In a TSEB circuit, the base voltage is approximat- 123docz.net

c. Stay the same d. Double

27. If the emitter resistance decreases with TSEB, the collector voltage will a. Decrease

b. Stay the same c. Increase

d. Equal the collector supply voltage

28. If the base resistor opens with TSEB, the collector voltage will

a. Decrease b. Stay the same c. Increase slightly

d. Equal the collector supply voltage

29. In TSEB, the base current must be very

a. Small b. Large c. Unstable d. Stable

30. The Q point of TSEB does not depend on the

a. Emitter resistance b. Collector resistance c. Current gain d. Emitter voltage

31. The majority carriers in the emitter of a pnp transistor are a. Holes

b. Free electrons c. Trivalent atoms d. Pentavalent atoms 32. The current gain of a pnp

transistor is

a. The negative of the npn current gain

b. The collector current divided by the emitter current c. Near zero

d. The ratio of collector current to base current

33. Which is the largest current in a pnp transistor?

a. Base current b. Emitter current c. Collector current d. None of these

34. The currents of a pnp transistor are

a. Usually smaller than npn currents

b. Opposite npn currents c. Usually larger than npn

currents d. Negative

35. With pnp voltage-divider bias, you must use

a. Negative power supplies b. Positive power supplies c. Resistors

d. Grounds

36. With a TSEB pnp circuit using a negative VCC supply, the emitter voltage is

a. Equal to the base voltage b. 0.7 V higher than the base

voltage

c. 0.7 V lower than the base voltage

d. Equal to the collector voltage

37. In a well-designed VDB circuit, the base current is

a. Much larger than the voltage divider current

b. Equal to the emitter current c. Much smaller than the voltage

divider current d. Equal to the collector

current

38. In a VDB circuit, the base input resistance RIN is

a. Equal to bdc RE

b. Normally smaller than RTH

c. Equal to bdc RC

d. Independent of bdc

39. In a TSEB circuit, the base voltage is approximately zero when

a. The base resistor is very large

b. The transistor is saturated c. bdc is very small

d. RB , 0.01 bdc RE

274 Chapter 7 SEC. 7-1 EMITTER BIAS

7-1 What is the collector voltage in Fig. 7-30a? The emitter voltage?

(a) VCC +20 V

RC 10 kΩ

RE 1.8 kΩ VBB

+2.5 V

(b) VCC +10 V

RC 910 Ω

RE 180 Ω +VBB

RE 100 Ω VBB

Figure 7-30

7-2 If the emitter resistor is doubled in Fig. 7-30a, what is the collector-emitter voltage?

7-3 If the collector supply voltage is decreased to 15 V in Fig. 7-30a, what is the collector voltage?

7-4 What is the collector voltage in Fig. 7-30b if VBB 5 2 V?

7-5 If the emitter resistor is doubled in Fig. 7-30b, what is the collector- emitter voltage for a base supply voltage of 2.3 V?

7-6 If the collector supply voltage is increased to 15 V in Fig. 7-30b, what

is the collector-emitter voltage for VBB 5 1.8 V?

SEC. 7-2 LED DRIVERS

7-7 If the base supply voltage is 2 V in Fig. 7-30c, what is the current through the LED?

7-8 If VBB 5 1.8 V in Fig. 7-30c, what is the LED current? The approximate VC?

(a) RB 1 MΩ

RC 4.7 kΩ VBB

+10 V

VCC +10 V

hFE ⫽ 100

(b) RE 1 kΩ RC 3.6 kΩ VBB

+1.8 V

VCC +10 V

Figure 7-31

SEC. 7-3 TROUBLESHOOTING EMITTER BIAS CIRCUITS

7-9 A voltmeter reads 10 V at the collector of Fig. 7-31a.

What are some of the troubles that can cause this high reading?

7-10 What if the ground on the emitter is open in Fig. 7-31a? What will a voltmeter read for the base voltage? For the collector voltage?

7-11 A dc voltmeter measures a very low voltage at the collector of Fig. 7-31a. What are some of the possible troubles?

7-12 A voltmeter reads 10 V at the collector of Fig. 7-31b. What are some of the troubles that can cause this high reading?

7-13 What if the emitter resistor is open in Fig. 7-31b?

