a. Increases b. Decreases c. Remains constant d. Cannot be determined
Problems
SEC. 5-1 THE ZENER DIODE
5-1 An unloaded zener regulator has a source voltage of 24 V, a series resistance of 470 V, and a zener voltage of 15 V. What is the zener current?
5-2 If the source voltage in Prob. 5-1 varies from 24 to 40 V, what is the maximum zener current?
5-3 If the series resistor of Prob. 5-1 has a tolerance of 65 percent, what is the maximum zener current?
SEC. 5-2 THE LOADED ZENER REGULATOR 5-4 If the zener diode is disconnected in
Fig. 5-44, what is the load voltage?
5-5 Calculate all three currents in Fig. 5-44.
5-6 Assuming a tolerance of 65 percent in both resistors of Fig. 5-44, what is the maximum zener current?
5-7 Suppose the supply voltage of Fig. 5-44 can vary from 24 to 40 V. What is the maximum zener current?
5-8 The zener diode of Fig. 5-44 is replaced with a 1N4742A. What are the load voltage and the zener current?
5-9 Draw the schematic diagram of a zener regulator with a supply voltage of 20 V, a series resistance of 330 V, a zener voltage of 12 V, and a load resis- tance of 1 kV. What are the load voltage and the zener current?
SEC. 5-3 SECOND APPROXIMATION OF A ZENER DIODE
5-10 The zener diode of Fig. 5-44 has a zener resistance of 14 V. If the power supply has a ripple of 1 Vp-p, what is the ripple across the load resistor?
POWER
SUPPLY 15 V RL
1.5 kΩ RS
470 Ω +
–
+ – 24 V
Figure 5-44
1000 mF 115 V ac
1N5314 1N753 TIL221
R1 R2 V130LA2
Figure 5-45
1000 mF 115 V ac
1N5314 1N753 TIL221
R1 R2 V130LA2
Figure 5-45
1000 mF 115 V ac
1N5314 1N753 TIL221
R1 R2 V130LA2
Figure 5-45
5-11 During the day, the ac line voltage changes. This causes the unregulated 24-V output of the power supply to vary from 21.5 to 25 V. If the zener resis- tance is 14 V, what is the voltage change over the foregoing range?
SEC. 5-4 ZENER DROP-OUT POINT
5-12 Assume the supply voltage of Fig. 5-44 decreases from 24 to 0 V. At some point along the way, the zener diode will stop regulating. Find the supply voltage where regulation is lost.
5-13 In Fig. 5-44, the unregulated voltage out of the power supply may vary from 20 to 26 V and the load resistance may vary from 500 V to 1.5 kV. Will the zener regulator fail under these conditions? If so, what value should the series resistance be?
5-14 The unregulated voltage in Fig. 5-44 may vary from 18 to 25 V, and the load current may vary from 1 to 25 mA. Will the zener regulator stop regulating under these conditions? If so, what is the maximum value for RS?
5-15 What is the minimum load resistance that may be used in Fig. 5-44 without losing zener regulation?
SEC. 5-5 READING A DATA SHEET
5-16 A zener diode has a voltage of 10 V and a current of 20 mA. What is the power dissipation?
5-17 A 1N5250B has 5 mA through it. What is the power?
5-18 What is the power dissipation in the resistors and zener diode of Fig. 5-44?
5-19 The zener diode of Fig. 5-44 is a 1N4744A. What is the minimum zener voltage? The maximum?
5-20 If the lead temperature of a 1N4736A zener diode rises to 100°C, what is the diode’s new power rating?
SEC. 5-6 TROUBLESHOOTING
5-21 In Fig. 5-44, what is the load voltage for each of these conditions?
a. Zener diode shorted b. Zener diode open
c. Series resistor open d. Load resistor shorted
5-22 If you measure approximately 18.3 V for the load voltage of Fig. 5-44, what do you think the trouble is?
5-23 You measure 24 V across the load of Fig. 5-44.
An ohmmeter indicates the zener diode is open.
Before replacing the zener diode, what should you check for?
5-24 In Fig. 5-45, the LED does not light. Which of the following are possible troubles?
a. V130LA2 is open.
b. Ground between two left bridge diodes is open.
c. Filter capacitor is open.
d. Filter capacitor is shorted.
e. 1N5314 is open.
f. 1N5314 is shorted.
