Teach Yourself Electricity and Electronics, 5th edition

Stan Gibilisco

Explanations for Quiz Answers in Chapter 24

1. The Colpitts, Clapp, and Hartley oscillators all employ inductance-capacitance (LC) tuned circuits to determine their operating frequencies. The Pierce oscillator employs a quartz crystal for that purpose. Crystal oscillators have better frequency stability, in general, than LC-tuned oscillators do. Therefore, the correct choice is (d).

2. We can solve this problem using the formula for decibel gain in terms of relative RMS voltages. Here's that formula again, for reference:

Gain (dB) = 20 log (E_out / E_in)

where E_out represents the RMS signal output voltage and E_in represents the RMS signal input voltage. We're told that the RMS voltage increases by a factor of 10,000, so we can substitute 10,000 for the ratio (E_out / E_in). When make that substitution in the above formula, we get

Gain (dB) = 20 log 10,000
= 20 x 4.0
= 80 dB

The correct choice is (b).

3. To solve this problem, we must use the formula for decibel gain in terms of relative RMS power levels. Here's the formula for reference:

Gain (dB) = 10 log (P_out / P_in)

where P_out represents the RMS signal output power and P_in represents the RMS signal input power. We know that the gain equals 33.0 dB while P_in equals 1.00 W. When we substitute these numbers into the above formula, we obtain

33.0 = 10 log (P_out / 1.00)

When we simplify the power ratio and divide each side by 10, we get

3.30 = log P_out

Taking the antilogarithm of both sides gives us

antilog 3.30 = antilog (log P_out)

which simplifies to

1995 = P_out

We can convert this figure to 1.995 kW rms and then round off to 2.00 kW RMS. The correct choice is (a).

4. Let's use the formula for the decibel gain in terms of RMS voltages, setting the output voltage E_out = 10 and the input voltage E_in = 30. Then we have

Gain (dB) = 20 log (E_out / E_in)
= 20 log (10/30)
= 20 log (1/3)
= 20 x (-0.4771)
= -9.5 dB

rounded off to two significant digits. That's the voltage gain, remember! We can also call it a voltage loss of 9.5 dB. The correct choice is therefore (a).

5. If we want to let DC pass while blocking high-frequency AC, we'd choose a radio-frequency (RF) choke. The correct choice is (c), with one cautionary note: We must make certain that the choke has the appropriate inductance for the impedance and AC frequency involved. Choices (a), (b), and (d) are completely wrong. None of the other three components (varactor, blocking capacitor, or Gunn diode) would do the job here.

6. Figure 24-20 is a schematic diagram of an Armstrong oscillator using a P-channel MOSFET and a transformer with a powdered-iron core. The correct choice is (c). In this particular example, we take the output from the drain circuit. We could also take the output from the source circuit for improved reliability, as is commonly done with other types of oscillators such as the Hartley, Colpitts, and Pierce designs.

7. The circuit of Fig. 24-20 contains to technical errors, as long as we ensure that we connect the transformer to cause feedback in the proper phase. Because we have a common-source circuit configuration, the AC signal at the gate should have the opposite phase from the AC signal at the drain. The correct choice is (a).

8. If we want an Armstrong circuit such as the one diagrammed in Fig. 24-20 to function properly, we must connect the transformer to ensure that the gate receives an AC signal in phase opposition with respect to the AC signal at the drain. The answer is (a).

9. The collector or drain carries current for less than half of the AC input signal cycle in a class-C bipolar-transistor amplifier. The correct choice is (a).

10. When we intend to design and build a class-B push-pull amplifier, we should select two bipolar transistors or FETs with the same manufacturer's part number, and with characteristics that match as closely as possible. The correct response is (d).

11. The output signal frequency of a Hartley oscillator depends on the component values in a tuned inductance-capacitance (LC) circuit. This type of oscillator does not have a quartz crystal, nor does it take advantage of a phase-locked loop. The correct choice is (b).

12. Class-A or class-B push-pull FET amplifier circuits introduce essentially no distortion into the input signal wave. However, drain current flows for the entire signal cycle only in the class-A system. The correct choice is (a). We might be tempted to answer (c), but we can reject this choice because it doesn't specify "push-pull."

13. We can make a class-B amplifier linear with respect to the AC input signal by connecting two bipolar or field-effect transistors in a push-pull configuration. The correct choice is (c). We must remember, however, to choose two transistors with the same manufacturer's part number, and to test them (if possible) to ensure that their operating characteristics are essentially identical.

14. If we connect a low-impedance load to an oscillator, we might have trouble obtaining or maintaining oscillation. The correct choice is (d).

15. We're told that an FET-based amplifier has an efficiency of 60%, and that its drain input power equals 90 W. We can calculate its RF output power using the formula

eff_% = 100 P_out / P_in

where eff_% represents the efficiency in percent, P_out represents the output power in watts, and P_in represents the input power in watts. Plugging in the values eff_% = 60 and P_in = 90, we obtain

60 = 100 P_out / 90

Multiplying through by 90, we get

5400 = 100 P_out

We can divide each side by 100 to get the answer as

P_out = 5400 / 100
= 54 W

The correct choice is (a).

16. In the circuit of Fig. 24-21, capacitor V functions as a blocking capacitor. It allows the input signal to reach the base of the transistor, but keeps it isolated for DC from preceding components, circuits, or systems. The correct answer is (b).

17. In the circuit of Fig. 24-21, resistor W provides bias for the transistor. The correct choice is (a). Incidentally, this resistor's ohmic value should roughly match the (purely resistive) output impedance of the circuit or system that drives the PA.

18. In the circuit of Fig. 24-21, the RF choke labeled X keeps the high-frequency AC signal from taking a shortcut through the power supply instead of going to the output terminals. The correct choice is (d). Choke X doesn't limit the current through the transistor, nor does it provide DC isolation for or from anything, nor does it have any effect on the resonant frequency, so we can reject choices (a), (b), and (c).

19. In the circuit of Fig. 24-21, the inductors labeled Y constitute part of an LC network designed to optimize the transfer of signal to the output. The correct choice is (a).

20. In the circuit of Fig. 24-21, the capacitors labeled Z, along with the inductors labeled Y, form an LC network that optimizes the transfer of signal to the output. The correct choice, once again, is (a). To obtain ideal system performance, we should adjust the values of the capacitors to obtain maximum RF output as indicated by an RF wattmeter, while providing the amplifier with an input signal at the desired frequency and at the correct power level.