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A NSWERS ◆^1
ANSWERS Chapter 1
SECTION CHECKUPS
Section 1–1 The Atom
1. The Bohr model has a central nucleus consisting of protons and neutrons orbited by electrons at varying distances from the nucleus. 2. An electron is the smallest particle of negative electrical charge. 3. Protons and neutrons. A proton is a particle of positive charge and a neutron has no net charge. 4. The atomic number is the number of electrons in the nucleus of an atom. 5. An electron shell contains orbiting electrons at a certain energy level. Each shell of a given atom is at a different energy level. 6. A valence electron is one that is in the outer shell of an atom. 7. A free electron is a valence electron that has broken free of its parent atom. 8. A positive ion is a previously neutral atom that has lost a valence electron and has a net positive charge. A negative ion is one that has gained an extra electron and has a net negative charge. 9. The quantum model is based on the uncertainty principle and wave-particle duality.
Section 1–2 Materials Used in Electronics
1. Conductors have many free electrons and easily conduct current. Insulators have essentially no free electrons and do not conduct current. 2. Semiconductors do not conduct current as well as conductors do. In terms of conductivity, they are between conductors and insulators. 3. Conductors such as copper have one valence electron. 4. Semiconductors have four valence electrons. 5. Gold, silver, and copper are the best conductors. 6. Silicon is the most widely used semiconductor. 7. The valence electrons of a semiconductor are more tightly bound to the atom than those of conductors. 8. Covalent bonds are formed by the sharing of valence electrons with neighboring atoms. 9. An intrinsic material is one that is in a pure state. 10. A crystal is a solid material formed by atoms bonding together in a symmetrical pattern.
Section 1–3 Current in Semiconductors
1. Free electrons are in the conduction band. 2. Free (conduction) electrons are responsible for electron current in silicon. 3. A hole is the absence of an electron in the valence band. 4. Hole current occurs at the valence level.
Section 1–4 N -Type and P -Type Semiconductors
1. Doping is the process of adding impurity atoms to a semiconductor in order to modify its con- ductive properties. 2. A pentavalent atom has five valence electrons and a trivalent atom has three valence electrons. 3. A pentavalent atom is called a donor atom and a trivalent atom is called an acceptor atom. 4. An n -type material is formed by the addition of pentavalent impurity atoms to the intrinsic semiconductive material. 5. A p- type material is formed by the addition of trivalent impurity atoms to the intrinsic semi- conductive material. 6. The majority carrier in an n -type semiconductor is the free electron. 7. The majority carrier in a p -type semiconductor is the hole.
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8. Majority carriers are produced by doping. 9. Minority carriers are thermally produced when electron-hole pairs are generated. 10. A pure semiconductor is intrinsic. A doped (impure) semiconductor is extrinsic.
Section 1–5 The PN Junction
1. A pn junction is the boundary between p -type and n -type semiconductors in a diode. 2. Diffusion is the movement of the free electrons (majority carriers) in the n -region across the pn junction and into the p region. 3. The depletion region is the thin layers of positive and negative ions that exist on both sides of the pn junction. 4. The barrier potential is the potential difference of the electric field in the depletion region and is the amount of energy required to move electrons through the depletion region. 5. The barrier potential for a silicon diode is approximately 0.7 V. 6. The barrier potential for a germanium diode is approximately 0.3 V.
