in Fig. 3-21, how low must RL
be before the diode’s bulk re- sistance must be considered?
a. 1 V c. 23 V b. 10 V d. 100 V
SEC. 3-1 BASIC IDEAS
3-1 A diode is in series with 220 V. If the voltage across the resistor is 6 V, what is the current through the diode?
3-2 A diode has a voltage of 0.7 V and a current of 100 mA. What is the diode power?
3-3 Two diodes are in series. The fi rst diode has a volt- age of 0.75 V and the second has a voltage of 0.8 V. If the current through the fi rst diode is 400 mA, what is the current through the second diode?
SEC. 3-2 THE IDEAL DIODE
3-4 In Fig. 3-22a, calculate the load current, load voltage, load power, diode power, and total power.
3-5 If the resistor is doubled in Fig. 3-22a, what is the load current?
Problems
3-6 In Fig. 3-22b, calculate the load current, load volt- age, load power, diode power, and total power.
3-7 If the resistor is doubled in Fig. 3-22b, what is the load current?
3-8 If the diode polarity is reversed in Fig. 3-22b, what is the diode current? The diode voltage?
SEC. 3-3 THE SECOND APPROXIMATION 3-9 In Fig. 3-22a, calculate the load current, load volt-
age, load power, diode power, and total power.
3-10 If the resistor is doubled in Fig. 3-22a, what is the load current?
3-11 In Fig. 3-22b, calculate the load current, load volt- age, load power, diode power, and total power.
3-12 If the resistor is doubled in Fig. 3-22b, what is the load current?
3-13 If the diode polarity is reversed in Fig. 3-22b, what is the diode current? The diode voltage?
SEC. 3-4 THE THIRD APPROXIMATION 3-14 In Fig. 3-22a, calculate the load current, load
voltage, load power, diode power, and total power. (RB 5 0.23 V)
3-15 If the resistor is doubled in Fig. 3-22a, what is the load current? (RB 5 0.23 V)
3-16 In Fig. 3-22b, calculate the load current, load voltage, load power, diode power, and total power. (RB 5 0.23 V)
3-17 If the resistor is doubled in Fig. 3-22b, what is the load current? (RB 5 0.23 V)
3-18 If the diode polarity is reversed in Fig. 3-22b, what is the diode current? The diode voltage?
RL 1 kΩ
(a)
(b) VS
12 V VS 20 V
RL 470 Ω –
+
– +
Figure 3-22
the junction of R1 and R2. (Remember, potentials are always with respect to ground.) Next you measure 0 V at the junction of the diode and the 5-kV resistor. Name some possible troubles.
3-23 The forward and reverse DMM diode test reading is 0.7 V and 1.8 V. Is this diode good?
SEC. 3-6 READING A DATA SHEET
3-24 Which diode would you select in the 1N4000 se- ries if it has to withstand a peak repetitive reverse voltage of 300 V?
3-25 The data sheet shows a band on one end of the diode. What is the name of this band? Does the diode arrow of the schematic symbol point toward or away from this band?
3-26 Boiling water has a temperature of 100°C. If you drop a 1N4001 into a pot of boiling water, will it be destroyed or not? Explain your answer.
100 kΩ +5 V
R
10 kΩ
(a) (b)
R1
R2 R3
5 kΩ 30 kΩ +12 V
Figure 3-23
3-20 Something causes R to short in Fig. 3-23a. What will the diode voltage be? What will happen to the diode?
Calculate the forward and the reverse resistance for each of these diodes.
3-28 In Fig. 3-23a, what value should R be to get a diode current of approximately 20 mA?
3-29 What value should R2 be in Fig. 3-23b to set up a diode current of 0.25 mA?
3-30 A silicon diode has a forward current of 500 mA at 1 V. Use the third approximation to calculate its bulk resistance.
Critical Thinking
3-27 Here are some diodes and their worst-case specifi cations:
Diode IF IR
1N914 10 mA at 1 V 25 nA at 20 V 1N4001 1 A at 1.1 V 10 A at 50 V 1N1185 10 A at 0.95 V 4.6 mA at 100 V
3-31 Given a silicon diode with a reverse current of 5 a at 25°C and 100 A at 100°C, calculate the surface leakage current.
3-32 The power is turned off and the upper end of R1 is grounded in Fig. 3-23b. Now you use an ohmme- ter to read the forward and reverse resistance of the diode. Both readings are identical. What does the ohmmeter read?
