a. Low b. High c. Unchanged d. Unknown
SEC. 6-3 TRANSISTOR CURRENTS
6-1 A transistor has an emitter current of 10 mA and a collector current of 9.95 mA. What is the base current?
6-2 The collector current is 10 mA, and the base cur- rent is 0.1 mA. What is the current gain?
6-3 A transistor has a current gain of 150 and a base current of 30 A. What is the collector current?
6-4 If the collector current is 100 mA and the current gain is 65, what is the emitter current?
SEC. 6-5 THE BASE CURVE
6-5 What is the base current in Fig. 6-33?
Problems
VCC 10 V bdc = 200
VBB 10 V
RB 470 kΩ
RC 820 Ω
–
+ –
+
Figure 6-33
6-6 If the current gain decreases from 200 to 100 in Fig. 6-33, what is the base current?
236 Chapter 6 6-7 If the 470 kV of Fig. 6-33 has a tolerance of 65
percent, what is the maximum base current?
SEC. 6-6 COLLECTOR CURVES
6-8 A transistor circuit similar to Fig. 6-33 has a collector supply voltage of 20 V, a collector resistance of 1.5 kV, and a collector current of 6 mA. What is the collector-emitter voltage?
6-9 If a transistor has a collector current of 100 mA and a collector-emitter voltage of 3.5 V, what is its power dissipation?
SEC. 6-7 TRANSISTOR APPROXIMATIONS 6-10 What are the collector-emitter voltage and the tran-
sistor power dissipation in Fig. 6-33? (Give answers for the ideal and the second approximation.) 6-11 Figure 6-34a shows a simpler way to draw a
transistor circuit. It works the same as the circuits already discussed. What is the collector- emitter voltage? The transistor power dissipation?
(Give answers for the ideal and the second approximation.)
6-12 When the base and collector supplies are equal, the transistor can be drawn as shown in Fig. 6-34b.
What is the collector-emitter voltage in this circuit?
The transistor power? (Give answers for the ideal and the second approximation.)
SEC. 6-8 READING DATA SHEETS
6-13 What is the storage temperature range of a 2N3904?
6-14 What is the minimum hFE for a 2N3904 for a collec- tor current of 1 mA and a collector-emitter voltage of 1 V?
6-15 A transistor has a power rating of 1 W. If the collector-emitter voltage is 10 V and the collector current is 120 mA, what happens to the transistor?
6-16 A 2N3904 has a power dissipation of 625 mW without a heat sink. If the ambient temperature is 65°C, what happens to the power rating?
SEC. 6-10 VARIATIONS IN CURRENT GAIN 6-17 Refer to Fig. 6-19. What is the current gain of a
2N3904 when the collector current is 100 mA and the junction temperature is 125°C?
6-18 Refer to Fig. 6-19. The junction temperature is 25°C, and the collector current is 1.0 mA. What is the current gain?
SEC. 6-11 THE LOAD LINE
6-19 Draw the load line for Fig. 6-35a. What is the collector current at the saturation point? The collector-emitter voltage at the cutoff point?
6-20 If the collector supply voltage is increased to 25 V in Fig. 6-35a, what happens to the load line?
6-21 If the collector resistance is increased to 4.7 kV in Fig. 6-35a, what happens to the load line?
6-22 If the base resistance of Fig. 6-35a is reduced to 500 kV, what happens to the load line?
6-23 Draw the load line for Fig. 6-35b. What is the collector current at the saturation point? The collector-emitter voltage at the cutoff point?
6-24 If the collector supply voltage is doubled in Fig. 6-35b, what happens to the load line?
6-25 If the collector resistance is increased to 1 kV in Fig. 6-35b, what happens to the load line?
SEC. 6-12 THE OPERATING POINT
6-26 In Fig. 6-35a, what is the voltage between the col- lector and ground if the current gain is 200?
6-27 The current gain varies from 25 to 300 in Fig. 6-35a. What is the minimum voltage from the collector to ground? The maximum?
(a)
(b) RB
330 kΩ
RC 1.2 kΩ
bdc = 150
RB 680 kΩ
RC 1.5 kΩ
bdc = 175 VBB
+5 V
VCC +15 V
VCC +12 V
Figure 6-34
(a) RC 3.3 kΩ VBB
+10 V
VCC +20 V
(b) RC 470 Ω VBB
+5 V
VCC +5 V
RB 1 MΩ
RB 680 kΩ
Figure 6-35
6-28 The resistors of Fig. 6-35a have a tolerance of 65 percent. The supply voltages have a tolerance of 610 percent. If the current gain can vary from 50 to 150, what is the minimum possible voltage from the collector to ground? The maximum?
