by Earl Boysen
22 Now, calculate the values of R1, R2, and R3 for this circuit. The process is similar to the one you used before, but you must expand it to deal with the second transistor. This is similar to the steps you used in problem 8. Follow these steps to calculate R1, R2, and R3:
1. Determine the load current IC2.
2. Determine β for Q2. Call this β2.
3. Calculate IB2 for Q2. Use IB2 = IC2/β2.
4. Calculate R3 to provide this base current. Use R3 = VS/IB2.
5. R3 is also the load for Q1 when Q1 is ON. Therefore, the collector current for Q1 (IC1) has the same value as the base current for Q2, as calculated in step 3.
6. Determine β1, the β for Q1.
7. Calculate the base current for Q1. Use IB1 = IC1/β1.
8. Find R1. Use R1 = Vs/IB1.
9. Choose R2. For convenience, let R2 = R1.
Continue to work with the same circuit shown in Figure 4.18. Use the following values:
Figure 4.18
A 10-volt lamp that draws 1 ampere; therefore VS = 10 volts, IC2 = 1 A.
β2 = 20, β1 = 100
Ignore any voltage drops across the transistors.
Questions
Calculate the following:
A. Find IB2 as in step 3.
IB2 = _____
B. Find R3 as in step 4.
R3 = _____
C. Calculate the load current for Q1 when it is ON, as shown in Step 5.
IC1 = _____
D. Find the base current for Q1.
IB1 = _____
E. Find R1 as in step 8.
R1 = _____
F. Choose a suitable value for R2. R2 = _____
Answers
The following answers correspond to the steps.
A.
1. IC2 is given as 1 ampere.
2. β2 = 20 (given). This is a typical value for a transistor that would handle 1 ampere.
3.
B.
4.
Note that the 0.7 volt base-emitter drop has been ignored.
C.
5. IC1 = IB2 = 50 mA
D.
6. β1 = 100
7.
E.
8.
Again, the 0.7-volt drop is ignored.
F.
9. For convenience, choose a value for R2 that is the same as R1, or 20 kΩ. This reduces the number of different components in the circuit. The fewer different components you have in a circuit, the less components you must keep in your parts bin. You could, of course, choose any value between 1 kΩ and 1 MΩ.
23 Following the same procedure, and using the same circuit shown in Figure 4.18, work through this example. Assume that you are using a 28-volt lamp that draws 560 mA, and that β2 = 10 and β1 = 100.
Questions
Calculate the following:
A. IB2 = _____
B. R3 = _____
C. IC1 = _____
D. IB1 = _____
E. R1 = _____
F. R2 = _____
Answers
A. 56 mA
B. 500 ohms
C. 56 mA
D. 0.56 mA
E. 50 kΩ
F. 50 kΩ by choice
The Three-Transistor Switch
24 The circuit shown in Figure 4.19 uses three transistors to switch a load on and off. In this circuit, Q1 is used to turn Q2 ON and OFF, and Q2 is used to turn Q3 ON and OFF. The calculations are similar to those you performed in the last few problems, but a few additional steps are required to deal with the third transistor. Use the circuit diagram in Figure 4.19 to determine the answers to the following questions.
Figure 4.19
Questions
Assume that the switch is in position A.
A. Is Q1 ON or OFF? _____
B. Is Q2 ON or OFF? _____
C. Where is current through R4 flowing? _____
D. Is Q3 ON or OFF? _____
Answers
A. ON
B. OFF
C. Into the base of Q3
D. ON
25 Now use the same circuit as in problem 24.
Questions
Assume that the switch is in position B.
A. Is Q1 ON or OFF? _____
B. Is Q2 ON or OFF? _____
C. Where is the current through R4 flowing? _____
D. Is Q3 ON or OFF? _____
E. Which switch position turns on the lamp? _____
F. How do the ON/OFF positions for the switch in the three-transistor switch differ from the ON/OFF positions for the switch in the two-transistor switch circuit? _____
Answers
A. OFF.
B. ON.
C. Through Q2 to ground.
D. OFF.
E. Position A.
F. The positions are opposite. Therefore, if a circuit controls lamps with two transistors and another circuit controls lamps with three transistors, flipping the switch that controls both circuits would change which lamps (or which other loads) are on.
26 Work through this example using the same equations you used for the two-transistor switch in problem 22. The steps are similar but with a few added steps, as shown here:
1. Find the load current. This is often given.
2. Determine the current gain of Q3. This is β3 and usually it is a given value.
3. Calculate IB3. Use IB3 = IC3/β3.
4. Calculate R4. Use R4 = VS/IB3.
5. Assume IC2 = IB3.
6. Find β2. Again this is a given value.
7. Calculate IB2. Use IB2 = IC2/β2.
8. Calculate R3. Use R3 = VS/IB2.
9. Assume IC1 = IB2.
10. Find β1.
11. Calculate IB1. Use IB1 = IC1/β1.
12. Calculate R1. Use R1 = Vs/IB1.
13. Choose R2.
For this example, use a 10-volt lamp that draws 10 amperes. Assume that the βs of the transistors are given in the manufacturer's data sheets as β1 = 100, β2 = 50, and β3 = 20. Now, work through the steps, checking the answers for each step as you complete it.
