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ACMA Advanced Amateur Radio Theory study materials
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ACMA Advanced Amateur Radio Theory practice questions

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Question 1Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

How many milliamperes are equivalent to 0.025 amperes?

  1. 2.5 mA
  2. 25 mA Correct answer.
  3. 2500 mA
  4. 250 mA

Explanation

The prefix milli means one thousandth, so one ampere equals 1000 milliamperes. Multiplying 0.025 A by 1000 gives 25 mA. Moving the decimal three places to the right converts amperes to milliamperes.

Question 2Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A power gain of 3 dB corresponds to what ratio of output power to input power?

  1. About 10 to 1
  2. About 1.4 to 1
  3. About 2 to 1 Correct answer.
  4. About 3 to 1

Explanation

A 3 dB power change represents a factor of about 2, since 10 log10(2) equals approximately 3.01 dB. Doubling power is the standard reference point in decibel work. A factor of 10 corresponds instead to 10 dB.

Question 3Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

In a purely resistive AC circuit, how do voltage and current relate in phase?

  1. They are 180 degrees out of phase
  2. Voltage leads current by 90 degrees
  3. Current leads voltage by 90 degrees
  4. They are exactly in phase Correct answer.

Explanation

A resistor has no reactance, so the voltage across it and the current through it rise and fall together. The phase angle between them is zero. Reactance, not resistance, is what produces phase shift.

Question 4Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the total resistance of two 100-ohm resistors connected in parallel?

  1. 50 ohms Correct answer.
  2. 25 ohms
  3. 200 ohms
  4. 100 ohms

Explanation

For two equal resistors in parallel the equivalent resistance is half of one value, since the reciprocal formula gives 1/100 + 1/100 = 1/50. Two equal 100-ohm resistors therefore present 50 ohms. Parallel paths always give a value lower than the smallest branch.

Question 5Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

Which property of a capacitor mainly determines its capacitance?

  1. The resistance of its dielectric
  2. The voltage rating printed on its case
  3. The length of its connecting leads
  4. The plate area, plate spacing, and dielectric material Correct answer.

Explanation

Capacitance depends on plate area, the distance between plates, and the permittivity of the dielectric. Larger area or closer spacing increases capacitance. Voltage rating and lead length do not set the capacitance value.

Question 6Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What happens to the resonant frequency of an LC circuit if the inductance is increased while capacitance stays the same?

  1. The circuit stops resonating
  2. The resonant frequency increases
  3. The resonant frequency stays the same
  4. The resonant frequency decreases Correct answer.

Explanation

Resonance occurs where inductive and capacitive reactances cancel, giving f = 1/(2 pi sqrt(LC)). Because inductance appears in the denominator, raising L lowers the resonant frequency. The square root means the change is not linear.

Question 7Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the impedance of an ideal capacitor at DC?

  1. Equal to its reactance at 1 kHz
  2. Equal to its ESR
  3. Infinite Correct answer.
  4. Zero

Explanation

Capacitive reactance is 1/(2 pi f C), which grows without limit as frequency approaches zero. At steady DC the capacitor therefore behaves as an open circuit with infinite impedance. This is why capacitors block direct current.

Question 8Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A resistor has bands of brown, black, red, and gold. What is its nominal value?

  1. 10 000 ohms
  2. 1000 ohms Correct answer.
  3. 100 ohms
  4. 10 ohms

Explanation

Brown is 1, black is 0, and red as a multiplier means times 100. The first two digits form 10, and 10 times 100 gives 1000 ohms. The gold band then indicates a tolerance of plus or minus 5 percent.

Question 9Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the reactance of a 1 microhenry inductor at 10 MHz?

  1. About 6300 ohms
  2. About 63 ohms Correct answer.
  3. About 6.3 ohms
  4. About 630 ohms

Explanation

Inductive reactance is 2 pi f L. Substituting 10 MHz and 1 microhenry gives 2 pi times 10, which is about 63 ohms. Reactance rises in direct proportion to frequency.

Question 10Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A signal power drops from 100 watts to 1 watt. What is the change in decibels?

  1. -20 dB Correct answer.
  2. -30 dB
  3. -40 dB
  4. -10 dB

Explanation

Power decibels equal 10 log10(Pout/Pin). The ratio here is 0.01, and the log of 0.01 is -2, so the result is -20 dB. Each 10 dB step corresponds to a factor of ten in power.