What will a voltmeter read for the base voltage?

For the collector voltage?

7-14 A dc voltmeter measures 1.1 V at the collector of Fig. 7-31b. What are some of the possible troubles?

Problems

SEC. 7-5 VOLTAGE-DIVIDER BIAS 7-15 What is the emitter voltage in

Fig. 7-32? The collector voltage?

7-16 What is the emitter voltage in Fig. 7-33? The collector voltage?

7-17 What is the emitter voltage in Fig. 7-34? The collector voltage?

7-18 What is the emitter voltage in Fig. 7-35? The collector voltage?

7-19 All resistors in Fig. 7-34 have a tolerance of 65 percent. What is the lowest possible value of the collector voltage? The highest?

7-20 The power supply of Fig. 7-35 has a tolerance of 610 percent. What is the lowest possible value of the collector voltage? The highest?

SEC. 7-7 VDB LOAD LINE AND Q POINT 7-21 What is the Q point for Fig. 7-32?

7-22 What is the Q point for Fig. 7-33?

7-23 What is the Q point for Fig. 7-34?

7-24 What is the Q point for Fig. 7-35?

7-25 All resistors in Fig. 7-34 have a tolerance of 65 per cent. What is the lowest value of the collector current? The highest?

7-26 The power supply in Fig. 7-35 has a tolerance of 610 percent. What is the lowest possible value of the collector current? The highest?

SEC. 7-8 TWO-SUPPLY EMITTER BIAS 7-27 What is the emitter current in Fig. 7-36? The

collector voltage?

7-28 If all resistances are doubled in Fig. 7-36, what is the emitter current? The collector voltage?

7-29 All resistors in Fig. 7-36 have a tolerance of 65 percent. What is the lowest possible value of the collector voltage? The highest?

SEC. 7-9 OTHER TYPES OF BIAS

7-30 Does the collector voltage increase, decrease, or remain the same in Fig. 7-35 for small changes in each of the following?

a. R1 increases d. RC decreases b. R2 decreases e. VCC increases c. RE increases f. bdc decreases

7-31 Does the collector voltage increase, decrease, or remain the same in Fig. 7-37 for small increases in each of the following circuit values?

a. R1 d. RC

b. R2 e. VEE

c. RE f. bdc

VCC +12 V

VEE –12 V RC 4.7 kΩ

10 kΩ RE

10 kΩ

Figure 7-36

VEE +10 V

RE 1 kΩ R2

2.2 kΩ

10 kΩ RC

3.6 kΩ 2N3906

Figure 7-37

SEC. 7-10 TROUBLESHOOTING VDB CIRCUITS 7-32 What is the approximate value of the collector

voltage in Fig. 7-35 for each of these troubles?

a. R1 open d. RC open

b. R2 open e. Collector-emitter open c. RE open

VCC +25 V

RC 3.6 kΩ R1

10 kΩ

2.2 kΩ RE

1 kΩ

Figure 7-32

VCC +15 V RC 2.7 kΩ R1

10 kΩ

2.2 kΩ RE

1 kΩ

Figure 7-33

VCC +10 V

RC 150 kΩ R1

330 kΩ

100 kΩ RE

51 kΩ

Figure 7-34

VCC +12 V RC 39 Ω R1

150 Ω

33 Ω RE

10 Ω

Figure 7-35

276 Chapter 7 7-33 What is the approximate value of the collector

voltage in Fig. 7-37 for each of these troubles?

a. R1 open b. R2 open c. RE open d. RC open

e. Collector-emitter open SEC. 7-11 PNP TRANSISTORS

7-34 What is the collector voltage in Fig. 7-37?

7-35 What is the collector-emitter voltage in Fig. 7-37?

7-36 What is the collector saturation current in Fig. 7-37? The collector-emitter cutoff voltage?

7-37 What is the emitter voltage in Fig. 7-38? The collector voltage?

Critical Thinking

7-38 Somebody has built the circuit in Fig. 7-35, except for changing the voltage divider as follows: R1 5 150 kV and R2 5 33 kV. The builder cannot understand why the base voltage is only 0.8 V instead of 2.16 V (the ideal output of the voltage divider). Can you explain what is happening?

7-39 Somebody builds the circuit in Fig. 7-35 with a 2N3904. What do you have to say about that?