Questions
Calculate the following:
A. IB3 = _____
B. R4 = _____
C. IB2 = _____
D. R3 = _____
E. IB1 = _____
F. R1 = _____
G. R2 = _____
Answers
The answers here correspond to the steps.
A.
1. The load current is given as 10 amperes.
2. β3 is given as 20.
3.
B.
4.
C.
5.
6.β2 is given as 50.
7.
D.
8.
E.
9.
10.β1 is given as 100.
11.
F.
12.
G.
13.R2 can be chosen to be 100 kΩ also.
27 Determine the values in the same circuit for a 75-volt lamp that draws 6 amperes. Assume that β3 = 30, β2 = 100, and β1 = 120.
Questions
Calculate the following values using the steps in problem 26:
A. IB3 = _____
B. R4 = _____
C. IB2 = _____
D. R3 = _____
E. IB1 = _____
F. R1 = _____
G. R2 = _____
Answers
A. 200 mA
B. 375 Ω
C. 2 mA
D. 37.5 kΩ
E. 16.7 μA
F. 4.5 MΩ
G. Choose R2 = 1 MΩ
Alternative Base Switching
28 In the examples of transistor switching, the actual switching was performed using a small mechanical switch placed in the base circuit of the first transistor. This switch has three terminals and switches from position A to position B. (This is a single-pole, double-throw switch.) This switch does not have a definite ON or OFF position, as does a simple ON-OFF switch.
Question
Why
couldn't a simple ON-OFF switch with only two terminals have been used with these examples? _____
Answer
An ON-OFF switch is either open or closed, and cannot switch between position A and position B, as shown earlier in Figure 4.19.
29 If you connect R1, R2, and a switch together, as shown in Figure 4.20, you can use a simple ON-OFF switch with only two terminals. (This is a single-pole, single-throw switch.)
Figure 4.20
Questions
A. When the switch is open, is Q1 ON or OFF? _____
B. When the switch is closed, is the lamp ON or Off? _____
Answers
A. OFF
B. ON
30 When the switch is closed, current flows through R1. However, at point A in Figure 4.20, the current divides into two paths. One path is the base current IB, and the other is marked I2.
Question
How could you calculate the total current I1? _____
Answer
31 The problem now is to choose the values of both R1 and R2 so that when the current divides, there is sufficient base current to turn Q1 ON.
Question
Consider this simple example. Assume the load is a 10-volt lamp that needs 100 mA of current and β = 100. Calculate the base current required.
1. IB = _____
Answer
32 After the current I1 flows through R1, it must divide, and 1 mA of it becomes IB. The remainder of the current is I2. The difficulty at this point is that there is no unique value for either I1 or I2. In other words, you could assign them almost any value. The only restriction is that both must permit 1 mA of current to flow into the base of Q1.
You must make an arbitrary choice for these two values. Based on practical experience, it is common to set I2 to be 10 times greater than IB. This split makes the circuits work reliably and keeps the calculations easy:
Question
In problem 31 you determined that IB = 1 mA. What is the value of I2? _____
Answer
I2 = 10 mA
33 Now you can calculate the value of R2. The voltage across R2 is the same as the voltage drop across the base-emitter junction of Q1. Assume that the circuit uses a silicon transistor, so this voltage is 0.7 volt.
Questions
A. What is the value of R2? _____
B. What is the value R1? _____
Answers
A.
B.
You can ignore the 0.7 volt in this case, which would give R1 = 910 ohms.
34 The resistor values you calculated in problem 33 ensure that the transistor turns ON and that the 100 mA current (IC) you need to illuminate the lamp flows through the lamp and the transistor. Figure 4.21 shows the labeled circuit.
Figure 4.21
Questions
For each of the following lamps, perform the same calculations you used in the last few problems to find the values of R1 and R2.
A. A 28-volt lamp that draws 56 mA. β = 100 _____
B. A 12-volt lamp that draws 140 mA. β = 50 _____
Answers
A.
B.
35 The arbitrary decision to make the value of I2 10 times the value of IB is obviously subject to considerable discussion, doubt, and disagreement. Transistors are not exact devices; they are not carbon copies of each other.