Question 11Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

Two 10 microfarad capacitors are connected in series. What is the total capacitance?

  1. 20 microfarads
  2. 40 microfarads
  3. 10 microfarads
  4. 5 microfarads Correct answer.

Explanation

Capacitors in series combine like resistors in parallel, so two equal values give half of one value. The reciprocal sum 1/10 + 1/10 yields 1/5, giving 5 microfarads. Series connection always reduces total capacitance below the smallest part.

Question 12Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

Which statement correctly describes a band-pass filter?

  1. It attenuates frequencies between two cutoff points and passes others
  2. It passes only a single frequency with zero bandwidth
  3. It passes all frequencies above a single cutoff
  4. It passes frequencies between two cutoff points and attenuates others Correct answer.

Explanation

A band-pass filter has a lower and an upper cutoff frequency, and it passes signals lying between them. Frequencies outside that band are attenuated. The width between cutoffs is the passband bandwidth.

Question 13Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A circuit draws 2 amperes from a 12 volt supply. How much power is dissipated?

  1. 24 watts Correct answer.
  2. 6 watts
  3. 14 watts
  4. 48 watts

Explanation

DC power equals voltage multiplied by current, so 12 volts times 2 amperes gives 24 watts. The product of potential and current gives the rate of energy transfer. Dividing or adding the two quantities does not yield power.

Question 14Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the quality factor Q of a series resonant circuit with 100 ohms of resistance and 500 ohms of inductive reactance at resonance?

  1. 5 Correct answer.
  2. 0.2
  3. 2
  4. 50

Explanation

For a series resonant circuit, Q equals the reactance at resonance divided by the resistance. Dividing 500 ohms by 100 ohms gives 5. A higher Q means a narrower, sharper resonance.

Question 15Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

In a series RLC circuit at resonance, what is the impedance?

  1. Zero in all cases
  2. Equal to the sum of R, XL, and XC
  3. Equal to the inductive reactance
  4. Equal to the resistance only Correct answer.

Explanation

At resonance the inductive and capacitive reactances are equal and opposite, so they cancel. The remaining impedance is purely resistive and equals R. This is the point of minimum impedance for a series circuit.

Question 16Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

How does the capacitive reactance of a fixed capacitor change as frequency increases?

  1. It stays constant
  2. It decreases in inverse proportion to frequency Correct answer.
  3. It increases in proportion to frequency
  4. It rises to a peak then falls

Explanation

Capacitive reactance is given by 1/(2 pi f C), so frequency appears in the denominator. Doubling the frequency halves the reactance. This inverse relationship lets a capacitor pass higher frequencies more readily.

Question 17Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A 12 volt source feeds two 100 ohm resistors in series. What current flows?

  1. 0.24 A
  2. 0.06 A Correct answer.
  3. 0.12 A
  4. 1.2 A

Explanation

Series resistors add, giving 200 ohms total. Ohm's law then gives 12 volts divided by 200 ohms, which is 0.06 amperes. Using only one resistor would overstate the current.

Question 18Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the main effect of a low-pass filter on a square wave?

  1. It amplifies the higher harmonics
  2. It removes the high-frequency harmonics and rounds the edges Correct answer.
  3. It shifts the waveform phase by exactly 90 degrees
  4. It converts the waveform to a pure sine at twice the frequency

Explanation

A square wave is built from a fundamental plus many odd harmonics. A low-pass filter attenuates those higher harmonics, which softens the fast transitions and rounds the corners. The fundamental remains largely intact.

Question 19Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A transformer has 100 turns on the primary and 500 turns on the secondary. What is the voltage step-up ratio?

  1. 5 to 1
  2. 1 to 2
  3. 25 to 1
  4. 1 to 5 Correct answer.

Explanation

The voltage ratio equals the turns ratio, so 100 to 500 reduces to 1 to 5. The secondary voltage is five times the primary voltage. Current transforms in the opposite direction.

Question 20Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

What is the phase relationship between voltage and current in an ideal inductor?

  1. Voltage leads current by 90 degrees Correct answer.
  2. Current leads voltage by 90 degrees
  3. Voltage lags current by 45 degrees
  4. Voltage and current are in phase

Explanation

In an inductor the voltage is proportional to the rate of change of current, which places voltage a quarter cycle ahead. This is often remembered with the phrase ELI. A capacitor shows the opposite relationship.

Question 21Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A resistor carries 0.5 amperes and has a value of 200 ohms. What is the voltage across it?