RELATED PROBLEM FOR EXAMPLE
1–1 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^10 4 s^2 4 p^2
TRUE/FALSE QUIZ
1. F 2. T 3. T 4. F 5. T 6. T 7. F 8. T 9. F
SELF-TEST
1. (c) 2. (d) 3. (a) 4. (b) 5. (a) 6. (d) 7. (d) 8. (c) 9. (d) 10. (d) 11. (d) 12. (d) 13. (c) 14. (b) 15. (a) 16. (d) 17. (e) 18. (a) 19. (b) 20. (c) 21. (c) 22. (a) 23. (c) 24. (d) 25. (d)
ANSWERS Chapter 2
SECTION CHECKUPS
Section 2–1 Diode Operation
1. When forward-biased, a diode conducts current. The free electrons in the n region move across the pn junction and combine with the holes in the p region. 2. To forward-bias a diode, the positive side of an external bias voltage is applied to the p region and the negative side to the n region. 3. When reverse-biased, a diode does not conduct current except for an extremely small reverse current. 4. To reverse-bias a diode, the positive side of an external bias voltage is applied to the n region and the negative side to the p region. 5. The depletion region for forward bias is much narrower than for reverse bias. 6. Majority carrier current is produced by forward bias. 7. Reverse current is produced by the minority carriers. 8. Reverse breakdown occurs when the reverse-bias voltage equals or exceeds the breakdown voltage of the pn junction of a diode. 9. Avalanche effect is the rapid multiplication of current carriers in reverse breakdown.
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Section 2–9 The Diode Datasheet
1. 1N4002: 100 V; 1N4003: 200 V; 1N4004: 400 V; 1N4005: 600 V; 1N4006: 800 V 2. No 3. Approximately 0.65 A 4. 12 A
Section 2–10 Troubleshooting
1. 0.5 V to 0.9 V 2. OL 3. An open diode results in no output voltage. 4. An open diode produces a half-wave output voltage. 5. The shorted diode may burn open. Transformer will be damaged. Fuse will blow. 6. The amplitude of the ripple voltage increases with a leaky filter capacitor. 7. There will be no output voltage when the primary opens. 8. The problem may be a partially shorted secondary winding.
RELATED PROBLEMS FOR EXAMPLES
2–1 V D 5 V; V LIMIT 0 V
2–2 3.82 V
2–3 (a) 2.3 V (b) 49.3 V 2–4 (a) 623.3 V (b) 624 V (c) negative half-cycles rather than positive half cycles 2–5 98.7 V 2–6 79.3 V including diode drop 2–7 41.0 V; 41.7 V 2–8 26.9 mV 2–9 3.7% 2–10 A positive peak of 9.9 V and clipped at 2–11 Limited at +10.7 V and 2–12 Change R 3 to or R 2 to 2–13 Same voltage waveform as Figure 2– 2–14 Verify C is shorted and replace it.
TRUE/FALSE QUIZ
1. F 2. F 3. T 4. F 5. T 6. T 7. F 8. T 9. F
10. F 11. T 12. T 13. T 14. F 15. T 16. F 17. F 18. T
19. F 20. T
CIRCUIT-ACTION QUIZ
1. (a) 2. (c) 3. (c) 4. (a) 5. (a) 6. (c) 7. (b) 8. (a) 9. (a) 10. (b) 11. (c) 12. (c) 13. (b) 14. (a) 15. (a) 16. (c) 17. (b) 18. (c) 19. (b) 20. (a)
SELF-TEST
1. (c) 2. (d) 3. (d) 4. (b) 5. (b) 6. (b) 7. (d) 8. (b) 9. (a) 10. (c) 11. (a) 12. (d) 13. (a) 14. (c) 15. (d) 16. (a) 17. (b) 18. (a) 19. (d) 20. (b) 21. (c) 22. (a) 23. (b) 24. (c) 25. (a) 26. (d) 27. (b) 28. (d) 29. (c) 30. (b) 31. (a) 32. (b) 33. (c) 34. (b) 35. (a) 36. (c) 37. (b)
100 Æ 220 Æ.
- 10.7 V
- 0.7 V
A NSWERS ◆^5
ANSWERS Chapter 3
SECTION CHECKUPS
Section 3–1 The Zener Diode
1. Zener diodes are operated in the reverse-breakdown region. 2. The test current, I Z 3. The zener impedance causes the voltage to vary slightly with current. 4. The zener voltage increases (or decreases) 0.05% for each degree centigrade increase (or decrease). 5. Power derating is the reduction in the power rating of a device as a result of an increase in temperature.