7-40 A student wants to measure the collector-emitter voltage in Fig. 7-35, and so connects a voltmeter between the collector and the emitter. What does it read?

7-41 You can vary any circuit value in Fig. 7-35. Name all the ways you can think of to destroy the transistor.

7-42 The power supply in Fig. 7-35 has to supply current to the transistor circuit. Name all the ways you can think of to ﬁ nd this current.

7-43 Calculate the collector voltage for each transistor in Fig. 7-39. (Hint: Capacitors are open to direct current.)

7-44 The circuit in Fig. 7-40a uses silicon diodes. What is the emitter current? The collector voltage?

7-45 What is the output voltage in Fig. 7-40b?

7-46 How much current is there through the LED in Fig. 7-41a?

7-47 What is the LED current in Fig. 7-41b?

7-48 We want the voltage divider in Fig. 7-34 to be stiff . Change R1 and R2 as needed without changing the Q point.

VCC +15 V 1 kΩ

240 Ω 1.8 kΩ

300 Ω Q1

510 Ω

120 Ω 910 Ω

150 Ω Q2

620 Ω

150 Ω 1 kΩ

180 Ω Q3

GND

vin vout

Figure 7-39

VCC –10 V

RC 3.6 kΩ R1

10 kΩ

2.2 kΩ RE

1 kΩ 2N3906

Figure 7-38

R1 10 kΩ

VCC +20 V RC 8.2 kΩ

RE 1 kΩ

(a)

RC1 1 kΩ

VCC +16 V

RE2 1 kΩ

(b) RE1 200 Ω VBB

+2 V vout

Figure 7-40

VEE +12 V

RC 200 Ω

(a) R2 680 Ω

RC 200 Ω

R1 620 Ω

(b) 6.2 V + –

Figure 7-41

Troubleshooting

Use Fig. 7-42 for the remaining problems.

7-49 Find Trouble 1.

7-50 Find Trouble 2.

7-51 Find Troubles 3 and 4.

7-52 Find Troubles 5 and 6.

7-53 Find Troubles 7 and 8.

7-54 Find Troubles 9 and 10.

7-55 Find Troubles 11 and 12.

Figure 7-42

R2 2.2 kΩ R1 10 kΩ

RC 3.6 kΩ

RE 1 kΩ B

+VCC (10 V)

1.8 1.1 6 OK

10 9.3 9.4 OK

0.7 0 0.1 OK

1.8 1.1 10 OK

0 0 10 OK

0 0 1 00

1.1 0.4 0.5 OK

1.1 0.4 10 OK

0 0 0 O K

1.83 0 10 OK

2.1 2.1 2.1 OK

3.4 2.7 2.8 1.83 1.212 10 OK VB (V)

Trouble

MEASUREMENTS VE (V) VC (V) R2 (Ω) OK

T1 T2 T3 T4

T6 T7 T8 T9 T10 T11 T12 T5

278 Chapter 7

Job Interview Questions

Multisim Troubleshooting Problems

1. Draw a VDB circuit. Then, tell me all the steps in calculating the collector-emitter voltage. Why does this circuit have a very stable Q point?

2. Draw a TSEB circuit and tell me how it works. What happens to the collector current when the transistor is replaced or the temperature changes?

3. Describe a few other kinds of bias. What can you tell me about their Q points?

4. What are the two types of feedback biasing, and why were they developed?

5. What is the primary type of biasing used with discrete bipolar transistor circuits?

6. Should transistors being used as switching circuits be biased in the active region? If not, what two points

associated with the load line are important with switching circuits?

7. In a VDB circuit, the base current is not small compared to the current through the voltage divider. What is the shortcoming of this circuit? What should be changed to correct it?

8. What is the most commonly used transistor biasing conﬁ guration? Why?

9. Draw a VDB circuit using an npn transistor. Label directions of divider, base, emitter, and collector currents.

10. What is wrong with a VDB circuit in which R1 and R2 are 100 times greater than RE?

The Multisim troubleshooting ﬁ les are found on the Instructor Resources section of Connect for Electronic Principles, in a folder named Multisim Troubleshooting Circuits (MTC). See page XVI for more details. For this chapter, the ﬁ les are labeled MTC07-56 through MTC07-60 and are based on the circuit of Figure 7-42.