  1. 50 volts
  2. 100 volts Correct answer.
  3. 400 volts
  4. 200 volts

Explanation

Ohm's law states that voltage equals current times resistance. Multiplying 0.5 amperes by 200 ohms gives 100 volts. Dividing current by resistance would give the wrong quantity entirely.

Question 22Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A resonant circuit has a bandwidth of 20 kHz at a center frequency of 1000 kHz. What is its loaded Q?

  1. 20
  2. 2000
  3. 50 Correct answer.
  4. 200

Explanation

Loaded Q equals the resonant frequency divided by the bandwidth. Dividing 1000 kHz by 20 kHz gives 50. Higher Q therefore corresponds to a narrower bandwidth.

Question 23Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

An amplifier has a gain of 20 dB followed by a loss of 3 dB. What is the overall change in power?

  1. About 50 times
  2. About 17 times
  3. About 23 times
  4. About 5 times Correct answer.

Explanation

Decibel gains in cascade add, so the net change is 20 dB minus 3 dB, or 17 dB. That equals 10^(1.7), which is about 50; however, 20 dB alone is 100 times and 3 dB is about half, so 100 divided by 2 gives about 50. Rechecking: 17 dB corresponds to roughly 50 times.

Question 24Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A parallel resonant circuit is used as the load in an amplifier. Compared with a series resonant circuit, how does its impedance behave at resonance?

  1. It equals the reactance of the inductor
  2. It reaches a maximum rather than a minimum Correct answer.
  3. It falls to zero at resonance
  4. It reaches a minimum rather than a maximum

Explanation

In a parallel tank the circulating currents cancel at resonance, leaving only the losses, so the impedance peaks. A series circuit does the opposite and reaches minimum impedance. This high impedance is why parallel tanks suit amplifier loads.

Question 25Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A coil has an inductance of 100 microhenries and a series resistance of 10 ohms. What is its Q at 1 MHz?

  1. About 63 Correct answer.
  2. About 6.3
  3. About 6300
  4. About 630

Explanation

First find the reactance: 2 pi times 1 MHz times 100 microhenries equals about 628 ohms. Dividing by the 10 ohm series resistance gives a Q near 63. Coil Q rises with frequency until parasitic effects dominate.

Question 26Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A signal is attenuated by a factor of 1000 in power. Expressed in decibels, what is this attenuation?

  1. 30 dB Correct answer.
  2. 20 dB
  3. 10 dB
  4. 3 dB

Explanation

Power in decibels is computed as ten times the base-10 logarithm of the power ratio. A ratio of 1000 corresponds to a logarithm of 3, so the attenuation is 30 dB. Because the ratio is a loss, the change is often stated as minus 30 dB.

Question 27Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

Three resistors of 220 ohms, 330 ohms, and 470 ohms are connected in parallel. Which value is closest to the total resistance?

  1. 1020 ohms
  2. 340 ohms
  3. 220 ohms
  4. 104 ohms Correct answer.

Explanation

For parallel resistors the reciprocals add: 1/R = 1/220 + 1/330 + 1/470, which equals about 0.00966 siemens. Inverting gives roughly 104 ohms. A parallel combination is always smaller than the smallest branch resistance.

Question 28Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

An inductor and a capacitor are connected in parallel and the combination is driven well below its resonant frequency. How does the network behave?

  1. It behaves as a pure resistance equal to the coil resistance.
  2. It behaves as an open circuit with no current flowing.
  3. It behaves inductively, with current lagging the applied voltage. Correct answer.
  4. It behaves capacitively, with current leading the applied voltage.

Explanation

Below resonance the inductive reactance is smaller than the capacitive reactance, so the inductor dominates the parallel combination. The net current therefore lags the applied voltage, giving an inductive character. As frequency rises through resonance the behavior reverses and the network becomes capacitive.

Question 29Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A coil has an inductive reactance of 200 ohms and a series resistance of 25 ohms at a given frequency. What is its quality factor at that frequency?

  1. 0.125
  2. 5000
  3. 175
  4. 8 Correct answer.

Explanation

The quality factor of a coil at a specified frequency is the ratio of its inductive reactance to its series resistance. Dividing 200 ohms by 25 ohms gives 8. A higher ratio means lower relative loss in the coil.

Question 30Radio and electronics fundamentals (units, decibels, AC and DC circuits, components, resonance, filters)

A series resonant circuit is driven at a frequency slightly above resonance. Which statement best describes the resulting phase and magnitude of the current?

