous figures

Used to determine the harmonic relationship between signals at different frequencies.

``````  100 : 150
2 : 3``````

A-3-5-1 (A) If a 100 Hz signal is fed to the horizontal input of an oscilloscope and a 150 Hz signal is fed to the vertical input, what type of pattern should be displayed on the screen?
A A looping pattern with 3 horizontal loops, and 2 vertical loops
B A rectangular pattern 100 mm wide and 150 mm high
C An oval pattern 100 mm wide and 150 mm high
D A looping pattern with 100 horizontal loops and 150 vertical loops

Using the horizontal and vertical inputs on an oscilloscope, it is possible to compare the frequency of two sine waves or the phase relationship of two signals of equal frequency. The resulting pattern is called a “Lissajous Figure”. Apply a known signal to the horizontal input. The ratio of the number of loops on the horizontal edge to the number on the vertical edge equals the ratio of the vertical frequency Y divided by horizontal frequency X: the number of cycles the Y frequency covers during one horizontal cycle. [French mathematician Jules-Antoine Lissajous]

``````      5 : 2
100 kHz : 40 kHz``````

A-3-5-8 (D) A 100-kHz signal is applied to the horizontal channel of an oscilloscope. A signal of unknown frequency is applied to the vertical channel. The resultant wave form has 5 loops displayed vertically and 2 loops horizontally. The unknown frequency is:
A 20 kHz
B 50 kHz
C 30 kHz
D 40 kHz

Using the horizontal and vertical inputs on an oscilloscope, it is possible to compare the frequency of two sine waves or the phase relationship of two signals of equal frequency. The resulting pattern is called a “Lissajous Figure”. Apply a known signal to the horizontal input. The ratio of the number of loops on the horizontal edge to the number on the vertical edge equals the ratio of the vertical frequency Y divided by horizontal frequency X: the number of cycles the Y frequency covers during one horizontal cycle. In this example, the unknown frequency is two fifths the known frequency. [French mathematician Jules-Antoine Lissajous]

Also used to determine the phase relationship between signals of the same frequency.

Signals that are in-phase display as a diagonal line, and approach a circle as the move out of phase.

A-3-5-7 (A) When using Lissajous figures to determine phase differences, an indication of zero or 180 degrees is represented on the screen of an oscilloscope by:
A a diagonal straight line
B a horizontal straight line
C an ellipse
D a circle

Two in-phase signals of the same frequency applied to the X and Y amplifiers will grow exactly at the same rate: equal deflection on the X and Y create a trace at a 45 degree angle. Ponder these X and Y values: 1 and 1, 2 and 2, 3 and 3, etc. A precise 180 degree phase difference has a similar effect except the trace appears in the other two quadrants. Ponder these X and Y values: 1 and -1, 2 and -2, 3 and -3, etc. [other angles correspond to the sine function of the ratio of Y as the ellipse crosses the vertical centre axis to the maximum Y value.]

oscilloscopes

Time-base accuracy and amplifier bandwidth determine scope performance.

A-3-5-2 (B) What factors limit the accuracy, frequency response and stability of an oscilloscope?
A Tube face voltage increments and deflection amplifier voltages
B Accuracy of the time base and the linearity and bandwidth of the deflection amplifiers
C Deflection amplifier output impedance and tube face frequency increments
D Accuracy and linearity of the time base and tube face voltage increments

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

Frequency response (range) can be increased by using a faster timebase, and increasing the deflection amplifier

A-3-5-3 (B) How can the frequency response of an oscilloscope be improved?
A By using triggered sweep and a crystal oscillator for the timebase
B By increasing the horizontal sweep rate and the vertical amplifier frequency response
C By using a crystal oscillator as the time base and increasing the vertical sweep rate
D By increasing the vertical sweep rate and the horizontal amplifier frequency response

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

Dual trace is a thing

A-3-5-4 (D) You can use an oscilloscope to display the input and output of a circuit at the same time by:
A measuring the input on the X axis and the output on the Y axis
B measuring the input on the X axis and the output on the Z axis
C measuring the input on the Y axis and the output on the X axis
D utilizing a dual trace oscilloscope

A dual-trace oscilloscope has two distinct vertical channels. They can be used to take simultaneous measurements at different points in a circuit.

Not good for measuring FM deviation.

A-3-5-5 (C) An oscilloscope cannot be used to:
A measure DC voltage
B determine the amplitude of complex voltage wave forms
C determine FM carrier deviation directly
D measure frequency

Measuring frequency is a matter of using the horizontal sweep rate and number of divisions to determine the period of one cycle of the waveform. Measuring deviation requires a deviation meter.

Scope “bandwidth” is the highest frequency it can display.

A-3-5-6 (B) The bandwidth of an oscilloscope is:
A a function of the time-base accuracy
B the highest frequency signal the scope can display
C directly related to gain compression
D indirectly related to screen persistence

Similarly to the frequency counter, the accuracy and stability of the time base responsible for sweeping on the horizontal axis are paramount. The top frequency is limited by the highest horizontal sweep rate and bandwidth of the horizontal and vertical deflection amplifiers.

A-3-5-9 (A) An oscilloscope probe must be compensated:
A every time the probe is used with a different oscilloscope
B when measuring a sine wave
C through the addition of a high-value series resistor
D when measuring a signal whose frequency varies

“Probe compensation is the process of adjusting the probe’s RC network divider so that the probe maintains its attenuation ratio over the probe’s rated bandwidth. Most scopes have a square wave reference signal available on the front panel to use for compensating the probe. You can attach the probe tip to the probe compensation terminal and connect the probe to an input of the scope. Viewing the square wave reference signal, make the proper adjustments on the probe using a small screw driver so that the square waves on the scope screen look square” (Jae-yong Chang, Agilent Technologies). Given that the input circuitry cannot be identical between instruments, adjustment must be made every time a high-impedance probe is used with a different scope.

Best way to check CW & sideband signals, by displaying waveform.

A-3-5-10 (C) What is the best instrument to use to check the signal quality of a CW or single-sideband phone transmitter?
A A signal tracer and an audio amplifier
B A field-strength meter
C An oscilloscope
D A sidetone monitor

An oscilloscope of sufficient bandwidth permits visualizing the transmitter’s output. The sidetone monitor merely produces a tone when a CW transmitter is keyed. The field-strength meter reports relative field strength in proximity to antennas. The signal tracer permits troubleshooting audio circuitry.

Transmitter output checked using a sampler

A-3-5-11 (C) What is the best signal source to connect to the vertical input of an oscilloscope for checking the quality of a transmitted signal?
A The IF output of a monitoring receiver
B The audio input of the transmitter
C The RF output of the transmitter through a sampling device
D The RF signals of a nearby receiving antenna

The most accurate representation of a transmitted signal can be viewed at the RF output of a transmitter. The correct method supposes the use of a “sampler” in the transmission line. That way, proper impedance match is maintained, instruments are protected from large voltages and extraneous signals are minimized.

Voltmeters & Ammeters

Meters are based on a d’Arsonval movement. A moving coil is in series with two fixed coils on either side. Passing a current through all the coils, the fields tend to align - deflecting the moving coil.

The sensitivity of an ammeter is the amount of current that causes full-scale deflection.

A-3-6-4 (A) The sensitivity of an ammeter is an expression of:
A the amount of current causing full-scale deflection
B the resistance of the meter
C the loading effect the meter will have on a circuit
D the value of the shunt resistor

Ammeter sensitivity is the current needed for a full-scale deflection.

To decrease the sensitivity (increase the range), shunt resistors in parallel with the meter’s internal resistance are used.

Ammeters have a low overall resistance to reduce the impact on the circuit under test.

A-3-6-7 (D) The range of a DC ammeter can easily be extended by:
A connecting an external resistance in series with the internal resistance
B changing the internal inductance of the meter
C changing the internal capacitance of the meter to resonance
D connecting an external resistance in parallel with the internal resistance

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one.

A-3-6-1 (B) A meter has a full-scale deflection of 40 microamperes and an internal resistance of 96 ohms. You want it to read 0 to 1 mA. The value of the shunt to be used is:
A 40 ohms
B 4 ohms
C 24 ohms
D 16 ohms

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one. For example, making a 40 microamperes movement, with a 96 ohms internal resistance, read 1000 microamperes at full scale is a factor of 25: the shunt = 96 ohms divided by 24 (i.e., 25 minus 1) = 4 ohms. Method B: you could have computed voltage across the meter as 40 times 96 = 3840 microvolts ; and then, computed the shunt resistor as 3840 microvolts divided by 960 microamperes = 4 ohms.

A-3-6-9 (B) How can the range of an ammeter be increased?
A By adding resistance in series with the meter
B By adding resistance in parallel with the meter
C By adding resistance in series with the circuit under test
D By adding resistance in parallel with the circuit under test

Extending the range of an ammeter supposes placing a shunt resistor across the meter movement to divert part of the current: the shunt resistor = internal resistance divided by the multiplication factor from which we will first have deducted a quantity of one.

The sensitivity of voltmeters are expressed in ohms per volt, where the resistance refers to .

A-3-6-5 (A) Voltmeter sensitivity is usually expressed in ohms per volt. This means that a voltmeter with a sensitivity of 20 kilohms per volt would be a:
A 50 microampere meter
B 1 milliampere meter
C 50 milliampere meter
D 100 milliampere meter

Voltmeter sensitivity in ohms per volt: i.e., the full-scale reading times the sensitivity yields total instrument resistance. Given sensitivity, the current needed for full-scale voltmeter reading follows Ohm’s Law: e.g., a voltmeter with a sensitivity of 20 000 ohms per volt uses a 50 microampere movement ( I = 1 volt divided by 20 000 = 50 microamperes ).

A-3-6-6 (B) The sensitivity of a voltmeter, whose resistance is 150 000 ohms on the 150-volt range, is:
A 150 ohms per volt
B 1000 ohms per volt
C 100 000 ohms per volt
D 10 000 ohms per volt

Voltmeter sensitivity in ohms per volt: i.e., the full-scale reading times the sensitivity yields total instrument resistance. Given total resistance, sensitivity can be computed as resistance divided by full-scale voltage reading: e.g., a total of 150 000 ohms for a voltmeter reading 150 volts is a sensitivity of 1000 ohms per volt..

Voltmeters use large bias resistors in series with the movement to reduce the load of the meter on the circuit under test.

To increase the range of a voltmeter, a greater bias resistance is used to reduce the current flow to a value within the meter’s range.

A-3-6-3 (D) A voltmeter having a range of 150 volts and an internal resistance of 150 000 ohms is to be extended to read 750 volts. The required multiplier resistor would have a value of:
A 1 500 ohms
B 750 000 ohms
C 1 200 000 ohms
D 600 000 ohms

Extending the range of a voltmeter supposes placing a suitable multiplier resistor in series with the instrument: the multiplier resistance = total internal resistance times the multiplication factor from which we will first have deducted a quantity of one. For example, to turn a 150 volts meter, with an internal resistance of 150 000 ohms, to read 750 volts full scale is an increase of 5 times: the multiplier resistor = 150 000 times 4 (i.e., 5 minus 1) = 600 000 ohms. Method B: current for full-scale reading = 150 volts divided by 150 000 ohms = 1 milliampere. The total resistance that would allow 1 milliampere under 750 volts = 750 000 ohms. Deduct the internal resistance from this value to determine the multiplier resistor.

A-3-6-2 (C) A moving-coil milliammeter having a full-scale deflection of 1 mA and an internal resistance of 0.5 ohms is to be converted to a voltmeter of 20 volts full-scale deflection. It would be necessary to insert a:
A shunt resistance of 19 999.5 ohms
B shunt resistance of 19.5 ohms
C series resistance of 19 999.5 ohms
D series resistance of 1 999.5 ohms

Turning an ammeter into a voltmeter supposes inserting a suitable resistor in series with the instrument. For a current of 1 milliampere to flow under 20 volts, Ohm’s Law tells us that a total resistance of 20 000 ohms is required. Subtract the internal resistance from that number and the actual series resistance needed is 19 999.5 ohms.

A-3-6-8 (B) What happens inside a multimeter when you switch it from a lower to a higher voltage range?
A Resistance is added in parallel with the meter
B Resistance is added in series with the meter
C Resistance is reduced in series with the meter
D Resistance is reduced in parallel with the meter

Extending the range of a voltmeter supposes placing a suitable multiplier resistor in series with the instrument; the higher the series resistance, the higher the voltage needed to obtain full-scale deflection on the meter.

Wattmeters measure power.

Wattmeters inductively with a torroidal coil surroundnign the feedline. The sampled signal is rectified down to DC, and the resulting voltage/current shown on a meter.

RF wattmeters are connected at the transmitter output.

A-3-6-10 (D) Where should an RF wattmeter be connected for the most accurate readings of transmitter output power?
A One-half wavelength from the transmitter output
B One-half wavelength from the antenna feed point
C At the antenna feed point
D At the transmitter output connector

Measuring right at the transmitter output connector removes any line losses from the measurement.

They must have the same impedance as the rest of the signal chain; typically 50 ohms.

A-3-6-11 (D) At what line impedance do most RF wattmeters usually operate?
A 25 ohms
B 100 ohms
C 300 ohms
D 50 ohms

In-line RF wattmeters are designed to work with the most common Characteristic Impedance in radio work: 50 ohms.

{L07} Oscillators.

Oscillators depend on positive feedback to reinforce a signal at a single frequency.

A-5-1-9 (C) A circuit depending on positive feedback for its operation would be a:
A detector
B audio amplifier
C variable-frequency oscillator
D mixer

An oscillator, fixed or variable, is an amplifier with a positive feedback path from output to input used to start and maintain oscillation.

The exam deals with three types of oscillators. These are:

• The Hartley oscillator, which is distinguished by a tapped coil (inductive divider),
• the Colpitts oscillator, distinguished by a capacitive divider, and also
• the Pierce oscillator, which uses a crystal with capacitive feedback.

The Hartley and Colpitts oscillators both use a resonant LC tank circuit.

• The Hartley oscillator uses a tapped coil in the tank circuit.
• The Colpitts oscillator uses capacitive divider for feedback coupling.

In a Hartley oscillator, the feedback is coupled back in to the input through a tapped coil. A-5-1-1 (C) How is the positive feedback coupled to the input in a Hartley oscillator?
A Through link coupling
B Through a neutralizing capacitor
C Through a tapped coil
D Through a capacitive divider

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A Colpitts oscillator uses a capacitive divider. A-5-1-6 (D) Positive feedback from a capacitive divider indicates the oscillator type is:
A Pierce
B Hartley
C Miller
D Colpitts

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A-5-1-2 (B) How is positive feedback coupled to the input in a Colpitts oscillator?
A Through a link coupling
B Through a capacitive divider
C Through a tapped coil
D Through a neutralizing capacitor

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A Pierce oscillator replaces the LC tank circuit with a crystal. It uses series capacitive feedback coupling. A-5-1-11 (A) In an oscillator where positive feedback is provided through a capacitor in series with a crystal, that type of oscillator is a:
A Pierce
B Colpitts
C Hartley
D Franklin

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A-5-1-8 (D) In an oscillator circuit where positive feedback is obtained through a single capacitor in series with the crystal, the type of oscillator is:
A Colpitts
B Hartley
C Miller
D Pierce

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A-5-1-3 (A) How is positive feedback coupled to the input in a Pierce oscillator?
A Through capacitive coupling
B Through a neutralizing capacitor
C Through link coupling
D Through a tapped coil

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

Crystal oscillators are very stable narrow bandwidth, and high Q

Very frequency stable.

A-2-11-4 (C) The main advantage of a crystal oscillator over a tuned LC oscillator is:
A freedom from harmonic emissions
B simplicity
C much greater frequency stability
D longer life under severe operating use

Piezoelectric crystals behave like tuned circuits with an extremely high “Q” (“Quality Factor”, in excess of 25 000). Their accuracy and stability are outstanding.

Multiples of the crystal frequency can be exploited as overtone oscilator

A-2-11-8 (B) Crystals are sometimes used in a circuit which has an output close to an integral multiple of the crystal frequency. This circuit is called:
A a crystal ladder
B an overtone oscillator
C a crystal multiplier
D a crystal lattice

Key word: MULTIPLE. Crystals are capable of resonance either at a fundamental frequency depending on their physical dimensions or at overtone frequencies near odd-integer multiples (3rd, 5th, 7th, etc.) of the fundamental. In a filter, crystals are used at their fundamental frequencies; the crystal lattice filter and the crystal ladder filter are two typical configurations. Crystal oscillators can be designed to work on a fundamental or overtone resonance; crystals are manufactured accordingly.

Not a source of high power signals.

A-2-11-9 (B) Which of the following properties does not apply to a crystal when used in an oscillator circuit?
A Good frequency accuracy
B High power output
C Good frequency stability
D Very low noise because of high Q

The piezoelectric property of quartz is two-fold: apply mechanical stress to a crystal and it produces a small electrical field; subject quartz to an electrical field and the crystal changes dimensions slightly. Crystals are capable of resonance either at a fundamental frequency depending on their physical dimensions or at overtone frequencies near odd-integer multiples (3rd, 5th, 7th, etc.). Piezoelectric crystals can serve as filters because of their extremely high “Q” (> 25 000) or as stable, noise-free and accurate frequency references.

The purpose of an oscillator is to provide a stable signal that is free of frequency drift.

• Colpitts oscillators use large capacitors in the capacitance divider that exhibit comparatively little variation over temperature.

A-5-1-4 (D) Why is the Colpitts oscillator circuit commonly used in a VFO?
A It can be used with or without crystal lock-in
B The frequency is a linear function with load impedance
C It has high output power
D It is stable

Hartley = a tap on the inductor of the tuned circuit permits inserting positive feedback from output to input. Colpitts = the inductor tap of the Hartley is replaced by two series capacitors in a capacitive divider configuration. Relatively large capacitor values, when compared to the Hartley, mean less influence from internal capacitance changes in the device, hence stability. Pierce = derived from the Colpitts, a piezoelectric crystal replaces the tuned circuit. Capacitive coupling maintains oscillation.

A-5-1-5 (C) Why must a very stable reference oscillator be used as part of a phase-locked loop (PLL) frequency synthesizer?
A Any amplitude variations in the reference oscillator signal will prevent the loop from changing frequency
B Any amplitude variations in the reference oscillator signal will prevent the loop from locking to the desired signal
C Any phase variations in the reference oscillator signal will produce phase noise in the synthesizer output
D Any phase variations in the reference oscillator signal will produce harmonic distortion in the modulating signal

Key words: PHASE-LOCKED. As the name implies, a PLL synthesizer includes a voltage-controlled oscillator (VCO) whose output is constantly compared to a stable crystal reference. If the phase of the output signal begins to lead or lag the reference (a phase error), a correction is applied to the oscillator tuning. If the reference is noisy or subject to phase jitter, the output is similarly corrupted.

