Electronics Demystified, 2nd edition Stan Gibilisco Explanations for Quiz Answers in Chapter 7 1. Class-A, class-AB, and class-C amplifiers will all work well for either AM or FM signals. As long as they're biased properly and we do not overdrive them, these amplifiers will not cause serious distortion of the modulation envelope in an AM signal. Class-C amplifiers will distort an AM signal regardless of the bias or drive. Class-C amplifiers work okay for FM, because FM signals maintain constant amplitude. The correct choice is D. 2. If we don't set the tuning and loading controls properly in a tuned RF power amplifier, we'll get excessive collector or drain current but greatly reduced RF power output. That state of affairs translates to low efficiency and excessive heat dissipation in the transistor. The correct choice is A. 3. Of all the amplifier configurations discussed in this chapter, the class-A scheme provides the best signal-waveform linearity (although some other classes work okay in regards to modulation-envelope linearity). The correct choice is A. 4. The op-amp circuit of Fig. 7-14 contains no technical errors. The correct choice is D. 5. Note that the potentiometer connects between the op-amp output and the inverting input. Therefore, the signal path through the potentiometer produces negative feedback. As we reduce the resistance in this path, we increase the amount of negative feedback and thereby reduce the circuit gain. The correct choice is B. 6. To solve this problem, we must calculate the voltage gain in decibels (dB). The formula isGain (dB) = 20 log10 (Eout / Ein) where Eout represents the RMS output voltage, Ein represents the RMS input voltage, and log10 stands for the base-10 logarithm function. In this case, we don't know the actual output and input voltages, but we do know that the output-to-input voltage ratio equals 20:1, or 20. Therefore Gain (dB) = 20 log10 20 = 20 x 1.3 = 26 dB Because this figure represents positive gain (that is, an increase in the amplitude), we can say that the change, in terms of RMS voltage, equals +26 dB. The correct choice is D. 7. Before we do any calculations here, we must take note of the fact that the question asks us about power gain, not voltage gain as the previous question specified! We want to find the power gain in decibels, so we should use the formulaGain (dB) = 10 log10 (Pout / Pin) where Pout represents the output power, Pin represents the input power, and log10 stands for the base-10 logarithm function. We don't know the actual output and input power levels, but we do know that the output-to-input power ratio equals 1:20, or 0.050. Therefore Gain (dB) = 10 log10 0.050 = 10 x -1.3 = -13 dB The figure comes out negative because we have a loss, which translates to negative gain. The correct choice is B. 8. The circuit described in this question employs tuned-circuit transformer coupling. When we use this type of arrangement, we can get excellent selectivity and gain. However, these advantages come along with an increased risk of oscillation at the resonant frequency of the tuned transformer windings. The correct choice is C. 9. In a class-A, class-AB1, or class-AB2 amplifier, we'll always see some collector or drain current, even in the complete absence of any input signal. That current translates into nonzero DC power input. However, in a class-B amplifier, we don't see any collector or drain current unless we apply a signal. Under no-signal conditions, the bipolar or field-effect transistor operates in a state of cutoff or pinchoff in class B. We must apply an input signal in order to generate any DC power input. The correct choice is A. 10. We should use a weak-signal audio amplifier if we want a microphone to detect faint sounds. The best design for weak-signal amplification at any frequency employs the class-A mode. We can generally expect to get better sensitivity with a JFET than we can with a bipolar transistor. The correct choice is C.