enable signal is zero volts? When the enable signal is 15.0 V?
LAMP 4 Ω
MTV10N100E
R2 1 MΩ R1 1 MΩ
20 mF
VDD +20 V
Figure 12-57
12-33 When the enable signal of Fig. 12-58 is 15.0 V, what is the output voltage across the load if Q2 has a RDS(on) value of 100 mV?
12-34 With a RDS(on) value of 100 mV, what is the power loss across Q2 and output load power when the enable signal is 15.0 V?
SEC. 12-11 E-MOSFET AMPLIFIERS
12-35 Find the constant k value and ID of Fig. 12-59 using the minimum values of ID(on), VGS(on), and VGS(th) for the 2N7000.
12-36 Determine the gm, Av, and vout values for
Fig. 12-59 using the minimum rated specifi cations.
12-37 In Fig. 12-59, change RD to 50 V. Find the con- stant k value and ID using the typical values of ID(on), VGS(on), and VGS(th) for the 2N7000.
12-38 Determine the gm, Av, and vout values for Fig. 12-59 using the typical rated specifi cations, VDD at 112 V and RD 5 15 V.
Q1RDS(on) = 1 Ω R1
Q2
Vout
Vin 15 V
EN
p-CHANNEL LOAD SWITCH
RL 5 Ω
+5.0 V 0 V
10 kΩ
Figure 12-58
vin
50 mV R2 1 MΩ R1 2 MΩ
+VDD
vout 150 Ω
RD
Q1
2N7000
1 kΩ RL 9 V
Figure 12-59
Critical Thinking
12-39 In Fig. 12-50c, the gate input voltage is a square wave with a frequency of 1 kHz and a peak volt- age of 15 V. What is the average power dissipa- tion in the load resistor?
12-40 The gate input voltage of Fig. 12-50d is a series of rectangular pulses with a duty cycle of 25 per- cent. This means that the gate voltage is high for 25 percent of the cycle and low the rest of the time. What is the average power dissipation in the load resistor?
12-41 The CMOS inverter of Fig. 12-53 uses MOSFETs with RDS(on) 5 100 V and RDS(off ) 5 10 MV. What is the quiescent power consumption of the circuit?
When a square wave is the input, the average
current through Q1 is 50 A. What is the power consumption?
12-42 If the gate voltage is 3 V in Fig. 12-55, what is the photodiode current?
12-43 The data sheet of an MTP16N25E shows a nor- malized graph of RDS(off ) versus temperature. The normalized value increases linearly from 1 to 2.25 as the junction temperature increases from 25 to 125°C. If RDS(on) 5 0.17 V at 25°C, what does it equal at 100°C?
12-44 In Fig. 12-29, Vin 5 12 V. If the transformer has a turns ratio of 4:1 and the output ripple is very small, what is the dc output voltage Vout?
522 Chapter 12
Job Interview Questions
1. Draw an E-MOSFET showing the p and n regions.
Then, explain the off -on action.
2. Describe how active-load switching works. Use circuit diagrams in your explanation.
3. Draw a CMOS inverter and explain the circuit action.
4. Draw any circuit that shows a power FET controlling a large load current. Explain the off -on action. Include RDS(on) in your discussion.
5. Some people say that MOS technology revolution- ized the world of electronics. Why?
6. List and compare the advantages and disadvantages of BJT and FET amplifi ers.
7. Explain what happens when drain current starts to increase through a power FET.
8. Why must an E-MOSFET be handled with care?
9. Why is a thin metal wire connected around all the leads of a MOSFET during shipment?
10. What are some precautionary measures that are taken while working with MOS devices?
11. Why would a designer generally select a MOSFET over a BJT for a power-switching function in a switch- ing power supply?
Self-Test Answers
1. c 2. d 3. d 4. c 5. c 6. d 7. d 8. c 9. b
10. d 11. a 12. b 13. d 14. c 15. a 16. b 17. d 18. d
19. c 20. d 21. a 22. b 23. d 24. d 25. c 26. d
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 MTC12-45 through MTC12- 49 and are based on the circuit of Fig. 12-59.
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.
