injection
c. Taking resistance measurements d. Replacing components
b = 100 b = 100
vg 1 mV p-p
RG 600 Ω
R1 10 kΩ
RC1 3.6 kΩ
R3 10 kΩ
RC2 3.6 kΩ
VCC +10 V
R2 2.2 kΩ
R4 2.2 kΩ
Q1 Q2
RE1 1 kΩ
RE2 1 kΩ
RL 10 kΩ
Figure 9-27
SEC. 9-1 MULTISTAGE AMPLIFIERS
9-1 In Fig. 9-27, what is the ac voltage at the fi rst base? At the second base? Across the load resistor?
9-2 If the supply voltage is increased to 112 V in Fig. 9-27, what is the output voltage?
9-3 If 5 300 in Fig. 9-27, what is the output voltage?
Problems
SEC. 9-2 TWO-STAGE FEEDBACK
9-4 A feedback amplifi er like Fig. 9-4 has rf 5 5 kV and re 5 50 V. What is the voltage gain?
9-5 In a feedback amplifi er like Fig. 9-5, re 5 125 V. If you want a voltage gain of 100, what value should rf be?
SEC. 9-3 CC AMPLIFIER
9-6 In Fig. 9-28, what is the input impedance of the base if 5 200? The input impedance of the stage?
9-8 What is the voltage gain in Fig. 9-28? If 5 175, what is the ac load voltage?
9-9 What is the input voltage in Fig. 9-28 if varies over a range of 50 to 300?
9-10 All resistors are doubled in Fig. 9-28. What happens to the input impedance of the stage if 5 150? To the input voltage?
9-11 What is the input impedance of the base if 5 200 in Fig. 9-29? The input impedance of the stage?
9-12 In Fig. 9-29, what is the ac input voltage to the emitter follower if 5 150 and vg 5 1 V?
9-13 What is the voltage gain in Fig. 9-29? If 5 175, what is the ac load voltage?
SEC. 9-4 OUTPUT IMPEDANCE
9-14 What is the output impedance in Fig. 9-28 if 5 200?
9-15 What is the output impedance in Fig. 9-29 if 5 100?
SEC. 9-5 CASCADING CE AND CC
9-16 What is the voltage gain of the CE stage in Fig. 9-30 if the second transistor has a dc and ac current gain of 200?
9-17 If both transistors in Fig. 9-30 have a dc and ac cur- rent gain of 150, what is the output voltage when vg 5 10 mV?
9-18 If both transistors have a dc and ac current gain of 200 in Fig. 9-30, what is the voltage gain of the CE stage if the load resistance drops to 125 V?
9-19 In Fig. 9-30, what would happen to the voltage gain of the CE amplifi er if the emitter follower stage were removed and a capacitor were used to cou- ple the ac signal to the 150 V load?
vg 1 V
RG 50 Ω
R2 2.2 kΩ R1 2.2 kΩ
RL 3.3 kΩ RE
1 kΩ VCC +15 V
Figure 9-28
R2 200 Ω R1 100 Ω
vg RE
30 Ω VCC +20 V
RG 50 Ω
RL 10 Ω
Figure 9-29
R2 1 kΩ R1 4.7 kΩ
Q1
Q2
RL 150 Ω RC
1.5 kΩ
VCC +15 V
RE 330 Ω vg
270 Ω RG
Figure 9-30
9-7 If 5 150 in Fig. 9-28, what is the ac input voltage to the emitter follower?
362 Chapter 9 R1
1 kΩ
R2 2 kΩ vg
1 Vp-p
RE 10 Ω
VCC +20 V
RG 600 Ω
RL 8 Ω Q1
Q2
Figure 9-32
+ – 1N958
RL 33 Ω +
–
Vout Vin 15 V
+
– RS
1 kΩ
Figure 9-33
SEC. 9-6 DARLINGTON CONNECTIONS 9-20 If the Darlington pair of Fig. 9-31 has an overall
current gain of 5000, what is the input impedance of the Q1 base?
9-21 In Fig. 9-31, what is the ac input voltage to the Q1 base if the Darlington pair has an overall current gain of 7000?
