by Earl Boysen
VB = _____
Answers
Following are the values you should have found:
1. 220 ohms
2. 0.27 volt
3. 0.97 volt (You can use 1 volt if you want.)
4. 1.2 mA
5. 0.012 mA
6. 0.12 mA
7. 8.3 kΩ
8. 68.2 kΩ
9. These are close to the standard values of 8.2 kΩ and 68 kΩ.
10. 1.08 volts using the standard values. This is close enough to the value of VB calculated in question 3.
Project 8.1: The Transistor Amplifier
Objective
The objective of this project is to demonstrate how AC voltage gain changes when you use resistors of different values and transistors with different current gain in a transistor amplifier circuit.
General Instructions
When the circuit is set up, you measure Vout for each set of resistors, and find AV, using the ratio Vout/Vin. You also determine a calculated AV using the ratio RC/RE in each case to determine how close the calculated AV is to the measured AV. You repeat this measurement with a second transistor for each set of resistors.
Parts List
You need the following equipment and supplies:
One 1 kΩ, 0.25-watt resistor
One 100 Ω, 0.25-watt resistor
One 15 kΩ, 0.25-watt resistor
One 2.2 kΩ, 0.25-watt resistor
One 3.3 kΩ, 0.25-watt resistor
One 220 Ω, 0.25-watt resistor
One 68 kΩ, 0.25-watt resistor
One 8.2 kΩ, 0.25-watt resistor
One 0.1 μF capacitor
One lab type power supply or 9-volt battery
One function generator.
One oscilloscope
One breadboard
One 2N3904 transistor
One PN2222 transistor
Figure 8.11 shows the pinout diagram for 2N3904 and PN2222 transistors.
Figure 8.11
Step-by-Step Instructions
Set up the circuit shown in Figure 8.12 using the components listed for Circuit # 1 in the following table. 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 help building the circuit, look at the photos of the completed circuit in the “Expected Results” section. (If you don't have a lab type power supply to provide 10 volts as indicated on the schematic, use a 9-volt battery.)
Figure 8.12
Carefully check your circuit against the diagram.
After you check your circuit, follow these steps, and record your measurements in the blank table following the steps.
1. Connect the oscilloscope probe for channel 2 to a jumper wire connected to Vin, and then connect the ground clip to a jumper wire attached to the ground bus.
2. Connect the oscilloscope probe for channel 1 to a jumper wire connected to Vout, and then connect the ground clip to a jumper wire attached to the ground bus.
3. Set the function generator to generate a 10 kHz sine wave with approximately 0.2 Vpp.
4. Measure and record Vin and Vout.
5. Change the components to those listed in the next row of the table (Circuit # 2 in this case.) You should turn off the power to the circuit before changing components to avoid shorting leads together.
6. Measure and record Vin and Vout.
7. Repeat steps 5 and 6 until you have recorded Vin and Vout in the last row of the table.
8. Determine β for each of the transistors used in this project. Insert the transistors one at a time into the circuit you built in Project 3-1 to take this measurement.
9. For each transistor, record β in the following table.
Expected Results
Figure 8.13 shows the breadboarded Circuit # 1.
Figure 8.13
Figure 8.14 shows the breadboarded Circuit # 3.
Figure 8.14
Figure 8.15 shows a function generator and oscilloscope attached to the circuit.
Figure 8.15
The input signal is represented by the upper sine wave shown in Figure 8.16, and the output signal is represented by the lower sine wave. Read the number of divisions for the peak-to-peak output sine wave, and multiply it by the corresponding VOLTS/DIV setting to determine Vout.
Figure 8.16
As you measure Vin and Vout for each circuit, you may need to adjust the TIME/DIV control, VOLTS/DIV control, and vertical POSITION controls on the oscilloscope. The controls shown in Figure 8.17 are adjusted to measure Vin and Vout for Circuit # 2.
Figure 8.17
Your values should be close to those shown in the following table.
The measured values of AV are quite close to the calculated values of AV, well within variations that could be caused by the ± 5 percent tolerance specified for the resistor values. Also, the variation in transistor β had no effect on the measured values of AV.
19 The AC voltage gain for the circuit discussed in problem 18 was 15. Earlier, you learned that the maximum practical gain of the amplifier circuit shown in Figure 8.9 is approximately 50.
However, in problem 10, you learned that AC voltage gains of up to 500 are possible for the amplifier circuit shown in Figure 8.4. Therefore, by ensuring the stability of the DC bias point, the amplifier has much lower gain than is possible with the transistor amplifier circuit shown in Figure 8.4.
You can make an amplifier with stable bias points without giving up high AC voltage gain by placing a capacitor in parallel with the emitter resistor, as shown in Figure 8.18.