6-29 In Fig. 6-35b, what is the voltage between the collector and ground if the current gain is 150?
6-30 The current gain varies from 100 to 300 in Fig. 6-35b. What is the minimum voltage from the collector to ground? The maximum?
6-31 The resistors of Fig. 6-35b have a tolerance of 65 percent. The supply voltages have a tolerance of 610 percent. If the current gain can vary from 50 to 150, what is the minimum possible voltage from the collector to ground? The maximum?
SEC. 6-13 RECOGNIZING SATURATION 6-32 In Fig. 6-35a, use the circuit values shown
unless otherwise indicated. Determine whether the transistor is saturated for each of these changes:
a. RB 5 33 kV and hFE 5 100 b. VBB 5 5 V and hFE 5 200 c. RC 5 10 kV and hFE 5 50 d. VCC 5 10 V and hFE 5 100
6-33 In Fig. 6-35b, use the circuit values shown unless otherwise indicated. Determine whether the tran- sistor is saturated for each of these changes:
a. RB 5 51 kV and hFE 5 100 b. VBB 5 10 V and hFE 5 500 c. RC 5 10 kV and hFE 5 100 d. VCC 5 10 V and hFE 5 100
SEC. 6-14 THE TRANSISTOR SWITCH
6-34 The 680 kV in Fig. 6-35b is replaced by 4.7 kV and a series switch. Assuming an ideal transistor, what is the collector voltage if the switch is open? What is the collector voltage if the switch is closed?
SEC. 6-15 TROUBLESHOOTING
6-35 In Fig. 6-33, does the collector-emitter voltage increase, decrease, or remain the same for each of these troubles?
a. 470 kV is shorted b. 470 kV is open c. 820 V is shorted d 820 V is open
e. No base supply voltage f. No collector supply
Critical Thinking
6-36 What is the dc alpha of a transistor that has a current gain of 200?
6-37 What is the current gain of a transistor with a dc alpha of 0.994?
6-38 Design a CE circuit to meet these specifi cations:
VBB 5 5 V, VCC 5 15 V, hFE 5 120, IC 510 mA, and VCE 5 7.5 V.
6-39 In Fig. 6-33, what value base resistor would be needed so VCE 5 6.7 V?
6-40 A 2N3904 has a power rating of 350 mW at room temperature (25°C). If the collector-emitter voltage is 10 V, what is the maximum current that the tran- sistor can handle for an ambient temperature of 50°C?
6-41 Suppose we connect an LED in series with the 820 V of Fig. 6-33. What does the LED current equal?
6-42 What is the collector-emitter saturation voltage of a 2N3904 when the collector current is 50 mA? Use the data sheet.
Multisim Troubleshooting Problems
The Multisim troubleshooting fi les are found on the Instructor Resources section of Connect for Electronic Principles, in a folder named Multisim Troubleshoot- ing Circuits (MTC). See page XVI for more details. For this chapter, the fi les are labelled MTC06-43 through MTC06-47 and are based on the circuit of Figure 6-33.
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.
6-43 Open up and troubleshoot fi le MTC06-43.
6-44 Open up and troubleshoot fi le MTC06-44.
6-45 Open up and troubleshoot fi le MTC06-45.
6-46 Open up and troubleshoot fi le MTC06-46.
6-47 Open up and troubleshoot fi le MTC06-47.
238 Chapter 6
Digital/Analog Trainer System
The following questions, 6-48 through 6-52, are directed toward the schematic diagram of the Digital/Analog Trainer System found on the Instructor Resources section of Connect for Electronic Principles. A full Instruction Manual for the Model XK-700 trainer can be found at www.elenco.com.
6-48 Q3 is a 2N3904 npn transistor. What type of bias confi guration does this circuit use?
6-49 What is the input signal applied to the 2N3904 called? What is the output of the 2N3904, which is found at the transistor’s collector called?
6-50 If the input to R10 is +5 V, determine the level of base current and collector current using the sec- ond approximation of this transistor.
6-51 With +5 V dc applied to R10, is Q3 operating in the active, cutoff , or saturated region?
6-52 With 0 V applied to R10, what is the approximate voltage at the collector of Q3?
Job Interview Questions
1. I want you to draw an npn transistor showing the n and p regions. Then I want you to bias the transistor properly and tell me how it works.