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.

7-56 Open up and troubleshoot ﬁ le MTC07-56.

7-57 Open up and troubleshoot ﬁ le MTC07-57.

7-58 Open up and troubleshoot ﬁ le MTC07-58.

7-59 Open up and troubleshoot ﬁ le MTC07-59.

7-60 Open up and troubleshoot ﬁ le MTC07-60.

Self-Test Answers

1. b 2. b 3. d 4. b 5. a 6. a 7. c 8. a 9. c 10. d 11. a 12. a 13. d

14. b 15. b 16. b 17. a 18. c 19. a 20. b 21. a 22. c 23. c 24. c 25. a 26. b

27. a 28. d 29. a 30. c 31. a 32. d 33. b 34. b 35. c 36. b 37. c 38. a 39. d

Practice Problem Answers

7-1 VCE 5 8.1 V 7-2 RE 5 680 V 7-5 VB 5 2.7 V;

VE 5 2 mA;

VC 5 7.78 V;

VCE 5 5.78 V 7-6 VCE 5 5.85 V;

Very close to the predicted value

7-8 RE 5 1 kV;

RC 5 4 kV;

R2 5 700 V (680);

R1 5 3.4 kV (3.3k) 7-9 VCE 5 6.96 V 7-10 VCE 5 7.05 V

7-11 For 7-29a:

VB 5 2.16 V;

VE 5 21.46 V;

VC 5 26.73 V;

VCE 5 25.27 V For 7-29b:

VB 5 9.84 V;

VE 5 10.54 V;

VC 5 5.27 V;

VCE 5 25.27 V

280 chapter

8 Basic BJT

Ampliﬁ ers

After a transistor has been biased with the Q point near the middle of the load line, we can couple a small ac voltage into the base. This will produce an ac collector voltage. The ac

collector voltage looks like the ac base voltage, except that it’s a lot bigger. In other words, the ac collector voltage is an ampliﬁ ed version of the ac base voltage.

This chapter will show you how to calculate the voltage gain and the ac voltages from the circuit values. This is important when troubleshooting because you can measure the ac voltages to see whether they are in reasonable agreement with theoretical values. This chapter also discusses ampliﬁ er input/output impedances and negative feedback.

© Arthur S. Aubry/Getty Images

ac collector resistance ac current gain ac emitter feedback ac emitter resistance ac-equivalent circuit ac ground

ac short

bypass capacitor

CB amplifi er CC amplifi er CE amplifi er coupling capacitor dc-equivalent circuit distortion

Ebers-Moll model feedback resistor

␲ model

small-signal ampliﬁ ers superposition theorem swamped ampliﬁ er swamping

T model voltage gain

Vocabulary

Objectives

After studying this chapter, you should be able to:

■ Draw a transistor ampliﬁ er and explain how it works.

■ Describe what coupling and bypass capacitors are supposed to do.

■ Give examples of ac shorts and ac grounds.

■ Use the superposition theorem to draw the dc- and ac-equivalent circuits.

■ Deﬁ ne small-signal operation and why it may be desirable.

■ Draw an ampliﬁ er that uses VDB.

Then, draw its ac-equivalent circuit.

■ Discuss the important

characteristics of the CE ampliﬁ er.

■ Show how to calculate and predict the voltage gain of a CE ampliﬁ er.

■ Explain how the swamped ampliﬁ er works and list three of its advantages.

■ Describe two capacitor-related problems that can occur in the CE ampliﬁ er.

■ Troubleshoot CE ampliﬁ er circuits.

Chapter Outline

8-1 Base-Biased Ampliﬁ er 8-2 Emitter-Biased Ampliﬁ er 8-3 Small-Signal Operation 8-4 AC Beta

8-5 AC Resistance of the Emitter Diode

8-6 Two Transistor Models 8-7 Analyzing an Ampliﬁ er 8-8 AC Quantities on the Data

Sheet 8-9 Voltage Gain

8-10 The Loading Eff ect of Input Impedance

8-11 Swamped Ampliﬁ er 8-12 Troubleshooting

282 Chapter 8

8-1 Base-Biased Ampliﬁ er

In this section, we will discuss a base-biased amplifi er. A base-biased amplifi er has instructional value because its basic ideas can be used to build more compli- cated amplifi ers.