A-5-1-7 (D) In an RF oscillator circuit designed for high stability, the positive feedback is drawn from two capacitors connected in series. These two capacitors would most likely be:
A ceramic
B electrolytics
C Mylar
D silver mica

In an oscillator, stability is paramount. Silver Mica, NP0 (N-P-Zero, negative-positive-zero) ceramic and polystyrene capacitors are temperature-stable. The electrolytic is suitable as a filter or audio bypass. The plain ceramic capacitor is good for coupling or RF bypass.

{L09} Transmitters.

In the grounded grid triode amplifier

final tubes

A-5-2-2 (B) The purpose of using a centre-tap return connection on the secondary of transmitting tube’s filament transformer is to:
A obtain optimum power output
B prevent modulation of the emitted wave by the alternating current filament supply
C reduce the possibility of harmonic emissions
D keep the output voltage constant with a varying load

When the cathode is simply the filament (a “directly-heated cathode”), the voltage drop across the filament (e.g., 6.3 volts AC) is in series with the cathode DC reference voltage: as an example, while one side of the filament might be at a certain DC voltage, the other extremity is at some other value, a value influenced by the AC voltage impressed on the filament. Electron flow is affected by an AC variation, hum results.

In the grounded grid triode amplifier

As the name suggests, the grid (screen) is connected to ground. Similar to a common-base BJT configuration. enlarge

The plate is connected to the high voltage supply.

A-5-2-5 (C) In a grounded grid amplifier using a triode vacuum tube, the plate is connected to a radio frequency choke. The other end of the radio frequency choke connects to the:
A ground
B B- (bias)
C B+ (high voltage)
D filament voltage

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The input is applied to the Cathode

A-5-2-3 (A) In a grounded grid amplifier using a triode vacuum tube, the input signal is applied to:
A the cathode
B the plate
C the control grid
D the filament leads

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The plate (output) is connected via a (DC) blocking capacitor to the Pi network.

A-5-2-4 (B) In a grounded grid amplifier using a triode vacuum tube, the plate is connected to the pi-network through a:
A electrolytic capacitor
B blocking capacitor
C by-pass capacitor
D tuning capacitor

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The filament is bridged by an RF bypass caps.

A-5-2-9 (C) In a grounded grid amplifier using a triode vacuum tube, each side of the filament is connected to a capacitor whose other end is connected to ground. These are:
A electrolytic capacitors
B blocking capacitors
C by-pass capacitors
D tuning capacitors

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The plate voltage can be determined using the relation shiop betwwen voltage, current, and power; taking into account efficiency.

For example, P = V I so V = P / I

A-5-2-8 (D) In a grounded grid amplifier using a triode vacuum tube, what would be the approximate B+ voltage required for an output of 400 watts at 400 mA with approximately 50 percent efficiency?
A 500 volts
B 3000 volts
C 1000 volts
D 2000 volts

400 watts out at 50% efficiency supposes that 800 watts DC are needed. Power is voltage times current ; thus, voltage is power divided by current ; 800 watts divided by 0.4 ampere = 2000 volts.

The filament (heater) voltage is provided by a transformer.

A-5-2-7 (B) In a grounded grid amplifier using a triode vacuum tube, the secondary winding of a transformer is connected directly to the vacuum tube. This transformer provides:
A Screen voltage
B filament voltage
C B- (bias)
D B+ (high voltage)

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The cathode is negative-biased using the B- supply.

A-5-2-6 (A) In a grounded grid amplifier using a triode vacuum tube, the cathode is connected to a radio frequency choke. The other end of the radio frequency choke connects to the:
A B- (bias)
B ground
C filament voltage
D B+ (high voltage)

A grounded-grid amplifier runs with the grid at ground potential. The cathode is above RF ground and serves as the input. A DC bias is applied to the cathode via an RF choke. Positive voltage (B+) is supplied to the plate via an RF choke. The plate is the output, a blocking capacitor passes the RF out to the matching network. A transformer provides AC filament voltage. The heater (in an indirectly-heated cathode tube) is bypassed to ground so radiofrequency does not stray out on the filament supply lines. [ If a tube is directly-heated (no separate cathode), filament voltage is brought through a filament choke. The side of the choke connected to the transformer is bypassed to ground with two capacitors. ]

The final matching network matches the impedance of the tubes to the impedance of the antenna system for efficient power transfer.

A-5-2-1 (D) The output tuning controls on a transmitter power amplifier with an adjustable PI network:
A allow switching to different antennas
B reduce the possibility of cross-modulation in adjunct receivers
C are involved with frequency multiplication in the previous stage
D allow efficient transfer of power to the antenna

Tube power amplifiers always include a matching network to match the high impedance of the tube to the antenna system impedance. As always, impedance match is all about maximum power transfer.

simple 2-stage CW transmitter

A CW transmimtter consists of an oscillator stage and a power amplifier stage.

A-5-1-10 (A) An apparatus with an oscillator and a class C amplifier would be:
A a two-stage CW transmitter
B a fixed-frequency single-sideband transmitter
C a two-stage frequency-modulated transmitter
D a two-stage regenerative receiver

A simple two-stage CW transmitter comprises an oscillator and a Class-C power amplifier. A transformer at the output of the oscillator serves the dual purpose of tuned circuit and coupling to the next stage. The DC supply to the final amplifier is bypassed to ground with a capacitor and decoupled through an RF choke so RF is kept out of the supply. The first stage is an oscillator.

The second stage is a power amplifier.

A-5-3-3 (C) In a simple 2 stage CW transmitter, the transistor in the second stage would act as:
A the master oscillator
B an audio oscillator
C a power amplifier
D a frequency multiplier

A simple two-stage CW transmitter comprises an oscillator and a Class-C power amplifier. A transformer at the output of the oscillator serves the dual purpose of tuned circuit and coupling to the next stage. The DC supply to the final amplifier is bypassed to ground with a capacitor and decoupled through an RF choke so RF is kept out of the supply.

Note the LC pi-network after the amplifer stage. This is a low-pass filter.

A-5-3-2 (A) In a simple 2 stage CW transmitter, current to the collector of the transistor in the class C amplifier stage flows through a radio frequency choke (RFC) and a tapped inductor. The RFC, on the tapped inductor side, is also connected to grounded capacitors. The purpose of the RFC and capacitors is to:
A form a low-pass filter
B provide negative feedback
C form a key-click filter
D form a RF-tuned circuit

A simple two-stage CW transmitter comprises an oscillator and a Class-C power amplifier. A transformer at the output of the oscillator serves the dual purpose of tuned circuit and coupling to the next stage. The DC supply to the final amplifier is bypassed to ground with a capacitor and decoupled through an RF choke so RF is kept out of the supply.

there is a tuned circuit that is inductively coupled

A-5-3-1 (C) In a simple 2 stage CW transmitter circuit, the oscillator stage and the class C amplifier stage are inductively coupled by a RF transformer. Another role of the RF transformer is to:
A provide the necessary feedback for oscillation
B act as part of a balanced mixer
C be part of a tuned circuit
D act as part of a pi filter

A simple two-stage CW transmitter comprises an oscillator and a Class-C power amplifier. A transformer at the output of the oscillator serves the dual purpose of tuned circuit and coupling to the next stage. The DC supply to the final amplifier is bypassed to ground with a capacitor and decoupled through an RF choke so RF is kept out of the supply.

A buffer stage may be incorporated between the oscillator and PA stages to eliminate chirp, which is caused by perturbation of the oscillator frequency as its loaded down by the PA.

A-5-3-4 (A) An advantage of keying the buffer stage in a transmitter is that:
A changes in oscillator frequency are less likely
B key clicks are eliminated
C the radiated bandwidth is restricted
D high RF voltages are not present

Keying a subsequent stage provides the oscillator with a fairly constant load (isolation) and allows it to run continuously for better stability.

Furthermore, tuned circuit filters may be used between stages to prevent the passage of unwanted harmonics.

A-5-2-11 (C) Harmonics produced in an early stage of a transmitter may be reduced in a later stage by:
A greater input to the final stage
B transistors instead of tubes
C tuned circuit coupling between stages
D larger value coupling capacitors

Key words: STAGES. Resonant circuits in the coupling between stages help convey only the operating frequency. Larger coupling capacitors would pass the harmonics more readily.

In a power amplifier, there may be some feedback from the output to the input. This feedback may cause “parasitic” oscillation in the absence of a neutralizing circuit to cancel it out.

Unwanted, parasitic oscillations may occur if the feedback is resonant at these frequencies.

A-5-3-9 (A) Parasitic oscillations would tend to occur mostly in:
A RF power output stages
B high gain audio output stages
C high voltage rectifiers
D mixer stages

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A-5-3-8 (B) Parasitic oscillations are usually generated due to:
A a mismatch between power amplifier and transmission line
B accidental resonant frequencies in the power amplifier
C harmonics from some earlier multiplier stage
D excessive drive or excitation to the power amplifier

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A-5-3-11 (B) Parasitic oscillations in an RF power amplifier may be caused by:
A excessive harmonic production
B lack of neutralization
C overdriven stages
D poor voltage regulation

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A neutralizing circuit cancels unwanted feedback signals coupled from the output of power amplifier to the input.

A-5-3-7 (C) What is the reason for neutralizing the final amplifier stage of a transmitter?
A To cut off the final amplifier during standby periods
B To keep the carrier on frequency
C To eliminate parasitic oscillations
D To limit the modulation index

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A-5-3-6 (D) What does a neutralizing circuit do in an RF amplifier?
A It eliminates AC hum from the power supply
B It reduces incidental grid modulation
C It controls differential gain
D It cancels the effects of positive feedback

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A-5-3-5 (D) As a power amplifier is tuned, what reading on its grid current meter indicates the best neutralization?
A Minimum grid current
B Maximum grid current
C A maximum change in grid current as the output circuit is changed
D A minimum change in grid current as the output circuit is changed

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

A-5-3-10 (D) Why is neutralization necessary for some vacuum-tube amplifiers?
A To reduce grid-to-cathode leakage
B To cancel AC hum from the filament transformer
C To reduce the limits of loaded Q
D To cancel oscillation caused by the effects of interelectrode capacitance

Undesired positive feedback in an RF amplifier causes parasitic oscillations: the amplifier becomes an oscillator. Inter-electrode capacitance (e.g., plate-to-grid), coupling from output to input, stray inductance or capacitance can start up oscillations. Neutralization is the process of cancelling positive-feedback paths. To test a tube amplifier for parasitic oscillations, connect nothing to the input and output terminals, apply DC power, monitor grid and plate current while slowly varying the controls on the output tuning network; if grid current develops or plate current changes, oscillations are present.

safety

A-5-2-10 (B) After you have opened a VHF power amplifier to make internal tuning adjustments, what should you do before you turn the amplifier on?
A Remove all amplifier shielding to ensure maximum cooling
B Be certain all amplifier shielding is fastened in place
C Make sure that the power interlock switch is bypassed so you can test the amplifier
D Be certain no antenna is attached so that you will not cause any interference

Running a VHF or UHF power amplifier without shielding presents a safety risk in terms of RF exposure.

SSB

balanced modulator

The balmod suppresses the carrier, and produces double-sideband.

A-5-4-1 (C) What type of signal does a balanced modulator produce?
A Full carrier
B Single sideband, suppressed carrier
C Double sideband, suppressed carrier
D FM with balanced deviation

One method of producing SSB is a Balanced Modulator followed by a filter. The modulator takes in a fixed-frequency RF signal and mixes it with audio from the speech amplifier. The modulator is said to be balanced because the two original inputs are not present at the output: carrier suppression has taken place. Present, however, are a lower and upper sideband. A subsequent filter selects one of the sidebands to complete the creation of a single sideband suppressed-carrier signal. Note that there is no RF output when no audio is applied.

A-5-4-3 (B) Carrier suppression in a single-sideband transmitter takes place in:
A the frequency multiplier stage
B the balanced modulator stage
C the carrier decouple stage
D the mechanical filter

One method of producing SSB is a Balanced Modulator followed by a filter. The modulator takes in a fixed-frequency RF signal and mixes it with audio from the speech amplifier. The modulator is said to be balanced because the two original inputs are not present at the output: carrier suppression has taken place. Present, however, are a lower and upper sideband. A subsequent filter selects one of the sidebands to complete the creation of a single sideband suppressed-carrier signal. Note that there is no RF output when no audio is applied.

A following filter remoevs the unwanted sideband.

A-5-4-2 (C) How can a single-sideband phone signal be produced?
A By using a loop modulator followed by a mixer
B By using a reactance modulator followed by a mixer
C By using a balanced modulator followed by a filter
D By driving a product detector with a DSB signal

One method of producing SSB is a Balanced Modulator followed by a filter. The modulator takes in a fixed-frequency RF signal and mixes it with audio from the speech amplifier. The modulator is said to be balanced because the two original inputs are not present at the output: carrier suppression has taken place. Present, however, are a lower and upper sideband. A subsequent filter selects one of the sidebands to complete the creation of a single sideband suppressed-carrier signal. Note that there is no RF output when no audio is applied.

ssb power

Going from AM to SSB results in improvements both in terms of effective transmit gain (power more concentrated) and receive gain (narrower bandwidth signal includes less noise)

A-5-4-4 (C) Transmission with SSB, as compared to conventional AM transmission, results in:
A a greater bandpass requirement in the receiver
B 3 dB gain in the transmitter
C 6 dB gain in the transmitter and 3 dB gain in the receiver
D 6 dB gain in the receiver

Under noisy conditions, SSB can bring up to a 9 dB improvement over an AM signal of the same peak power. In AM, the Peak Envelope Power present in one of the two sidebands equals one fourth the carrier power: e.g., a 100-watt AM transmitter only packs 25 watts PEP in each sideband. SSB concentrates all the available power in one sideband alone: a 4 to 1 improvement or 6 dB. Using half the bandwidth on SSB reception, permits taking in only half of the noise at the receiver, an additional 3 dB improvement.

The carrier must be suppressed, which in practice means 40 dB below (a factor of 10,000).

A-5-4-10 (B) How much is the carrier suppressed below peak output power in a single-sideband phone transmission?
A At least 60 dB
B At least 40 dB
C No more than 20 dB
D No more than 30 dB

“Most well-designed balanced modulators can provide between 30 and 50 dB of carrier suppression. …The filter roll-off can be used to obtain an additional 20 dB of carrier suppression.” (ARRL Handbook 1985)

two-tone test

Two-tone test is a test of linearity for SSB transmissions.

A-5-4-6 (C) What kind of input signal is used to test the amplitude linearity of a single-sideband phone transmitter while viewing the output on an oscilloscope?
A An audio-frequency square wave
B Normal speech
C Two audio-frequency sine waves
D An audio-frequency sine wave

A two-tone test permits verifying the linearity of an SSB transmitter. The test requires a generator producing two low-distortion non-harmonically related audio sine waves of equal amplitude. The frequencies must fall within the normal transmitter audio passband: e.g., 700 and 1900 Hz. A sample of the transmitter’s output is observed on an oscilloscope while the tones are fed into the microphone input. Feeding an SSB transmitter with two equal-amplitude steady audio tones produces two equal-amplitude RF signals: total power is thus twice the power present in each RF signal.

A-5-4-9 (A) What measurement can be made of a single-sideband phone transmitter’s amplifier by performing a two-tone test using an oscilloscope?
A Its linearity
B Its frequency deviation
C Its percent of carrier phase shift
D Its percent of frequency modulation

A two-tone test permits verifying the linearity of an SSB transmitter. The test requires a generator producing two low-distortion non-harmonically related audio sine waves of equal amplitude. The frequencies must fall within the normal transmitter audio passband: e.g., 700 and 1900 Hz. A sample of the transmitter’s output is observed on an oscilloscope while the tones are fed into the microphone input. Feeding an SSB transmitter with two equal-amplitude steady audio tones produces two equal-amplitude RF signals: total power is thus twice the power present in each RF signal.

The peak power of the output transmission is the integral of power spectrum. So, in the context of the two-tone test, where both input signals are the same power, the output is twice the power of either input signal.

A-5-4-5 (C) The peak power output of a single-sideband transmitter, when being tested by a two-tone generator is:
A one-half of the RF peak output power of any of the tones
B one-quarter of the RF peak output power of any of the tones
C twice the RF power output of any of the tones
D equal to the RF peak output power of any of the tones

A two-tone test permits verifying the linearity of an SSB transmitter. The test requires a generator producing two low-distortion non-harmonically related audio sine waves of equal amplitude. The frequencies must fall within the normal transmitter audio passband: e.g., 700 and 1900 Hz. A sample of the transmitter’s output is observed on an oscilloscope while the tones are fed into the microphone input. Feeding an SSB transmitter with two equal-amplitude steady audio tones produces two equal-amplitude RF signals: total power is thus twice the power present in each RF signal.

The test uses two non-harmonically-related tones. The output is observed in the time domain, on an oscilloscope.

A-5-4-7 (D) When testing the amplitude linearity of a single-sideband transmitter what audio tones are fed into the microphone input and on what kind of kind of instrument is the output observed?
A Two harmonically related tones are fed in, and the output is observed on an oscilloscope
B Two harmonically related tones are fed in, and the output is observed on a distortion analyzer
C Two non-harmonically related tones are fed in, and the output is observed on a distortion analyzer
D Two non-harmonically related tones are fed in, and the output is observed on an oscilloscope

A two-tone test permits verifying the linearity of an SSB transmitter. The test requires a generator producing two low-distortion non-harmonically related audio sine waves of equal amplitude. The frequencies must fall within the normal transmitter audio passband: e.g., 700 and 1900 Hz. A sample of the transmitter’s output is observed on an oscilloscope while the tones are fed into the microphone input. Feeding an SSB transmitter with two equal-amplitude steady audio tones produces two equal-amplitude RF signals: total power is thus twice the power present in each RF signal.