SEC. 5-8 LIGHT-EMITTING DIODES (LEDS) 5-25 What is the current through the LED of
Fig. 5-46?
5-26 If the supply voltage of Fig. 5-46 increases to 40 V, what is the LED current?
5-27 If the resistor is decreased to 1 kV, what is the LED current in Fig. 5-46?
5-28 The resistor of Fig. 5-46 is decreased until the LED current equals 13 mA. What is the value of the resistance?
POWER SUPPLY
+ – 15 V
RS 2.2 kΩ
Figure 5-46
186 Chapter 5 1000 mF
6 V ac
#27 BULB GEN
Figure 5-47 5-29 The zener diode of Fig. 5-44 has a zener resis-
tance of 14 V. What is the load voltage if you in- clude RZ in your calculations?
5-30 The zener diode of Fig. 5-44 is a 1N4744A. If the load resistance varies from 1 to 10 kV, what is the minimum load voltage? The maximum load volt- age? (Use the second approximation.)
5-31 Design a zener regulator to meet these specifi ca- tions: Load voltage is 6.8 V, source voltage is 20 V, and load current is 30 mA.
5-32 A TIL312 is a seven-segment indicator. Each seg- ment has a voltage drop between 1.5 and 2 V at 20 mA. The supply voltage is 15 V. Design a seven- segment display circuit controlled by on-off switches that has a maximum current drain of 140 mA.
5-33 The secondary voltage of Fig. 5-45 is 12.6 Vrms
when the line voltage is 115 Vrms. During the day, the power line varies by 610 percent. The resistors have tolerances of 65 percent. The 1N4733A has a tolerance of 65 percent and a zener resistance of 7 V. If R2 equals 560 V, what is the maximum possible value of the zener current at any instant during day?
Critical Thinking
5-34 In Fig. 5-45, the secondary voltage is 12.6 Vrms, and diode drops are 0.7 V each. The 1N5314 is a constant-current diode with a current of 4.7 mA. The LED current is 15.6 mA, and the zener current is 21.7 mA. The fi lter capacitor has a tolerance of 620 per- cent. What is the maximum peak-to-peak ripple?
5-35 Figure 5-47 shows part of a bicycle lighting system.
The diodes are Schottky diodes. Use the second approximation to calculate the voltage across the fi lter capacitor.
RL 1 kΩ C
VL RS
270 Ω A +18 V
B D
E D1
1N5240B
+
–
18 10.3 10.3 10.3 OK
18 0 0 0 OK
18 14.2 14.2 0 OK 18 14.2 14.2 14.2
0 0 0 0 OK
18 0 0 0 0
18 18 18 18
18 10.5 10.5 10.5 OK 18 14.2 14.2 14.2 OK VA VB VC VD D1 OK
T1 T2 T3 T4
T6 T7 T8 T5
Figure 5-48 Troubleshooting.
The troubleshooting table shown in Fig. 5-48 lists the voltage values at each respective circuit point and the condition of the diode D1, for circuit troubles T1 through T8. The fi rst row displays what values would be found under normal operating conditions.
Troubleshooting
5-36 Find Troubles 1 to 4 in Fig. 5-48.
5-37 Find Troubles 5 to 8 in Fig. 5-48.
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 MTC05-38 through MTC05-42 and are based on the circuit of Figure 5-48.
Multisim Troubleshooting Problems
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.
5-38 Open up and troubleshoot fi le MTC05-38.
The following questions, 5-43 through 5-47, 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.
5-43 Are the LEDs D18 through D25 connected as com- mon anode or common cathode?
5-44 If the output voltage of the 74HC04 U8A is +5 V, is D18 biased on or off ?
Digital/Analog Trainer System
5-45 If the output voltage of U8E is 0 V, what is the ap- proximate current through D22?
5-46 What resistor value would R42 need to be so the current through D24 was approximately 15 mA?
5-47 What resistor value and wattage is needed to add an LED 12 V indicator light to U2’s 12 V DC output?
Design the LED current for approximately 21 mA.
1. Draw a zener regulator. Then explain to me how it works and what its purpose is.
2. I have a power supply that produces an output of 25 V dc. I want three regulated outputs of approximately 15 V, 15.7 V, and 16.4 V. Show me a circuit that will produce these outputs.