12-45 Open up and troubleshoot fi le MTC12-45.
12-46 Open up and troubleshoot fi le MTC12-46.
12-47 Open up and troubleshoot fi le MTC12-47.
12-48 Open up and troubleshoot fi le MTC12-48.
12-49 Open up and troubleshoot fi le MTC12-49.
Practice Problem Answers
12-1 VGS ID
21 V 2.25 mA 22 V 1 mA
0 V 4 mA
11 V 6.25 mA 12 V 9 mA 12-2 vout 5 105.6 mV
12-3 ID(sat) 5 10 mA; Vout(off ) 5 20 V;
Vout(on) 5 0.06 V
12-4 ILED 5 32 mA
12-6 vout 5 20 V and 198 mV
12-7 RDS(on) > 222 V
12-8 If vin . VGS(th); vout 5 115 V pulse 12-9 ID 5 0.996 A
12-10 IL 5 2.5 A
12-13 Vload 5 4.76 V; Pload 5 4.76 W;
Ploss 5 238 mW
12-14 k 5 5.48 3 1023 A/V2; ID 5 26 mA 12-15 RD 5 4 kV
12-16 VGS 5 2.98 V; ID 5 80 mA;
gm 5 183 mS; Av 5 11.7;
vout 5 1.17 V
524 chapter
13
The word thyristor comes from the Greek and means “door,”
as in opening a door and letting something pass through it. A thyristor is a semiconductor device that uses internal feedback to produce switching action. The most important thyristors are the silicon controlled rectifi er (SCR) and the triac. Like power FETs, the SCR and the triac can switch large currents on and off . Because of this, they can be used for overvoltage protection, motor controls, heaters, lighting systems, and other heavy- current loads. Insulated-gate bipolar transistors (IGBTs) are not included in the thyristor family, but are covered in this chapter as an important power-switching device.
Thyristors
© Jason Reed/Getty Images
breakover conduction angle diac
fi ring angle four-layer diode gate trigger current IGT
gate trigger voltage VGT
holding current insulated-gate bipolar transistor (IGBT) low-current drop-out programmable unijunction transistor (PUT)
sawtooth generator
Schockley diode SCR
silicon unilateral switch (SUS)
thyristor triac
unijunction transistor (UJT)
Vocabulary
bchob_ha
bchop_ha
bchop_ln
Objectives
After studying this chapter, you should be able to:
■ Describe the four-layer diode, how it is turned on, and how it is turned off .
■ Explain the characteristics of SCRs.
■ Demonstrate how to test SCRs.
■ Calculate the fi ring and
conduction angles of RC phase control circuits.
■ Explain the characteristics of triacs and diacs.
■ Compare the switching control of IGBTs to power MOSFETs.
■ Describe the major
characteristics of the photo-SCR and silicon controlled switch.
■ Explain the operation of UJT and PUT circuits.
bchop_haa
Chapter Outline
13-1 The Four-Layer Diode 13-2 The Silicon Controlled
Rectifi er
13-3 The SCR Crowbar 13-4 SCR Phase Control 13-5 Bidirectional Thyristors 13-6 IGBTs
13-7 Other Thyristors 13-8 Troubleshooting
526 Chapter 13
13-1 The Four-Layer Diode
Thyristor operation can be explained in terms of the equivalent circuit shown in Fig. 13-1a. The upper transistor Q1 is a pnp device, and the lower transistor Q2 is an npn device. The collector of Q1 drives the base of Q2. Similarly, the collector of Q2 drives the base of Q1.
Positive Feedback
The unusual connection of Fig. 13-1a uses positive feedback. Any change in the base current of Q2 is amplifi ed and fed back through Q1 to magnify the original change. This positive feedback continues changing the base current of Q2 until both transistors go into either saturation or cutoff.
For instance, if the base current of Q2 increases, the collector current of Q2 increases. This increases the base current of Q1 and the collector current of Q1. More collector current in Q1 will further increase the base current of Q2. This amplify-and-feedback action continues until both transistors are driven into saturation. In this case, the overall circuit acts like a closed switch (Fig. 13-1b).