9-22 Both transistors have a of 150 in Fig. 9-32. What is the input impedance of the fi rst base?
9-23 In Fig. 9-32, what is the ac input voltage to the Q1
base if the Darlington pair has an overall current gain of 2000?
SEC. 9-7 VOLTAGE REGULATION
9-24 The transistor in Fig. 9-33 has a current gain of 150.
If the 1N958 has a zener voltage of 7.5 V, what is the output voltage? The zener current?
9-25 If the input voltage in Fig. 9-33 changes to 25 V, what is the output voltage? The zener current?
9-26 The potentiometer in Fig. 9-34 can vary from 0 to 1 kV. What is the output voltage when the wiper is at the center?
9-27 What is the output voltage in Fig. 9-34 if the wiper is all the way up? If it is all the way down?
R3 1 kΩ R4
1 kΩ RL R2
1.5 kΩ R1
1 kΩ
R5 1 kΩ
Vout
Vz
– + Q2
Q1
7.5 V – + –
+ Vin 25 V
Figure 9-34
VCC +15 V
Q2 Q1
vg 10 mV
RG 5.1 kΩ
R1 150 kΩ
R2 150 kΩ
RE 470Ω
RL 1 kΩ vout
Figure 9-31
Figure 9-35
RE 2 kΩ vg
2 mVp-p
R2 2 kΩ
R1 10 kΩ
RC 3.3 kΩ
1 mF 47 mF
47 mF VCC
12 V 50 Ω
RL 10 kΩ RG
vout
9-33 In Fig. 9-33, what is the power dissipation of the tran- sistor if the current gain is 100 and the zener voltage is 7.5 V?
9-34 In Fig. 9-36a, the transistor has a dc of 150. Cal- culate the following dc quantities: VB, VE, VC, IE, IC, and IB.
9-35 If an input signal with a peak-to-peak value of 5 mV drives the circuit of Fig. 9-36a, what are the two ac output voltages? What do you think is the purpose of this circuit?
Critical Thinking
9-36 Figure 9-36b shows a circuit in which the control voltage can be 0 V or 15 V. If the audio input voltage is 10 mV, what is the audio output voltage when the control voltage is 0 V? When the control voltage is 15 V? What do you think this circuit is supposed to do?
9-37 In Fig. 9-33, what would the output voltage be if the zener diode opened? (Use bdc 5 200)
9-38 In Fig. 9-33, if the 33 V load shorts, what is the transistor’s power dissipation? (use bdc 5 100)
(a)
vin vout(1)
VCC +15 V
vout(2) R1
4.7 kΩ
R2 2 kΩ
RE 1 kΩ RC 1 kΩ
(b)
VCC +12 V
R1 33 kΩ
RE 2.2 kΩ RC 4.7 kΩ
AUDIO OUT AUDIO
IN CONTROL VOLTAGE
R3 1 kΩ
R2 10 kΩ
10 mF
10 mF 10 mF
Figure 9-36
SEC. 9-8 COMMON-BASE AMPLIFIER
9-28 In Fig. 9-35, what is the Q point emitter current?
9-29 What is the approximate voltage gain in Fig. 9-35?
9-30 In Fig. 9-35, what is the input impedance looking into the emitter? What is the input impedance of the stage?
9-31 In Fig. 9-35, with an input of 2 mV from the genera- tor, what is the value of vout?
9-32 In Fig. 9-35, if the VCC supply voltage were increased to 15 V, what would vout equal?