3. I have a zener regulator that stops regulating during the day. The ac line voltage in my area varies from 105 to 125 Vrms. Also, the load resistance on the zener regulator varies from 100 V to 1 kV. Tell me some of the possible reasons why the zener regulator fails during the day.
4. This morning, I was breadboarding an LED indicator.
After I connected the LED and turned on the power, the LED did not light up. I checked the LED and dis- covered that it was open. I tried another LED and got
Job Interview Questions
the same results. Tell me some of the possible rea- sons why this happened.
5. I have heard that a varactor can be used to tune a television receiver. Tell me the basic idea of how it tunes a resonant circuit.
6. Why would an optocoupler be used in an electronic circuit?
7. Given a standard plastic-dome LED package, name two ways to identify the cathode.
8. Explain the diff erences, if any, between a rectifi er diode and a Schottky diode.
9. Draw a circuit like Fig. 5-4a, except replace the dc source with an ac source with a peak value of 40 V.
Draw the graph of output voltage for a zener voltage of 10 V.
1. d 2. b 3. b 4. a 5. a 6. c
7. c 8. a 9. c 10. b 11. c 12. a
13. b 14. d 15. d 16. a 17. c 18. c
19. b 20. b 21. a 22. c 23. c 24. c
25. b 26. d 27. a 28. c 29. b 30. b
31. d 32. a
Self-Test Answers
5-1 IS 5 24.4 mA 5-3 IS 5 18.5 mA;
IL 5 10 mA;
IZ 5 8.5 mA
5-5 VRL 5 8 Vp-p square-wave
5-7 VL 5 10.1 V 5-8 VR(out) 5 94 mVp-p 5-10 RS(max) 5 65 V 5-11 RS(max) 5 495 V
5-13 RS 5 330 V 5-14 IS 5 27 mA;
P 5 7.2 W
Practice Problem Answers
5-39 Open up and troubleshoot fi le MTC05-39.
5-40 Open up and troubleshoot fi le MTC05-40.
5-41 Open up and troubleshoot fi le MTC05-41.
5-42 Open up and troubleshoot fi le MTC05-42.
188 chapter
6
In 1951, William Schockley invented the fi rst junction transistor, a semiconductor device that can amplify (enlarge) electronic signals such as radio and television signals. The transistor has led to many other semiconductor inventions, including the integrated circuit (IC), a small device that contains thousands of miniaturized transistors. Because of the IC, modern computers and other electronic miracles are possible.
This chapter introduces the fundamental operation of a bipolar junction transistor (BJT), the kind that uses both free electrons and holes. The word bipolar is an abbreviation for
“two polarities.” This chapter will also explore how this BJT can be properly biased to act as a switch.
BJT Fundamentals
© Jason Reed/Getty Images
active region amplifying circuit base
base bias
bipolar junction transistor breakdown region collector
collector diode common emitter current gain cutoff point
cutoff region dc alpha dc beta emitter emitter diode h parameters heat sink
integrated circuit junction transistor load line
power transistors
quiescent point saturation point saturation region small-signal transistors soft saturation
surface-mount transistors switching circuit
thermal resistance two-state circuit
Vocabulary
bchob_ha
bchop_ha
bchop_ln
Objectives
After studying this chapter, you should be able to:
■ Describe the relationships among the base, emitter, and collector currents of a bipolar junction transistor.
■ Draw a diagram of the CE circuit and label each terminal, voltage, and resistance.
■ Draw a hypothetical base curve and a set of collector curves, labeling both axes.
■ Label the three regions of operation on a bipolar junction transistor collector curve.
■ Calculate the respective CE transistor current and voltage values using the ideal transistor and the second transistor approximation.
■ List several bipolar junction transistor ratings that might be used by a technician.
■ State why base bias does not work well in amplifying circuits.
■ Identify the saturation point and the cutoff point for a given base- biased circuit.
■ Calculate the Q point for a given base-biased circuit.
bchop_haa
Chapter Outline
6-1 The Unbiased Transistor 6-2 The Biased Transistor 6-3 Transistor Currents 6-4 The CE Connection 6-5 The Base Curve 6-6 Collector Curves
6-7 Transistor Approximations 6-8 Reading Data Sheets 6-9 Surface-Mount Transistors 6-10 Variations in Current Gain 6-11 The Load Line
6-12 The Operating Point 6-13 Recognizing Saturation 6-14 The Transistor Switch 6-15 Troubleshooting
190 Chapter 6
6-1 The Unbiased Transistor
A transistor has three doped regions, as shown in Fig. 6-1. The bottom region is called the emitter, the middle region is the base, and the top region is the collector. In an actual transistor, the base region is much thinner as compared to the collector and emitter regions. The transistor of Fig. 6-1 is an npn device be- cause there is a p region between two n regions. Recall that the majority carriers are free electrons in n-type material and holes in p-type material.