Section 3–2 Zener Diode Applications
1. An infinite resistance (open) 2. With no load, there is no current to a load. With full load, there is maximum current to the load. 3. Approximately 0.7 V, just like a rectifier diode
Section 3–3 The Varactor Diode
1. A varactor exhibits variable capacitance. 2. A varactor is operated in reverse bias. 3. The depletion region 4. Capacitance decreases with more reverse bias. 5. The capacitance ratio is the ratio of a varactor’s capacitance at a specified minimum voltage to the capacitance at a specified maximum voltage.
Section 3–4 Optical Diodes
1. Infrared and visible light 2. Infrared has the greater wavelength. 3. An LED operates in forward bias. 4. Light emission increases with forward current. 5. False, V F of an LED is usually greater than 1.2 V. 6. A tiny grouping of red, green, and blue LEDs. 7. A photodiode operates in reverse bias. 8. The internal resistance decreases. 9. Dark current is the reverse photodiode current when there is no light.
Section 3–5 Other Types of Diodes
1. l ight a mplification by s timulated e mission of r adiation 2. Coherent light has only a single wavelength, but incoherent light has a wide band of wave- lengths. A laser diode produces coherent light. 3. High-frequency and fast-switching circuits 4. Hot carrier is another name for Schottky diodes. 5. Tunnel diodes have negative resistance. 6. Oscillators 7. p region, n region, and intrinsic ( i ) region 8. A current regulator operates between V L (limiting voltage) and POV (peak operating voltage).
Section 3–6 Troubleshooting
1. The output voltage is too high and equal to the rectifier output. 2. More
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Section 4–3 BJT Characteristics and Parameters
1. bDC I C I B; aDC I C I E; h FE is bDC. 2. bDC 100; aDC 100 (100 1) 0. 3. I C is plotted versus V CE. 4. Forward-reverse bias is required for amplifier operation. 5. bDC increases with temperature. 6. No. bDC generally varies some from one device to the next for a given type.
Section 4–4 The BJT as an Amplifier
1. Amplification is the process where a smaller signal is used to produce a larger identical signal. 2. Voltage gain is the ratio of output voltage to input voltage. 3. R C and determine the voltage gain. 4. Av 5 V250 mV 20 5. Av 1200 Æ 20 Æ 60
Section 4–5 The BJT as a Switch
1. A transistor switch operates in cutoff and saturation. 2. The collector current is maximum in saturation. 3. The collector current is approximately zero in cutoff. 4. V CE V CC in cutoff. 5. V CE is minimum in saturation.
Section 4–6 The Phototransistor
1. The base current of a phototransistor is light induced. 2. Base 3. The collector current depends on bDC and I l. 4. Current transfer ratio
Section 4–7 Transistor Categories and Packaging
1. Three categories of BJTs are small signal/general purpose, power, and RF. 2. Emitter is the lead closest to the tab. 3. The metal mounting tab or case in power transistors is the collector.
Section 4–8 Troubleshooting
1. First, test it in-circuit. 2. If R B opens, the transistor is in cutoff. 3. The base voltage is +3 V and the collector voltage is +9 V.