3-33 Some systems, like burglar alarms and comput- ers, use battery backup just in case the main source of power should fail. Describe how the circuit in Fig. 3-24 works.
15–V
SOURCE LOAD
– +
12 V
Figure 3-24
Multisim Troubleshooting Problems
The Multisim troubleshooting fi les are found on the In- structor 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 MTC03-34 through MTC03- 38 and are based on the circuit of Fig. 3-23b.
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.
3-34. Open up and troubleshoot fi le MTC03-34.
84 Chapter 3
Digital/Analog Trainer System
The following questions, 3-39 through 3-43, are directed towards 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.
3-39 What type of diode is D1?
3-40 Could D1 be replaced with a 1N4002 diode? Explain why or why not.
3-41 Which side of diodes D5 and D6 are connected together? Anodes or cathodes?
3-42 Is D14 normally forward biased or reverse biased?
3-43 Is D15 normally forward biased or reverse biased?
Job Interview Questions
For the following questions, whenever possible, draw circuits, graphs, or any fi gures that will help illustrate your answers. If you can combine words and pictures in your explanations, you are more likely to understand what you are talking about. Also, if you have privacy, pretend that you are at an interview and speak out loud. This practice will make it easier later, when the interview actually takes place.
1. Have you ever heard of an ideal diode? If so, tell me what it is and when you would use it.
2. One of the approximations for a diode is the second approximation. Tell me what the equivalent circuit is and when a silicon diode conducts.
3. Draw the diode curve and explain the diff erent parts of it.
4. A circuit on my lab bench keeps destroying a diode every time I connect a new one. If I have a data sheet for the diode, what are some of the quantities I need to check?
5. In the most basic terms, describe what a diode acts like when it is forward biased and when it is reverse biased.
6. What is the diff erence between the typical knee volt- age of a germanium diode and a silicon diode?
7. What would be a good technique for a technician to use to determine the current through a diode without breaking the circuit?
8. If you suspect that there is a defective diode on a circuit board, what steps would you take to determine whether it is actually defective?
9. For a diode to be useful, how much larger should the reverse resistance be than the forward resistance?
10. How might you connect a diode to prevent a second battery from discharging in a recreational vehicle, and yet still allow it to charge from the alternator?
11. What instruments can you use to test a diode in or out of a circuit?
12. Describe the operation of a diode in detail. Include majority and minority carriers in your discussion.
Self-Test Answers
1. b 2. b 3. c 4. d 5. a 6. b 7. c 8. c
9. a 10. a 11. b 12. b 13. a 14. d 15. a 16. c
17. b 18. b 19. a 20. b 21. a 22. c
3-2 PD 5 2.2 W 3-3 IL 5 5 mA 3-4 VL 5 2 V;
IL 5 2 mA
D
3-6 IL 5 1.77 mA;
VL 5 1.77 V;
PD 5 1.24 mW
L
PD 5 335 mW
86
4 Diode Circuits
Most electronic systems, like HDTVs, audio power amplifi ers, and computers, need a dc voltage to work properly. Since the power-line voltage is alternating and normally too high of a value, we need to reduce the ac line voltage and then convert it to a relatively constant dc output voltage. The section of the electronic system that produces this dc voltage is called the power supply. Within the power supply are circuits that allow current to fl ow in only one direction. These circuits are called rectifi ers. Other circuits will fi lter and regulate the dc output.
This chapter discusses rectifi er circuits, fi lters, an introduction to voltage regulators, clippers, clampers, and voltage multipliers.
© Arthur S. Aubry/Getty Images
Transformer Rectifier
AC Input
Filter Regulator
DC
RL
bridge rectifi er capacitor-input fi lter choke-input fi lter clamper
clipper
dc value of a signal fi lter
full-wave rectifi er
half-wave rectifi er IC voltage regulator integrated circuit passive fi lter peak detector peak inverse voltage polarized capacitor power supply
rectifi ers ripple
surge current surge resistor switching regulator unidirectional load current voltage multiplier
Vocabulary
■ Draw a diagram of a half-wave rectifi er and explain how it works.
■ Describe the role of the input transformer in power supplies.
■ Draw a diagram of a full-wave rectifi er and explain how it works.
■ Draw a diagram of a bridge rectifi er and explain how it works.