Transistors are also manufactured as pnp devices. A pnp transistor has an n region between two p regions. To avoid confusion between the npn and the pnp transistors, our early discussions will focus on the npn transistor.
Doping Levels
In Fig. 6-1, the emitter is heavily doped. On the other hand, the base is lightly doped. The doping level of the collector is intermediate, between the heavy dop- ing of the emitter and the light doping of the base. The collector is physically the largest of the three regions.
Emitter and Collector Diodes
The transistor of Fig. 6-1 has two junctions: one between the emitter and the base, and another between the collector and the base. Because of this, a transistor is like two back-to-back diodes. The lower diode is called the emitter-base diode, or simply the emitter diode. The upper diode is called the collector-base diode, or the collector diode.
Before and After Diff usion
Figure 6-1 shows the transistor regions before diffusion has occurred. Because of their repulsion for each other, the free electrons in the n regions will spread in all directions. Some of the free electrons in the n region will diffuse across the junction and recombine with the holes in the p region. Visualize the free electrons in each n region crossing the junction and recombining with holes.
The result is two depletion layers, as shown in Fig. 6-2a. For each of these depletion layers, the barrier potential is approximately 0.7 V at 25°C for a silicon
N
P
N
COLLECTOR
BASE
EMITTER –
+ – +
– +
– +
– +
– +
– + –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
– +
– +
– +
– +
– +
– +
– + –
+ – +
– +
– +
– +
– +
– + –
+ – +
– +
– +
– +
– +
– + –
+ – +
– +
– +
– +
– +
– + +
– + –
+ –
+ –
+ –
+ –
+ – +
– + –
+ –
+ –
+ –
+ –
+ – +
– + –
+ –
+ –
+ –
+ –
+ –
Figure 6-1 Structure of a transistor.
GOOD TO KNOW
During the afternoon of December 23, 1947, Walter H.
Brattain and John Bardeen demonstrated the amplifying action of the first transistor at the Bell Telephone Laboratories.
The first transistor was called a point-contact transistor, which was the predecessor to the junction transistor invented by Schockley.
GOOD TO KNOW
The transistor in Fig. 6-1 is sometimes referred to as a bipolar junction transistor, or BJT. However, most people in the electronics industry still use the word transistor, with the understanding that a bipolar junction transistor is meant.
transistor (0.3 V at 25°C for a germanium transistor). As before, we emphasize silicon devices because they are now more widely used than germanium devices.
6-2 The Biased Transistor
An unbiased transistor is like two back-to-back diodes, as shown in Fig. 6-2b.
Each diode has a barrier potential of approximately 0.7 V. Keep this diode equiv- alent in mind when testing an npn transistor with a DMM. When you connect ex- ternal voltage sources to the transistor, you will get currents through the different parts of the transistor.
Emitter Electrons
Figure 6-3 shows a biased transistor. The minus signs represent free electrons.
The heavily doped emitter has the following job: to emit or inject its free electrons into the base. The lightly doped base also has a well-defi ned purpose: to pass emitter-injected electrons on to the collector. The collector is so named because it collects or gathers most of the electrons from the base.
Figure 6-3 is the usual way to bias a transistor. The left source VBB of Fig. 6-3 forward-biases the emitter diode, and the right source VCC reverse-biases the collector
– + VBB
n
n p RB B
C
E
– +
VCE – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
RC
VCC
VBE
– +
– +
Figure 6-3 Biased transistor.
B
C
E (n) (n)
(p)
(b) N
P
N –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+
+ + + + + + +
+ + + + + + +
–
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –+ –
+ –
+ –
+ –
+ –
+ –
+
– – – – – – –
+
– +
– +
– + – +
– + – +
–
– – – – – – –
DEPLETION LAYER
DEPLETION LAYER
(a)
Figure 6-2 Unbiased transistor. (a) Depletion layers; (b) diode equivalent.