  1. The circuit is capacitive, so the current leads the source voltage and its magnitude is above the resonant value.
  2. The circuit is inductive, so the current lags the source voltage and its magnitude is below the resonant value. Correct answer.
  3. The circuit is capacitive, so the current lags the source voltage and its magnitude is below the resonant value.
  4. The circuit is resistive, so the current is in phase with the source voltage at its maximum value.

Explanation

Above resonance the inductive reactance exceeds the capacitive reactance, so the net reactance is inductive and the current lags the applied voltage. The net reactance also raises the total impedance above the purely resistive value at resonance, so the current magnitude falls below its peak.

Question 31Semiconductor devices, amplifiers, oscillators and power supplies

In a bipolar junction transistor operating in its normal active region, which charge carriers are the majority carriers that travel from the emitter to the collector?

  1. Electrons drifting from the collector to the emitter
  2. Holes injected from the base into the emitter
  3. Electrons injected from the emitter into the base, in an NPN device Correct answer.
  4. Holes generated thermally inside the collector depletion region

Explanation

In an NPN transistor the emitter is heavily doped N-type material, so it injects electrons into the thin base. Because the base is thin and lightly doped, most of these electrons diffuse across it and are swept into the collector by the reverse-biased collector junction, giving transistor action. The base current is a small fraction due to recombination.

Question 32Semiconductor devices, amplifiers, oscillators and power supplies

Which property best describes the gate-to-source junction of a junction field-effect transistor (JFET) when it is used as an amplifier?

  1. It is forward biased to draw gate current
  2. It is biased into avalanche breakdown for gain
  3. It is unbiased and floats at the source potential
  4. It is always reverse biased so the gate draws almost no current Correct answer.

Explanation

A JFET controls drain current by the width of the depletion region around a reverse-biased gate junction. Reverse bias keeps gate current extremely small, so the device has very high input impedance at low frequencies. Forward biasing the gate would cause conduction and destroy the high-impedance control action.

Question 33Semiconductor devices, amplifiers, oscillators and power supplies

What is the primary function of a Zener diode in a regulated power supply?

  1. Amplifying the error signal in the feedback loop
  2. Providing a stable reference voltage when operated in reverse breakdown Correct answer.
  3. Filtering ripple by storing large amounts of charge
  4. Rectifying the AC mains before the transformer

Explanation

A Zener diode is designed to operate in controlled reverse breakdown, where its voltage stays nearly constant over a range of currents. That constant voltage serves as a shunt reference or regulator. Rectification, amplification and bulk filtering are done by other components.

Question 34Semiconductor devices, amplifiers, oscillators and power supplies

In a class B push-pull audio amplifier, what is the main purpose of forward-biasing the output devices slightly into class AB?

  1. To reduce crossover distortion at the zero crossing Correct answer.
  2. To eliminate the need for an output transformer
  3. To increase the maximum supply voltage rating
  4. To raise the efficiency above the theoretical class B limit

Explanation

Pure class B operation leaves a dead zone near zero volts where neither device conducts, producing crossover distortion. A small forward bias keeps both devices slightly on so the transfer characteristic is continuous. This trades a little efficiency for much lower distortion.

Question 35Semiconductor devices, amplifiers, oscillators and power supplies

Which oscillator topology uses a tapped inductor or tapped capacitor divider to feed back a fraction of the resonant tank voltage to the amplifier input?

  1. Wien bridge oscillator
  2. Pierce oscillator
  3. Colpitts oscillator
  4. Hartley oscillator Correct answer.

Explanation

In the Hartley oscillator the feedback is taken from a tap on the inductor, forming an inductive divider across the resonant tank. The Colpitts and Pierce circuits use capacitive dividers or crystals instead. The Wien bridge is an RC oscillator with no tapped inductor.

Question 36Semiconductor devices, amplifiers, oscillators and power supplies

What is the effect of connecting a large filter capacitor across the output of a full-wave rectifier?

  1. It converts the output back to AC
  2. It lowers the peak inverse voltage seen by the diodes
  3. It increases the ripple frequency
  4. It reduces ripple by charging near the peak and discharging between peaks Correct answer.

Explanation

The capacitor charges to the peak rectified voltage and then supplies the load while the rectified waveform falls, holding the output up between peaks. This smooths the ripple, with larger capacitance or lighter load giving less ripple. Ripple frequency itself is set by the rectifier configuration and mains frequency.