RELATED PROBLEMS FOR EXAMPLES
4–1 10 mA 4–2 I B 241 mA; I C 21.7 mA; I E 21.94 mA; V CE 4.23 V; V CB 3.53 V 4–3 Along the horizontal axis 4–4 Not saturated 4–5 10 V 4–6 V CC(max) 44.5 V; V CE(max) is exceeded first. 4–7 4.55 W 4–8 P D(max) 500 mW @ 50°C 4–9 2.5 kÆ
r ¿ e
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4–10 78.4 mA 4–11 Reduce R C to 140 Æ and R B to 2.2 kÆ. 4–12 R B open
TRUE/FALSE QUIZ
1. T 2. F 3. T 4. T 5. F 6. F
7. F 8. T 9. F 10. T 11. F 12. T
CIRCUIT-ACTION QUIZ
1. (a) 2. (a) 3. (c) 4. (b) 5. (c) 6. (b) 7. (c) 8. (a) 9. (b) 10. (a) 11. (a) 12. (a)
SELF-TEST
1. (d) 2. (c) 3. (a) 4. (d) 5. (a) 6. (c) 7. (b) 8. (a) 9. (c) 10. (c) 11. (b) 12. (b) 13. (f) 14. (c) 15. (b) 16. (b) 17. (b) 18. (a) 19. (d) 20. (a) 21. (c)
ANSWERS Chapter 5
SECTION CHECKUPS
Section 5–1 The DC Operating Point
1. The upper load line limit is I C(sat) and V CE(sat). The lower limit is I C 0 and V CE(cutoff). 2. The Q-point is the dc point at which a transistor is biased and is specified by V CE and I C. 3. Ideally, saturation occurs at the intersection of the load line and the y -axis ( V CE 0 V). Cutoff occurs at the intersection of the load line and the I B 0 curve. 4. The Q-point must be centered on the load line for maximum Vce.
**Section 5–2 Voltage-Divider Bias
2.**
3. V B 5 V 4. Voltage-divider bias is stable and requires only one supply voltage.
Section 5–3 Other Bias Methods
1. Emitter bias is much less dependent on the value of beta than is base bias. 2. Emitter bias requires two separate supply voltages. 3. I C increases with causing a reduction in V C and, therefore, less voltage across R B, thus less I B. 4. Base bias is beta-dependent. 5. The Q-point changes due to changes in and V CE over temperature. 6. Emitter-feedback improves stability.
Section 5–4 Troubleshooting
1. A transistor is saturated when V CE 0 V. A transistor is in cutoff when V CE V CC. 2. R E is open because the BE junction of the transistor is still forward-biased. 3. If R C is open, V C is about 0.7 V less than V B.
bDC
bDC,
R IN(BASE) = (bDC V B)> I E = [(190)(2 V)] > 2 mA = 190 kÆ
R IN(BASE) = V IN> I IN = 5 V> 5 mA = 1 MÆ
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Section 6–4 The Common-Collector Amplifier
1. A common-collector amplifier is an emitter-follower. 2. The maximum voltage gain of a common-collector amplifier is 1. 3. A common-collector amplifier has a high input resistance.
Section 6–5 The Common-Base Amplifier
1. Yes 2. The common-base amplifier has a low input resistance. 3. The maximum current gain is 1 in a CB amplifier.
Section 6–6 Multistage Amplifiers
1. A stage is one amplifier in a cascaded arrangement. 2. The overall voltage gain is the product of the individual gains. 3. 4. At lower frequencies, XC becomes large enough to affect the gain.
Section 6–7 The Differential Amplifier
1. Double-ended differential input is between two input terminals. Single-ended differential input is from one input terminal to ground (with other input grounded). 2. Common-mode rejection is the ability of an op-amp to produce very little output when the same signal is applied to both inputs. 3. A higher CMRR results in a lower common-mode gain.
Section 6–8 Troubleshooting
1. If C 4 opens, the gain drops. The dc level would not be affected. 2. Q 2 would be biased in cutoff. 3. The collector voltage of Q 1 and the base, emitter, and collector voltages of Q 2 would change.