Figure 8.18
If the reactance of this capacitor for an AC signal is significantly smaller than RE, the AC signal passes through the capacitor rather than the resistor. Therefore, the capacitor is called an emitter bypass capacitor. The AC signal “sees” a different circuit from the DC, which is blocked by the capacitor and must flow through the resistor. Figure 8.19 shows the different circuits seen by AC and DC signals.
Figure 8.19
The AC voltage gain is now close to that of the amplifier circuit discussed in problems 1–10.
Questions
A. What effect does the emitter bypass capacitor have on an AC signal? _____
B. What effect does the emitter bypass capacitor have on the AC voltage gain? _____
C. What is the AC voltage gain formula with an emitter bypass capacitor included in the circuit? _____
Answers
A. It makes the emitter look like a ground and effectively turns the circuit into the circuit shown in Figure 8.4.
B. It increases the gain.
C. The same formula used in problem 10:
20 You can use the circuit shown in Figure 8.18 when you need as much AC voltage gain as possible. When high AC voltage gain is your priority, predicting the actual amount of gain is usually not important, so the fact that the equation is inexact is unimportant. If you need an accurate amount of gain, then you must use a different type of amplifier circuit that produces lower amounts of gain.
You can find the value of the capacitor CE using the following steps:
1. Determine the lowest frequency at which the amplifier must operate.
2. Calculate XC with the following formula:
3. Calculate CE with the following formula using the lowest frequency at which the amplifier must operate (determined in step 1):
For the following question, use the circuit shown in Figure 8.10, with an emitter bypass capacitor added, as shown in Figure 8.20.
Figure 8.20
Questions
Follow the previous steps to calculate the value of CE required if the lowest operating frequency of the amplifier is 50 Hz.
1. 50 Hz is the lowest frequency at which the amplifier must operate.
2. XC = _____
3. CE = _____
Answers
XC = 10 ohms
CE = 320 μF (approximately)
The AC voltage gain
formula for an amplifier with an emitter bypass capacitor (Circuit 2 in Figure 8.21) is the same as the AC voltage gain formula for the amplifiers discussed in problems 1–10, where the emitter is directly connected to ground (Circuit 1 in Figure 8.21).
Figure 8.21
The AC voltage gain formula for an amplifier is as follows:
(RC is used instead of RL because the collector resistor is the total load on the amplifier.)
Circuit 1—Here, RE = zero, so the AC voltage gain formula is as follows:
Circuit 2—Here, RE = zero for an AC signal because the AC signal is grounded by the capacitor, and RE is out of the AC circuit. Thus, the AC voltage gain formula is as follows:
21 To obtain even larger voltage gains, two transistor amplifiers can be cascaded. That is, you can feed the output of the first amplifier into the input of the second amplifier. Figure 8.22 shows a two-transistor amplifier circuit, also called a two-stage amplifier.
Figure 8.22
You find the total AC voltage gain by multiplying the individual gains. For example, if the first amplifier has an AC voltage gain of 10, and the second has an AC voltage gain of 10, then the overall AC voltage gain is 100.
Questions
A. Suppose you cascade an amplifier with a gain of 15 with one that has a gain of 25. What is the overall gain? _____
B. What is the overall gain if the individual gains are 13 and 17? _____
Answers
A. 375
B. 221
22 Two-stage amplifiers can achieve large AC voltage gains if each amplifier uses an emitter bypass capacitor.
Question
What is the total AC voltage gain if each stage of a two-transistor amplifier has a gain of 100? _____
Answer
10,000
The Emitter Follower
23 Figure 8.23 shows another type of amplifier circuit.
Figure 8.23
Question
How is the circuit shown in Figure 8.23 different from the amplifier circuit discussed in problems 11–18? _____
Answer
There is no collector resistor, and the output signal is taken from the emitter.
24 The circuit shown in Figure 8.23 is called an emitter follower amplifier. (In some cases, it is also called the common collector amplifier.)
The output signal has some interesting features:
The peak-to-peak value of the output signal is almost the same as the input signal. In other words, the circuit gain is slightly less than 1; although in practice it is often considered to be 1.
The output signal has the same phase as the input signal. It is not inverted; the output is simply considered to be the same as the input.
The amplifier has a high input resistance. Therefore, it draws little current from the signal source.
The amplifier has a low output resistance. Therefore, the signal at the emitter appears to be emanating from a battery or signal generator with a low internal resistance.
Questions
A. What is the voltage gain of an emitter follower amplifier? _____
B. Is the output signal inverted? _____
C. What is the input resistance of the emitter follower amplifier? _____
D. What is its output resistance? _____
Answers
A. 1
B. No
C. High
D. Low
25 The example in this problem demonstrates the importance of the emitter follower circuit. The circuit shown in Figure 8.24 contains a small AC motor with 100 ohms resistance that is driven by a 10 Vpp signal from a generator. The 50-ohm resistor labeled RG is the internal resistance of the generator. In this circuit, only 6.7 Vpp is applied to the motor; the rest of the voltage is dropped across RG.