Question 37Semiconductor devices, amplifiers, oscillators and power supplies

A silicon diode is forward biased and conducting. Approximately what forward voltage drop should be expected?

  1. 0.7 V Correct answer.
  2. 0.2 V
  3. 1.5 V
  4. 5.0 V

Explanation

A forward-biased silicon PN junction has a barrier potential of roughly 0.6 to 0.7 volts at normal currents. Germanium junctions are nearer 0.3 volts and Schottky devices lower still. Values like 1.5 or 5 volts far exceed a single silicon junction drop.

Question 38Semiconductor devices, amplifiers, oscillators and power supplies

An NPN transistor has a base current of 0.1 mA and a collector current of 8 mA in the active region. What is its approximate DC current gain?

  1. 0.0125
  2. 8
  3. 800
  4. 80 Correct answer.

Explanation

DC current gain beta equals collector current divided by base current. Dividing 8 mA by 0.1 mA gives 80, a typical value for a small-signal transistor. The result is dimensionless and should be in the tens to hundreds range.

Question 39Semiconductor devices, amplifiers, oscillators and power supplies

A common-emitter amplifier stage is required to present a high input impedance and moderate voltage gain. Which configuration best meets this need?

  1. Common emitter with a fully bypassed emitter resistor
  2. Common base
  3. Common collector
  4. Common emitter with an unbypassed emitter resistor Correct answer.

Explanation

Leaving the emitter resistor unbypassed introduces local negative feedback, which raises input impedance and stabilizes gain at roughly the collector-to-emitter resistor ratio. A bypassed emitter resistor gives higher gain but lower input impedance. Common base has low input impedance and common collector has near-unity voltage gain.

Question 40Semiconductor devices, amplifiers, oscillators and power supplies

A MOSFET is used as a switching device in a DC-DC converter. Why is it driven hard between fully on and fully off states?

  1. To minimize conduction and switching losses by avoiding the active region Correct answer.
  2. To increase its gate threshold voltage permanently
  3. To allow the body diode to conduct continuously
  4. To keep it in its linear region for smooth output

Explanation

In switching applications the device spends little time in the high-dissipation active region. Fully on gives low drain-source resistance and fully off gives negligible current, so power loss is minimized. Remaining in the linear region would dissipate large amounts of heat.

Question 41Semiconductor devices, amplifiers, oscillators and power supplies

A shunt Zener regulator has a 12 V Zener, a series resistor of 470 ohms from a 20 V supply, and a load drawing 10 mA. What current flows through the Zener, approximately?

  1. About 10 mA
  2. About 17 mA
  3. About 27 mA
  4. About 7 mA Correct answer.

Explanation

The series resistor drops 20 minus 12, or 8 volts, giving 8/470, about 17 mA total current. Subtracting the 10 mA load leaves roughly 7 mA through the Zener. The Zener must carry the difference between supplied and consumed current.

Question 42Semiconductor devices, amplifiers, oscillators and power supplies

Which condition must be satisfied for a feedback oscillator to sustain steady oscillation?

  1. Loop gain magnitude greater than one with any phase shift
  2. Loop gain magnitude equal to one with total loop phase shift of 180 degrees
  3. Loop gain magnitude equal to one with total loop phase shift of 360 degrees Correct answer.
  4. Loop gain magnitude less than one with zero phase shift

Explanation

The Barkhausen criterion requires unity loop gain magnitude and a total loop phase shift of a full 360 degrees, equivalent to zero, so feedback is regenerative. In practice the loop gain is made slightly greater than one at start-up and then limited to unity by nonlinearity. A 180 degree loop shift would give negative feedback and suppress oscillation.

Question 43Semiconductor devices, amplifiers, oscillators and power supplies

Why is a quartz crystal preferred over a simple LC tank in a frequency-standard oscillator?

  1. It produces a larger output amplitude than any LC circuit
  2. It can be tuned over a wider range than an LC tank
  3. It operates without any active amplifier
  4. It has a much higher Q and far greater frequency stability Correct answer.

Explanation

The piezoelectric resonance of quartz has an extremely high quality factor, so the resonant frequency is sharply defined and drifts little with temperature and component variation. LC tanks have much lower Q and drift more. Crystals trade wide tunability for this stability.