RELATED PROBLEMS FOR EXAMPLES
6–1 I C 5 mA; V CE 1.5 V 6–2 3.13 mA 6–3 9.3 mV 6– 6–5 97. 6–6 83 6–7 5; 165 6–8 9. 6–9 Increases 6–10 71. A single transistor loads the CE amplifier much more than the Darlington pair. 6–11 64. 6– 6–13 34,000; 90.6 dB 6–14 C 3 open
TRUE/FALSE QUIZ
1. T 2. T 3. F 4. T 5. F 6. F 7. T 8. T
9. T 10. F 11. T 12. F 13. T 14. T 15. F
A ¿ v = 1500; Av 1(dB) = 27.96 dB; Av 2(dB) = 13.98 dB; Av 3(dB) = 21.58 dB; A ¿ v (db) = 63.52 dB
C 2 = 28.4 mF
20 log (500) = 54.0 dB
A NSWERS ◆^11
CIRCUIT-ACTION QUIZ
1. (a) 2. (b) 3. (a) 4. (b) 5. (a) 6. (c) 7. (c) 8. (c) 9. (c) 10. (c)
SELF-TEST
1. (a) 2. (b) 3. (c) 4. (b) 5. (a) 6. (b) 7. (b) 8. (d) 9. (d) 10. (d) 11. (d) 12. (c) 13. (a) 14. (c) 15. (a) 16. (d) 17. (b) 18. (a) 19. (c)
ANSWERS Chapter 7
SECTION CHECKUPS
Section 7–1 The Class A Power Amplifier
1. To dissipate excessive heat 2. The collector 3. Cutoff and saturation clipping 4. 25% 5. The ratio of input resistance to output resistance
Section 7–2 The Class B and Class AB Push-Pull Amplifiers
1. The class B Q-point is at cutoff. 2. The barrier potential of the base-emitter junction causes crossover distortion. 3. Maximum efficiency of a class B amplifier is 79%. 4. Push-pull reproduces both positive and negative alternations of the input signal with greater efficiency. 5. Both transistors in class AB are biased slightly above cutoff. In class B they are biased at cutoff.
Section 7–3 The Class C Amplifier
1. Class C is biased well into cutoff. 2. The purpose of the tuned circuit is to produce a sinusoidal voltage output. 3. [1 W (1 W 0.1 W)]100 90.9%
Section 7–4 Troubleshooting
1. Excess input signal voltage 2. Open bypass capacitor, C 2
RELATED PROBLEMS FOR EXAMPLES
7–1 Power gain increases. 7–2 The efficiency goes up because no power is wasted in R E3. The problem is that the speaker has direct current in the coil and may burn open. 7–3 15 V, 0.1 A 7–4 7.5 V; 0.25 A 7–5 Pout 1.76 W; P DC 2.24 W 7–6 Rin would increase. 7–7 3 mW 7–8 The efficiency decreases. 7–9 Halve the input frequency.
h = >
A NSWERS ◆^13
Section 8–8 The IGBT
1. IGBT stands for insulated-gate bipolar transistor. 2. High-voltage switching applications 3. The IGBT has a lower output saturation voltage than the MOSFET. 4. The IGBT has a very high input resistance compared to a BJT. 5. Latch-up is a condition in which the IGBT is in the on state and cannot be turned off by the gate voltage.
Section 8–9 Troubleshooting
1. R S open, no ground connection 2. Because V GS remains at approximately zero 3. The device is off and V D V DD.
RELATED PROBLEMS FOR EXAMPLES
8–1 I D remains at approximately 12 mA. 8– 8–3 I D 3.52 mA 8– 8– 8– 8– 8– 8 8– 8– 8– 8– 8–13 I D does not change because of the constant-current source. 8–14 R DS values would not change because slopes are constant. 8– 8–16 I D 25 mA 8–17 (a) p channel (b) 6.48 mA (c) 35.3 mA 8–18 V GS 3.13 V; V DS 21.4 V 8–19 I D 2.13 mA 8–20 V DS 5.6 V
TRUE/FALSE QUIZ
1. T 2. F 3. T 4. F 5. F 6. T 7. T
8. F 9. T 10. T 11. F 12. T 13. F 14. T
CIRCUIT-ACTION QUIZ
1. (b) 2. (c) 3. (b) 4. (b) 5. (a) 6. (a) 7. (c) 8. (b)
SELF-TEST
1. (e) 2. (b) 3. (a) 4. (c) 5. (d) 6. (c) 7. (a) 8. (c) 9. (b) 10. (d) 11. (a) 12. (c) 13. (d) 14. (b) 15. (d) 16. (c) 17. (d) 18. (c) 19. (d) 20. (b) 21. (a) 22. (a) 23. (d) 24. (c) 25. (c)
1.07 kÆ
I D 1.25 mA, V GS - 2.25 V
I D = 1.81 mA, V GS = -2.44 V
V GS - 1.8 V, I D 1.8 mA
R S = 294 Æ; R D = 3 kÆ
R S = 889 Æ
R S = 245 Æ
V DS = 2 V; V GS = -3.12 V
R IN = 25,000 MÆ
g (^) m = 1800 mS; I D = 4.32 mA
V P = - 4 V
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ANSWERS Chapter 9
SECTION CHECKUPS
Section 9–1 The Common-Source Amplifier
1. The FET with gm 3.5 mS can produce the higher gain. 2. 3. Id is at its positive peak and Vds is at its negative peak when Vgs is at its positive peak. 4. Vgs is an ac quantity, V GS is a dc quantity. 5. Voltage gain of a CS amplifier is determined by gm and Rd. 6. The gain is halved because Rd R D 2.