364 Chapter 9 1 mV
vg
A
C D
E
B Q1
C3 C2 C1
RG 600 Ω
R2 2.2 kΩ R1 10 kΩ
RC1 3.6 kΩ
RE1 1 kΩ
VCC +10 V
G
H I
F Q2
R4 39 kΩ R3 15 kΩ
RE2 4.3 kΩ
RL 10 kΩ C4
0.6 0.6 0.6 1
0.75 0.75
0.6 0.6 VA
0.6 0.6 0.6 0
0.75 0.75
0.6 0.6 VB
0.6 0.6 0.6 0
0 0.75
0.6 0.6 VC
70 70 70 0
0 2
95 70 VD
0 0 0 0
0 0.75
0 0 VE
70 70 70 0
0 0 2
70 VF
70 70 0 0
0 0 2
70 VG
70 70 0 0
0 0 2
0 VH
70 0 0 0
0 0 2
0 VI OK
Trouble
Ac Millivolts
T1 T2 T3 T4
T6 T7 T5
(a)
(b)
Figure 9-37
The Multisim troubleshooting fi les are found on the Instructor Resources section of Connect for Electronic Principles, in a folder named Multisim Troubleshooting
Multisim Troubleshooting Problems
Circuits (MTC). See page XVI for more details. For this chapter, the fi les are labeled MTC09-45 through MTC09-49 and are based on the circuit of Figure 9-37.
Use Fig. 9-37 for the remaining problems. The table la- beled “Ac Millivolts” contains the measurements of the ac voltages expressed in millivolts. For this exercise, all resistors are OK. The troubles are limited to open capac- itors, open connecting wires, and open transistors.
Troubleshooting
9-43 Find Troubles T1 to T3.
9-44 Find Troubles T4 to T7.
9-39 In Fig. 9-34, what is the power dissipation of Q2 when the wiper is at the center and the load resist- ance is 100 V?
9-40 Using Fig. 9-31, if both transistors have a b of 100, what is the approximate output impedance of the amplifi er?
9-41 In Fig. 9-30, if the input voltage from the generator were 100 mVp-p and the emitter-bypass capacitor opened, what would the output voltage across the load be?
9-42 In Fig. 9-35, what would be the output voltage if the base-bypass capacitor shorted?
1. Draw the schematic diagram of an emitter follower.
Tell me why this circuit is widely used in power ampli- fi ers and voltage regulators.
2. Tell me all that you know about the output imped- ance of an emitter follower.
3. Draw a Darlington pair and explain why the overall current gain is the product of the individual current gains.
4. Draw a zener follower and explain why it regulates the output voltage against changes in the input voltage.
5. What is the voltage gain of an emitter follower? This being the case, in what applications would such a circuit be useful?
6. Explain why a Darlington pair has a higher power gain than a single transistor.
Job Interview Questions
7. Why are “follower” circuits so important in acoustic circuits?
8. What is the approximate ac voltage gain for a CC amplifi er?
9. What is another name for a common-collector amplifi er?
10. What is the relationship between an ac signal phase (output to input) and a common-collector amplifi er?
11. If a technician measures unity voltage gain (output voltage divided by input voltage) from a CC amplifi er, what is the problem?
12. The Darlington amplifi er is used in the fi nal power amplifi er (FPA) in most higher-quality audio amplifi ers because it increases the power gain. How does a Darlington amplifi er increase the power gain?
Self-Test Answers
1. a 2. b 3. c 4. b 5. c 6. b
7. c 8. d 9. c 10. a 11. a 12. d
13. c 14. a 15. c 16. d 17. a 18. c
19. c 20. a 21. c 22. c 23. a 24. a
25. d 26. a 27. d 28. a 29. c 30. d
31. c 32. b 33. d 34. b 35. d 36. b
Practice Problem Answers
9-1 vout 5 2.24 V 9-3 rf 5 4.9 kV 9-5 zin(base) 5 303 kV;
zin(stage) 5 4.92 kV 9-6 vin < 0.893 V 9-7 vin 5 0.979 V;
vout 5 0.974 V
9-8 zout 5 3.33 V 9-9 zout 5 2.86 V 9-10 Av 5 222 9-11 Av 5 6.28 9-12 5 5625;
IB1 5 14.3 A;
zin(base) 5 112.5 kV
9-13 Vout 5 7.5 V;
Iz 5 5 mA 9-14 Vout 5 18.9 V 9-15 vout 5 76.9 mVp-p
Open up and troubleshoot each of the respec- tive fi les. Take measurements to determine if there is a fault and, if so, determine the circuit fault.
9-45 Open up and troubleshoot fi le MTC09-45.