Any two audio tones may be used, but they must be within the transmitter audio passband, and should not be harmonically related

A-5-4-8 (A) What audio frequencies are used in a two-tone test of the linearity of a single-sideband phone transmitter?
A Any two audio tones may be used, but they must be within the transmitter audio passband, and should not be harmonically related
B 20 Hz and 20 kHz tones must be used
C 1200 Hz and 2400 Hz tones must be used
D Any two audio tones may be used, but they must be within the transmitter audio passband, and must be harmonically related

A two-tone test permits verifying the linearity of an SSB transmitter. The test requires a generator producing two low-distortion non-harmonically related audio sine waves of equal amplitude. The frequencies must fall within the normal transmitter audio passband: e.g., 700 and 1900 Hz. A sample of the transmitter’s output is observed on an oscilloscope while the tones are fed into the microphone input. Feeding an SSB transmitter with two equal-amplitude steady audio tones produces two equal-amplitude RF signals: total power is thus twice the power present in each RF signal.

ALC is a type of RF signal processing that prevents “flat-topping” of the transmitted signal due to over-driving the power amplifier.

A-5-4-11 (B) What is meant by “flat topping” in a single-sideband phone transmission?
A The transmitter’s carrier is properly suppressed
B Signal distortion caused by excessive drive
C Signal distortion caused by insufficient collector current
D The transmitter’s automatic level control is properly adjusted

Flattening of the peaks is an extreme form of distortion where the output of the transmitter is incapable of reproducing the original pattern of the audio input on voice peaks. This is generally caused by excessive audio input to the transmitter: too much audio causes the amplifier stage to exceed its linear operation range. The purpose of the ALC (Automatic Level Control) is to prevent overdrive.

A-5-7-1 (A) Maintaining the peak RF output of a SSB transmitter at a relatively constant level requires a circuit called the:
A automatic level control (ALC)
B automatic gain control (AGC)
C automatic output control (AOC)
D automatic volume control (AVC)

Automatic Level Control (ALC) serves to prevent overdriving an amplifier. The ALC circuit samples the envelope (peak) of the RF output to develop a DC control voltage used to control the gain of an earlier stage. AGC and AVC are receiver circuits.

A-5-7-11 (B) Automatic Level Control (ALC) is another name for:
A AF clipping
B RF compression
C AF compression
D RF clipping

Clipping places a hard limit on voltage swings. Compression is a reduction in gain when signal exceed a certain threshold. The ALC circuit samples the envelope (peak) of the RF output and produces a DC control voltage used to control the gain of an earlier stage when the output reaches a certain level.

Analog Signal processing

Audio signal processing takes place after the audio amp, and before the modulator.

A-5-6-9 (A) In most modern FM transmitters, to produce a better sound, a compressor and a clipper are placed:
A between the audio amplifier and the modulator
B between the multiplier and the PA
C between the modulator and the oscillator
D in the microphone circuit, before the audio amplifier

In this context, compression and clipping are AUDIO processes aimed at maintaining high average deviation without exceeding a given peak value. Two answers can be readily scratched as they pertain to radiofrequency (RF) paths. The microphone circuit is not suitable as the audio level at that point is too low for a simple clipper.

The consequence of

Clipping, or peak-limiting is a simple example of AF signal processing.

peak limiting

A-5-7-8 (B) Which of the following is not a method used for peak limiting, in a signal processor?
A AF clipping
B Frequency clipping
C RF clipping
D Compression

The expression “peak limiting” entails limiting the amplitude. Compression, AF clipping and RF clipping are valid operations. There is no such thing as frequency clipping.

A-5-7-9 (B) What is the undesirable result of AF clipping in a speech processor?
A Reduction in peak amplitude
B Increased harmonic distortion
C Reduced average power
D Increased average power

Audio frequency clipping abruptly stops voltage excursions at a certain level. This gives the audio a square wave appearance; square waves are rich in harmonics. AF clippers need to be followed by a low-pass filter to prevent harmonics from entering modulation stages. You may also eliminate the bad answers: “reduction in peak amplitude” is the object of clipping, “increased average power” is a result of clipping (softer passages are no longer dwarfed by the peaks), “increased average power” simply contradicts the previous answer.

A-5-7-10 (C) Which description is not correct? You are planning to build a speech processor for your transceiver. Compared to AF clipping, RF clipping:
A is more expensive to implement
B is more difficult to implement
C is easier to implement
D has less distortion

Working at radio-frequencies is evidently more difficult and thus more expensive than dealing with audio frequencies. RF clipping is generally presumed to induce less distortion because any harmonics generated through clipping automatically end up outside the passband of subsequent filters. At audio frequencies, harmonics of the lower speech frequencies fall within the audio passband and can muddle the audio signal.

Speech compression makes the amplitude response non-linear; bringing up the bottom rather than limiting the top.

A-5-7-2 (C) Speech compression associated with SSB transmission implies:
A a lower signal-to-noise ratio
B circuit level instability
C full amplification of low level signals and reducing or eliminating amplification of high level signals
D full amplification of high level signals and reducing or eliminating signals amplification of low level

Audio compression maintains a high voice level despite variations in the voice signal incoming from a microphone. To produce a high average output without exceeding a certain peak value, low level signals need to be amplified while high level signals are passed along with little or no gain. [ compression is the automatic reduction of gain as the signal level increases beyond a pre-set level known as the threshold. ]

A-5-7-7 (B) Which principle is not associated with analog signal processing?
A Clipping
B Frequency division
C Compression
D Bandwidth limiting

Key words: NOT ASSOCIATED WITH ANALOG. Compression, bandwidth limiting and clipping can all be performed as analog processes. Frequency division requires a numerical computation.

FM

For the transmitted RF signal, the three important parameters to optimize together are

• power,
• deviation, and
• stability.

A-5-6-10 (D) Three important parameters to be verified in an FM transmitter are:
A distortion, bandwidth and sideband power
B modulation, pre-emphasis and carrier suppression
C frequency stability, de-emphasis and linearity
D power, frequency deviation and frequency stability

Stability is paramount in all transmitters, frequency deviation ultimately determines bandwidth while linearity (absence of distortion) minimizes out-of-channel emissions. Carrier Suppression is a concern with SSB, pre-emphasis (FM transmitter) and de-emphasis (FM receiver) are simple resistor-capacitor networks.

When there is no modulation, the carrier sits at the at the Center frequency.

A-5-5-5 (B) When the transmitter is not modulated, or the amplitude of the modulating signal is zero, the frequency of the carrier is called its:
A modulating frequency
B centre frequency
C frequency deviation
D frequency shift

Centre Frequency is the transmitter output frequency in the absence of modulation. Frequency deviation and frequency shift both are synonyms for the offset in carrier frequency caused by modulation at a given instant. Modulating frequency relates to the audio frequency used for modulation.

Deviation meters multiply the modulation frequency by something called the modulation index.

A-5-5-8 (A) Some types of deviation meters work on the principle of:
A a carrier null and multiplying the modulation frequency by the modulation index
B detecting the frequencies in the sidebands
C the amplitude of power in the sidebands
D a carrier peak and dividing by the modulation index

Certain Modulation Index values cause nulls at the centre carrier frequency: e.g., the Bessel function returns zero for the carrier component at Modulation Indices of 2.4048, 5.5201 or 8.6537 . Detecting a carrier null permits determining deviation as Modulation Index times modulating frequency. For example, with a tone of 905 hertz and deviation set at 4996 hertz (nearly 5 kHz), a null in the carrier will be observed because 4996 Hz deviation for a tone of 905 Hz is a Modulation Index of 5.52 . An all-mode receiver with a sharp filter permits observing the carrier component. The procedure could be used to set a transmitter or calibrate a home-made deviation meter.
‘ Expressed to 8 decimal places, the first three indices producing Bessel nulls for J(0) are 2.40482556, 5.52007811 and 8.65372791.

A-5-5-9 (B) When using some deviation meters, it is important to know:
A pass-band of the IF filter
B modulating frequency and the modulation index
C modulation index
D modulating frequency

Certain Modulation Index values cause nulls at the centre carrier frequency: e.g., the Bessel function returns zero for the carrier component at Modulation Indices of 2.4048, 5.5201 or 8.6537 . Detecting a carrier null permits determining deviation as Modulation Index times modulating frequency. For example, with a tone of 905 hertz and deviation set at 4996 hertz (nearly 5 kHz), a null in the carrier will be observed because 4996 Hz deviation for a tone of 905 Hz is a Modulation Index of 5.52 . An all-mode receiver with a sharp filter permits observing the carrier component. The procedure could be used to set a transmitter or calibrate a home-made deviation meter.

modulation index is the proportion by which the modulating frequency displaces the carrier frequency. It is given by:

``````    f_dev
M = -----
f_mod``````

For example, if a deviation of 3000 Hz results from a modulating frequency of 1000 Hz, this is a modulation index of
`3000/1000 = 3`

A-5-5-1 (D) In an FM phone signal having a maximum frequency deviation of 3000 Hz either side of the carrier frequency, what is the modulation index, when the modulating frequency is 1000 Hz?
A 0.3
B 3000
C 1000
D 3

Deviation = the amount of frequency shift, at a given instant, from the centre carrier frequency (e.g., plus or minus 5 kHz). Modulation Index = the ratio of deviation to modulating frequency for a particular audio frequency (both being expressed in the same units): e.g., 3 kHz deviation with 1 kHz audio represents a Modulation Index of 3. Deviation Ratio = the ratio of maximum deviation to the maximum modulating frequency: e.g. maximum deviation of 5 kHz with a highest modulating frequency of 3 kHz is a Deviation Ratio of 1.66 .

For example, if there’s a deviation of 6 kHz resulting from a modulating frequency of 2 kHz, the modulation index is

`6/2 = 3` times greater than than the modulating frequency

A-5-5-2 (A) What is the modulation index of an FM phone transmitter producing an instantaneous carrier deviation of 6 kHz when modulated with a 2 kHz modulating frequency?
A 3
B 0.333
C 2000
D 6000

Deviation = the amount of frequency shift, at a given instant, from the centre carrier frequency (e.g., plus or minus 5 kHz). Modulation Index = the ratio of deviation to modulating frequency for a particular audio frequency (both being expressed in the same units): e.g., 3 kHz deviation with 1 kHz audio represents a Modulation Index of 3. Deviation Ratio = the ratio of maximum deviation to the maximum modulating frequency: e.g. maximum deviation of 5 kHz with a highest modulating frequency of 3 kHz is a Deviation Ratio of 1.66 .

Deviation ratio is basically the same thing, but calculated at the maximum extent of either:

``````     dev_max
DR = -------
mod_max``````

For example:

If the maximum deviation is 5 kHz, and the maximum modulating signal is 3 kHz, then the deviation ratio is

5/3 = about 1.5

A-5-5-3 (A) What is the deviation ratio of an FM phone transmitter having a maximum frequency swing of plus or minus 5 kHz and accepting a maximum modulation rate of 3 kHz?
A 1.66
B 60
C 0.16
D 0.6

Deviation = the amount of frequency shift, at a given instant, from the centre carrier frequency (e.g., plus or minus 5 kHz). Modulation Index = the ratio of deviation to modulating frequency for a particular audio frequency (both being expressed in the same units): e.g., 3 kHz deviation with 1 kHz audio represents a Modulation Index of 3. Deviation Ratio = the ratio of maximum deviation to the maximum modulating frequency: e.g. maximum deviation of 5 kHz with a highest modulating frequency of 3 kHz is a Deviation Ratio of 1.66 .

The deviation ratio if the maximum deviation is 7.5 kHz and the maximum modulating frequency is 3.5 kHz, then the deviation ratio is

7.5/3.5 = about 2

A-5-5-4 (D) What is the deviation ratio of an FM phone transmitter having a maximum frequency swing of plus or minus 7.5 kHz and accepting a maximum modulation rate of 3.5 kHz?
A 0.47
B 47
C 0.214
D 2.14

Deviation = the amount of frequency shift, at a given instant, from the centre carrier frequency (e.g., plus or minus 5 kHz). Modulation Index = the ratio of deviation to modulating frequency for a particular audio frequency (both being expressed in the same units): e.g., 3 kHz deviation with 1 kHz audio represents a Modulation Index of 3. Deviation Ratio = the ratio of maximum deviation to the maximum modulating frequency: e.g. maximum deviation of 5 kHz with a highest modulating frequency of 3 kHz is a Deviation Ratio of 1.66 .

The deviation is dependent on the amplitude of the modulating frequency.

A-5-5-6 (C) In an FM transmitter system, the amount of deviation from the centre frequency is determined solely by the:
A amplitude and the frequency of the modulating frequency
B modulating frequency and the amplitude of the centre frequency
C amplitude of the modulating frequency
D frequency of the modulating frequency

In Frequency Modulation, the amplitude of the modulation is translated into the importance of the deviation, the modulation frequency is reflected in the rhythm of the deviation.

FM produces sidebands

A-5-5-7 (A) Any FM wave with single-tone modulation has:
A an infinite number of sideband frequencies
B two sideband frequencies
C four sideband frequencies
D one sideband frequency

Unlike AM where a single modulating frequency creates only a pair of side frequencies (one on each side of the carrier), FM creates an infinite number of side frequency pairs; the Modulation Index influences the amplitude of the side frequencies through a mathematical relation known as a Bessel Function. The number of side frequencies with significant amplitude determines the required bandwidth. For certain Modulation Index values, there is zero energy at the centre frequency; the energy is then totally found in the side frequencies.

Counterintuitively, a phase modulator is used to achieve frequency modulation of a power amplifier.

A-5-6-5 (A) What type of circuit varies the tuning of an amplifier tank circuit to produce FM signals?
A A phase modulator
B A balanced modulator
C A double balanced mixer
D An audio modulator

Two methods exist to produce Frequency Modulation. The Direct Method supposes forcing deviation directly on the oscillator; deviation is then multiplied along with the oscillator frequency up to the operating frequency. Phase Modulation is an indirect method whereby the phase of the signal is affected (i.e., retarding/advancing) in step with the modulation by varying a reactance on a stage other than the oscillator.

emphasis and de-emphasis

A-6-4-1 (B) What audio shaping network is added at an FM receiver to restore proportionally attenuated lower audio frequencies?
A A heterodyne suppressor
B A de-emphasis network
C A pre-emphasis network
D An audio prescaler

With “true” FM, deviation is independent of modulating frequency, actual deviation is determined solely by the modulating amplitude. With Phase Modulation, deviation depends on the amount of phase shift and its rapidity, increasing modulating frequency results in proportionally more deviation even if amplitude is held constant. Because commercial standards were based on Phase Modulation, an FM transmitter requires an artificial boost in high frequency response so that PM and FM sound the same at the receiver. A pre-emphasis network tailors the frequency response in the FM transmitter. De-emphasis is employed in the receiver to restore a flat audio response.

FM bandwidth, deviation

A-5-5-10 (D) What is the significant bandwidth of an FM-phone transmission having a +/- 5-kHz deviation and a 3-kHz modulating frequency?
A 8 kHz
B 5 kHz
C 3 kHz
D 16 kHz

Carson’s Rule permits estimating the bandwidth of an FM signal: bandwidth equals twice the sum of deviation + modulating frequency, in this example, 5 + 3 = 8, twice 8 = 16. [ Mathematician and engineer John R. Carson (1887-1940) had predicted the approximate bandwidth of an FM signal circa 1922. ]

A-5-5-11 (D) What is the frequency deviation for a 12.21-MHz reactance-modulated oscillator in a +/- 5-kHz deviation, 146.52-MHz FM-phone transmitter?
A +/- 12 kHz
B +/- 5 kHz
C +/- 41.67 Hz
D +/- 416.7 Hz

In this example, the frequency multiplication ratio between oscillator and output is 12 ( 146.52 divided by 12.21 = 12 ). Hence, the oscillator needs only be shifted by 416.7 Hz, i.e., 5000 Hz divided by 12.

‘ — — — —

Audio shaping called pre-emphasis is used on FM transmitters.

A-5-6-6 (A) What audio shaping network is added at an FM transmitter to attenuate the lower audio frequencies?
A A pre-emphasis network
B An audio prescaler
C A heterodyne suppressor
D A de-emphasis network

With direct FM, deviation is independent of modulating frequency, actual deviation is determined solely by the modulating amplitude.

With Phase Modulation, deviation depends on the amount of phase shift and its rapidity, increasing modulating frequency results in proportionally more deviation even if amplitude is held constant. Because commercial standards were based on Phase Modulation, an FM transmitter requires an artificial boost in high frequency response so that PM and FM sound the same at the receiver. A pre-emphasis network tailors the frequency response in the FM transmitter. De-emphasis is employed in the receiver to restore a flat audio response.

A-5-6-8 (D) The characteristic difference between a phase modulator and a frequency modulator is:
A the centre frequency
B de-emphasis
C frequency inversion
D pre-emphasis

With direct FM, deviation is independent of modulating frequency, actual deviation is determined solely by the modulating amplitude. With Phase Modulation, deviation depends on the amount of phase shift and its rapidity, increasing modulating frequency results in proportionally more deviation even if amplitude is held constant. Because commercial standards were based on Phase Modulation, an FM transmitter requires an artificial boost in high frequency response so that PM and FM sound the same at the receiver. A pre-emphasis network tailors the frequency response in the FM transmitter. De-emphasis is employed in the receiver to restore a flat audio response.

The simplest receiver consists of a tuned front-end and a detector. An example is a crystal receiver.

A tuned-RF receiver - old - cascaded amplifiers, all operating at the input RF frequency

A heterodyne receiver (also called a direct conversion design) couples a local oscillator into the front end of a basic receiver; this allows unmodulated (continuous wave) carriers to be detected the their resulting AF beat tone.

All transmitters are now continuous wave; do spark-gap transmitters behave differently when received by crystal sets?

The word “heterodyne” is an archaic 1920s marketing term involving the roots hetero, meaning different; and dyne, evoking automatic transformation. In this context, the word alludes to the two different frequencies present, and the mixed between them.

In this context, the oscillatoris called a beat frequency oscillator or BFO.

A-6-2-2 (A) A superheterodyne receiver designed for SSB reception must have a beat-frequency oscillator (BFO) because:
A the suppressed carrier must be replaced for detection
B it phases out the unwanted sideband signal
C it reduces the pass-band of the IF stages
D it beats with the receiver carrier to produce the missing sideband

The Beat Frequency Oscillator feeds the Product Detector for CW and SSB detection. Mixing the Intermediate Frequency with the BFO signal in the Product Detector produces an audio output. In Single Sideband, it is said to “reinsert the carrier” as it recreates a reference at the exact frequency at which the carrier, suppressed at the transmitter, would have appeared out of the Intermediate Frequency chain.