Section 9–2 The Common-Drain Amplifier
1. The ideal maximum voltage gain of a CD amplifier is 1. 2. The voltage gain of a CD amplifier is determined by gm and Rs.
Section 9–3 The Common-Gate Amplifier
1. The CG amplifier has a low input resistance (1 gm ). 2. gm affects both voltage gain and input resistance. 3. High input resistance, high gain, and excellent high-frequency response
Section 9–4 The Class D Amplifier
1. Modulator, push-pull switching amplifier, and low-pass filter 2. The pulse width is proportional to the amplitude of the input signal. 3. It is passed through a low-pass filter.
Section 9–5 MOSFET Analog Switching
1. When it is turned off 2. When it is turned on 3. A pulse or dc voltage 4. R depends inversely on frequency and capacitance.
Section 9–6 MOSFET Digital Switching
1. A CMOS inverter consists of a p -channel and an n -channel MOSFET connected in a cascade arrangement. 2. NAND gate 3. NOR gate
Section 9–7 Troubleshooting
1. To be a good troubleshooter, you must understand the circuit. 2. There would be a lower than normal first-stage gain if C 2 opens. 3. No, but there would be a loss of signal to the second stage.
RELATED PROBLEMS FOR EXAMPLES
9–2 See Figure ANS9–1. 9–3 2.319 mA 9–4 See Figure ANS9–2. 9–5 350 mV 9–6 Rin = 9.99 MÆ
Av = gm Rd = (2500 mS)(10 kÆ) = 25
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Section 10–2 The Decibel (a) 12 dB corresponds to a voltage gain of approximately 4. (b) (c) 0 dBm corresponds to 1 mW.
Section 10–3 Low-Frequency Amplifier Response (a) fcl 2 167 Hz is dominant. (b) (c) attenuation at one decade below fcl. (d) ( e)
Section 10–4 High-Frequency Amplifier Response (a) The internal transistor capacitances determine the high-frequency response. (b) Cin ( tot ) Cin ( Miller ) Cce (4 pF)(81) 8 pF 342 pF (c) The input RC circuit dominates. (d) Ciss and C (^) rss are usually specified on a FET datasheet. (e) C (^) in ( tot ) (3 pF)(26) 4 pF 82 pF
Section 10–5 Total Amplifier Frequency Response (a) The gain is 1 at fT. (b) (c) Av 130 MHz50 MHz 2.
Section 10–6 Frequency Response of Multistage Amplifiers (a) (b) (c) BW decreases.