9-46 Open up and troubleshoot fi le MTC09-46.
9-47 Open up and troubleshoot fi le MTC09-47.
9-48 Open up and troubleshoot fi le MTC09-48.
9-49 Open up and troubleshoot fi le MTC09-49.
366
chapter
10
In most electronic systems applications, the input signal is small. After several stages of voltage gain, however, the signal becomes large and uses the entire load line. In these later stages of a system, the collector currents are much larger
because the load impedances are much smaller. Stereo amplifi er speakers, for example, may have an impedance of 8 V or less.
Small-signal transistors have a power rating of less than 1 W, whereas power transistors have a power rating of more than 1 W. Small-signal transistors are typically used at the front end of systems where the signal power is low, and power transistors are used near the end of systems because the signal power and current are high.
Power
Amplifi ers
© Jason Reed/Getty Images
ac output compliance ac load line
audio amplifi er bandwidth (BW) capacitive coupling Class-A operation Class-AB operation Class-B operation Class-C operation compensating diodes
crossover distortion current drain direct coupling driver stage duty cycle effi ciency harmonics
large-signal operation narrowband amplifi er power amplifi er
power gain preamp
push-pull circuit radio-frequency (RF) amplifi er
thermal runaway transformer coupling tuned RF amplifi er wideband amplifi er
Vocabulary
bchob_ha
bchop_ha
bchop_ln
Objectives
After studying this chapter, you should be able to:
■ Show how the dc load line, ac load line, and Q point are determined for CE and CC power amplifi ers.
■ Calculate the maximum peak-to- peak (MPP) unclipped ac voltage that is possible with CE and CC power amplifi ers.
■ Describe the characteristics of amplifi ers, including classes of operation, types of coupling, and frequency ranges.
■ Draw a schematic of Class-B/AB push-pull amplifi er and explain its operation.
■ Determine the effi ciency of transistor power amplifi ers.
■ Discuss the factors that limit the power rating of a transistor and what can be done to improve the power rating.
bchop_haa
Chapter Outline
10-1 Amplifi er Terms 10-2 Two Load Lines 10-3 Class-A Operation 10-4 Class-B Operation 10-5 Class-B Push-Pull Emitter
Follower
10-6 Biasing Class-B/AB Amplifi ers
10-7 Class-B/AB Driver 10-8 Class-C Operation 10-9 Class-C Formulas 10-10 Transistor Power Rating
368 Chapter 10
10-1 Amplifi er Terms
There are different ways to describe amplifi ers. For instance, we can describe them by their class of operation, by their interstage coupling, or by their frequency range.
Classes of Operation
Class-A operation of an amplifi er means that the transistor operates in the active region at all times. This implies that collector current fl ows for 360° of the ac cycle, as shown in Fig. 10-1a. With a Class-A amplifi er, the designer usually tries to locate the Q point somewhere near the middle of the load line. This way, the signal can swing over the maximum possible range without saturating or cutting off the transistor, which would distort the signal.
Class-B operation is different. It means that collector current fl ows for only half the cycle (180°), as shown in Fig. 10-1b. To have this kind of operation, a designer locates the Q point at cutoff. Then, only the positive half-cycle of ac base voltage can produce collector current. This reduces the wasted heat in power transistors.
Class-C operation means that collector current fl ows for less than 180°
of the ac cycle, as shown in Fig. 10-1c. With Class-C operation, only part of the positive half-cycle of ac base voltage produces collector current. As a result, we get brief pulses of collector current like those of Fig. 10-1c.
Types of Coupling
Figure 10-2a shows capacitive coupling. The coupling capacitor transmits the amplifi ed ac voltage to the next stage. Figure 10-2b illustrates transformer coupling. Here, the ac voltage is coupled through a transformer to the next stage.
Capacitive coupling and transformer coupling are both examples of ac coupling, which blocks the dc voltage.
Direct coupling is different. In Fig. 10-2c, there is a direct connection between the collector of the fi rst transistor and the base of the second transistor.
IC
ICQ
t (a)
IC
t (b)
IC
t (c)
Figure 10-1 Collector current: (a) Class-A; (b) Class-B; (c) Class-C.