On the other hand, if something causes the base current of Q2 to decrease, the collector current of Q2 decreases, the base current of Q1 decreases, the col- lector current of Q1 decreases, and the base current of Q2 decreases further. This action continues until both transistors are driven into cutoff. Then, the circuit acts like an open switch (Fig. 13-1c).
The circuit of Fig. 13-1a is stable in either of two states: open or closed.
It will remain in either state indefi nitely until acted on by an outside force. If the circuit is open, it stays open until something increases the base current of Q2. If the circuit is closed, it stays closed until something decreases the base current of Q2. Because the circuit can remain in either state indefi nitely, it is called a latch.
Closing a Latch
Figure 13-2a shows a latch connected to a load resistor with a supply voltage of VCC. Assume that the latch is open, as shown in Fig. 13-2b. Because there is no current through the load resistor, the voltage across the latch equals the supply voltage. So, the operating point is at the lower end of the dc load line (Fig. 13-2d ).
The only way to close the latch of Fig. 13-2b is by breakover. This means using a large enough supply voltage VCC to break down the Q1 collector diode.
Figure 13-1 Transistor latch.
Q2 Q1
(a) (b) (c)
Figure 13-2 Latching circuit.
RL +VCC
RL
v +
– VCC
+ –
RL
0 V + – +VCC
IC(sat) +VCC
(a) (b) (c) (d )
V VCC I
LATCH CLOSED
LATCH OPEN Q2
Q1
Since the collector current of Q1 increases the base current of Q2, the positive feedback will start. This drives both transistors into saturation, as previously described. When saturated, both transistors ideally look like short-circuits, and the latch is closed (Fig. 13-2c). Ideally, the latch has zero voltage across it when it is closed and the operating point is at the upper end of the load line (Fig. 13-2d ).
In Fig. 13-2a, breakover can also occur if Q2 breaks down fi rst. Although breakover starts with the breakdown of either collector diode, it ends with both transistors in the saturated state. This is why the term breakover is used instead of breakdown to describe this kind of latch closing.
Opening a Latch
How do we open the latch of Fig. 13-2a? By reducing the VCC supply to zero.
This forces the transistors to switch from saturation to cutoff. We call this type of opening low-current drop-out because it depends on reducing the latch current to a value low enough to bring the transistors out of saturation.
The Schockley Diode
Figure 13-3a was originally called a Schockley diode after the inventor. Several other names are also used for this device: four-layer diode, pnpn diode, and sili- con unilateral switch (SUS). The device lets current fl ow in only one direction.
The easiest way to understand how it works is to visualize it separated into two halves, as shown in Fig. 13-3b. The left half is a pnp transistor, and the right half is an npn transistor. Therefore, the four-layer diode is equivalent to the latch of Fig. 13-3c.
Figure 13-3d shows the schematic symbol of a four-layer diode. The only way to close a four-layer diode is by breakover. The only way to open it is by low-current drop-out, which means reducing the current to less than the holding current (given on data sheets). The holding current is the low value of current where the transistors switch from saturation to cutoff.
After a four-layer diode breaks over, the voltage across it ideally drops to zero. In reality, there is some voltage across the latched diode. Figure 13-3e shows current versus voltage for a 1N5158 that is latched on. As you can see, the voltage across the device increases when the current increases: 1 V at 0.2 A, 1.5 V at 0.95 A, 2 V at 1.8 A, and so on.
GOOD TO KNOW
The four-layer diode is rarely, if ever, used in modern circuit design. In fact, most device manufacturers no longer make them. In spite of the fact that the device is nearly obsolete, it is covered in detail here because most of the operating principles of the four-layer diode can be applied to many of the more commonly used thyristors. In fact, most thyristors are nothing more than slight variations of the basic four-layer diode.
Figure 13-3 Four-layer diode.
n p p n
(a)
n p p
n p n
(b) (c) (d ) (e)
10 7.0 5.0 2.0 1.0 0.7 0.5 0.2
0.10 1 2 3 4
VOLTAGE, V
5 6 7
0.3 3.0
CURRENT, A
TA = 25⬚ C [77⬚ F]
528 Chapter 13
Breakover Characteristic
Figure 13-4 shows the graph of current versus voltage of a four-layer diode. The device has two operating regions: cutoff and saturation. The dashed line is the transition path between cutoff and saturation. It is dashed to indicate that the device switches rapidly between the off and on states.