A-6-2-5 (B) The BFO is off-set slightly (500 - 1 500 Hz) from the incoming signal to the detector. This is required:
A to protect the incoming signal from interference
B to beat with the incoming signal
C to pass the signal without interruption
D to provide additional amplification

The Beat Frequency Oscillator feeds the Product Detector for CW and SSB detection. Mixing the Intermediate Frequency with the BFO signal in the Product Detector produces an audio output. In Single Sideband, it is said to “reinsert the carrier” as it recreates a reference at the exact frequency at which the carrier, suppressed at the transmitter, would have appeared out of the Intermediate Frequency chain.

In this context, direct refers to the intelligence carried by the RF signal being converted directly from RF to AF.

The super-heterodyne uses a (tunable) (somewhat-selective) RF filter stage called a per-selector, the output of which is mixed with the output of a local oscillator, to down-convert to a lower frequency called the intermediate frequency. This is followed by an even more selective IF filter.

A single-conversion superheterodyn incorporates a single IF stage; a double-conversion uses two IF stages.

The local oscillator mixes with the input frequency; the result must be near the IF frequency

``f_LO +/- f_in = f_IF``

For example:

``````16 +/- x = 9

so

-x = 16 - 9 = 7

but also
x = 9 - 16 = -7

or

x =

x = 16 - 9 = 7``````

A-6-1-9 (A) A single conversion receiver with a 9 MHz IF has a local oscillator operating at 16 MHz. The frequency it is tuned to is:
A 7 MHz
B 16 MHz
C 21 MHz
D 9 MHz

In a superheterodyne receiver, injection from the Local Oscillator can be above or below the operating frequency. There could be two answers: Local Oscillator minus Intermediate Frequency or Local Oscillator plus Intermediate Frequency.

image rejection

The process of mixing the wideband input signal with the local oscillator, and accepting a given IF passband, results in the reception of image signals; which are due to the nature of the mixing transformation.

A pre-selector is a passband filter in front of the RF amp that increases the empty gap between the IF band and the image band.

The location of the IF passband determines the location of the image band. Graphical methods are recommended here.

A-6-5-1 (C) What part of a superheterodyne receiver determines the image rejection ratio of the receiver?
A AGC loop
B IF filter
C RF amplifier pre-selector
D Product detector

The Image is the other frequency that can successfully mix with the Local Oscillator and produce an output out of the Mixer at the Intermediate frequency. Selectivity ahead of the Mixer must be employed to prevent that signal from reaching the Mixer.

A-6-5-6 (D) Which of the following is an important reason for using a VHF intermediate frequency in an HF receiver?
A To provide a greater tuning range
B To tune out cross-modulation distortion
C To prevent the generation of spurious mixer products
D To move the image response far away from the filter passband

Whether injection from the Local Oscillator is above or below the operating frequency, the Image Frequency is always separated from the operating frequency by twice the Intermediate Frequency. A very high Intermediate Frequency moves the Image well out of the preselector bandpass.

double-conversion

The incorporate two cascaded IF stages, each fed by a dedicated local oscillator.

A-6-1-10 (D) A double conversion receiver designed for SSB reception has a beat frequency oscillator and:
A one IF stage and one local oscillator
B two IF stages and three local oscillators
C two IF stages and one local oscillator
D two IF stages and two local oscillators

Key words: DOUBLE CONVERSION. Double conversion entails two Mixers, two Local Oscillators and two Intermediate Frequency chains.

A-6-1-11 (B) The advantage of a double conversion receiver over a single conversion receiver is that it:
A produces a louder audio signal
B suffers less from image interference for a given front end sensitivity
C does not drift off frequency
D is a more sensitive receiver

Two frequencies, one above and one below the Local Oscillator frequency, can produce a mixing result at the Intermediate Frequency. The one resulting in the unwanted product is the Image Frequency. With a low Intermediate Frequency, selectivity and gain are easier to achieve but image rejection suffers. With a high Intermediate Frequency, image rejection is facilitated but selectivity is more difficult to achieve. A Double-conversion receiver deals with image rejection with an initial conversion and restores selectivity with a subsequent down-conversion. Two conversions, however, expose the designer to twice the risk of spurious responses due to spurious oscillations in local oscillators.

The first IF stage is optimized for image rejection, and the second IF stage is optimized for selectivity.

A-6-1-6 (C) In a dual-conversion superheterodyne receiver what are the respective aims of the first and second conversion:
A selectivity and dynamic range
B image rejection and noise figure
C image rejection and selectivity
D selectivity and image rejection

The first conversion to a high IF places the image frequency far away from the operating frequency so it can be optimally rejected by the front-end filtering. The second conversion to a low IF performs the traditional function of ensuring selectivity to protect the receiver from adjacent channels.

Recall that mixing is a non-linear process, with many moving parts. With two mixing stages, the resulting mixing products may be all over the place, waiting to suck.

When it works, we call it mixing; when it does not work, we call it intermodulation.

A-6-1-5 (A) A multiple conversion superheterodyne receiver is more susceptible to spurious responses than a single-conversion receiver because of the:
A additional oscillators and mixing frequencies involved in the design
B poorer selectivity in the IF caused by the multitude of frequency changes
C greater sensitivity introducing higher levels of RF to the receiver
D AGC being forced to work harder causing the stages concerned to overload

Two frequencies, one above and one below the Local Oscillator frequency, can produce a mixing result at the Intermediate Frequency. The one resulting in the unwanted product is the Image Frequency. With a low Intermediate Frequency, selectivity and gain are easier to achieve but image rejection suffers. With a high Intermediate Frequency, image rejection is facilitated but selectivity is more difficult to achieve. A Double-conversion receiver deals with image rejection with an initial conversion and restores selectivity with a subsequent down-conversion. Two conversions, however, expose the designer to twice the risk of spurious responses due to spurious oscillations in local oscillators.

A-6-1-3 (D) One of the greatest advantages of the double-conversion over the single-conversion receiver is that it:
A is much more stable
B is much more sensitive
C produces a louder signal at the output
D greater reduction of image interference for a given front end selectivity

Two frequencies, one above and one below the Local Oscillator frequency, can produce a mixing result at the Intermediate Frequency. The one resulting in the unwanted product is the Image Frequency. With a low Intermediate Frequency, selectivity and gain are easier to achieve but image rejection suffers. With a high Intermediate Frequency, image rejection is facilitated but selectivity is more difficult to achieve. A Double-conversion receiver deals with image rejection with an initial conversion and restores selectivity with a subsequent down-conversion. Two conversions, however, expose the designer to twice the risk of spurious responses due to spurious oscillations in local oscillators.

The front-end consists of an (optional) rf amplifier, and the first mixer

A-6-1-7 (B) Which stage of a receiver has its input and output circuits tuned to the received frequency?
A The detector
B The RF amplifier
C The local oscillator
D The audio frequency amplifier

Key word: TUNED. Of all the stages listed, only one runs at the operating frequency: the radio-frequency amplifier.

A-6-2-8 (C) In a superheterodyne receiver without an RF amplifier, the input to the mixer stage has a variable capacitor in parallel with an inductance. The variable capacitor is for:
A tuning the beat-frequency oscillator
B tuning both the antenna and the local oscillator
C tuning the receiver preselector to the reception frequency
D tuning both the antenna and the beat-frequency oscillator

In the absence of a radio-frequency amplifier, the input to the Mixer must be an antenna tuning circuit.

A-6-2-10 (D) Which two stages in a superheterodyne receiver have input tuned circuits tuned to the same frequency?
A IF and local oscillator
B RF and IF
C RF and local oscillator
D RF and first mixer

Through an elimination process, only one answer makes sense. The input and the output of the radio-frequency amplifier run at the operating frequency. The output of the RF amplifier constitutes the input to the Mixer.

A-6-3-3 (A) How much gain should be used in the RF amplifier stage of a receiver?
A Sufficient gain to allow weak signals to overcome noise generated in the first mixer stage
B As much gain as possible, short of self-oscillation
C It depends on the amplification factor of the first IF stage
D Sufficient gain to keep weak signals below the noise of the first mixer stage

The Radio-Frequency Amplifier should only introduce enough gain to override the internal noise of the subsequent Mixer. Too much gain will degrade Dynamic Range; Dynamic Range is broadly defined as a ratio between the strongest signals that can be tolerated near the passband and the “Minimum Discernible Signal”.

A-6-3-4 (B) What is the primary purpose of an RF amplifier in a receiver?
A To provide most of the receiver gain
B To improve the receiver noise figure
C To vary the receiver image rejection by using the AGC
D To develop the AGC voltage

The Intermediate Frequency chain is responsible for the selectivity and a large part of the gain. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

IF stage

A-6-3-2 (A) Which of the following is a purpose of the first IF amplifier stage in a receiver?
A To improve selectivity and gain
B To tune out cross-modulation distortion
C To increase dynamic response
D To improve noise figure performance

The Intermediate Frequency chain is responsible for the selectivity and a large part of the gain. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

A-6-1-4 (A) In a communications receiver, a crystal filter would be located in the:
A IF circuits
B local oscillator
C audio output stage
D detector

The Intermediate Frequency chain is responsible for most of the selectivity. Crystal filters or mechanical filters can be used at the Intermediate Frequency. Digital Signal Processing (DSP) is used in modern receivers.

A-6-1-2 (C) What factors should be considered when selecting an intermediate frequency?
A Interference to other services
B Cross-modulation distortion and interference
C Image rejection and responses to unwanted signals
D Noise figure and distortion

Two frequencies, one above and one below the Local Oscillator frequency, can produce a mixing result at the Intermediate Frequency. The one resulting in the unwanted product is the Image Frequency. With a low Intermediate Frequency, selectivity and gain are easier to achieve but image rejection suffers. With a high Intermediate Frequency, image rejection is facilitated but selectivity is more difficult to achieve. A Double-conversion receiver deals with image rejection with an initial conversion and restores selectivity with a subsequent down-conversion. Two conversions, however, expose the designer to twice the risk of spurious responses due to spurious oscillations in local oscillators.

A-6-2-4 (C) If the incoming signal to the mixer is 3 600 kHz and the first IF is 9 MHz, at which one of the following frequencies would the local oscillator (LO) operate?
A 10 600 kHz
B 21 600 kHz
C 5 400 kHz
D 3 400 kHz

In a superheterodyne receiver, injection from the Local Oscillator can be above or below the operating frequency. There could be two answers: Intermediate Frequency plus operating frequency or Intermediate Frequency minus operating frequency.

A-6-2-3 (D) The first mixer in the receiver mixes the incoming signal with the local oscillator to produce:
A an audio frequency
B a radio frequency
C a high frequency oscillator (HFO) frequency
D an intermediate frequency

The superheterodyne concept is based on converting the operating frequency to a fixed Intermediate Frequency: the Mixer performs that function by combining the output of the tuneable RF amplifier with the Local Oscillator signal to feed the fixed-tuned Intermediate Frequency chain.

mixer

A-6-2-9 (A) What receiver stage combines a 14.25-MHz input signal with a 13.795-MHz oscillator signal to produce a 455-kHz intermediate frequency (IF) signal?
A Mixer
B BFO
C VFO
D Multiplier

The superheterodyne concept is based on converting the operating frequency to a fixed Intermediate Frequency: the Mixer performs that function by combining the output of the tuneable RF amplifier with the Local Oscillator signal to feed the fixed-tuned Intermediate Frequency chain.

A-6-2-11 (C) The mixer stage of a superheterodyne receiver:
A acts as a buffer stage
B demodulates SSB signals
C produces an intermediate frequency
D produces spurious signals

The superheterodyne concept is based on converting the operating frequency to a fixed Intermediate Frequency: the Mixer performs that function by combining the output of the tuneable RF amplifier with the Local Oscillator signal to feed the fixed-tuned Intermediate Frequency chain.

Local oscillator

A-6-2-7 (C) In a superheterodyne receiver, a stage before the IF amplifier has a variable capacitor in parallel with a trimmer capacitor and an inductance. The variable capacitor is for:
A tuning of the beat-frequency oscillator (BFO)
B tuning both the antenna and the LO
C tuning of the local oscillator (LO)
D tuning both the antenna and the BFO

Two stages may require tuning ahead of the Intermediate Frequency amplifier: the preselector and the high-frequency oscillator, commonly known as the Local Oscillator. As the question alludes to one trimmer and one inductance, only one circuit can be tuned.

A-6-2-6 (D) It is very important that the oscillators contained in a superheterodyne receiver are:
A sensitive and selective
B stable and sensitive
C selective and spectrally pure
D stable and spectrally pure

Oscillators need to be free of drift regardless of voltage and temperature variations or mechanical vibrations. Spectral purity is the absence of harmonics, other spurious oscillations or noise; purity limits spurious responses in the subsequent mixing processes.

superhet

A-6-2-1 (A) The mixer stage of a superheterodyne receiver is used to:
A change the frequency of the incoming signal to that of the IF
B allow a number of IF frequencies to be used
C remove image signals from the receiver
D produce an audio frequency for the speaker

The superheterodyne concept is based on converting the operating frequency to a fixed Intermediate Frequency: the Mixer performs that function by combining the output of the tuneable RF amplifier with the Local Oscillator signal to feed the fixed-tuned Intermediate Frequency chain.

A-6-1-8 (B) Which stage of a superheterodyne receiver lies between a tuneable stage and a fixed tuned stage?
A Local oscillator
B Mixer
C Radio frequency amplifier
D Intermediate frequency amplifier

The superheterodyne concept is based on converting the operating frequency to a fixed Intermediate Frequency: the Mixer performs that function by combining the output of the tuneable RF amplifier with the Local Oscillator signal to feed the fixed-tuned Intermediate Frequency chain.

A-6-1-1 (C) What are the advantages of the frequency conversion process in a superheterodyne receiver?
A Automatic soft-limiting and automatic squelching
B Automatic squelching and increased sensitivity
C Increased selectivity and optimal tuned circuit design
D Automatic detection in the RF amplifier and increased sensitivity

Down-converting the operating frequency to a lower Intermediate Frequency facilitates selectivity: for example, 0.6% of 455 kilohertz is 2.7 kHz, 0.6% of 3.7 megahertz is 22 kHz.

noise floor

A-6-3-1 (A) What is meant by the noise floor of a receiver?
A The weakest signal that can be detected above the receiver internal noise
B The weakest signal that can be detected under noisy atmospheric conditions
C The minimum level of noise that will overload the receiver RF amplifier stage
D The amount of noise generated by the receiver local oscillator

A receiver’s “Noise Floor” is the power level at which an incoming signal exhibits a Signal-To-Noise ratio of zero decibel: that is, the signal power equals the internal noise power level. Noise Floor is evaluated while measuring “Minimum Discernible Signal (MDS)”. The “Noise Figure” of a receiver is a comparison of Signal-to-Noise ratio at the input and Signal-to-Noise ratio at the output; it assesses the degradation in Signal-to-Noise ratio caused by added noise. A low Noise Figure suggests that little noise was added internally and is a hallmark of sensitivity. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

A-6-3-7 (A) The lower the receiver noise figure becomes, the greater will be the receiver’s _:
A sensitivity
B rejection of unwanted signals
C selectivity
D stability

The “Noise Figure” of a receiver is a comparison of Signal-to-Noise ratio at the input and Signal-to-Noise ratio at the output; it assesses the degradation in Signal-to-Noise ratio caused by added noise. A low Noise Figure suggests that little noise was added internally and is a hallmark of sensitivity. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

A-6-3-8 (D) The noise generated in a receiver of good design originates in the:
A detector and AF amplifier
B BFO and detector
C IF amplifier and detector
D RF amplifier and mixer

The “Noise Figure” of a receiver is a comparison of Signal-to-Noise ratio at the input and Signal-to-Noise ratio at the output; it assesses the degradation in Signal-to-Noise ratio caused by added noise. A low Noise Figure suggests that little noise was added internally and is a hallmark of sensitivity. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

A-6-3-9 (C) Why are very low noise figures relatively unimportant for a high frequency receiver?
A The use of SSB and CW on the HF bands overcomes the noise
B Regardless of the front end, the succeeding stages when used on HF are very noisy
C External HF noise, man-made and natural, are higher than the internal noise generated by the receiver
D Ionospheric distortion of the received signal creates high noise levels

Below 30 megahertz, the antenna picks-up atmospheric noise and man-made noise at levels far more important than internal noise. As the frequency of operation rises, those types of noise become less prevalent. On Ultra High Frequencies (UHF) and above, the internal noise becomes the limiting factor in receiving weak signals. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver: for weak signal work on 2 metres and up, more attention must be placed on reducing internal noise in the front-end.

dynamic range

A-6-3-6 (C) What is the term used for the decibel difference (or ratio) between the largest tolerable receiver input signal (without causing audible distortion products) and the minimum discernible signal (sensitivity)?
A Stability
B Noise figure
C Dynamic range
D Design parameter

Dynamic Range is broadly defined as a ratio between the strongest signals that can be tolerated near the passband and the “Minimum Discernible Signal”. The “Noise Figure” of a receiver is a comparison of Signal-to-Noise ratio at the input and Signal-to-Noise ratio at the output; it assesses the degradation in Signal-to-Noise ratio caused by added noise. A low Noise Figure suggests that little noise was added internally and is a hallmark of sensitivity. The front-end of the receiver, where signals are weakest, is responsible for the noise performance of a receiver.

A-6-3-10 (B) The term which relates specifically to the amplitude levels of multiple signals that can be accommodated during reception is called:
A noise figure
B dynamic range
C AGC
D cross-modulation index

Dynamic Range is broadly defined as a ratio between the strongest signals that can be tolerated near the passband and the “Minimum Discernible Signal”. Other related notions include Blocking Dynamic Range and Intermodulation Dynamic Range. Blocking Dynamic Range measures how much of a single strong off-channel signal can be tolerated while receiving a weak signal, it is a measure of desensitization or immunity to overload. Intermodulation Dynamic Range verifies how strong two off-channel signals can be without spurious responses being generated in the receiver, it is a measure of resistance to intermodulation.

A-6-5-10 (B) Poor dynamic range of a receiver can cause many problems when a strong signal appears within or near the front-end bandpass. Which of the following is NOT caused as a direct result?
A Cross-modulation
B Feedback
C Desensitization
D Intermodulation

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

selectivity

A-6-3-11 (B) Normally, front-end selectivity is provided by the resonant networks both before and after the RF stage in a superheterodyne receiver. This whole section of the receiver is often referred to as the:
A pass-selector
B preselector
C preamble
D preamplifier

A preselector is a tuned stage which passes a desired range of signals to a receiver: it ensures a certain preliminary selection. It may or may not be amplified; in other words, active or passive.

product detector

A-6-4-2 (C) What does a product detector do?
A It amplifies and narrows band-pass frequencies
B It detects cross-modulation products
C It mixes an incoming signal with a locally generated carrier
D It provides local oscillations for input to a mixer

The Beat Frequency Oscillator feeds the Product Detector for CW and SSB detection. Mixing the Intermediate Frequency with the BFO signal in the Product Detector produces an audio output. In Single Sideband, it is said to “reinsert the carrier” as it recreates a reference at the exact frequency at which the carrier, suppressed at the transmitter, would have appeared out of the Intermediate Frequency chain.

distortion

A-6-4-3 (B) Distortion in a receiver that only affects strong signals usually indicates a defect in or mis-adjustment of the:
A RF amplifier
B automatic gain control (AGC)
C IF amplifier
D AF amplifier

“Distortion that affects only strong signals is the normal symptom of AGC (Automatic Gain Control) failure.” (ARRL Handbook 1985)

AGC

A-6-4-4 (C) In a superheterodyne receiver with automatic gain control (AGC), as the strength of the signal increases, the AGC:
A distorts the signal
B introduces limiting
C reduces the receiver gain
D increases the receiver gain

The AGC (Automatic Gain Control) circuit reduces receiver gain as signal strength increases. AGC can be “IF-derived”, some say “RF-Derived” (a slight misnomer), by sampling the output of the last Intermediate Frequency stage or “AF-derived” by sampling the output of the detector. The resulting control voltage is applied to the Intermediate Frequency amplifiers and, sometimes, Radio-Frequency amplifier.

A-6-4-7 (B) The overall output of an AM/CW/SSB receiver can be adjusted by means of manual controls on the receiver or by use of a circuit known as:
A automatic load control
B automatic gain control
C automatic frequency control
D inverse gain control

The AGC (Automatic Gain Control) circuit reduces receiver gain as signal strength increases. AGC can be “IF-derived”, some say “RF-Derived” (a slight misnomer), by sampling the output of the last Intermediate Frequency stage or “AF-derived” by sampling the output of the detector. The resulting control voltage is applied to the Intermediate Frequency amplifiers and, sometimes, Radio-Frequency amplifier.

A-6-4-8 (A) AGC voltage is applied to the:
A RF and IF amplifiers
B AF and IF amplifiers
C RF and AF amplifiers
D detector and AF amplifiers

The AGC (Automatic Gain Control) circuit reduces receiver gain as signal strength increases. AGC can be “IF-derived”, some say “RF-Derived” (a slight misnomer), by sampling the output of the last Intermediate Frequency stage or “AF-derived” by sampling the output of the detector. The resulting control voltage is applied to the Intermediate Frequency amplifiers and, sometimes, Radio-Frequency amplifier.

A-6-4-9 (C) AGC is derived in a receiver from one of two circuits. Depending on the method used, it is called:
A IF derived or RF derived
B detector derived or audio derived
C IF derived or audio derived
D RF derived or audio derived

The AGC (Automatic Gain Control) circuit reduces receiver gain as signal strength increases. AGC can be “IF-derived”, some say “RF-Derived” (a slight misnomer), by sampling the output of the last Intermediate Frequency stage or “AF-derived” by sampling the output of the detector. The resulting control voltage is applied to the Intermediate Frequency amplifiers and, sometimes, Radio-Frequency amplifier.

A-6-4-10 (D) Which two variables primarily determine the behaviour of an automatic gain control (AGC) loop?
A Blanking level and slope
B Slope and bandwidth
C Clipping level and hang time
D Threshold and decay time

The AGC threshold is the level in the monitored circuit at which the AGC circuit begins to reduce gain. The AGC decay time determines how quickly gain is restored once the strong signal disappears.

detector

A-6-4-5 (B) The amplified IF signal is applied to the __ stage in a superheterodyne receiver:
A LO
B detector
C RF amplifier
D audio output

Remember your Basic Qualification? The Detector follows the Intermediate Frequency amplifier.

A-6-4-6 (C) The low-level output of a detector is:
A fed directly to the speaker
B applied to the RF amplifier
C applied to the AF amplifier
D grounded via the chassis

Remember your Basic Qualification? The Audio Amplifier follows the Detector.

A-6-4-11 (B) What circuit combines signals from an IF amplifier stage and a beat-frequency oscillator (BFO), to produce an audio signal?
A A VFO circuit
B A product detector circuit
C An AGC circuit
D A power supply circuit

The Beat Frequency Oscillator feeds the Product Detector for CW and SSB detection. Mixing the Intermediate Frequency with the BFO signal in the Product Detector produces an audio output. In Single Sideband, it is said to “reinsert the carrier” as it recreates a reference at the exact frequency at which the carrier, suppressed at the transmitter, would have appeared out of the Intermediate Frequency chain.

desensitization

A-6-5-2 (A) What is the term for the reduction in receiver sensitivity caused by a strong signal near the received frequency?
A Desensitization
B Cross-modulation interference
C Squelch gain rollback
D Quieting

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

A-6-5-3 (D) What causes receiver desensitization?
A Squelch gain adjusted too high
B Squelch gain adjusted too low
C Audio gain adjusted too low
D Strong near frequency signals

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

A-6-5-4 (B) What is one way receiver desensitization can be reduced?
A Increase the transmitter audio gain
B Use a cavity filter
C Decrease the receiver squelch gain
D Increase the receiver bandwidth

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

intermodulation

A-6-5-5 (A) What causes intermodulation in an electronic circuit?
A Nonlinear circuits or devices
B Too little gain
C Positive feedback
D Lack of neutralization

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

A-6-5-7 (B) Intermodulation interference is produced by:
A the mixing of more than one signal in the first or second intermediate frequency amplifiers of a receiver
B the mixing of two or more signals in the front-end of a superheterodyne receiver
C the interaction of products from high-powered transmitters in the area
D the high-voltage stages in the final amplifier of an amplitude or frequency-modulated transmitter

Desensitization is a symptom of front-end overload where a strong adjacent off-channel signal provokes a drop in receiver sensitivity. The only cure for desensitization is to keep the offending signal out of the receiver. Other manifestations of front-end overload are intermodulation and cross-modulation where strong signals push the RF amplifier or Mixer into non-linear operation resulting in spurious responses.

A-6-5-11 (B) Which of these measurements is a good indicator of VHF receiver performance in an environment of strong out-of-band signals?
A Intermediate frequency rejection ratio
B Two-tone Third-Order IMD Dynamic Range, 10 MHz spacing
C Third-Order Intercept Point
D Blocking Dynamic Range

“The FM Two-tone, third-order dynamic range, 10-MHz offset … is a wide-band dynamic-range test on VHF equipment, using two strong signals just outside the amateur band (usually the abode of nearby pager transmitters). (…) This test is a good indicator of relative IMD performance”. (RFI - Intermodulation, ARRL, Ed Hare, W1RFI)

frequency stability

A-6-5-8 (A) Which of the following is NOT a direct cause of instability in a receiver?
A Dial display accuracy
B Mechanical rigidity
C Feedback components
D Temperature variations

Key words: NOT A DIRECT CAUSE. Temperature variations, voltage variations and movements due to mechanical stresses will cause changes in frequency. The selection of feedback components, notably their temperature coefficient, is paramount for stability. Dial accuracy is not instability per se.

A-6-5-9 (B) Poor frequency stability in a receiver usually originates in the:
A mixer
B local oscillator and power supply
C detector
D RF amplifier

Stability is the ability to stay on frequency despite other variations. The Local Oscillator indirectly sets the operating frequency. Temperature variations, voltage variations and movements due to mechanical stresses will cause changes in frequency.

{L13} Digital Signal Processing.

Digital Signal Processors consist of an analog to digital converter, a mathematical transform, a digital to analog converter and a low pass filter.

A-5-7-6 (D) What do you call the circuit which employs an analog to digital converter, a mathematical transform, a digital to analog converter and a low pass filter?
A Digital formatter
B Mathematical transformer
C Digital transformer
D Digital signal processor

Digital Signal Processing first transforms an analog signal to digital via an analog-to-digital converter (ADC), performs mathematical operations to filter or demodulate, for example, and recreates an analog signal through a digital-to-analog converter (DAC). Aliasing is a situation where input frequencies above the sampling rate get erroneously reproduced as lower frequencies; the situation is prevented through the use of an anti-aliasing filter (a low-pass filter).

There is no such thing as an aliasing amplifier.

A-5-7-3 (B) Which of the following functions is not included in a typical digital signal processor?
A Mathematical transform
B Aliasing amplifier
C Analog to digital converter
D Digital to analog converter

Digital Signal Processing first transforms an analog signal to digital via an analog-to-digital converter (ADC), performs mathematical operations to filter or demodulate, for example, and recreates an analog signal through a digital-to-analog converter (DAC). Aliasing is a situation where input frequencies above the sampling rate get erroneously reproduced as lower frequencies; the situation is prevented through the use of an anti-aliasing filter (a low-pass filter).

In a DAC or ADC, 8 bits gives you 2^8 = 256 levels.

A-5-7-4 (D) How many bits are required to provide 256 discrete levels, or a ratio of 256:1?
A 6 bits
B 16 bits
C 4 bits
D 8 bits