Figure 8.24
Figure 8.25 shows the same circuit, with a transistor connected between the generator and the motor in an emitter follower configuration.
Figure 8.25
You can use the following formula to calculate the approximate input resistance of the transistor:
The 10 Vpp from the generator is divided between the 10,000-ohm input resistance of the transistor and the 50-ohm internal resistance of the generator. Therefore, there is no significant voltage drop across RG, and the full 10 Vpp is applied to the base of the transistor. The emitter voltage remains at 10 Vpp.
Also, the current through the motor is now produced by the power supply and not the generator, and the transistor looks like a generator with a low internal resistance.
This internal resistance (RO) is called the output impedance of the emitter follower. You can calculate it using this formula:
For the circuit shown in Figure 8.25, if RG = 50 ohms and β = 100, RO = 0.5 ohms. Therefore, the circuit shown in Figure 8.25 is effectively a generator with an internal resistance of only 0.5 ohms driving a motor with a resistance of 100 ohms. Therefore, the output voltage of 10 Vpp is maintained across the motor.
Questions
A. What is the emitter follower circuit used for in this example? _____
B. Which two properties of the emitter follower are useful in circuits? _____
Answers
A. To drive a load that could not be driven directly by a generator
B. High input resistance and its low output resistance
26 The questions in this problem apply to the emitter follower circuit discussed in problems 23–25.
Questions
A. What is the approximate gain of an emitter follower circuit? _____
B. What is the phase of the output signal compared to the phase of the input signal? _____
C. Which has the higher value, the input resistance or the output resistance? _____
D. Is the emitter follower more effective at amplifying signals or at isolating loads? _____
Answers
A. 1
B. The same phase
C. The input resistance
D. Isolating loads
27 You can design an emitter follower circuit using the following steps:
1. Specify VE. This is a DC voltage level, which is usually specified as half the supply voltage.
2. Find VB. Use VB = VE + 0.7 volt.
3. Specify RE. Often this is a given factor, especially if it is a motor or other load that is being driven.
4. Find IE by using the following formula:
5. Find IB by using the following formula:
6. Find I2 by using I2 = 10IB.
7. Find R2 by using the following formula:
8. Find R1 by using the following formula:
Usually, IB is small enough to be dropped from this formula.
9. Choose the nearest standard values for R1 and R2.
10. Check that these standard values give a voltage close to VB. Use the voltage divider formula.
A simple design example illustrates this procedure. Use the values shown in the circuit in Figure 8.26 for this problem.
Figure 8.26
Questions
Work through Steps 1–10 to find the values of the two bias resistors.
1. VE = _____
2. VB = _____
3. RE = _____
4. IE = _____
5. IB = _____
6. I2 = _____
7. R2 = _____
8. R1 = _____
9. The nearest standard values are as follows:
R1 = _____
R2 = _____
10. VB = _____
Answers
Your answers should be close to the following values:
1. 5 volts (This was given in Figure 8.26.)
2. 5.7 volts
3. 1 kΩ (This was given in Figure 8.26.)
4. 5 mA
5. 0.05 mA
6. 0.5 mA
7. 11.4 kΩ
8. 7.8 kΩ
9. The nearest standard values are 8.2 kΩ and 12 kΩ.
10. The standard resistor values result in VB = 5.94 volts. This is a little higher than the VB calculated in Step 2, but it i
s acceptable.
VE is set by the biasing resistors. Therefore, it is not dependent upon the value of RE. Almost any value of RE can be used in this circuit. The minimum value for RE is obtained by using this simple equation:
Analyzing an Amplifier
28 Up to now, the emphasis has been on designing a simple amplifier and an emitter follower. This section shows how to “analyze” a circuit that has already been designed. In this case, to “analyze” means to calculate the collector DC voltage (the bias point) and find the AC gain. This procedure is basically the reverse of the design procedure.
Start with the circuit shown in Figure 8.27.
Figure 8.27
Following are the steps you use to analyze a circuit:
1. Find VB by using the following equation:
2. Find VE by using VE = VB − 0.7 volt.
3. Find IC by using the following equation:
Note that IC = IE.
4. Find VR by using VR = RC × IC.
5. Find VC by using VC = VS2VR. This is the bias point.
6. Find AV by using the following equation:
When you use the second formula, you must find the value of Rin (or hie) on the data sheets for the transistor from the manufacturer.
Use the circuit shown in Figure 8.28 for the following questions. For these questions, use β = 100, Rin = 2 kΩ and the values given in the circuit drawing.
Figure 8.28
Questions
Calculate VB, VE, IC, VR, VC, and AV using Steps 126 of this problem.