When the device is at cutoff, it has zero current. If the voltage across diode tries to exceed VB, the device breaks over and moves rapidly along the dashed line to the saturation region. When the diode is in saturation, it is operat- ing on the upper line. As long as the current through it is greater than the holding current IH, the diode remains latched in the on state. If the current becomes less than IH, the device switches into cutoff.
The ideal approximation of a four-layer diode is an open switch when cut off and a closed switch when saturated. The second approximation includes the knee voltage VK, approximately 0.7 V in Fig. 13-4. For higher approximations, use computer simulation software or refer to the data sheet of the four-layer diode.
Figure 13-4 Breakover characteristic.
The diode of Fig. 13-5 has a breakover voltage of 10 V. If the input voltage of Fig. 13-5 is increased to 115 V, what is the diode current?
SOLUTION Since an input voltage of 15 V is more than the break over voltage of 10 V, the diode breaks over. Ideally, the diode is like a closed switch, so the current is:
I 5 15 V ______
100 V 5 150 mA To a second approximation:
I 5 15 V ____________ 2 0.7 V 100 V 5 143 mA
For a more accurate answer, look at Fig. 13-3e and you will see that the voltage is 0.9 V when the current is around 150 mA. Therefore, an improved answer is:
I 5 15 V ____________ 2 0.9 V 100 V 5 141 mA
PRACTICE PROBLEM 13-1 In Fig. 13-5, determine the diode current if the input voltage V is 12 V, to a second approximation.
Example 13-1
V VB
VK IH
I
Figure 13-5 Example.
1N5158 VB = 10 V
IH = 4 mA V
– +
RS 100 Ω
Example 13-2
The diode of Fig. 13-5 has a holding current of 4 mA. The input voltage is increased to 15 V to latch the diode and then decreased to open the diode. What is the input voltage that opens the diode?
SOLUTION The diode opens when the current is slightly less than the holding current, given as 4 mA. At this small cur- rent, the diode voltage is approximately equal to the knee voltage, 0.7 V. Since 4 mA fl ows through 100 V, the input voltage is:
Vin 5 0.7 V 1 (4 mA)(100 V) 5 1.1 V
So, the input voltage has to be reduced from 15 V to slightly less than 1.1 V to open the diode.
PRACTICE PROBLEM 13-2 Repeat Example 13-2 using a diode with a holding current of 10 mA.
Application Example 13-3
Figure 13-6a shows a sawtooth generator. The capacitor charges toward the supply voltage, as shown in Fig. 13-6b. When the capacitor voltage reaches 110 V, the diode breaks over. This discharges the capacitor, producing the fl yback (sudden voltage drop) of the output waveform. When the voltage is ideally zero, the diode opens and the capacitor begins to charge again. In this way, we get the ideal sawtooth shown in Fig. 13-6b.
What is the RC time constant for capacitor charging? What is the frequency of the sawtooth wave if its period is approximately 20 percent of the time constant?
SOLUTION The RC time constant is:
RC 5 (2 kV)(0.02 F) 5 40 s
The period is approximately 20 percent of the time constant. The fi rst 20 percent of one time constant is basically linear.
Therefore:
T 5 0.2(40 s) 5 8 s The frequency is:
f 5 1 ____
8 s 5 125 kHz
PRACTICE PROBLEM 13-3 Using Fig. 13-6, change the resistor value to 1 kV and solve for the sawtooth frequency.
Figure 13-6 Sawtooth generator.
VB = 10 V 10 V R1 55 V
2 kΩ +55 V
C1 0.02 mF
(a) (b)
t vout
530 Chapter 13
13-2 The Silicon Controlled Rectifi er
The SCR is the most widely used thyristor. It can switch very large currents on and off. Because of this, it is used to control motors, ovens, air conditioners, and induction heaters.