■ Analyze a capacitor input fi lter and its surge current.
■ List three important specifi cations found on a rectifi er data sheet.
■ Explain how a clipper works and draw waveforms.
■ Explain how a clamper works and draw waveforms.
■ Describe the action of voltage multipliers.
Chapter Outline
4-1 The Half-Wave Rectifi er 4-2 The Transformer 4-3 The Full-Wave Rectifi er 4-4 The Bridge Rectifi er 4-5 The Choke-Input Filter 4-6 The Capacitor-Input Filter 4-7 Peak Inverse Voltage and
Surge Current
4-8 Other Power-Supply Topics 4-9 Troubleshooting
4-10 Clippers and Limiters 4-11 Clampers
4-12 Voltage Multipliers
88 Chapter 4
positive half-cycle of source voltage will appear across the load resistor. On the negative half-cycle, the diode is reverse biased. In this case, the ideal diode will appear as an open switch, as shown in Fig. 4-1c, and no voltage appears across the load resistor.
Ideal Waveforms
Figure 4-2a shows a graphical representation of the input voltage waveform. It is a sine wave with an instantaneous value of vin and a peak value of Vp(in). A pure sinusoid like this has an average value of zero over one cycle because each instantaneous voltage has an equal and opposite voltage half a cycle later. If you measure this voltage with a dc voltmeter, you will get a reading of zero because a dc voltmeter indicates the average value.
In the half-wave rectifi er of Fig. 4-2b, the diode is conducting during the positive half-cycles but is nonconducting during the negative half-cycles. Because of this, the circuit clips off the negative half-cycles, as shown in Fig. 4-2c. We call a waveform like this a half-wave signal. This half-wave voltage produces a unidirectional load current. This means that it fl ows in only one direction. If the diode were reversed, it would become forward biased when the input voltage was negative. As a result, the output pulses would be negative. This is shown in Fig. 4-2d. Notice how the negative peaks are offset from the positive peaks and follow the negative alternations of the input voltage.
A half-wave signal like the one in Fig. 4-2c is a pulsating dc voltage that increases to a maximum, decreases to zero, and then remains at zero during the negative half-cycle. This is not the kind of dc voltage we need for electron- ics equipment. What we need is a constant voltage, the same as you get from a
(a)
(b)
(c) CLOSED
– +
– +
RL 0 V OPEN
+ –
RL
RL
(b)
(c)
IDEAL
vin
t t
Vp(out) vout
RL vout
(a) Vp(in)
vin
(d)
t –Vp(out)
–vout
Figure 4-2 (a) Input to half-wave rectifi er; (b) circuit; (c) output of positive half-wave rectifi er; (d) output of negative half- wave rectifi er.
Ideal half wave: Vp(out) 5 Vp(in) (4-1)
DC Value of Half-Wave Signal
The dc value of a signal is the same as the average value. If you measure a signal with a dc voltmeter, the reading will equal the average value. In basic courses, the dc value of a half-wave signal is derived. The formula is:
Half wave: Vdc 5 ___ Vp (4-2)
The proof of this derivation requires calculus because we have to work out the average value over one cycle.
Since 1/ < 0.318, you may see Eq. (4-2) written as:
Vdc < 0.318Vp
When the equation is written in this form, you can see that the dc or average value equals 31.8 percent of the peak value. For instance, if the peak voltage of the half- wave signal is 100 V, the dc voltage or average value is 31.8 V.
Output Frequency
The output frequency is the same as the input frequency. This makes sense when you compare Fig. 4-2c with Fig. 4-2a. Each cycle of input voltage produces one cycle of output voltage. Therefore, we can write:
Half wave: fout 5 fin (4-3)
We will use this derivation later with fi lters.
Second Approximation
We don’t get a perfect half-wave voltage across the load resistor. Because of the barrier potential, the diode does not turn on until the ac source voltage reaches approximately 0.7 V. When the peak source voltage is much greater than 0.7 V, the load voltage will resemble a half-wave signal. For instance, if the peak source voltage is 100 V, the load voltage will be close to a perfect half-wave voltage. If the peak source voltage is only 5 V, the load voltage will have a peak of only 4.3 V.