Section 10–7 Frequency Response Measurements (a) fcl 125 Hz; fcu 500 kHz (b) Rise time is between the 10% and 90% points and fall time is between the 90% and 10% points. (c) tr 150 ns (d) tf 2.8 ms (e) Since
RELATED PROBLEMS FOR EXAMPLES
10–1 (a) 61.6 dB (b) 17 dB (c) 102 dB 10–2 (a) 50 V (b) 6.25 V (c) 1.56 V 10– 10–4 212 @ 400 Hz; 30 @ 40 Hz; 3 @ 4 Hz 10–5 It will increase the gain and reduce the lower critical frequency. 10–6 C 2 “sees” a smaller resistance. 10–7 fcl changes from 16.2 Hz to 16.1 Hz. 10–8 Ideally, the low-frequency response is not affected because an infinite load makes fc of the output stage even lower, so the input stage determines the lower cutoff frequency of the amplifier. 10–9 The resistance of the input will be higher, so the critical frequency is lower.
0.22 mF
fcu 77 fcl , BW fcu = 2.5 MHz.
f (^) cu ¿( dom ) = 49 kHz
f (^) cl ¿( dom ) = 1 kHz
BW = 25 kHz - 100 Hz = 24.9 kHz
fcl = 1 >(2p(6500 Æ)(0.0022 mF)) = 11.1 kHz
u = tan-^1 (0.5) = 26.6°
Av (dB) = 50 dB - 3 dB = 47 dB
Ap = 10 log (25) = 13.98 dB
A NSWERS ◆^17
10–10 Change 10–11 in series with 215 pF, fc 2.31 MHz 10–12 28.7 MHz 10–13 1 pF 10–14 fc decreases to 83.8 MHz. 10–15 48.2 MHz 10–16 BW decreases; BW increases 10–17 20 MHz 10–18 980 Hz 10–19 39.8 kHz 10–20 26 dB
TRUE/FALSE QUIZ
1. T 2. F 3. F 4. T 5. T 6. F
7. T 8. T 9. T 10. F 11. T 12. T
CIRCUIT-ACTION QUIZ
1. (a) 2. (a) 3. (a) 4. (b) 5. (a) 6. (b) 7. (c) 8. (a) 9. (b) 10. (b)
SELF-TEST
1. (d) 2. (c) 3. (b) 4. (d) 5. (c) 6. (b) 7. (a) 8. (a) 9. (d) 10. (b) 11. (c) 12. (c) 13. (d) 14. (b) 15. (c) 16. (c) 17. (a) 18. (c) 19. (b) 20. (d) 21. (b) 22. (a) 23. (a)
ANSWERS Chapter 11
SECTION CHECKUPS
Section 11–1 The Four-Layer Diode
1. The 4-layer diode is a thyristor because it has four semiconductor layers in a pnpn configuration. 2. A region of 4-layer diode operation in which the device is nonconducting 3. The device turns on and conducts when exceeds the forward-breakover voltage. 4. When the anode current is reduced below the holding current value, the device turns off.
Section 11–2 The Silicon-Controlled Rectifier (SCR)
1. An SCR (silicon-controlled rectifier) is a three-terminal thyristor. 2. The SCR terminals are anode, cathode, and gate. 3. A positive gate pulse turns the SCR on. 4. Reduce the anode current below (holding current) to turn a conducting SCR off. 5. Open the series switch
Section 11–3 SCR Applications
1. The SCR will conduct for more than 90° but less than 180°. 2. To block discharge of the battery through that path
Section 11–4 The Diac and Triac
1. The diac is like two parallel 4-layer diodes connected in opposite directions. 2. A triac is like two parallel SCRs having a common gate and connected in opposite directions. 3. A triac has a gate terminal, but a diac does not.
I H
V AK
320 Æ
C 1 to 0.68 mF.
A NSWERS ◆^19
Section 12–2 Op-Amp Input Modes and Parameters
1. Double-ended input is between two input terminals. Single-ended input is from one input terminal to ground (with other input grounded). 2. Common-mode rejection is the ability of an op-amp to produce very little output when the same signal is applied to both inputs. 3. A higher common-mode gain results in a lower CMRR. 4. Input bias current, input offset voltage, drift, input offset current, input impedance, output impedance, maximum output voltage swing, CMRR, open-loop voltage gain, slew rate, frequency response. 5. A fast pulse is used as the input and the rate of change of the output is measured.