192 Chapter 6
diode. Although other biasing methods are possible, forward-biasing the emitter diode and reverse-biasing the collector diode produce the most useful results.
Base Electrons
At the instant that forward bias is applied to the emitter diode of Fig. 6-3, the electrons in the emitter have not yet entered the base region. If VBB is greater than the emitter-base barrier potential in Fig. 6-3, emitter electrons will enter the base region, as shown in Fig. 6-4. Theoretically, these free electrons can fl ow in either of two directions. First, they can fl ow to the left and out of the base, passing through RB on the way to the positive source terminal. Second, the free electrons can fl ow into the collector.
Which way will the free electrons go? Most will continue on to the col- lector. Why? Two reasons: The base is lightly doped and very thin. The light dop- ing means that the free electrons have a long lifetime in the base region. The very thin base means that the free electrons have only a short distance to go to reach the collector. For these two reasons, almost all the emitter-injected electrons pass through the base to the collector.
Only a few free electrons will recombine with holes in the lightly doped base of Fig. 6-4. Then, as valence electrons, they will fl ow through the base resis- tor to the positive side of the VBB supply.
Collector Electrons
Almost all the free electrons go into the collector, as shown in Fig. 6-5. Once they are in the collector, they feel the attraction of the VCC source voltage. Because of
GOOD TO KNOW
In a transistor, the emitter-base depletion layer is narrower than the collector-base deple- tion layer. The reason can be attributed to the different doping levels of the emitter and collec- tor regions. With much heavier doping in the emitter region, the penetration into the n material is minimal because of the availabil- ity of many more free electrons.
However, on the collector side, fewer free electrons are availa- ble and the depletion layer must penetrate more deeply in order to set up the barrier potential.
– + VBB
n
n p RB B
C
E
– +
VCE – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
RC
VCC
VBE
– +
– +
Figure 6-4 Emitter injects free electrons into base.
– + VBB
n
n p RB B
C
E
– +
VCE – – – – – – –
– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –
RC
VCC
VBE
– +
– +
Figure 6-5 Free electrons from base fl ow into collector.
this, the free electrons fl ow through the collector and through RC until they reach the positive terminal of the collector supply voltage.
Here’s a summary of what’s going on: In Fig. 6-5, VBB forward-biases the emitter diode, forcing the free electrons in the emitter to enter the base. The thin and lightly doped base gives almost all these electrons enough time to diffuse into the collector. These electrons fl ow through the collector, through RC, and into the positive terminal of the VCC voltage source.
6-3 Transistor Currents
Figures 6-6a and 6-6b show the schematic symbol for an npn transistor. If you prefer conventional fl ow, use Fig. 6-6a. If you prefer electron fl ow, use Fig. 6-6b.
In Fig. 6-6, there are three different currents in a transistor: emitter current IE, base current IB, and collector current IC.
How the Currents Compare
Because the emitter is the source of the electrons, it has the largest current. Since most of the emitter electrons fl ow to the collector, the collector current is almost as large as the emitter current. The base current is very small by comparison, often less than 1 percent of the collector current.
Relation of Currents
Recall Kirchhoff’s current law. It says that the sum of all currents into a point or junction equals the sum of all currents out of the point or junction. When applied to a transistor, Kirchhoff’s current law gives us this important relationship:
IE 5 IC 1 IB (6-1)
This says that the emitter current is the sum of the collector current and the base current. Since the base current is so small, the collector current approximately equals the emitter current:
IC < IE
and the base current is much smaller than the collector current:
IB ,, IC
(Note: ,, means much smaller than.)
Figure 6-6c shows the schematic symbol for a pnp transistor and its cur- rents. Notice that the current directions are opposite that of the npn. Again notice that Eq. (6-1) holds true for the pnp transistor currents.
Alpha
The dc alpha (symbolized dc) is defi ned as the dc collector current divided by the dc emitter current:
adc 5 I__ C
IE (6-2)
Since the collector current almost equals the emitter current, the dc alpha is slightly less than 1. For instance, in a low-power transistor, the dc alpha is typi- cally greater than 0.99. Even in a high-power transistor, the dc alpha is typically greater than 0.95.
IB
IC
IE
(a)
(b) IB
IC
IE
IB
IC
IE
(c) B
B
B
C
E
C
E
C
E
Figure 6-6 Three transistor currents. (a) Conventional fl ow;
(b) electron fl ow; (c) pnp currents.