Coupling Capacitor

Figure 8-1a shows an ac voltage source connected to a capacitor and a resistor. Since the impedance of the capacitor is inversely proportional to frequency, the capacitor effectively blocks dc voltage and transmits ac voltage. When the frequency is high enough, the capacitive reactance is much smaller than the resistance. In this case, almost all the ac source voltage appears across the resistor. When used in this way, the capacitor is called a coupling capacitor because it couples or transmits the ac signal to the resistor. Coupling capacitors are important because they allow us to couple an ac signal into an amplifi er without disturbing its Q point.

For a coupling capacitor to work properly, its reactance must be much smaller than the resistance at the lowest frequency of the ac source. For instance, if the frequency of the ac source varies from 20 Hz to 20 kHz, the worst case occurs at 20 Hz. A circuit designer will select a capacitor whose reactance at 20 Hz is much smaller than the resistance.

How small is small? As a defi nition:

Good coupling: XC 0.1R (8-1)

In words: The reactance should be at least 10 times smaller than the resistance at the lowest frequency of operation.

When the 10:1 rule is satisfi ed, Fig. 8-1a can be replaced by the equivalent circuit in Fig. 8-1b. Why? The magnitude of impedance in Fig. 8-1a is given by:

Z 5 ẽ—R2 1 XC2

When you substitute the worst case into this, you get:

Z 5 ẽ——R2 1 (0.1R)2 5 ẽ——R2 10.01R2 5 ẽ—1.01R2 5 1.005R

Since the impedance is within half of a percent of R at the lowest frequency, the current in Fig. 8-1a is only half a percent less than the current in Fig. 8-1b. Since any well-designed circuit satisfi es the 10:1 rule, we can approximate all coupling capacitors as an ac short (Fig. 8-1b).

A fi nal point about coupling capacitors: Since dc voltage has a frequency of zero, the reactance of a coupling capacitor is infi nite at zero frequency. There- fore, we will use these two approximations for a capacitor:

1. For dc analysis, the capacitor is open.

2. For ac analysis, the capacitor is shorted.

Figure 8-1 (a) Coupling capacitor; (b) capacitor is an ac short; (c) dc open and ac short.

(a) (b)

SHORT

AC (c) C

V R V R

Figure 8-1c summarizes these two important ideas. Unless otherwise stated, all the circuits we analyze from now on will satisfy the 10:1 rule, so that we can visualize a coupling capacitor as shown in Fig. 8-1c.

Example 8-1

Using Fig. 8-1a, if R 5 2 kV and the frequency range is from 20 Hz to 20 kHz, fi nd the value of C needed to act as a good coupling capacitor.

SOLUTION Following the 10 :1 rule, XC should be 10 times smaller than R at the lowest frequency.

Therefore:

XC , 0.1 R at 20 Hz XC , 200 V at 20 Hz Since XC 5 1 _____

2␲fC

by rearrangement, C 5 ______ 1

2␲f XC 5 _________________ 1 (2␲)(20 Hz)(200 V) C 5 39.8 ␮F

PRACTICE PROBLEM 8-1 Using Example 8-1, fi nd the value of C when the lowest frequency is 1 kHz and R is 1.6 kV.

DC Circuit

Figure 8-2a shows a base-biased circuit. The dc base voltage is 0.7 V. Because 30 V is much greater than 0.7 V, the base current is approximately 30 V divided by 1 MV, or:

IB 5 30 ␮A

With a current gain of 100, the collector current is:

IC 5 3 mA and the collector voltage is:

VC 5 30 V 2 (3 mA)(5 kV) 5 15 V So, the Q point is located at 3 mA and 15 V.

Amplifying Circuit

Figure 8-2b shows how to add components to build an amplifi er. First, a coupling capacitor is used between an ac source and the base. Since the coupling capacitor is open to direct current, the same dc base current exists, with or without the capacitor and ac source. Similarly, a coupling capacitor is used between the collector and the load resistor of 100 kV. Since this capacitor is open to direct

284 Chapter 8

current, the dc collector voltage is the same, with or without the capacitor and load resistor. The key idea is that the coupling capacitors prevent the ac source and load resistance from changing the Q point.