2. Draw a set of collector curves. Then, using these curves, show me where the four operating regions of a transistor are.
3. Draw the two equivalent circuits (ideal and second approximation) to represent a transistor that is operat- ing in the active region. Then, tell me when and how you would use these circuits to calculate the transis- tor currents and voltages.
4. Draw a transistor circuit with a CE connection. Now, what kind of troubles can you get with a circuit like this and what measurements would you make to iso- late each trouble?
5. When looking at a schematic diagram that shows npn and pnp transistors, how can you identify each type?
How can you tell the direction of electron (or conven- tional) fl ow?
6. Name a test instrument that can display a set of col- lector curves, IC versus VCE, for a transistor.
7. What is the formula for transistor power dissipa- tion? Knowing this relationship, where on the load
line would you expect the power dissipation to be maximum?
8. What are the three currents in a transistor, and how are they related?
9. Draw an npn and a pnp transistor. Label all currents and show directions of fl ow.
10. Transistors may be connected in any of the following confi gurations: common emitter, common collec- tor, and common base. Which is the most common confi guration?
11. Draw a base-biased circuit. Then, tell me how to cal- culate the collector-emitter voltage. Why is this circuit likely to fail in mass production if a precise value of current gain is needed?
12. Draw another base-biased circuit. Draw a load line for the circuit and tell me how to calculate the saturation and cutoff points. Discuss the eff ects of a changing current gain on the location of the Q point.
13. Tell me how you would test a transistor out of a circuit. What kind of tests can you do while the transistor is in a circuit with the power on?
14. What eff ect does temperature have on current gain?
Self-Test Answers
1. b 2. a 3. c 4. a 5. b 6. b 7. b 8. b
9. d 10. b 11. b 12. d 13. b 14. a 15. a 16. b
17. b 18. d 19. c 20. a 21. a 22. a 23. b 24. d
25. d 26. c 27. c 28. c
29. b 30. d 31. c 32. b
33. d 34. c 35. b
Practice Problem Answers
6-1 dc 5 200 6-2 IC 5 10 mA 6-3 IB 5 74.1 A 6-4 VB 5 0.7 V;
IB 5 33 A;
IC 5 6.6 mA 6-5 IB 5 13.7 A;
IC 5 4.11 mA;
VCE 5 1.78 V;
PD 5 7.32 mW
6-6 IB 5 16.6 A;
IC 5 5.89 mA;
dc 5 355 6-10 Ideal: IB 5 14.9 A;
IC 5 1.49 mA;
VCE 5 9.6 V
Second: IB 5 13.4 A;
IC 5 1.34 mA;
VCE 5 10.2 V
6-12 PD (max) 5 375 mW. Not within the safety factor of 2.
6-14 IC (sat) 5 6 mA;
VCE (cutoff ) 5 12 V 6-16 IC (sat) 5 3 mA;
The slope would decrease.
6-17 VCE 5 8.25 V 6-19 RB 5 47 kV
6-20 VCE 5 11.999 V and 0.15 V
240 chapter
7 BJT Biasing
A prototype is a basic circuit design that can be modifi ed to get more advanced circuits. Base bias is a prototype used in the design of switching circuits. Emitter bias is a prototype used in the design of amplifying circuits. In this chapter, we emphasize emitter bias and the practical circuits that can be derived from it.
© Royalty-Free/CORBIS
collector-feedback bias correction factor emitter bias
emitter-feedback bias fi rm voltage divider
phototransistor prototype self-bias stage
stiff voltage divider
swamp out
two-supply emitter bias (TSEB)
voltage-divider bias (VDB)
Vocabulary
Objectives
After studying this chapter, you should be able to:
■ Draw an emitter bias circuit and explain why it works well in amplifying circuits.
■ Draw a diagram of a voltage- divider bias circuit.
■ Calculate the divider current, base voltage, emitter voltage, emitter current, collector voltage, and collector-emitter voltage for an npn VDB circuit.
■ Determine how to draw the load line and calculate the Q point for a given VDB circuit.
■ Design a VDB circuit using design guidelines.
■ Draw a two-supply emitter bias circuit and calculate VRE, IE, VC, and VCE.
■ Compare several diff erent types of bias and describe how well each works.
■ Calculate the Q point of pnp VDB circuits.
■ Troubleshoot transistor-biasing circuits.