Section 12–3 Negative Feedback
1. Negative feedback provides a stable controlled voltage gain, control of impedances, and wider bandwidth. 2. The open-loop gain is so high that a very small signal on the input will drive the op-amp into saturation.
Section 12–4 Op-Amps with Negative Feedback
1. The main purpose of negative feedback is to stabilize the gain. 2. False 3. Acl 1 0.02 50
Section 12–5 Effects of Negative Feedback on Op-Amp Impedances
1. The noninverting configuration has a higher Zin than the op-amp alone. 2. Zin increases in a voltage-follower. 3.
Section 12–6 Bias Current and Offset Voltage
1. Input bias current and input offset voltage are sources of output error. 2. Add a resistor in the feedback path equal to the input source resistance.
Section 12–7 Open-Loop Frequency and Phase Responses
1. Open-loop voltage gain is without feedback, and closed-loop voltage gain is with negative feedback. Open-loop voltage gain is larger. 2. BW 100 Hz 3. Aol decreases. 4. Av ( tot ) 20 dB 30 dB 50 dB 5.
Section 12–8 Closed-Loop Frequency Response
1. Yes, Acl is always less than Aol. 2. BW 3,000 kHz 60 50 kHz 3. fT 3,000 kHz 1 3 MHz
Section 12–9 Troubleshooting
1. Check the output null adjustment. 2. After a verification that there is power supply voltage to the op-amp, then the absence of an output signal probably indicates a bad op-amp.
u tot = -49° + (-5.2°) = -54.2°
Zin (I) Ri = 2 kÆ, Z (^) out (I) = Zout >(1 + Aol B ) = 25 mÆ
20 ◆^ A NSWERS
RELATED PROBLEMS FOR EXAMPLES
12–1 340,000; 111 dB 12– 12–3 32. 12– 12–5 (a) (b) 23 12–6 Zin increases, Zout decreases. 12– 12–8 (a) 79,996 (b) 79,900 (c) 6380 12–9 173 Hz 12–10 75 dB; 12–11 2 MHz 12–12 (a) 29.6 kHz (b) 42.6 kHz
TRUE/FALSE QUIZ
1. T 2. F 3. T 4. T 5. F 6. T 7. T
8. F 9. T 10. F 11. T 12. T 13. T 14. F
CIRCUIT-ACTION QUIZ
1. (b) 2. (a) 3. (b) 4. (a) 5. (a) 6. (b) 7. (c) 8. (b)
SELF-TEST
1. (c) 2. (b) 3. (d) 4. (b) 5. (a) 6. (c) 7. (b) 8. (d) 9. (c) 10. (d) 11. (a) 12. (b) 13. (c) 14. (d) 15. (c) 16. (c) 17. (a) 18. (c) 19. (d) 20. (b) 21. (b) 22. (b) 23. (d) 24. (a) 25. (d) 26. (b) 27. (b) 28. (a) 29. (c) 30. (d)
ANSWERS Chapter 13
SECTION CHECKUPS
Section 13–1 Comparators
1. (a) (b) 2. Hysteresis makes the comparator less susceptable to noise. 3. Bounding limits the output amplitude to a specified level.
Section 13–2 Summing Amplifiers
1. The summing point is the point where the input resistors are commonly connected. 2. Rf R 1 5 0. 3.
Section 13–3 Integrators and Differentiators
1. The feedback element in an ideal integrator is a capacitor. 2. The capacitor voltage is linear because the capacitor current is constant. 3. The feedback element in a differentiator is a resistor. 4. The output of a differentiator is proportional to the rate of change of the input.
5 kÆ
V = (22 kÆ> 69 kÆ)(- 12 V) = -3.83 V
V = (10 kÆ> 110 kÆ)15 V = 1.36 V
Zin (I) = 560 Æ; Zout (I) = 110 mÆ; Acl = - 146
20.6 GÆ, 14 mÆ
67.5 kÆ
20 V/ms