256 levels in binary data spans the range 0 to 255. Add the value of each bit in an 8-bit binary number, starting with the lowest: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255.

Adding a bit doubles the range; which, for amplitudes, is equivalent to 6 dB.

A-5-7-5 (D) Adding one bit to the word length, is equivalent to adding __ dB to the dynamic range of the digitizer:
A 1 dB
B 4 dB
C 3 dB
D 6 dB

The number of bits determines the maximum number of distinct values available to represent a signal. For each sample, the analog-to-digital converter selects the closest value to describe the instantaneous value of the input signal. The difference between the real value and its numerical representation is a quantization error: plus or minus half a bit, i.e., 1 bit total. In digital terms, Signal-to-Noise Ratio and Dynamic Range are synonyms specifying the ratio between the largest possible sample value to the quantization error. Dynamic Range for fixed-point processors can be approximated by the data word size: 6 dB times number of bits. Each bit doubles the value, doubling voltage is 6 dB.

{L06} Basic Digital Techniques.

An OR Gate gives you a 1 when any of the inputs are 1.

A-2-10-2 (C) What is an OR gate?
A A circuit that produces logic “1” at its output if all inputs are logic “0”
B A circuit that produces a logic “0” at its output if any input is logic “1”
C A circuit that produces a logic “1” at its output if any input is logic “1”
D A circuit that produces a logic “0” at its output if all inputs are logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

A NOR gate is an OR gate with the output reversed. You get a zero if any of the inputs are 1.

A-2-10-3 (B) What is a NOR gate?
A A circuit that produces a logic “1” at its output if some but not all of its inputs are logic “1”
B A circuit that produces a logic “0” at its output if any or all inputs are logic “1”
C A circuit that produces a logic “0” at its output only if all inputs are logic “0”
D A circuit that produces a logic “1” at its output only if all inputs are logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

An exclusive OR (XOR) gate gives you a 1 if one (and only one) input is 1.

A-2-10-5 (C) What is an EXCLUSIVE OR gate?
A A circuit that produces a logic “1” at its output when all of the inputs are logic “1”
B A circuit that produces a logic “1” at its output when all of the inputs are logic “0”
C A circuit that produces a logic “1” at its output when only one of the inputs is logic “1”
D A circuit that produces a logic “0” at its output when only one of the inputs is logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

An exclusive NOR gate is an XOR in reverse.

A-2-10-6 (D) What is an EXCLUSIVE NOR gate?
A A circuit that produces a logic “1” at its output when only one of the inputs is logic “0”
B A circuit that produces a logic “1” at its output when only one of the inputs are logic “1”
C A circuit that produces a logic “0” at its output when all of the inputs are logic “1”
D A circuit that produces a logic “1” at its output when all of the inputs are logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

A NAND gate is an AND gate, witht he output reversed. You only get a 1 when the inputs are all zero.

A-2-10-1 (D) What is a NAND gate?
A A circuit that produces a logic “1” at its output only when all inputs are logic “1”
B A circuit that produces a logic “0” at its output if some but not all of its inputs are logic “1”
C A circuit that produces a logic “0” at its output only when all inputs are logic “0”
D A circuit that produces a logic “0” at its output only when all inputs are logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

An AND gate gives you a one only if all the inputs are 1.

A-2-10-7 (C) What is an AND gate?
A A circuit that produces a logic “1” at its output only if one of its inputs is logic “1”
B A circuit that produces a logic “1” at its output if all inputs are logic “0”
C A circuit that produces a logic “1” at its output only if all its inputs are logic “1”
D A circuit that produces a logic “1” at the output if at least one input is a logic “0”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

A NOT gate inverts the input.

A-2-10-4 (A) What is a NOT gate (also known as an INVERTER)?
A A circuit that produces a logic “0” at its output when the input is logic “1”
B A circuit that does not allow data transmission when its input is high
C A circuit that allows data transmission only when its input is high
D A circuit that produces a logic “1” at its output when the input is logic “1”

Logic circuitry uses three main operations: “AND”, “OR” and “NOT”. The “NOT” is also called negation, complement or inversion. AND: true if both inputs are true. NAND: AND followed by a NOT, false when both inputs are true (AND, NAND: all inputs have a desired state). OR: true if either input is true. NOR: OR followed by a NOT, false if either input is true (OR,NOR: at least one input has desired state). XOR: true if both inputs are complementary. XNOR: XOR followed by a NOT, false if both inputs are complementary (XOR,XNOR: only one input has a desired state).

A flip-flop is a sequential logic element with two output states. They are also called bistable multivibrators, or latches.

A-2-10-8 (B) What is a flip-flop circuit?
A A binary sequential logic element with one stable state
B A binary sequential logic element with two stable states
C A binary sequential logic element with eight stable states
D A binary sequential logic element with four stable states

The “flip-flop” is a bistable multivibrator. The adjective “bistable” alludes to two possible stable states, set or reset. The circuit remains in one of the two states until a change is triggered. The terms “flip-flop” and “latch” are sometimes used interchangeably. Simple logic gates implement Combinational Logic: the output is determined only by the current inputs. In Sequential Logic, the output depends on current inputs and the exact sequence of prior events. [Nowadays, purists will tell you that a latch follows the input levels (transparency) before a final value is locked-in; a flip-flop captures data strictly on a clock transition (edge-triggered).]

A-2-10-9 (A) What is a bistable multivibrator?
A A flip-flop
B An OR gate
C An AND gate
D A clock

The “flip-flop” is a bistable multivibrator. The adjective “bistable” alludes to two possible stable states, set or reset. The circuit remains in one of the two states until a change is triggered. The terms “flip-flop” and “latch” are sometimes used interchangeably. Simple logic gates implement Combinational Logic: the output is determined only by the current inputs. In Sequential Logic, the output depends on current inputs and the exact sequence of prior events. [ Nowadays, purists will tell you that a latch follows the input levels (transparency) before a final value is locked-in; a flip-flop captures data strictly on a clock transition (edge-triggered). ]

A-2-10-10 (D) What type of digital logic is also known as a latch?
A A decade counter
B An OR gate
C An op-amp
D A flip-flop

The “flip-flop” is a bistable multivibrator. The adjective “bistable” alludes to two possible stable states, set or reset. The circuit remains in one of the two states until a change is triggered. The terms “flip-flop” and “latch” are sometimes used interchangeably. Simple logic gates implement Combinational Logic: the output is determined only by the current inputs. In Sequential Logic, the output depends on current inputs and the exact sequence of prior events. [ Nowadays, purists will tell you that a latch follows the input levels (transparency) before a final value is locked-in; a flip-flop captures data strictly on a clock transition (edge-triggered). ]

They consist of two transistors wired together. When one of them is in conduction, the other is in cutoff.

A-2-10-11 (D) In a multivibrator circuit, when one transistor conducts, the other is:
A saturated
B reverse-biased
C forward-biased
D cut off

The transistors in a multivibrator circuit alternate between two states: conduction and cut-off. The Astable multivibrator can be used as a square-wave generator. The Monostable multivibrator assumes the alternate state for a given period when triggered. The Bistable multivibrator is used in latch and flip-flop circuits.

{L10} Digital Transmission Techniques.

This topic consists mostly of arbitrary trivia.

Baudot and ASCII

ASCII included lower case. Baudot does not.

A-5-8-4 (A) What is one advantage of using ASCII rather than Baudot code?
A It includes both upper and lower case text characters in the code
B ASCII includes built-in error correction
C ASCII characters contain fewer information bits
D The larger character set allows store-and-forward

With 5 bits, Baudot can accommodate 32 combinations ( 2 exponent 5 ) per character set; one set represents uppercase letters A through Z, the other are figures 0 through 9 plus other symbols. The original ASCII (American Standard Code for Information Interchange) used seven bits to accommodate 128 combinations; enough for lowercase and uppercase letters, all digits and other symbols. So called “extended ASCII” uses 8 bits for a total of 256 combinations, this adds accented letters.

ASCII uses 8 bits.

A-5-8-11 (A) How many information bits are included in the ISO-8859 extension to the ASCII code?
A 8
B 7
C 6
D 5

With 5 bits, Baudot can accommodate 32 combinations ( 2 exponent 5 ) per character set; one set represents uppercase letters A through Z, the other are figures 0 through 9 plus other symbols. The original ASCII (American Standard Code for Information Interchange) used seven bits to accommodate 128 combinations; enough for lowercase and uppercase letters, all digits and other symbols. So called “extended ASCII” uses 8 bits for a total of 256 combinations, this adds accented letters.

Baudot code uses five bits

A-5-8-10 (A) How many information bits are included in the Baudot code?
A 5
B 7
C 8
D 6

With 5 bits, Baudot can accommodate 32 combinations ( 2 exponent 5 ) per character set; one set represents uppercase letters A through Z, the other are figures 0 through 9 plus other symbols. The original ASCII (American Standard Code for Information Interchange) used seven bits to accommodate 128 combinations; enough for lowercase and uppercase letters, all digits and other symbols. So called “extended ASCII” uses 8 bits for a total of 256 combinations, this adds accented letters.

As per the name, varicode consists of elements having unequal (variable) length.

A-5-8-1 (C) What digital code consists of elements having unequal length?
A Baudot
B ASCII
C Varicode
D AX.25

Varicode is an encoding method where ASCII characters are represented by bit patterns ranging from 1 to 10 bits in length. Varicode is used for PSK31 (“Phase Shift Keying, 31 Baud”). ASCII (American Standard Code for Information Interchange) or Baudot (RTTY) rely on a precise number of evenly-timed bits. AX.25 is a protocol used for packet transmission.

The OSI model : the base layer is the physical layer.

A-5-8-2 (A) Open Systems Interconnection (OSI) model standardizes communications functions as layers within a data communications system. Amateur digital radio systems often follow the OSI model in structure. What is the base layer of the OSI model involving the interconnection of a packet radio TNC to a computer terminal?
A The physical layer
B The link layer
C The network layer
D The transport layer

Key words: TNC to COMPUTER INTERCONNECTION. Physical layer (for example, RS-232): the electrical, mechanical, procedural and functional specifications for moving data across a physical medium, including modulation, establishing and terminating connections. The Data Link layer packages data bits into frames, provides error-free transfer of frames including physical addressing, network topology, error notification. The Network layer (for example, Internetwork Protocol) uses logical addresses to route frames through a network of links. The Transport layer (for example, TCP Transmission Control Protocol) provides the end-to-end control, ensures that data is complete and properly sequenced ( e.g., retransmissions ).

A CRC is an error checking code.

A-5-8-3 (D) What is the purpose of a Cyclic Redundancy Check (CRC)?
A Lossy compression
B Error correction
C Lossless compression
D Error detection

“A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data” (Wikipedia). A CRC is computed at the originating end and sent with the message; the receiving end does the same calculation, a mismatch in the CRC indicates that the payload was damaged. The CRC manages error detection, other mechanisms can provide error correction.

AMTOR ARQ - automatic repeat request

A-5-8-5 (A) What type of error control system is used in AMTOR ARQ (Mode A)?
A The receiving station automatically requests repeats when needed
B The receiving station checks the frame check sequence (FCS) against the transmitted FCS
C Each character is sent twice
D Mode A AMTOR does not include an error control system

AMTOR is similar to RTTY but with error correction added; a special 7-bit code is used. Amateur Teleprinting Over Radio (AMTOR) can run under two modes. Mode B ( FEC - Forward Error Correction ): characters are sent twice in groups of five in consecutive blocks. Mode A ( ARQ - Automatic Repeat reQuest ): characters are transmitted in blocks of 3, receiving station returns a positive acknowledgement or a request to resend. [ Frame Check Sequence is part of the AX.25 Packet protocol, no relation to AMTOR ]

AMTOR FEC - characters sent twice

A-5-8-6 (B) What error-correction system is used in AMTOR FEC (Mode B)?
A The receiving station checks the frame check sequence (FCS) against the transmitted FCS
B Each character is sent twice
C Mode B AMTOR does not include an error-correction system
D The receiving station automatically requests repeats when needed

AMTOR is similar to RTTY but with error correction added; a special 7-bit code is used. Amateur Teleprinting Over Radio (AMTOR) can run under two modes. Mode B ( FEC - Forward Error Correction ): characters are sent twice in groups of five in consecutive blocks. Mode A ( ARQ - Automatic Repeat reQuest ): characters are transmitted in blocks of 3, receiving station returns a positive acknowledgement or a request to resend. [ Frame Check Sequence is part of the AX.25 Packet protocol, no relation to AMTOR ]

APRS is for messaging, telemetry, and broadcast. It is not for ALE.

A-5-8-7 (A) APRS (Automatic Packet Reporting System) does NOT support which one of these functions?
A Automatic link establishment
B Two-way messaging
C Telemetry
D Amateur-specific local information broadcast

Key word: NOT. “APRS is a digital communications information channel for Ham radio. (…) As a single national channel (…), it gives the mobile ham a place to monitor for 10 to 30 minutes in any area, at any time to capture what is happening in ham radio in the surrounding area. Announcements, Bulletins, Messages, Alerts, Weather, and of course a map of all this activity including objects, satellites, nets, meetings, hamfests, etc. (…) APRS also supports global call sign-to-call sign messaging (…)” (www.aprs.org Bob Bruninga WB4APR). Automatic Link Establishment (ALE) is a standard for systems capable of automatically selecting a band and frequency from a list of channels for HF communications with a given similarly-equipped station.

CRC use a hash function

A-5-8-8 (C) Which algorithm may be used to create a Cyclic Redundancy Check (CRC)?
A Convolution code
B Lempel-Ziv routine
C Hash function
D Dynamic Huffman code

“A hash function is any algorithm that maps data of arbitrary length to data of a fixed length. (…) The values returned by a hash function are called hash values, hash codes, hash sums, checksums or simply hashes” (Wikipedia). A convolutional code is a type of error-correcting code (e.g., Viterbi or Reed-Solomon). Lempel-Ziv and Dynamic Huffman coding are lossless data compression algorithms.

AX.25 is used for packet

A-5-8-9 (A) The designator AX.25 is associated with which amateur radio mode?
A packet
B RTTY
C ASCII
D spread spectrum speech

Packet radio adheres to the AX.25 protocol. AX.25 is derived from the X.25 networking protocol: one notable difference is the use of call signs as addresses. AX.25 uses a Frame-Check Sequence for error detection. The Frame-Check sequence (FCS) is a sixteen-bit number calculated by both the sending and receiving stations of a frame. Comparing the received FCS with a locally computed one permits detecting corruption in transit.

Spread-Spectrum Frequency-Hopping is a fucking thing

FHSS is a thing

A-5-9-8 (C) What is frequency hopping spread spectrum?
A The carrier is frequency-companded
B The carrier is phase-shifted by a fast binary bit stream
C The carrier frequency is changed in accordance with a pseudo-random list of channels
D The carrier is amplitude-modulated over a wide range called the spread

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

Spead spectrum is a thing

A-5-9-1 (B) What term describes a wide-band communications system in which the RF carrier varies according to some predetermined sequence?
A Time domain frequency modulation
B Spread spectrum communication
C Amplitude-companded single sideband
D AMTOR

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

It uses a pseuo-random sequence

A-5-9-11 (A) How does the spread-spectrum technique of frequency hopping work?
A The frequency of an RF carrier is changed very rapidly according to a particular pseudo-random sequence
B If interference is detected by the receiver, it will signal the transmitter to change frequency
C If interference is detected by the receiver, it will signal the transmitter to wait until the frequency is clear
D A pseudo-random bit stream is used to shift the phase of an RF carrier very rapidly in a particular sequence

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

A-5-9-6 (D) Which type of signal is used to produce a predetermined alteration in the carrier for spread spectrum communication?
A Frequency-companded sequence
B Quantizing noise
C Random noise sequence
D Pseudo-random sequence

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

Frequency hopping is a thing

A-5-9-2 (C) What is the term used to describe a spread spectrum communications system where the centre frequency of a conventional carrier is changed many times per second in accordance with a pseudorandom list of channels?
A Time-domain frequency modulation
B Frequency companded spread spectrum
C Frequency hopping
D Direct sequence

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

A-5-9-4 (C) Frequency hopping is used with which type of transmission?
A Packet
B RTTY
D AMTOR

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

Direct sequence is a thing

A-5-9-3 (A) What term is used to describe a spread spectrum communications system in which a very fast binary bit stream is used to shift the phase of an RF carrier?
A Direct sequence
B Frequency hopping
C Phase companded spread spectrum
D Binary phase-shift keying

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

A-5-9-5 (D) Direct sequence is used with which type of transmission?
A AMTOR
B Packet
C RTTY

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

It is difficult to monitor

A-5-9-7 (C) Why is it difficult to monitor a spread spectrum transmission?
A It varies too quickly in amplitude
B The signal is too distorted for comfortable listening
C Your receiver must be frequency-synchronized to the transmitter
D It requires narrower bandwidth than most receivers have

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

direct-sequence is thing

A-5-9-9 (D) What is direct-sequence spread spectrum?
A The carrier is amplitude modulated over a range called the spread
B The carrier is frequency-companded
C The carrier is altered in accordance with a pseudo-random list of channels
D The carrier is phase-shifted by a fast binary bit stream

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

they are resistant to interference

A-5-9-10 (B) Why are received spread-spectrum signals so resistant to interference?
A The high power used by a spread-spectrum transmitter keeps its signal from being easily overpowered
B Signals not using the spectrum-spreading algorithm are suppressed in the receiver
C The receiver is always equipped with a special digital signal processor (DSP) interference filter
D If interference is detected by the receiver, it will signal the transmitter to change frequencies

Spread-spectrum transmission relies on a wide range of frequencies rather than a single one to reduce the effects of noise and interference. It requires several megahertz of bandwidth. Frequency-Hopping Spread-Spectrum changes the carrier frequency a number of times per second in a given pattern. Direct-Sequence Spread-Spectrum uses a pseudo-random bit pattern, many times faster than the data stream, to impress Phase Shift Keying on a carrier. For proper demodulation, the receiver must synchronize itself with the incoming stream. Noise and interference do not follow the same pre-agreed patterns and are thus effectively ignored.

{L14} Transmission Lines.

Tuners

there are

A-7-1-1 (D) For an antenna tuner of the “Transformer” type, which of the following statements is FALSE?
A The input is suitable for 50 ohm impedance
B The output is suitable for impedances from low to high
C The circuit is known as a transformer-type antenna tuner
D The circuit is known as a Pi-type antenna tuner

A tuning circuit using two inductively-coupled windings can readily effect an impedance transformation like a transformer. The coupling may be fixed or variable. Moving a tap on the output coil changes the turns ratio and permits raising or lowering impedance.

A-7-1-2 (C) For an antenna tuner of the “Series” type, which of the following statements is false?
A The output is suitable for impedances from low to high
B The input is suitable for impedance of 50 ohms
C The circuit is known as a Pi-type antenna tuner
D The circuit is known as a Series-type antenna tuner

A simple series L-C network is one of the ways to couple a random-length antenna, whose impedance can be quite unpredictable, directly to a transmitter. Other options include the L and Pi networks.

L-match

A-7-1-3 (B) For an antenna tuner of the “L” type, which of the following statements is false?
A The circuit is known as an L-type antenna tuner
B The circuit is suitable for matching to a vertical ground plane antenna
C The transmitter input is suitable for 50 ohms impedance
D The antenna output is high impedance

A resonant vertical ground plane antenna offers an impedance in the range 30 to 50 ohms. A low-pass “L” network (series inductor followed by parallel capacitor) is commonly used with a high impedance random wire. With only two variables components, an “L” network has a limited range of impedance transformation. [ In reality, four “L” configurations are possible, two of which can match to a lower impedance. ]

Pi-match

A-7-1-4 (B) For an antenna tuner of the “Pi” type, which of the following statements is false?
A The circuit is a Pi-type antenna tuner
B The circuit is a series-type antenna tuner
C The transmitter input is suitable for impedance of 50 ohms
D The antenna output is suitable for impedances from low to high

The “Pi” configuration, usually an input shunt capacitor, a series inductor and an output shunt capacitor, resembles two L networks back-to-back. The Pi has greater impedance transformation range than the L network.

Matching Networks

A-7-1-5 (A) What is a pi-network?
A A network consisting of one inductor and two capacitors or two inductors and one capacitor
B An antenna matching network that is isolated from ground
C A network consisting of four inductors or four capacitors
D A power incidence network

The “Pi” configuration, usually an input shunt capacitor, a series inductor and an output shunt capacitor, resembles two L networks back-to-back. The Pi has greater impedance transformation range than the L network.

A-7-1-6 (D) Which type of network offers the greatest transformation ratio?
A Chebyshev
B Butterworth
C L-network
D Pi-network

With only two variables components, an “L” network has a limited range of impedance transformation. The “Pi” configuration, usually an input shunt capacitor, a series inductor and an output shunt capacitor, resembles two L networks back-to-back. The Pi has greater impedance transformation range than the L network. The “Pi-L” network, where the Pi output capacitor doubles as an input capacitor to a subsequent L section, provides even more harmonic suppression and a greater transformation ratio.

A-7-1-7 (B) Why is an L-network of limited utility in impedance matching?
A It has limited power handling capability
B It matches only a small impedance range
C It is thermally unstable
D It is prone to self-resonance

With only two variables components, an “L” network has a limited range of impedance transformation. The “Pi” configuration, usually an input shunt capacitor, a series inductor and an output shunt capacitor, resembles two L networks back-to-back. The Pi has greater impedance transformation range than the L network. The “Pi-L” network, where the Pi output capacitor doubles as an input capacitor to a subsequent L section, provides even more harmonic suppression and a greater transformation ratio.

A-7-1-8 (A) How does a network transform one impedance to another?
A It cancels the reactive part of an impedance and changes the resistive part
B It produces transconductance to cancel the reactive part of an impedance
C It introduces negative resistance to cancel the resistive part of an impedance
D Network resistances substitute for load resistances

Within the context of matching the line to a transmitter, the goal is to present a suitable resistive impedance to the final amplifier. Impedance comprises a reactive value and a resistive value. To achieve matching, reactance must be cancelled and resistance transformed.

A-7-1-9 (B) What advantage does a pi-L network have over a pi-network for impedance matching between a vacuum tube linear amplifier and a multiband antenna?
A Greater transformation range
B Greater harmonic suppression
C Higher efficiency
D Lower losses

Key words: MULTIBAND ANTENNA. Such an antenna may radiate harmonics more readily. The added harmonic suppression of the “Pi-L” network is advantageous.

A-7-1-10 (D) Which type of network provides the greatest harmonic suppression?
A Inverse pi-network
B Pi-network
C L-network
D Pi-L network

With only two variables components, an “L” network has a limited range of impedance transformation. The “Pi” configuration, usually an input shunt capacitor, a series inductor and an output shunt capacitor, resembles two L networks back-to-back. The Pi has greater impedance transformation range than the L network. The “Pi-L” network, where the Pi output capacitor doubles as an input capacitor to a subsequent L section, provides even more harmonic suppression and a greater transformation ratio.

Stubs

A-7-2-1 (B) What kind of impedance does a quarter wavelength transmission line present to the source when the line is shorted at the far end?
A A very low impedance
B A very high impedance
C The same as the characteristic impedance of the transmission line
D The same as the output impedance of the source

Line lengths that are multiples of a half-wavelength replicate the load impedance at the input regardless of the Characteristic Impedance: i.e., the input impedance equals the load impedance. Line lengths that are odd multiples of a quarter-wavelength behave as impedance transformers. Quarter-wavelength line sections are said to invert impedance: an open is reflected as a short and vice-versa. When used for matching right at the antenna, a quarter-wavelength line section is called a “Q Section” or “Quarter-Wave Transformer”.

A-7-2-2 (D) What kind of impedance does a quarter wavelength transmission line present to the source if the line is open at the far end?
A A very high impedance
B The same as the output impedance of the source
C The same as the characteristic impedance of the transmission line
D A very low impedance

Line lengths that are multiples of a half-wavelength replicate the load impedance at the input regardless of the Characteristic Impedance: i.e., the input impedance equals the load impedance. Line lengths that are odd multiples of a quarter-wavelength behave as impedance transformers. Quarter-wavelength line sections are said to invert impedance: an open is reflected as a short and vice-versa. When used for matching right at the antenna, a quarter-wavelength line section is called a “Q Section” or “Quarter-Wave Transformer”.

A-7-2-3 (A) What kind of impedance does a half wavelength transmission line present to the source when the line is open at the far end?
A A very high impedance
B The same as the characteristic impedance of the transmission line
C The same as the output impedance of the source
D A very low impedance

Line lengths that are multiples of a half-wavelength replicate the load impedance at the input regardless of the Characteristic Impedance: i.e., the input impedance equals the load impedance. Line lengths that are odd multiples of a quarter-wavelength behave as impedance transformers. Quarter-wavelength line sections are said to invert impedance: an open is reflected as a short and vice-versa. When used for matching right at the antenna, a quarter-wavelength line section is called a “Q Section” or “Quarter-Wave Transformer”.

A-7-2-4 (B) What kind of impedance does a half wavelength transmission line present to the source when the line is shorted at the far end?
A The same as the output impedance of the source
B A very low impedance
C A very high impedance
D The same as the characteristic impedance of the transmission line

Line lengths that are multiples of a half-wavelength replicate the load impedance at the input regardless of the Characteristic Impedance: i.e., the input impedance equals the load impedance. Line lengths that are odd multiples of a quarter-wavelength behave as impedance transformers. Quarter-wavelength line sections are said to invert impedance: an open is reflected as a short and vice-versa. When used for matching right at the antenna, a quarter-wavelength line section is called a “Q Section” or “Quarter-Wave Transformer”.

Velocity Factor

For a given wavelength, the cable is shorter than the wave in free space.

A-7-2-9 (D) Why is the physical length of a coaxial cable shorter than its electrical length?
A The surge impedance is higher in the parallel transmission line
B Skin effect is less pronounced in the coaxial cable
C The characteristic impedance is higher in a parallel transmission line
D RF energy moves slower along the coaxial cable than in air

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

velocity factor nu

v = nu c

nu = v/c

A-7-2-6 (C) What is the term for the ratio of the actual velocity at which a signal travels through a transmission line to the speed of light in a vacuum?
A Surge impedance
B Standing wave ratio
C Velocity factor
D Characteristic impedance

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

A-7-2-5 (A) What is the velocity factor of a transmission line?
A The velocity of the wave on the transmission line divided by the velocity of light
B The velocity of the wave on the transmission line multiplied by the velocity of light in a vacuum
C The index of shielding for coaxial cable
D The ratio of the characteristic impedance of the line to the terminating impedance

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

A-7-2-11 (C) The velocity factor of a transmission line is the:
A speed at which the signal travels in free space
B speed to which the standing waves are reflected back to the transmitter
C ratio of the velocity of propagation in the transmission line to the velocity of propagation in free space
D impedance of the line, e.g. 50 ohm, 75 ohm, etc.

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

‘ — — — —

velocity factor is the factor by which a free-space wavelength is multiplied to obtain the length in a medium.

lambda_free * vf = lambda_medium

so