Chapter Outline
7-1 Emitter Bias 7-2 LED Drivers
7-3 Troubleshooting Emitter Bias Circuits
7-4 More Optoelectronic Devices
7-5 Voltage-Divider Bias 7-6 Accurate VDB Analysis 7-7 VDB Load Line and Q Point 7-8 Two-Supply Emitter Bias 7-9 Other Types of Bias 7-10 Troubleshooting VDB
Circuits
7-11 PNP Transistors
242 Chapter 7
7-1 Emitter Bias
Digital circuits are the type of circuits used in computers. In this area, base bias and circuits derived from base bias are useful. But when it comes to amplifi ers, we need circuits whose Q points are immune to changes in current gain.
Figure 7-1 shows emitter bias. As you can see, the resistor has been moved from the base circuit to the emitter circuit. That one change makes all the difference in the world. The Q point of this new circuit is now rock-solid. When the current gain changes from 50 to 150, the Q point shows almost no movement along the load line.
Basic Idea
The base supply voltage is now applied directly to the base. Therefore, a trou- bleshooter will read VBB between the base and ground. The emitter is no longer grounded. Now the emitter is above the ground and has a voltage given by:
VE 5 VBB 2 VBE (7-1)
If VBB is more than 20 times VBE, the ideal approximation will be accurate. If VBB
is less than 20 times VBE, you may want to use the second approximation. Other- wise, your error will be more than 5 percent.
Finding the Q Point
Let us analyze the emitter-biased circuit of Fig. 7-2. The base supply voltage is only 5 V, so we use the second approximation. The voltage between the base and ground is 5 V. From now on, we refer to this base-to-ground voltage as the base voltage, or VB. The voltage across the base-emitter terminals is 0.7 V. We refer to this voltage as the base-emitter voltage, or VBE.
VBB
VCC RC
RE –
+ –
+
Figure 7-1 Emitter bias.
VBB 5 V
RC 1 kΩ
RE 2.2 kΩ
VCC 15 V bdc = 100
–
+ –
+
Figure 7-2 Finding the Q point.
The voltage between the emitter and ground is called the emitter voltage.
It equals:
VE 55 V 2 0.7 V 5 4.3 V
This voltage is across the emitter resistance, so we can use Ohm’s law to fi nd the emitter current:
IE 5 ______4.3 V
2.2 kV 5 1.95 mA
This means that the collector current is 1.95 mA to a close approximation. When this collector current fl ows through the collector resistor, it produces a voltage drop of 1.95 V. Subtracting this from the collector supply voltage gives the voltage between the collector and ground:
VC 5 15 V 2 (1.95 mA)(1 kV) 5 13.1 V
From now on, we will refer to this collector-to-ground voltage as the collector voltage.
This is the voltage a troubleshooter would measure when testing a transis- tor circuit. One lead of the voltmeter would be connected to the collector, and the other lead would be connected to ground. If you want the collector-emitter voltage, you have to subtract the emitter voltage from the collector voltage as follows:
VCE 5 13.1 V 2 4.3 V 5 8.8 V
So, the emitter-biased circuit of Fig. 7-2 has a Q point with these coordinates:
IC 5 1.95 mA and VCE 5 8.8 V.
The collector-emitter voltage is the voltage used for drawing load lines and for reading transistor data sheets. As a formula:
VCE 5 VC 2 VE (7-2)
Circuit Is Immune to Changes in Current Gain
Here is why emitter bias excels. The Q point of an emitter-biased circuit is immune to changes in current gain. The proof lies in the process used to analyze the circuit.
Here are the steps we used earlier:
1. Get the emitter voltage.
2. Calculate the emitter current.
3. Find the collector voltage.
4. Subtract the emitter from the collector voltage to get VCE.
At no time do we need to use the current gain in the foregoing process. Since we don’t use it to fi nd the emitter current, collector current, and so on, the exact value of current gain no longer matters.
By moving the resistor from the base to the emitter circuit, we force the base-to-ground voltage to equal the base supply voltage. Before, almost all this supply voltage was across the base resistor, setting up a fi xed base current. Now, all this supply voltage minus 0.7 V is across the emitter resistor, setting up a fi xed emitter current.
Minor Eff ect of Current Gain
The current gain has a minor effect on the collector current. Under all operating conditions, the three currents are related by:
IE 5 IC 1 IB
which can be rearranged as:
IE 5 IC 1 ___ IC
dc
GOOD TO KNOW
Because the values of IC and VCE
are not affected by the value of beta in an emitter-biased circuit, this type of circuit is said to be beta-independent.