In general, any transistor of the same type has a different β from any other because of the variance in tolerances found in component manufacturing. This leads to a degree of inexactness in designing and analyzing transistor circuits. The truth is that if you follow exact mathematical procedures, it can complicate your life. In practice, a few “rules of thumb” have been developed to help you make the necessary assumptions. These rules lead to simple equations that provide workable values for components that you can use in designing circuits.
The choice of I2 = 10IB is one such rule of thumb. Is it the only choice that works? Of course not. Almost any value of I2 that is at least 5 times larger than IB can work. Choosing 10 times the value is a good option for three reasons:
It is a good practical choice. It always works.
It makes the arithmetic easy.
It's not overly complicated and doesn't involve unnecessary calculations.
Question
In the example from problem 32, IB = 1 mA and I2 = 10 mA. Which of the following values can also work efficiently for I2?
A. 5 mA
B. 8 mA
C. 175 mA
D. 6.738 mA
E. 1 mA
Answers
Choices A, B, and D. Value C is too high to be a sensible choice, and E is too low.
36 Before you continue with this chapter, answer the following review questions.
Questions
A. Which switches faster, the transistor or the mechanical switch? _____
B. Which can be more accurately controlled? _____
C. Which is the easiest to operate remotely? _____
D. Which is the most reliable? _____
E. Which has the longest life? _____
Answers
A. The transistor is much faster.
B. The transistor.
C. The transistor.
D. The transistor.
E. Because transistors have no moving parts, they have a much longer operating lifetime than a mechanical switch. A mechanical switch will fail after several thousand operations, whereas transistors can be operated several million times a second and can last for years.
Switching the JFET
37 The use of the junction field effect transistor (JFET) as a switch is discussed in the next few problems. You may want to review problems 28 through 31 in Chapter 3 where this book introduced the JFET.
The JFET is considered a “normally on” device, which means that with 0 volts applied to the input terminal (called the gate), it is ON, and current can flow through the transistor. When you apply a voltage to the gate, the device conducts less current because the resistance of the drain to the source channel increases. At some point, as the voltage increases, the value of the resistance in the channel becomes so high that the device “cuts off” the flow of current.
Questions
A. What are the three terminals for a JFET called, and which one controls the operation of the device? _____
B. What turns the JFET ON and OFF? _____
Answers
A. Drain, source, and gate, with the gate acting as the control.
B. When the gate voltage is zero (at the same potential as the source), the JFET is ON. When the gate to source voltage difference is high, the JFET is OFF.
Project 4.2: The JFET
Objective
The objective of this project is to determine the drain current that flows when a JFET is fully ON, and the gate voltage needed to fully shut the JFET OFF, using the circuit shown in Figure 4.22.
Figure 4.22
General Instructions
After the circuit is set up, change the gate voltage (VGS) by adjusting the potentiometer. Measure the drain current (ID) for each VGS value. As you work through the project, observe how the drain current drops toward zero as you increase VGS. When the JFET is OFF, ID is at zero; when the JFET is fully ON, ID is at its maximum (called IDSS).
Parts List
You need the following equipment and supplies:
One 6-volt battery pack (4 AA batteries)
One 12-volt battery pack (8 AA batteries)
One multimeter set to mA
One multimeter set to measure DC voltage
One 10 k potentiometer
One breadboard
Two terminal blocks
One 2N3819 JFET (Figure 4.23 shows the pinout for the 2N3819.)
Figure 4.23
Step-by-Step Instructions
Set up the circuit shown in Figure 4.22. If you have some experience in building circuits, this schematic (along with the previous parts list) should provide all the information you need to build the circuit. If you need a bit more hel
p building the circuit, look at the photos of the completed circuit in the “Expected Results” section.
Carefully check your circuit against the diagram, especially the orientation of the JFET to ensure that the drain, gate, and source leads are connected correctly. One unusual aspect of this circuit you may want to check is that the +V bus of the 6-volt battery pack should be connected to the ground bus of the 12-volt battery pack.
After you check your circuit, follow these steps, and record your measurements in the blank table following the steps:
1. Adjust the potentiometer to set VGS at 0 volts. (Your multimeter may indicate a few tenths of a millivolt; that's close enough.)
2. Measure and record VGS and ID.
3. Adjust the potentiometer slightly to give a higher value of VGS.
4. Measure and record the new values of VGS and ID.
5. Repeat steps 3 and 4 until ID drops to 0 mA.
VGS (Volts) ID (mA)
6. Graph the points recorded in the table, using the blank graph in Figure 4.24. Draw a curve through the points. Your curve should look like the one in Figure 4.22.
Figure 4.24
Expected Results
Figure 4.25 shows the breadboarded circuit for this project.
Figure 4.25
Figure 4.26 shows the test setup for this project.
Figure 4.26
Compare your measurements with the ones shown in the following table. You should see a similar trend in the measured values, not exactly the same values.
VGS (Volts) ID (mA)
0 12.7