``````      lambda_free
vf =  -------------
lambda_medium``````

for example, at 14.1 MHz `lambda = ~20 m`,
with `vf = 0.66` then
`20 m * 0.66 = about 12 m`;
a quarter of that is about `3 m`

A-7-3-4 (C) Assuming a velocity factor of 0.66 what would be the physical length of a typical coaxial stub that is electrically one quarter wavelength long at 14.1 MHz?
A 2.33 metres (7.64 feet)
B 0.25 metre (0.82 foot)
C 3.51 metres (11.5 feet)
D 20 metres (65.6 feet)

An electrical quarter wavelength can be computed in metres as one fourth of 300 divided by frequency in megahertz. The physical length equals the electrical length times the Velocity Factor. In this example, 300 divided by 14.1 divided by 4 times 0.66 = 3.51 metre.

velocity factor is determined by the dielectric material, with the following relationship:

nu = 1/sqrt(C_d)

A-7-2-8 (A) What determines the velocity factor in a transmission line?
A Dielectrics in the line
B The line length
C The centre conductor resistivity
D The terminal impedance

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

A-7-2-10 (C) The reciprocal of the square root of the dielectric constant of the material used to separate the conductors in a transmission line gives the __ of the line:
A impedance
B hermetic losses
C velocity factor
D VSWR

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

Typ nu=.66

A-7-2-7 (A) What is a typical velocity factor for coaxial cable with polyethylene dielectric?
A 0.66
B 0.33
C 0.1
D 2.7

Velocity Factor is a ratio of wave travel speed on a transmission line with respect to wave speed in vacuum. It is expressed as a percentage or a decimal fraction because waves travel slower on lines than in space. The dielectric constant of the insulator between the conductors determines the Velocity Factor per this formula: 1 over the square root of the dielectric constant. Lines using polyethylene have a Velocity Factor of 66%, foam polyethylene brings it above 80%. Actual Velocity Factor can vary by as much as plus or minus 10%. Because of that delay in propagation, a given physical length will always seem longer electrically.