When you need to get a better answer, use this derivation:
2d half wave: Vp(out) 5 Vp(in) 2 0.7 V (4-4)
Higher Approximations
Most designers will make sure that the bulk resistance is much smaller than the Thevenin resistance facing the diode. Because of this, we can ignore bulk resist- ance in almost every case. If you must have better accuracy than you can get with the second approximation, you should use a computer and a circuit simulator like Multisim.
signal can be determined with the following formula:
Vrms51.57 Vavg
where Vavg5Vdc50.318Vp Another formula that works is:
Vrms5___Vp
ẽ—2
For any waveform, the rms value corresponds to the equivalent dc value that will produce the same heating effect.
90 Chapter 4
values of peak load voltage and the dc load voltage. Then, compare these values to the readings on the oscilloscope and the multimeter.
SOLUTION Figure 4-3 shows an ac source of 10 V and 60 Hz. Schematic diagrams usually show ac source voltages as effective or rms values. Recall that the effective value is the value of a dc voltage that produces the same heating effect as the ac voltage.
Figure 4-3 Lab example of half-wave rectifi er.
4-2 The Transformer
Power companies in the United States supply a nominal line voltage of 120 Vrms and a frequency of 60 Hz. The actual voltage coming out of a power outlet may vary from 105 to 125 Vrms, depending on the time of day, the locality, and other factors. Line voltage is too high for most of the circuits used in electronics equip- ment. This is why a transformer is commonly used in the power-supply section of almost all electronics equipment. The transformer steps the line voltage down to safer and lower levels that are more suitable for use with diodes, transistors, and other semiconductor devices.
Basic Idea
Earlier courses discussed the transformer in detail. All we need in this chapter is a brief review. Figure 4-4 shows a transformer. Here, you see line voltage applied to the primary winding of a transformer. Usually, the power plug has a third prong to ground the equipment. Because of the turns ratio N1/N2, the secondary voltage is stepped down when N1 is greater than N2.
Phasing Dots
Recall the meaning of the phasing dots shown at the upper ends of the windings.
Dotted ends have the same instantaneous phase. In other words, when a positive half-cycle appears across the primary, a positive half-cycle appears across the Vp 5 _____ Vrms
0.707 5 10 V _____
0.707 5 14.1 V With an ideal diode, the peak load voltage is:
Vp(out) 5 Vp(in) 5 14.1 V The dc load voltage is:
Vdc 5 V___ p 5 14.1 V ______ 5 4.49 V
With the second approximation, we get a peak load voltage of:
Vp(out) 5 Vp(in) 2 0.7 V 5 14.1 V 2 0.7 V 5 13.4 V and a dc load voltage of:
Vdc 5 V___ p 5 13.4 V ______ 5 4.27 V
Figure 4-3 shows you the values that an oscilloscope and a multimeter will read. Channel 1 of the oscilloscope is set at 5 V per major division (5 V/Div). The half-wave signal has a peak value between 13 and 14 V, which agrees with the result from our second approximation. The multimeter also gives good agreement with theoretical values because it reads approximately 4.22 V.
PRACTICE PROBLEM 4-1 Using Fig. 4-3, change the ac source voltage to 15 V. Calculate the second approxima- tion dc load voltage Vdc.
92 Chapter 4
secondary. If the secondary dot were on the ground end, the secondary voltage would be 180° out of phase with the primary voltage.
On the positive half-cycle of primary voltage, the secondary winding has a positive half sine wave across it and the diode is forward biased. On the negative half-cycle of primary voltage, the secondary winding has a negative half-cycle and the diode is reverse biased. Assuming an ideal diode, we will get a half-wave load voltage.
Turns Ratio
Recall from your earlier course work the following derivation:
V2 5 ______ V1
N1/N2 (4-5)
This says that the secondary voltage equals the primary voltage divided by the turns ratio. Sometimes you will see this equivalent form:
V2 5 N___ 2
N1
V1
This says that the secondary voltage equals the inverse turns ratio times the pri- mary voltage.
You can use either formula for rms, peak values, and instantaneous volt- ages. Most of the time, we will use Eq. (4-5) with rms values because ac source voltages are almost always specifi ed as rms values.
The terms step up and step down are also encountered when dealing with transformers. These terms always relate the secondary voltage to the primary volt- age. This means that a step-up transformer will produce a secondary voltage that is larger than the primary, and a step-down transformer will produce a secondary voltage that is smaller than the primary.
RL
Example 4-2
What are the peak load voltage and dc load voltage in Fig. 4-5?