Question 44Semiconductor devices, amplifiers, oscillators and power supplies

A linear voltage regulator is dropping 8 volts at a load current of 1.5 A. How much power is dissipated as heat in the regulator?

  1. 8 W
  2. 12 W Correct answer.
  3. 1.5 W
  4. 18 W

Explanation

Power dissipated in a series pass element equals the voltage across it multiplied by the current through it. Here 8 volts times 1.5 amperes gives 12 watts, which requires an adequate heat sink. This loss is the main efficiency penalty of linear regulation.

Question 45Semiconductor devices, amplifiers, oscillators and power supplies

What advantage does a switching-mode power supply have over a linear supply of similar output rating?

  1. Higher efficiency because the pass device dissipates little power Correct answer.
  2. Complete freedom from electromagnetic interference
  3. Much lower output ripple with no filtering
  4. Simpler circuitry with fewer components

Explanation

A switching supply rapidly switches its pass device between low-loss on and off states, so little power is dissipated as heat and efficiency is high. Linear supplies burn the excess voltage as heat. The trade-off is greater circuit complexity, more ripple and more electromagnetic interference.

Question 46Semiconductor devices, amplifiers, oscillators and power supplies

In a class C RF power amplifier, the transistor conducts for only a portion of each cycle. What is the main consequence for the output?

  1. High efficiency with a strongly non-sinusoidal current that requires a tuned output network Correct answer.
  2. Low efficiency but perfectly linear amplification
  3. Very low distortion suitable for wideband audio
  4. Operation independent of any resonant load

Explanation

Class C conduction over less than half a cycle gives high efficiency because the device is off most of the time. The resulting current pulses are rich in harmonics, so a resonant tank at the output restores a sinusoidal waveform and suppresses unwanted components. This suits narrowband RF amplification, not wideband audio.

Question 47Semiconductor devices, amplifiers, oscillators and power supplies

An operational amplifier is configured as a non-inverting amplifier with a 2.2 kilohm feedback resistor and a 1 kilohm resistor from the inverting input to ground. What is the approximate closed-loop voltage gain?

  1. 2.2
  2. 0.31
  3. 3.2 Correct answer.
  4. 1.0

Explanation

For the non-inverting configuration the gain is one plus the feedback resistor divided by the ground resistor. One plus 2.2 divided by 1 equals about 3.2. The inverting configuration would instead use the negative ratio of the two resistors.

Question 48Semiconductor devices, amplifiers, oscillators and power supplies

A power MOSFET in a switching converter shows a measured drain current that briefly spikes far above the steady on-state value each time it turns on. Which factor most likely explains this spike?

  1. The body diode is permanently forward biased
  2. Parasitic capacitance and diode reverse recovery charge must be supplied at turn-on Correct answer.
  3. The gate threshold voltage has fallen to zero
  4. The device has entered thermal runaway

Explanation

At turn-on the device must charge parasitic capacitances and, in circuits with a freewheeling diode, absorb the reverse-recovery charge of that diode. Both draw a brief current spike beyond the steady load current. Good layout and appropriate diode choice reduce this stress.

Question 49Semiconductor devices, amplifiers, oscillators and power supplies

A Colpitts oscillator is to be designed to run at 7.0 MHz using a 1 microhenry inductor. Using the series combination of the two feedback capacitors, what total capacitance is required approximately?

  1. About 520 pF Correct answer.
  2. About 52 pF
  3. About 5.2 pF
  4. About 5.2 nF

Explanation

The resonant frequency is one over two pi times the square root of LC. Solving for C gives one over four pi squared times frequency squared times L, which for 7 MHz and 1 microhenry is about 5.2 times ten to the minus ten farads, or roughly 520 picofarads. This is the effective series value of the two divider capacitors.

Question 50Semiconductor devices, amplifiers, oscillators and power supplies

A push-pull class B stage runs from a 24 V supply into an 8 ohm load. Assuming ideal devices and a sinusoidal output swinging to the full supply rails, what is the approximate maximum output power?

  1. About 72 W
  2. About 36 W
  3. About 18 W
  4. About 9 W Correct answer.

Explanation

The peak output voltage is half the total supply, or 12 volts, so the peak power is 12 squared divided by twice 8, giving 9 watts. Equivalently, RMS voltage of 8.49 volts squared over 8 ohms gives about 9 watts. Using the full 24 volts as peak would double-count the push-pull swing.