GOOD TO KNOW
As we progress through the letters A, B, and C designating the various classes of operation, we can see that linear operation occurs for shorter and shorter intervals of time. A Class-D amplifier is one whose output is switched on and off; that is, it essentially spends zero time during each input cycle in the linear region of operation. A Class-D amplifier is often used as a pulse-width modulator, which is a circuit whose output pulses have widths that are pro- portional to the amplitude level of the amplifier’s input signal.
GOOD TO KNOW
Most integrated circuit amplifiers use direct coupling between stages.
TO NEXT STAGE RC
(a)
TO NEXT STAGE
(b)
(c)
Figure 10-2 Types of coupling: (a) capacitive; (b) transformer; (c) direct.
Because of this, both the dc and the ac voltages are coupled. Since there is no lower frequency limit, a direct-coupled amplifi er is sometimes called a dc amplifi er.
Ranges of Frequency
Another way to describe amplifi ers is by stating their frequency range. For instance, an audio amplifi er refers to an amplifi er that operates in the range of 20 Hz to 20 kHz. On the other hand, a radio-frequency (RF) amplifi er is one that amplifi es frequencies above 20 kHz, usually much higher. For instance, the RF amplifi ers in AM radios amplify frequencies between 535 and 1605 kHz, and the RF amplifi ers in FM radios amplify frequencies between 88 and 108 MHz.
Amplifi ers are also classifi ed as narrowband or wideband. A narrow- band amplifi er works over a small frequency range like 450 to 460 kHz. A wide- band amplifi er operates over a large frequency range like 0 to 1 MHz.
Narrowband amplifi ers are usually tuned RF amplifi ers, which means that their ac load is a high-Q resonant tank tuned to a radio station or television channel. Wideband amplifi ers are usually untuned; that is, their ac load is resistive.
Figure 10-3a is an example of a tuned RF amplifi er. The LC tank is res- onant at some frequency. If the tank has a high Q, the bandwidth is narrow. The output is capacitively coupled to the next stage.
Figure 10-3b is another example of a tuned RF amplifi er. This time, the narrowband output signal is transformer-coupled to the next stage.
Signal Levels
We have already described small-signal operation, in which the peak-to-peak swing in collector current is less than 10 percent of quiescent collector current. In large-signal operation, a peak-to-peak signal uses all or most of the load line. In a stereo system, the small signal from a radio tuner or compact disc player is used as the input to a preamp, a low-noise amplifi er that is designed with the proper
370 Chapter 10
input impedance to pick up the signal from the input source, provide some level of amplifi cation, and deliver the output to the next stage. Following the preamp, one or more stages of amplifi cation are used to produce a larger output suitable for driving tone and volume controls. The signal is then used as the input to a power amplifi er, which produces output power ranging from a few hundred milliwatts up to hundreds of watts.
In the remainder of this chapter, we will discuss power amplifi ers and related topics like the ac load line, power gain, and effi ciency.
10-2 Two Load Lines
Every amplifi er has a dc-equivalent circuit and an ac-equivalent circuit. Because of this, it has two load lines: a dc load line and an ac load line. For small-signal operation, the location of the Q point is not critical. But with large-signal ampli- fi ers, the Q point has to be at the middle of the ac load line to get the maximum possible output swing.
DC Load Line
Figure 10-4a is a voltage-divider-based (VDB) amplifi er. One way to move the Q point is by varying the value of R2. For very large values of R2, the transistor goes into saturation and its current is given by:
IC(sat) 5 ________ VCC
RC 1 RE (10-1)
Very small values of R2 will drive the transistor into cutoff, and its voltage is given by:
VCE(cutoff ) 5 VCC (10-2)
Figure 10-4b shows the dc load line with the Q point.
AC Load Line
Figure 10-4c is the ac-equivalent circuit for the VDB amplifi er. With the emitter at ac ground, RE has no effect on the ac operation. Furthermore, the ac collec- tor resistance is less than the dc collector resistance. Therefore, when an ac sig- nal comes in, the instantaneous operating point moves along the ac load line of
R1 C
L
R2
RE
TO NEXT STAGE INPUT
+VCC
(a)
R1
C L
R2
RE
TO NEXT STAGE +VCC
(b)
Figure 10-3 Tuned RF amplifi ers: (a) capacitive coupling; (b) transformer coupling.