120 V 60 Hz
5:1
V1 V2 RL
1 kΩ
Figure 4-5 Transformer example.
4-3 The Full-Wave Rectifi er
Figure 4-6a shows a full-wave rectifi er circuit. Notice the grounded center tap on the secondary winding. The full-wave rectifi er is equivalent to two half-wave rectifi ers. Because of the center tap, each of these rectifi ers has an input voltage equal to half the secondary voltage. Diode D1 conducts on the positive half-cycle, and diode D2 conducts on the negative half-cycle. As a result, the rectifi ed load current fl ows during both half-cycles. The full-wave rectifi er acts the same as two back-to-back half-wave rectifi ers.
Figure 4-6b shows the equivalent circuit for the positive half-cycle. As you see, D1 is forward biased. This produces a positive load voltage as indicated by the plus-minus polarity across the load resistor. Figure 4-6c shows the equiva- lent circuit for the negative half-cycle. This time, D2 is forward biased. As you can see, this also produces a positive load voltage.
During both half-cycles, the load voltage has the same polarity and the load current is in the same direction. The circuit is called a full-wave rectifi er because it has changed the ac input voltage to the pulsating dc output voltage shown in Fig. 4-6d. This waveform has some interesting properties that we will now discuss.
and the peak secondary voltage is:
Vp 5 24 V _____
0.707 5 34 V
With an ideal diode, the peak load voltage is:
Vp(out) 5 34 V The dc load voltage is:
Vdc 5 ___ V p 5 34 V ____ 5 10.8 V
With the second approximation, the peak load voltage is:
Vp(out) 5 34 V 2 0.7 V 5 33.3 V and the dc load voltage is:
Vdc 5 ___ V p 5 33.3 V ______ 5 10.6 V
PRACTICE PROBLEM 4-2 Using Fig. 4-5, change the transformer’s turns ratio to 2 :1 and solve for the ideal dc load voltage.
94 Chapter 4 RL
D1
D2 N1:N2
(a)
(b) D1 N1:N2
+ +
RL + – –
–
(c)
(d)
RL
D2
Vp(out) vout
t N1:N2
+
+ – +
–
–
IDEAL
Full wave: Vdc 5 ____ (4-6) Since 2/ 5 0.636, you may see Eq. (4-6) written as:
Vdc < 0.636Vp
In this form, you can see that the dc or average value equals 63.6 percent of the peak value. For instance, if the peak voltage of the full-wave signal is 100 V, the dc voltage or average value is 63.6 V.
Output Frequency
With a half-wave rectifi er, the output frequency equals the input frequency. But with a full-wave rectifi er, something unusual happens to the output frequency. The ac line voltage has a frequency of 60 Hz. Therefore, the input period equals:
Tin 5 1 __
f 5 ______ 1 60 Hz 5 16.7 ms
Because of the full-wave rectifi cation, the period of the full-wave signal is half the input period:
Tout 5 0.5(16.7 ms) 5 8.33 ms
(If there is any doubt in your mind, compare Fig. 4-6d to Fig. 4-2c.) When we calculate the output frequency, we get:
fout 5 1 ____
Tout 5 _______ 1 8.33 ms 5 120 Hz
The frequency of the full-wave signal is double the input frequency. This makes sense. A full-wave output has twice as many cycles as the sine-wave input has. The full-wave rectifi er inverts each negative half-cycle so that we get double the number of positive half-cycles. The effect is to double the frequency. As a derivation:
Full wave: fout 5 2fin (4-7)
Second Approximation
Since the full-wave rectifi er is like two back-to-back half-wave rectifi ers, we can use the second approximation given earlier. The idea is to subtract 0.7 V from the ideal peak output voltage. The following example will illustrate the idea.
signal is Vrms 5 0.707Vp, which is the same as Vrms for a full sine wave.
Application Example 4-3
Figure 4-7 shows a full-wave rectifi er that you can build on a lab bench or on a computer screen with Multisim. Channel 1 of the oscilloscope displays the primary voltage (the sine wave), and channel 2 displays the load voltage (the full-wave signal).
Set Channel 1 as your positive input trigger point. Most oscilloscopes will need a 103 probe to measure the higher input voltage level. Calculate the peak input and output voltages. Then compare the theoretical values to the measured values.
SOLUTION
The peak primary voltage is:
Vp(1) 5 _____ Vrms
0.707 5 120 V ______
0.707 5 170 V