Triggering the Latch
By adding an input terminal to the base of Q2, as shown in Fig. 13-7a, we can create a second way to close the latch. Here is the theory of operation: When the latch is open, as shown in Fig. 13-7b, the operating point is at the lower end of the dc load line (Fig. 13-7d ). To close the latch, we can couple a trigger (sharp pulse) into the base of Q2, as shown in Fig. 13-7a. The trigger momentarily increases the base current of Q2. This starts the positive feedback, which drives both transistors into saturation.
When saturated, both transistors ideally look like short-circuits, and the latch is closed (Fig. 13-7c). Ideally, the latch has zero voltage across it when it is closed, and the operating point is at the upper end of the load line (Fig. 13-7d ).
Gate Triggering
Figure 13-8a shows the structure of the SCR. The input is called the gate, the top is the anode, and the bottom is the cathode. The SCR is far more useful than a four-layer diode because the gate triggering is easier than breakover triggering.
Figure 13-7 Transistor latch with trigger input.
GOOD TO KNOW
SCRs are designed to handle higher values of current and voltage than any other type of thyristor. Presently, some SCRs are capable of controlling large currents up to 1.5 kA and volt- ages in excess of 2 kV.
(d) RL
+VCC
(a) (b)
RL
v +
– VCC
+ –
(c) RL
0 V + – +VCC
IC(sat) +VCC
V VCC I
LATCH CLOSED
LATCH OPEN Q2
Q1
Figure 13-8 Silicon controlled rectifi er (SCR).
n p
p n
n p p
n p n
ANODE
CATHODE CATHODE
CATHODE
CATHODE GATE
ANODE
GATE
ANODE
GATE
GATE
ANODE
(d) (c)
(b) (a)
Again, we can visualize the four doped regions separated into two tran- sistors, as shown in Fig. 13-8b. Therefore, the SCR is equivalent to a latch with a trigger input (Fig. 13-8c). Schematic diagrams use the symbol of Fig. 13-8d.
Whenever you see this symbol, remember that it is equivalent to a latch with a trigger input. Typical SCRs are shown in Fig. 13-9.
Since the gate of an SCR is connected to the base of an internal transis- tor, it takes at least 0.7 V to trigger an SCR. Data sheets list this voltage as the gate trigger voltage VGT. Rather than specify the input resistance of the gate, a manufacturer gives the minimum input current needed to turn on the SCR. Data sheets list this current as the gate trigger current IGT.
Figure 13-10 shows a data sheet for the 2N6504 series of SCRs. For this series, it shows typical trigger voltage and current values of:
VGT 5 1.0 V IGT 5 9.0 mA
This means that the source driving the gate of a typical 2N6504 series SCR has to supply 9.0 mA at 1.0 V to latch the SCR.
Also, the breakover voltage or blocking voltage is specifi ed as its peak repetitive off-state forward voltage VDRM and its peak repetitive off-state reverse voltage VRRM. Depending on which SCR of the series is used, the breakover volt- age ranges from 50 to 800 V.
Required Input Voltage
An SCR like the one shown in Fig. 13-11 has a gate voltage VG. When this volt- age is more than VGT, the SCR will turn on and the output voltage will drop from 1VCC to a low value. Sometimes, a gate resistor is used as shown here. This resistor limits the gate current to a safe value. The input voltage needed to trigger an SCR has to be more than:
Vin 5 VGT 1 IGTRG (13-1)
In this equation, VGT and IGT are the gate trigger voltage and current for the device. For instance, the data sheet of a 2N4441 gives VGT 5 0.75 V and IGT 5 10 mA. When you have the value of RG, the calculation of Vin is straightforward.
If a gate resistor is not used, RG is the Thevenin resistance of the circuit driving the gate. Unless Eq. (13-1) is satisfi ed, the SCR cannot turn on.
Resetting the SCR
After the SCR has turned on, it stays on even though you reduce the gate supply Vin to zero. In this case, the output remains low indefi nitely. To reset the SCR, you must reduce the anode to cathode current to a value less than its holding current IH. This can be done by reducing VCC to a low value. The data sheet for Figure 13-9 Typical SCRs.
532 Chapter 13
Figure 13-10 SCR data sheet (Used with permission from SCILLC dba ON Semiconductor).