Fig. 10-4d. In other words, the peak-to-peak sinusoidal current and voltage are determined by the ac load line.
As shown in Fig. 10-4d, the saturation and cutoff points on the ac load line differ from those on the dc load line. Because the ac collector and emitter resistance are lower than the respective dc resistance, the ac load line is much steeper. It’s important to note that the ac and dc load lines intersect at the Q point.
This happens when the ac input voltage is crossing zero.
Here’s how to determine the ends of the ac load line. Writing a collector voltage loop gives us:
vce 1 icrc 5 0 or
ic 5 2 v___ rcec (10-3)
The ac collector current is given by:
ic 5 DIC 5 IC 2 ICQ
and the ac collector voltage is:
vce 5 DVCE 5 VCE 2 VCEQ
When substituting these expressions into Eq. (10-3) and rearranging, we arrive at:
IC 5 ICQ 1 _____ VCEQrc 2 ____ VrCEc (10-4)
VCE VCC
VCC IC
IC
Q
Q
VCE
vce(cutoff) =VCEQ + ICQrc AC LOAD LINE
DC LOAD LINE
DC LOAD LINE +VCC
RC R1
vin R2
RE
RL
(a) (b)
(d)
vin R2
rc R1
(c)
VCC RC + RE
ic(sat) = ICQ + VCEQ rc
Figure 10-4 (a) VDB amplifi er; (b) dc load line; (c) ac-equivalent circuit; (d) ac load line.
372 Chapter 10
This is the equation of the ac load line. When the transistor goes into saturation, VCE is zero, and Eq. (10-4) gives us:
ic(sat) 5 ICQ 1 _____ VCEQrc (10-5) where ic(sat)5 ac saturation current
ICQ5 dc collector current VCEQ5 dc collector-emitter voltage
rc5 ac resistance seen by the collector When the transistor goes into cutoff, IC equals zero. Since
vce(cutoff) 5 VCEQ 1 DVCE
and
DVCE 5 (DIC)(rc) we can substitute to get:
DVCE 5 (ICQ 2 0A)(rc) resulting in:
vce(cutoff) 5 VCEQ 1 ICQrc (10-6)
Because the ac load line has a higher slope than the dc load line, the maximum peak-to-peak (MPP) output is always less than the supply voltage. As a formula:
MPP , VCC (10-7)
For instance, if the supply voltage is 10 V, the maximum peak-to-peak sinusoidal output is less than 10 V.
Clipping of Large Signals
When the Q point is at the center of the dc load line (Fig. 10-4d ), the ac signal cannot use all of the ac load line without clipping. For instance, if the ac signal increases, we will get the cutoff clipping shown in Fig. 10-5a.
If the Q point is moved higher, as shown in Fig. 10-5b, a large signal will drive the transistor into saturation. In this case, we get saturation clipping.
Both cutoff and saturation clipping are undesirable because they distort the signal.
When a distorted signal like this drives a loudspeaker, it sounds terrible.
A well-designed large-signal amplifi er has the Q point at the middle of the ac load line (Fig. 10-5c). In this case, we get a maximum peak-to-peak unclipped output. This maximum unclipped peak-to-peak ac voltage is also referred to its ac output compliance.
Maximum Output
When the Q point is below the center of the ac load line, the maximum peak (MP) output is ICQrc, as shown in Fig. 10-6a. On the other hand, if the Q point is above the center of the ac load line, the maximum peak output is VCEQ, as shown in Fig. 10-6b.
For any Q point, therefore, the maximum peak output is:
MP 5 ICQrc or VCEQ, whichever is smaller (10-8) and the maximum peak-to-peak output is twice this amount:
MPP 5 2MP (10-9)
Equations (10-8) and (10-9) are useful in troubleshooting to determine the largest unclipped output that is possible.