244 Chapter 7
Solve this for the collector current, and you get:
IC 5 _______ dc
dc 1 1 IE (7-3)
The quantity that multiplies IE is called a correction factor. It tells you how IC differs from IE. When the current gain is 100, the correction factor is:
dc _______
dc 1 1 5 _______100 100 1 1 5 0.99
This means that the collector current is equal to 99 percent of the emitter current.
Therefore, we get only a 1 percent error when we ignore the correction factor and say that the collector current equals the emitter current.
Example 7-1
What is the voltage between the collector and ground in the Multisim Fig. 7-3?
Between the collector and the emitter?
SOLUTION The base voltage is 5 V. The emitter voltage is 0.7 V less than this, or:
VE 55 V 2 0.7 V 5 4.3 V
Figure 7-3 Meter Values.
7-2 LED Drivers
You have learned that base-biased circuits set up a fi xed value of base current, and emitter-biased circuits set up a fi xed value of emitter current. Because of the problem with current gain, base-biased circuits are normally designed to switch between saturation and cutoff, whereas emitter-biased circuits are usually de- signed to operate in the active region.
In this section, we discuss two circuits that can be used as LED drivers.
The fi rst circuit uses base bias, and the second circuit uses emitter bias. This will give you a chance to see how each circuit performs in the same application.
Base-Biased LED Driver
The base current is zero in Fig. 7-4a, which means that the transistor is at cut- off. When the switch of Fig. 7-4a closes, the transistor goes into hard saturation.
Visualize a short between the collector-emitter terminals. Then the collector sup- ply voltage (15 V) appears across the series connection of the 1.5 kV and the LED. If we ignore the voltage drop across the LED, the collector current is ideally 10 mA. But if we allow 2 V across the LED, then there is 13 V across the 1.5 kV, and the collector current is 13 V divided by 1.5 kV, or 8.67 mA.
This voltage is across the emitter resistance, which is now 1 kV. Therefore, the emitter current is 4.3 V divided by 1 kV, or:
IE 5 _____4.3 V 1 k
V 5 4.3 mA
The collector current is approximately equal to 4.3 mA. When this current fl ows through the collector resistance (now 2 kV), it produces a voltage of:
ICRC 5 (4.3 mA)(2 kV) 5 8.6 V
When you subtract this voltage from the collector supply voltage, you get:
VC 5 15 V 2 8.6 V 5 6.4 V
This voltage value is very close to the value measured by the Multisim meter.
Remember, this is the voltage between the collector and ground. This is what you would measure when troubleshooting.
Unless you have a voltmeter with a high input resistance and a fl oating ground lead, you should not attempt to connect a voltmeter directly between the collector and the emitter because this may short the emitter to ground. If you want to know the value of VCE, you should measure the collector-to-ground voltage, then measure the emitter-to-ground voltage, and subtract the two. In this case:
VCE 5 6.4 V 2 4.3 V 5 2.1 V
PRACTICE PROBLEM 7-1 Decrease the base supply voltage of Fig. 7-3 to 3 V. Predict and measure the new value of VCE.
246 Chapter 7
There is nothing wrong with this circuit. It makes a fi ne LED driver because it is designed for hard saturation, where the current gain doesn’t mat- ter. If you want to change the LED current in this circuit, you can change either the collector resistance or the collector supply voltage. The base resistance is made 10 times larger than the collector resistance because we want hard satura- tion when the switch is closed.
Emitter-Biased LED Driver
The emitter current is zero in Fig. 7-4b, which means that the transistor is at cutoff. When the switch of Fig. 7-4b closes, the transistor goes into the active region. Ideally, the emitter voltage is 15 V. This means that we get an emitter current of 10 mA. This time, the LED voltage drop has no effect. It doesn’t mat- ter whether the exact LED voltage is 1.8, 2, or 2.5 V. This is an advantage of the emitter-biased design over the base-biased design. The LED current is indepen- dent of the LED voltage. Another advantage is that the circuit doesn’t require a collector resistor.
The emitter-biased circuit of Fig. 7-4b operates in the active region when the switch is closed. To change the LED current, you can change the base supply voltage or the emitter resistance. For instance, if you vary the base supply voltage, the LED current varies in direct proportion.
(a)
VCC 15 V VBB
15 V
RC 1.5 kΩ
RB 15 kΩ
RE 1.5 kΩ
(b)
VCC 20 V VBB
15 V – +
–
+ –
+ – +
Figure 7-4 (a) Base-biased; (b) emitter-biased.