In Fig. 8-2b, the ac source voltage is 100 ␮V. Since the coupling capacitor is an ac short, all the ac source voltage appears between the base and the ground. This ac voltage produces an ac base current that is added to the existing dc base current. In other words, the total base current will have a dc component and an ac component.

Figure 8-3a illustrates the idea. An ac component is superimposed on the dc component. On the positive half-cycle, the ac base current adds to the 30 ␮A of the dc base current, and on the negative half-cycle, it subtracts from it.

The ac base current produces an amplifi ed variation in collector current because of the current gain. In Fig. 8-3b, the collector current has a dc component of 3 mA. Superimposed on this is an ac collector current. Since this amplifi ed collector current fl ows through the collector resistor, it produces a varying voltage across the collector resistor. When this voltage is subtracted from the supply voltage, we get the collector voltage shown in Fig. 8-3c.

Again, an ac component is superimposed on a dc component. The collector voltage is swinging sinusoidally above and below the dc level of 115 V. Also, the ac collector voltage is inverted 180° out of phase with the input voltage. Why?

On the positive half-cycle of the ac base current, the collector current increases, producing more voltage across the collector resistor. This means that there is less voltage between the collector and ground. Similarly, on the negative half-cycle, Figure 8-2 (a) Base bias; (b) base-biased ampliﬁ er.

100 mV

(b) (a)

+15 V

+0.7 V bdc = 100

RB 1 MΩ

RC 5 kΩ

RL 100 kΩ RC

5 kΩ

VCC +30 V

the collector current decreases. Since there is less voltage across the collector resistor, the collector voltage increases.

Voltage Waveforms

Figure 8-4 shows the waveforms for a base-biased amplifi er. The ac source voltage is a small sinusoidal voltage. This is coupled into the base, where it is superimposed on the dc component of 10.7 V. The variation in base voltage produces sinusoidal variations in base current, collector current, and collector voltage. The total collector voltage is an inverted sine wave superimposed on the dc collector voltage of 115 V.

Notice the action of the output coupling capacitor. Since it is open to direct current, it blocks the dc component of collector voltage. Since it is shorted to alternating current, it couples the ac collector voltage to the load resistor. This is why the load voltage is a pure ac signal with an average value of zero.

Figure 8-3 DC and ac components. (a) Base current; (b) collector current;

t IB

30 mA

t IC

3 mA

t VC

15 V

(a)

(b)

(c)

286 Chapter 8

Voltage Gain

The voltage gain of an amplifi er is defi ned as the ac output voltage divided by the ac input voltage. As a defi nition:

AV 5 ____ vvoutin (8-2) For instance, if we measure an ac load voltage of 50 mV with an ac input voltage of 100 ␮V, the voltage gain is:

AV 5 50 mV _______

100 ␮V 5 500

This says that the ac output voltage is 500 times larger than the ac input voltage.

Calculating Output Voltage

We can multiply both sides of Eq. (8-2) by vin to get this derivation:

vout 5 AV vin (8-3)

This is useful when you want to calculate the value of vout, given the values of AV and vin.

For instance, the triangular symbol shown in Fig. 8-5a is used to indicate an amplifi er of any design. Since we are given an input voltage of 2 mV and a voltage gain of 200, we can calculate an output voltage of:

vout 5 (200)(2 mV) 5 400 mV

Calculating Input Voltage

We can divide both sides of Eq. (8-3) by AV to get this derivation:

vin 5 v____ out

AV (8-4)

This is useful when you want to calculate the value of vin, given the values vout and AV. For instance, the output voltage is 2.5 V in Fig. 8-5b. With a voltage gain of 350, the input voltage is:

vin 5 2.5 V _____

350 5 7.14 mV

Figure 8-4 Base-biased ampliﬁ er with waveforms.

Figure 8-5 (a) Calculating output voltage; (b) calculating input voltage.

+15 V 0

+0.7 V 0

vin

vout RB

1 MΩ

RC 5 kΩ

VCC +30 V

RL 100 kΩ

AV = 200

vin vout

2 mV

AV = 350

vin vout

2.5 V (a)

(b)

In a TSEB circuit, the base voltage is approximately zero

The surface-leakage current doubles when the reverse

Voltage multipliers are circuits best used to produce