Waveguides

Waveguides are used above 3 GHz.

A-7-9-1 (A) Waveguide is typically used:
A at frequencies above 3000 MHz
B at frequencies above 2 MHz
C at frequencies below 150 MHz
D at frequencies below 1500 MHz

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

Above the cutoff/critical frequency (e.g. for microwaves) waveguides have low loss.

A-7-9-6 (A) Which of the following is a major advantage of waveguide over coaxial cable for use at microwave frequencies?
A Very low losses
B Frequency response from 1.8 MHz to 24GHz
C Easy to install
D Inexpensive to install

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

Waveguides are low-loss.

A-7-9-3 (B) Which of the following is an advantage of waveguide as a transmission line?
A Heavy and difficult to install
B Low loss
C Frequency sensitive based on dimensions
D Expensive

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

There is no such thing as hysteresis loss.

A-7-9-2 (D) Which of the following is not correct? Waveguide is an efficient transmission medium because it features:
A low radiation loss
B low dielectric loss
C low copper loss
D low hysteresis loss

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

Waveguides are a half-wavelengh wide.

A-7-9-4 (D) For rectangular waveguide to transfer energy, the cross-section should be at least:
A three-eighths wavelength
B one-eighth wavelength
C one-quarter wavelength
D one-half wavelength

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

Waveguides only support transmission above the “critical” or “cutoff” frequency, so they are in effect a high-pass filter.

A-7-9-9 (D) A section of waveguide:
A operates like a low-pass filter
B operates like a band-stop filter
C is lightweight and easy to install
D operates like a high-pass filter

Waveguides, used as transmission lines, are hollow pipes through which signals propagate as waves. The width or diameter of the waveguide must be slightly larger than a half-wavelength at the operating frequency. Below one gigahertz, dimensions become prohibitive. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. Waveguides do not suffer the conduction, dielectric or radiation losses that normal transmission lines present at microwave frequencies.

A-7-9-5 (D) Which of the following statements about waveguide IS NOT correct?
A In the transverse electric mode, a component of the magnetic field is in the direction of propagation
B In the transverse magnetic mode, a component of the electric field is in the direction of propagation
C Waveguide has low loss at high frequencies, but high loss below cutoff frequency
D Waveguide has high loss at high frequencies, but low loss below cutoff frequency

Key words: IS NOT. Signals with wavelengths too large for the physical size of the waveguide are attenuated: waveguides behave like high-pass filters, in other words, attenuation below the cutoff frequency. In free space, waves are known as transverse-electromagnetic: the electric field, the magnetic field and the direction of travel are all perpendicular to one another. In a waveguide, waves bounce from wall to wall thus travelling in a zigzag manner. Only one of the electric or magnetic field can be truly perpendicular with the length of the waveguide; the mode, transverse electric or transverse magnetic, describes which field is purely perpendicular.

Because waveguides carry heating microwaves, be careful.

A-7-9-11 (C) What precautions should you take before beginning repairs on a microwave feed horn or waveguide?
A Be sure propagation conditions are unfavourable for tropospheric ducting
B Be sure to wear tight-fitting clothes and gloves to protect your body and hands from sharp edges
C Be sure the transmitter is turned off and the power source is disconnected
D Be sure the weather is dry and sunny

With the possibility of shorter wavelengths to reach deeper into the body or to produce resonances in smaller structures, such as the eye, be extra careful not to expose anyone to microwave radiation. The significant gain available from physically small antennas also turn low power levels into definite risks. Heating is one known effect of RF on body tissues, other effects are possible.

Stripline & microstripline

“Stripline” is basically a specially sized trace on a PCB surrounded by ground layers, above and below.

A-7-9-7 (B) What is printed circuit transmission line called?
A Ground plane
B Microstripline
C Dielectric substrate
D Dielectric imprinting

A Microstrip transmission line is a type of line consisting of a thin and flat strip of conductive material separated from a ground plane by a dielectric. A double-sided printed circuit board lends itself to the construction of microstrip lines: traces on top with a ground plane underneath. Characteristic Impedance is determined by trace width, dielectric thickness and dielectric constant. One side of the line is exposed to air, external shielding may be required when high isolation is required. Stripline uses a similar thin and flat conductor but sandwiched between two parallel ground planes.

A-7-9-10 (B) Stripline is a:
A family of fluids for removing coatings from small parts
B printed circuit transmission line
C small semiconductor family
D high power microwave antenna

A Microstrip transmission line is a type of line consisting of a thin and flat strip of conductive material separated from a ground plane by a dielectric. A double-sided printed circuit board lends itself to the construction of microstrip lines: traces on top with a ground plane underneath. Characteristic Impedance is determined by trace width, dielectric thickness and dielectric constant. External shielding may be required when high isolation is required. Stripline uses a similar thin and flat conductor but sandwiched between two parallel ground planes.

microstripline omits the second shield layer, and consequently has poorer shielding

A-7-9-8 (B) Compared with coaxial cable, microstripline:
A must have much higher characteristic impedance
B has poorer shielding
C has superior shielding
D must have much lower characteristic impedance

A Microstrip transmission line is a type of line consisting of a thin and flat strip of conductive material separated from a ground plane by a dielectric. A double-sided printed circuit board lends itself to the construction of microstrip lines: traces on top with a ground plane underneath. Characteristic Impedance is determined by trace width, dielectric thickness and dielectric constant. One side of the line is exposed to air, external shielding may be required when high isolation is required. Stripline uses a similar thin and flat conductor but sandwiched between two parallel ground planes.

Smith Charts

A-7-1-11 (C) A Smith Chart is useful:
A to solve problems in direct current circuits
B because it only works with complex numbers
C because it simplifies mathematical operations
D only to solve matching and transmission line problems

“The Smith chart, invented by Phillip H. Smith (1905–1987), is a graphical aid or nomogram designed for electrical and electronics engineers specializing in radio frequency (RF) engineering to assist in solving problems with transmission lines and matching circuits. (…) The Smith chart is most frequently used at or within the unity radius region. However, the remainder is still mathematically relevant, being used, for example, in oscillator design and stability analysis”. (Wikipedia)

{L16} FM Repeaters.

Intermod `!`

intermodulation refers to mixing of outside signals

A-5-6-1 (B) If the signals of two repeater transmitters mix together in one or both of their final amplifiers and unwanted signals at the sum and difference frequencies of the original signals are generated and radiated, what is this called?
A Amplifier desensitization
B Intermodulation interference
C Neutralization
D Adjacent channel interference

Intermodulation is the unwanted mixing of two or more signals that produce new signals (products). The fundamental or harmonic energy from strong nearby transmitters intermix to create intermodulation products. Mixing can take place in the front-end of the affected receiver, in the Power Amplifier of one of the transmitters or through “external rectification or passive intermodulation” (in some electronic device or some corroded junction between two metals acting as a diode mixer).

This happens when you have two tranmsitters next to each other

A-5-6-2 (A) How does intermodulation interference between two repeater transmitters usually occur?
A When they are in close proximity and the signals mix in one or both of their final amplifiers
B When the signals are reflected in phase by aircraft passing overhead
C When they are in close proximity and the signals cause feedback in one or both of their final amplifiers
D When the signals are reflected out of phase by aircraft passing overhead

Intermodulation is the unwanted mixing of two or more signals that produce new signals (products). The fundamental or harmonic energy from strong nearby transmitters intermix to create intermodulation products. Mixing can take place in the front-end of the affected receiver, in the Power Amplifier of one of the transmitters or through “external rectification or passive intermodulation” (in some electronic device or some corroded junction between two metals acting as a diode mixer).

It can be prevented by the use of a device called a terminated circulator, also called a line isolator

A-5-6-3 (B) How can intermodulation interference between two repeater transmitters in close proximity often be reduced or eliminated?
A By using a Class C final amplifier with high driving power
B By installing a terminated circulator or ferrite isolator in the transmission line to the transmitter and duplexer
C By installing a low-pass filter in the antenna transmission line
D By installing a high-pass filter in the antenna transmission line

A circulator is usually a three-port device; any energy coming in one port is forwarded in one direction only, towards the next port. An isolator is a circulator with one port terminated into a dummy load. With the transmitter on port 1, the antenna on port 2 and a dummy load on port 3, RF from the transmitter is routed to the antenna but any outside energy picked-up by the antenna is diverted to the dummy load and thus cannot enter the Power Amplifier where it might lead to intermodulation. Magnetized ferrite material is used in the fabrication of circulators.

`!` IMD results from the mixing of harmonics; involving the sum and difference of harmonics of two interfering signals.

for example, IMD on 146.70 due to signals on 146.52 and may be due to signals on 146.34 MHz or 146.61 MHz

A-5-6-4 (D) If a receiver tuned to 146.70 MHz receives an intermodulation product signal whenever a nearby transmitter transmits on 146.52, what are the two most likely frequencies for the other interfering signal?
A 146.88 MHz and 146.34 MHz
B 146.01 MHz and 147.30 MHz
C 73.35 MHz and 239.40 MHz
D 146.34 MHz and 146.61 MHz

Third Order Intermodulation products where the second harmonic of one signal mixes with the fundamental frequency of another are the most troublesome because of the relative frequency proximity of the involved signals: filtering them out is difficult. Given two signals, F1 and F2, the Third Order IMD product can be computed as (a) “twice F1 minus F2” or (b) “twice F2 minus F1”. In this example, IMD = 146.70, F1 = 146.52 ; solving for F2 in case “a” is (F1 times 2), minus IMD ; solving for F2 in case “b” is (IMD + F1), divided by 2.

146.70 is …?

double the interfering signal:
146.52 = 293.04

293.04 - 146.70 = 146.34 one possible interfering signal

because
146.70 = 2*146.520 - 146.34

Another possibility:
146.70 = 146.520 +/- 2 X

So (146.520 146.70 -

Cavity filters are used for 2 m repeaters.

A-5-6-7 (D) Which type of filter would be best to use in a 2-metre repeater duplexer?
A A DSP filter
B An L-C filter
C A crystal filter
D A cavity filter

In the context of a repeater installation, a duplexer is a specialized filter which allows operating the receiver and transmitter simultaneously on the same antenna. The duplexer is built with four or more quarter-wavelength cavity resonators. The duplexer provides isolation ( 90 dB or more on 2 m ) between the receive and transmit paths at the expense of insertion loss.

IMD occurs in final amplifers and receiver front-ends, but not IF stages (this is presumably due to filtering between stages).

A-5-6-11 (A) Intermodulation interference products are not typically associated with which of the following:
A intermediate frequency stage
B final amplifier stage
D passive intermodulation

“Undesired mixing of two or more frequencies in a non-linear device which produces additional sum and difference frequencies” (ARRL RFI Book). Final amplifier stage: signals from nearby transmitters may find their way into a power amplifier creating intermodulation. Receiver front-end: strong offending signals may enter the RF amplifier or mixer and create intermodulation. Passive intermodulation (PIM), sometimes called “rusty bolt effect”: passive devices subjected to strong signals cause the problem; cables, antennas, connectors, dissimilar metals, etc. due to oxidation, corrosion, metal flakes, dirty mating surfaces or poor contact.

{L15} Antennas.

antenna feed matches

[stub match uses a short section (stub) of feedline

A-7-3-3 (A) What term describes a method of antenna impedance matching that uses a short section of transmission line connected to the antenna transmission line near the antenna and perpendicular to the transmission line?
A The stub match
B The omega match
C The delta match
D The gamma match

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

For example, a quarter wave stub at 15 Mhz using vf = 0.8 cable

``````v = f l

l = c nu / f

l = 300 10^6 * .8 / 15 10^6
= 300 * .8 / 15 [m]
= 16 [m]

so 1/4 l = 4 [m]``````

A-7-3-6 (A) A quarter-wave stub, for use at 15 MHz, is made from a coaxial cable having a velocity factor of 0.8. Its physical length will be:
A 4 m (13.1 ft)
B 12 m (39.4 ft)
C 8 m (26.2 ft)
D 7.5 m (24.6 ft)

An electrical quarter wavelength can be computed in metres as one fourth of 300 divided by frequency in megahertz. The physical length equals the electrical length times the Velocity Factor. In this example, 300 divided by 15 divided by 4 times 0.80 = 4 metres.

T-match matches high-impedance balanced transmission line (ladder line) to a low-impedance element

A-7-3-1 (C) What term describes a method used to match a high-impedance transmission line to a lower impedance antenna by connecting the line to the driven element in two places, spaced a fraction of a wavelength on each side of the driven element centre?
A The omega match
B The stub match
C The T match
D The gamma match

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

The gamma match is basically half a t-match. gamma match is where you connect the center conductor of an unbalanced feedline in two special places on the driven element.

A-7-3-2 (C) What term describes an unbalanced feed system in which the driven element of an antenna is fed both at the centre and a fraction of a wavelength to one side of centre?
A The stub match
B The T match
C The gamma match
D The omega match

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

yagi-uda gamma match

A-7-3-11 (A) A Yagi antenna uses a gamma match. The variable capacitor connects to the:
A adjustable gamma rod
B an adjustable point on the director
C center of the driven element
D coaxial line braid

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

A-7-3-10 (C) A Yagi antenna uses a gamma match. The adjustable gamma rod connects to:
A an adjustable point on the reflector
B the centre of the driven element
C the variable capacitor
D the coaxial line centre conductor

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

A-7-3-9 (C) A Yagi antenna uses a gamma match. The centre of the driven element connects to:
A the adjustable gamma rod
B a variable capacitor
C the coaxial line braid
D the coaxial line centre conductor

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

A-7-3-8 (C) A Yagi antenna uses a gamma match. The coaxial braid connects to:
A the adjustable gamma rod
B the centre of the reflector
C the centre of the driven element
D the variable capacitor

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

A-7-3-7 (D) The matching of a driven element with a single adjustable mechanical and capacitive arrangement is descriptive of:
A a “T” match
B an “omega” match
C a “Y” match
D a “gamma” match

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

A-7-3-5 (B) The driven element of a Yagi antenna is connected to a coaxial transmission line. The coax braid is connected to the centre of the driven element and the centre conductor is connected to a variable capacitor in series with an adjustable mechanical arrangement on one side of the driven element. The type of matching is:
A zeta match
B gamma match
C lambda match
D T match

Gamma match: an unbalanced feed system, the coaxial braid attaches to the centre of the radiating element, the centre conductor connects via a series capacitor further along the radiating element. This second connection is done with an adjustable rod parallel to the radiating element. T match: a balanced feed system, can be thought as two mirrored gamma matches, the feed line is brought via conductors parallel to the radiating element at points further along the element. A Matching Stub is a short section of line, open or shorted, connected across the transmission line at a specific distance from the antenna feedpoint.

half-wave dipole

A-7-4-10 (D) In a half-wave dipole, where does the minimum current occur?
A At the centre
B It is equal at all points
C At the right end
D At both ends

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-11 (B) In a half-wave dipole, where does the minimum impedance occur?
A At both ends
B At the centre
C It is the same at all points
D At the right end

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-9 (B) In a half-wave dipole, where does minimum voltage occur?
A Both ends
B The centre
C At the right end
D It is equal at all points

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-8 (C) The impedance of a half-wave antenna at its centre is low, because at this point:
A voltage and current are both low
B voltage is high and current is low
C voltage is low and current is high
D voltage and current are both high

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-7 (C) At the ends of a half-wave dipole:
A voltage and current are both low
B voltage is low and current is high
C voltage is high and current is low
D voltage and current are both high

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-6 (D) A half-wave dipole antenna is normally fed at the point where:
A the voltage is maximum
B the resistance is maximum
C the antenna is resonant
D the current is maximum

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-5 (A) In a half-wave dipole, the highest distribution of __ occurs at the middle.
A current
B inductance
C voltage
D capacity

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-4 (A) In a half-wave dipole, the lowest distribution of _ occurs at the middle.
A voltage
B capacity
C inductance
D current

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-3 (C) The feed point in a centre-fed half-wave antenna is at the point of:
A minimum voltage and current
B maximum voltage
C maximum current
D minimum current

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-2 (B) In a half-wave dipole, the distribution of _ is lowest at each end.
A capacitance
B current
C voltage
D inductance

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

A-7-4-1 (D) In a half-wave dipole, the distribution of _ is highest at each end.
A current
B inductance
C capacitance
D voltage

On a centre-fed resonant half-wave dipole, current is high and voltage low at the feedpoint, the ends exhibit high voltage and low current. Low voltage and high current at the centre make for low impedance ( Z = E divided by I ).

circular polarization

A-7-5-1 (D) What is meant by circularly polarized electromagnetic waves?
A Waves with an electric field bent into circular shape
B Waves that circle the earth
C Waves produced by a circular loop antenna
D Waves with a rotating electric field

The polarization of an electromagnetic radio wave corresponds to the position of the electrical field with respect to the surface of the Earth: horizontal when the E field is parallel to ground and vertical when perpendicular to ground. The magnetic field is at 90 degrees (perpendicular) to the electrical field. Dipoles and Yagis are linearly polarized antennas (i.e., the electrical field has a constant orientation). Circular polarization, where the polarization rotates, can be obtained from helical beam antennas or with crossed linear antennas fed with the correct phase difference. “Sense” refers to the direction of the rotation: clockwise polarization for a receding wave is termed right-hand.

A-7-5-2 (A) What type of polarization is produced by crossed dipoles fed 90 degrees out of phase?
A Circular polarization
B Cross-polarization
C Perpendicular polarization
D None of the other answers, the two fields cancel out

Crossed dipoles fed 90 degrees out of phase are the active elements of a turnstile antenna and produce circular polarization. The turnstile antenna is used for satellite communications.

A-7-5-3 (C) Which of these antennas does not produce circular polarization?
B Axial-mode helical antenna
C Loaded helical-wound antenna
D Crossed dipoles fed 90 degrees out of phase

Key word: NOT. Antennas featuring a fine wire wound around a shaft (e.g., HF mobile antennas) are “loaded helical-wound antennas”; these produce a linear polarization, i.e., vertical or horizontal depending on their positions relative to ground. The axial-mode helical antenna, with its corkscrew look, is used in satellite work and produces circular polarization.

Cross-polarization loss is in the vicinity of 20 dB

A-7-5-5 (C) For VHF and UHF signals over a fixed path, what extra loss can be expected when linearly-polarized antennas are crossed-polarized (90 degrees)?
A 6 dB
B 10 dB
C 20 dB or more
D 3 dB

“A loss in signal strength of 20 dB or more can be expected with cross-polarization so it is important to use antennas with the same polarization as the stations with which you expect to communicate.” (ARRL Antenna Book, 22nd ed., section 21.10.5 Polarization)

ERP - relative to dipole

dipole is ~2 dBi (i.e., relative to isotropic radiator)

A-7-6-11 (A) A transmitter has an output of 1000 watts PEP. The coaxial cable, connectors and antenna tuner have a composite loss of 1 dB, and the antenna gain is 10 dBd. What is the Effective Radiated Power (ERP) in watts PEP?
A 8000
B 1009
C 10 000
D 9000

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 1 dB plus 10 dB yields a net increase of 9 dB or eight times the power.

A-7-6-10 (B) A transmitter has an output of 2000 watts PEP. The transmission line, connectors and antenna tuner have a composite loss of 1 dB, and the gain from the stacked Yagi antenna is 10 dBd. What is the Effective Radiated Power (ERP) in watts PEP?
A 2009
B 16 000
C 18 000
D 20 000

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 1 dB plus 10 dB yields a net increase of 9 dB or eight times the power.

A-7-6-8 (D) A transmitter has a power output of 125 watts. There is a loss of 0.8 dB in the transmission line, 0.2 dB in the antenna tuner, and a gain of 10 dBd in the antenna. The Effective Radiated Power (ERP) is:
A 1250
B 1125
C 134
D 1000

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 1 dB plus 10 dB yields a net increase of 9 dB or eight times the power.

A-7-6-6 (D) A transmitter has a power output of 100 watts. There is a loss of 1.30 dB in the transmission line, a loss of 0.2 dB through the antenna tuner, and a gain of 4.50 dBd in the antenna. The Effective Radiated Power (ERP) is:
A 800 watts
B 400 watts
C 100 watts
D 200 watts

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 1.5 dB plus 4.5 dB yields a net increase of 3 dB or twice the power.

A-7-6-5 (A) A transmitter has an output power of 200 watts. The coaxial and connector losses are 3 dB in total, and the antenna gain is 9 dBd. What is the approximate Effective Radiated Power of this system?
A 800 watts
B 3200 watts
C 1600 watts
D 400 watts

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 3 dB plus 9 dB yields a net increase of 6 dB or four times the power.

A-7-6-4 (A) Effective Radiated Power means the:
A transmitter output power, minus line losses, plus antenna gain relative to a dipole
B power supplied to the antenna before the modulation of the carrier
C power supplied to the transmission line plus antenna gain
D ratio of signal output power to signal input power

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain.

A-7-6-3 (A) What is the Effective Radiated Power of an amateur transmitter, if the transmitter output power is 200 watts, the transmission line loss is 5 watts, and the antenna power gain is 3 dBd?
A 390 watts
B 197 watts
C 228 watts
D 178 watts

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 5 watts plus 3 dB yields a net increase of 3 dB or twice the remaining power of 195 watts.

polarization chirality

A-7-5-8 (D) A helical-beam antenna with right-hand polarization will best receive signals with:
A left-hand polarization
B vertical polarization only
C horizontal polarization
D right-hand polarization

The helical beam antenna is circularly polarized. Although it will respond to horizontally or vertically polarized waves, the full gain of the antenna can only be realized with a circularly polarized wave of the same “sense”.

line losses consist of dielectric loss, and conductor loss

A-7-6-2 (D) As standing wave ratio rises, so does the loss in the transmission line. This is caused by:
A high antenna currents
B high antenna voltage
C leakage to ground through the dielectric
D dielectric and conductor heat losses

Voltage peaks on the standing wave increase losses through the dielectric ( P = E squared divided by R ), current peaks on the standing wave increase conductor losses ( P = I squared times R ).

ERP
ERP

A-7-6-9 (A) If a 3 dBd gain antenna is replaced with a 9 dBd gain antenna, with no other changes, the Effective Radiated Power (ERP) will increase by:
A 4
B 6
C 1.5
D 2

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, an added 6 dB yields four times the power.

A-7-6-1 (D) A transmitter has an output of 100 watts. The cable and connectors have a composite loss of 3 dB, and the antenna has a gain of 6 dBd. What is the Effective Radiated Power?
A 350 watts
B 400 watts
C 300 watts
D 200 watts

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, minus 3 dB plus 6 dB yields a net increase of 3 dB or twice the power.

A-7-6-7 (B) If the overall gain of an amateur station is increased by 3 dB the ERP (Effective Radiated Power) will:
A be cut in half
B double
C decrease by 3 watts
D remain the same

Effective Radiated Power (ERP) equals transmitter power minus line losses plus antenna gain. In this example, a net gain of 3 dB yields twice the power.

cross polarized antennas

A-7-5-9 (A) One antenna which will respond simultaneously to vertically- and horizontally-polarized signals is the:
A helical-beam antenna
B folded dipole antenna
C ground-plane antenna

The dipole, ground-plane and quad are all linearly polarized and thus respond optimally to waves polarized in a single given direction, horizontal or vertical as the case may be. Unless proper alignment is assured, a significant loss is incurred. The helical beam antenna is circularly polarized, it can deal with the rotating fields of a wave with circular polarization. Consequently, it can deal with any single polarization at any given angle.

polarization

A-7-7-2 (D) Most simple horizontally polarized antennas do not exhibit significant directivity unless they are:
A an eighth of a wavelength above the ground
B a quarter wavelength above the ground
C three-eighths of a wavelength above the ground
D a half wavelength or more above the ground

Below one half-wavelength in antenna height, there is little point in selecting a given broadside direction: for example, choosing a north-south orientation to favour east-west radiation will not pay significant dividends for antennas close to ground.

parabolic antennas

A-7-5-7 (B) A parabolic antenna is very efficient because:
A a horn-type radiator can be used to trap the received energy
B all the received energy is focused to a point where the pick-up antenna is located
C a dipole antenna can be used to pick up the received energy
D no impedance matching is required

A parabolic reflector dish provides significant gain because energy striking any point of the parabola is reflected to the focal point with the correct phase. On transmit, the inverse process takes place: all energy directed at the parabola from the feed antenna is reflected forward with the correct phase. High gain antennas used on UHF or microwave frequencies present a real risk to living tissues: never stand in front of a transmitting antenna.

A-7-5-6 (C) Which of the following is NOT a valid parabolic dish illumination arrangement?
A Offset feed
B Cassegrain
C Newtonian
D Front feed

Key word: NOT. Front feed (also known as focal feed or axial feed): circular reflector, the feed is centered in front of the reflector, very common on larger dish antennas. Offset feed (also known as off-axis): elliptical reflector, the feed is off to one side, out of the path of the radio waves, typical of domestic satellite receiving antennas. Cassegrain (based on the Cassegrain telescope): the feed is behind the dish and relies on a small convex secondary reflector in front of the dish. Newtonian: a bogus answer, valid for telescopes.

Surface error should not be greater than a small fraction of wavelength

A-7-5-10 (D) In amateur work, what is the surface error upper limit you should try not to exceed on a parabolic reflector?
A 0.25 lambda
B 5 mm (0.2 in) regardless of frequency
C 1% of the diameter
D 0.1 lambda

“Surface errors should not exceed 1/8 lambda in amateur operation. At 430 MHz, 1/8 lambda is 3.4 inches [8.6 cm], but at 10 GHz, it is 0.1476 inch [3.7 mm]! (…) Mesh can be used for the reflector surface to reduce weight and wind loading, but hole size should be less than 1/12 lambda.” (ARRL Antenna Book 22nd ed., sect. 15.6.2)

The gain of a parabolic dish depends on diameter

A-7-5-11 (B) You want to convert a surplus parabolic dish for amateur radio use, the gain of this antenna depends on:
A the material composition of the dish
B the diameter of the antenna in wavelengths
C the polarization of the feed device illuminating it
D the focal length of the antenna

Gain is primarily affected by the antenna aperture (reflector area) to wavelength ratio.

ground planes

A-7-7-3 (D) The plane from which ground reflections can be considered to take place, or the effective ground plane for an antenna is:
A as much as 6 cm below ground depending upon soil conditions
B as much as a meter above ground
C at ground level exactly
D several centimeters to as much as 2 meters below ground, depending upon soil conditions

Current penetration around an antenna depends first on frequency and then on soil conductivity and dielectric constant. At HF frequencies over salt water, penetration ranges from 5 to 18 centimetres. Over poor ground, penetration can exceed 10 metres.

takeoff angle

A-7-7-4 (C) Why is a ground-mounted vertical quarter-wave antenna in reasonably open surroundings better for long distance contacts than a half-wave dipole at a quarter wavelength above ground?
A It has an omnidirectional characteristic
B It uses vertical polarization
C The vertical radiation angle is lower
D The radiation resistance is lower

Key words: DIPOLE AT A QUARTER WAVELENGTH HEIGHT. At heights below three eights of a wavelength, ground reflections cause horizontal dipoles to direct more energy straight up. At a height of one half-wavelength, radiation at 90 degrees is minimized and two lobes form at 30 degrees. In this comparison, the ground-mounted vertical undoubtedly exhibits a lower radiation angle as it cannot possibly radiate upwards.

A-7-7-5 (B) When a half-wave dipole antenna is installed one-half wavelength above ground, the:
A radiation pattern is unaffected
B vertical or upward radiation is effectively cancelled
C radiation pattern changes to produce side lobes at 15 and 50 degrees
D side lobe radiation is cancelled

At heights below three eights of a wavelength, ground reflections cause horizontal dipoles to direct more energy straight up. At a height of one half-wavelength, radiation at 90 degrees is minimized and two lobes form at 30 degrees.

A-7-7-7 (B) For long distance propagation, the vertical radiation angle of the energy from the antenna should be:
A more than 30 degrees but less than 45 degrees
B less than 30 degrees
C more than 45 degrees but less than 90 degrees
D 90 degrees

Depending on band (for example, 10, 15 and 20 metre) and distance, preferred radiation angles range from 1 to 25 degrees for long distance communication. A low radiation angle permits hitting the ionosphere at a greater distance for longer skip distances.

A-7-7-8 (D) Greater distance can be covered with multiple-hop transmissions by decreasing the:
A power applied to the antenna
B main height of the antenna
C length of the antenna
D vertical radiation angle of the antenna

Depending on band (for example, 10, 15 and 20 metre) and distance, preferred radiation angles range from 1 to 25 degrees for long distance communication. A low radiation angle permits hitting the ionosphere at a greater distance for longer skip distances.

NVIS

A-7-7-11 (D) Which antenna system and operating frequency are most suitable for Near Vertical Incidence (NVIS) communications?
A A horizontal antenna at a height of half a wavelength and an operating frequency at the optimum working frequency
B A vertical antenna and a frequency below the maximum usable frequency
C A vertical antenna and a frequency above the lowest usable frequency
D A horizontal antenna less than 1/4 wavelength above ground and a frequency below the current critical frequency

Near-Vertical Incidence Sky wave (NVIS) — “The use of very low dipole antennas that radiate at very high elevation angles has become popular in emergency communications (“emcomm”) systems. This works at low frequencies (7 MHz and below) that are lower than the ionosphere’s critical frequency — the highest frequency for which a signal traveling vertically will be reflected.” (ARRL Handbook, 2012 ed., 21.2.12 NVIS Antennas)

A-7-7-1 (C) For a 3-element Yagi antenna with horizontally mounted elements, how does the main lobe takeoff angle vary with height above flat ground?
A It does not vary with height
B It depends on E-region height, not antenna height
C It decreases with increasing height
D It increases with increasing height

Greater antenna heights tend to lower the main radiation lobe where ground reflections end up in phase with direct radiation from the antenna.

A-7-7-10 (A) Why can a horizontal antenna closer to ground be advantageous for close range communications on lower HF bands?
A The ground tends to act as a reflector
B Lower antenna noise temperature
C Low radiation angle for closer distances
D The radiation resistance is higher

Near-Vertical Incidence Sky wave (NVIS) — “The use of very low dipole antennas that radiate at very high elevation angles has become popular in emergency communications (“emcomm”) systems. This works at low frequencies (7 MHz and below) that are lower than the ionosphere’s critical frequency — the highest frequency for which a signal traveling vertically will be reflected.” (ARRL Handbook, 2012 ed., 21.2.12 NVIS Antennas)

ground interactions distort the free-space radiation pattern.

A-7-7-6 (A) How does antenna height affect the horizontal (azimuthal) radiation pattern of a horizontal dipole HF antenna?
A If the antenna is less than one-half wavelength high, reflected radio waves from the ground significantly distort the pattern
B Antenna height has no effect on the pattern
C If the antenna is less than one-half wavelength high, radiation off the ends of the wire is eliminated
D If the antenna is too high, the pattern becomes unpredictable

Below one half-wavelength in antenna height, there is little point in selecting a given broadside direction: for example, choosing a north-south orientation to favour east-west radiation will not pay significant dividends for antennas close to ground.

feedpoint impedance

A-7-7-9 (C) The impedance at the centre of a dipole antenna more than 3 wavelengths above ground would be nearest to:
A 300 ohms
B 600 ohms
C 75 ohms
D 25 ohms

The impedance of a dipole in free space is known as 73 ohms. Below a height of a half-wavelength, impedance is greatly affected by ground proximity. At one wavelength and up, impedance begins to track the free-space value more closely.

A-7-8-1 (A) What is meant by the radiation resistance of an antenna?
A The equivalent resistance that would dissipate the same amount of power as that radiated from an antenna
B The resistance in the atmosphere that an antenna must overcome to be able to radiate a signal
C The specific impedance of an antenna
D The combined losses of the antenna elements and transmission line

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-2 (A) Why would one need to know the radiation resistance of an antenna?
A To match impedances for maximum power transfer
B To measure the near-field radiation density from a transmitting antenna
C To calculate the front-to-side ratio of the antenna
D To calculate the front-to-back ratio of the antenna

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-3 (C) What factors determine the radiation resistance of an antenna?
A Sunspot activity and time of day
B It is a physical constant and is the same for all antennas
C Antenna location with respect to nearby objects and the conductors length/diameter ratio
D Transmission line length and antenna height

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-4 (A) What is the term for the ratio of the radiation resistance of an antenna to the total resistance of the system?
A Antenna efficiency
B Beamwidth
C Effective Radiated Power
D Radiation conversion loss

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-5 (D) What is included in the total resistance of an antenna system?
A Radiation resistance plus transmission resistance
B Transmission line resistance plus radiation resistance
C Radiation resistance plus space impedance
D Radiation resistance plus ohmic resistance

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-8 (D) What is the term used for an equivalent resistance which would dissipate the same amount of energy as that radiated from an antenna?
A j factor
B Antenna resistance
C K factor

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

The half-power beam width is defined as the angle through which the directivity is fewer than 3 dB down from the peak.

A-7-8-6 (C) How can the approximate beamwidth of a beam antenna be determined?
A Measure the ratio of the signal strengths of the radiated power lobes from the front and side of the antenna
B Measure the ratio of the signal strengths of the radiated power lobes from the front and rear of the antenna
C Note the two points where the signal strength is down 3 dB from the maximum signal point and compute the angular difference
D Draw two imaginary lines through the ends of the elements and measure the angle between the lines

Beamwidth is defined as the width in degrees over which the major lobe is within 3 dB of maximum gain, this is equally described as the angle between the half-power points.

A-7-8-9 (B) Antenna beamwidth is the angular distance between:
A the 3 dB power points on the first minor lobe
B the points on the major lobe at the half-power points
C the maximum lobe spread points on the major lobe
D the 6 dB power points on the major lobe

Beamwidth is defined as the width in degrees over which the major lobe is within 3 dB of maximum gain, this is equally described as the angle between the half-power points.

antenna efficiency

A-7-8-7 (D) How is antenna percent efficiency calculated?
A (radiation resistance / transmission resistance) x 100
B (total resistance / radiation resistance) x 100
C (effective radiated power / transmitter output) x 100
D (radiation resistance / total resistance) x 100

The power delivered to an antenna can be transformed in two ways: some of it is lost through heat and dielectric losses, the rest is radiated. The part that is radiated can be imagined to have “disappeared” into a virtual resistance. Radiation Resistance is defined as an equivalent resistance that would have dissipated all the power radiated. The dimensions of the radiating element, particularly its length, and its immediate environment, the proximity to ground for instance, affect radiation resistance. Except for electrically short antennas, radiation resistance makes up most of the antenna impedance. Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100.

A-7-8-10 (C) If the ohmic resistance of a half-wave dipole is 2 ohms, and the radiation resistance is 72 ohms, what is the antenna efficiency?
A 72%
B 100%
C 97.3%
D 74%

Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100. In this example, 72 divided by 74 is 97.3%

A-7-8-11 (A) If the ohmic resistance of a miniloop antenna is 2 milliohms and the radiation resistance is 50 milliohms, what is the antenna efficiency?
A 96.15%
B 52%
C 25%
D 50%

Antenna efficiency in percentage can be computed as Radiation Resistance over total resistance times 100. In this example, 50 divided by 52 is 96.2%

miscellaney

doppler shift is relevant in the context of satellite work

A-7-5-4 (C) On VHF/UHF frequencies, Doppler shift becomes of consequence on which type of communication?
A Simplex line-of-sight contact between hand-held transceivers
B Contact with terrestrial mobile stations
C Contact via satellite
D Contact through a hilltop repeater

“Doppler shift: the change in frequency of a received signal due to the motion of the satellite. This requires adjustment of the transmit or receive frequency, with the common practice being to change the higher of the two frequencies in use” (http://www.amsat.org/). Low Earth orbiting satellites travel at speeds around 28 000 km/h. The higher the operating frequency, the higher the possible shift: for example, +/- 600 Hz on 10 m, +/- 3 kHz on 2 m and +/- 9 kHz on 70 cm.