by Earl Boysen
1. Find IC by using the following equation:
2. Find IB by using the following equation:
3. Find RB by using the following equation:
Questions
Find the value of RB that will set the collector voltage to 5 volts, using steps 123 and the following values for the circuit:
A. IC = _____
B. IB = _____
C. RB = _____
Answers
A.
B.
C.
2 You have seen that using a 200 kΩ resistor for RB gives an output level of 5 volts at the collector. This procedure of setting the output DC level is called biasing. In problem 1, you biased the transistor to a 5-volt DC output.
Use the circuit shown in Figure 8.1 and the formulas given in problem 1 to answer the following questions.
Questions
A. If you decrease the value of RB, how do IB, IC, VR, and the bias point VC change? _____
B. If you increase the value of RB, how do IB, IC, VR, and VC change? _____
Answers
A. IB increases, IC increases, VR increases, and so the bias point VC decreases.
B. IB decreases, IC decreases, VR decreases, and so the bias point VC increases.
3 In problem 2, you found that changing the value of RB in the circuit shown in Figure 8.1 changes the value of IB.
The transistor amplifies slight variations in IB. Therefore, the amount IC fluctuates is β times the change in value in IB.
The variations in IC cause changes in the voltage drop VR across RC. Therefore, the output voltage measured at the collector also changes.
Questions
For the circuit shown in Figure 8.1, calculate the following parameters when RB = 168 kΩ and VS = 10 volts:
A. _____
B. IC = βIB = _____
C. VR = ICRC = _____
D. VC = VS2VR = _____
Answers
A.
B. IC = 100 × 0.059 = 5.9 mA
C. VR = 1 kΩ × 5.9 mA = 5.9 volts
D. VC = 10 volts 2 5.9 volts = 4.1 volts
4 Use the circuit shown in Figure 8.1 to answer the following questions when VS = 10 volts.
Questions
Calculate VC for each of the following values of RB:
A. 100 kΩ _____
B. 10 MΩ _____
C. 133 kΩ _____
D. 400 kΩ _____
Answers
A. IB = 0.1 mA, IC = 10 mA, VC = 0 volts
B. IB = 1 μA, IC = 0.1 mA, VC = 10 volts (approximately)
C. IB = 0.075 mA, IC = 7.5 mA, VC = 2.5 volts
D. IB = 0.025 mA, IC = 2.5 mA, VC = 7.5 volts
5 The values of IC and VC that you calculated in problems 1 and 4 are plotted on the graph on the left side of Figure 8.2. The straight line connecting these points on the graph is called the load line.
Figure 8.2
The axis labeled VC represents the voltage between the collector and the emitter of the transistor, and not the voltage between the collector and ground. Therefore, this axis should correctly be labeled VCE, as shown in the graph on the right of the figure. (For this circuit, VCE = VC because there is no resistor between the emitter and ground.)
Questions
A. At point A in the graph on the right, is the transistor ON or OFF? _____
B. Is it ON or OFF at point B? _____
Answers
A. ON because full current flows, and the transistor acts like a short circuit. The voltage drop across the transistor is 0 volts.
B. OFF because essentially no current flows, and the transistor acts like an open circuit. The voltage drop across the transistor is at its maximum (10 volts, in this case).
6 Point A on the graph shown in Figure 8.2 is called the saturated point (or the saturation point) because it is at that point that the collector current is at its maximum.
Point B on the graph shown in Figure 8.2 is often called the cutoff point because, at that point, the transistor is OFF and no collector current flows.
In regions X and Y, the gain (β) is not constant, so these are called the nonlinear regions. Note that β = IC/IB. Therefore, β is the slope of the line shown in the graph.
As a rough guide, V1 is approximately 1 volt, and V2 is approximately 1 volt less than the voltage at point B.
Question
What is the value of VCE at point B? _____
Answer
VCE = VS, which is 10 volts in this case.
7 In region Z of the graph shown in Figure 8.2, β (that is, the slope of the graph) is constant. Therefore, this is called the linear region. Operating the transistor in the linear region results in an output signal that is free of distortion.
Question
Which values of IC and VC would result in an undistorted output in the circuit shown in Figure 8.1?
A. IC = 9 mA, VC = 1 volt
B. IC = 1 mA, VC = 9 volts
C. IC = 6 mA, VC = 4.5 volts _____
Answer
C is the only one. A and B fall into nonlinear regions.
8 If you apply a small AC signal to the base of the transistor after it has been biased, the small voltage variations of the AC signal (shown in Figure 8.3 as a sine wave) cause small variations in the base current.
Figure 8.3
These variations in the base current will be amplified by a factor of β and will cause corresponding variations in the collector current. The variations in the collector current, in turn, will cause similar variations in the collector voltage.
The β used for AC gain calculations is different from the β used in calculating DC variations. The AC β is the value of the common emitter AC forward current transfer ratio, which is listed as hfe in the manufacturer's data sheets for the transistor. Use the AC β whenever you need to calculate the AC output for a given AC input, or to determine an AC current variation. Use the DC β to calculate the base or collector DC current values. You must know which β to use, and remember that one is used for DC, and the other is used for AC variations. The DC β is sometimes called hFE or βdc.
As Vin increases, the base current increases, which causes the collector current to increase. An increase in the collector current increases the voltage drop across RC, which causes VC to decrease.
Note The capacitor shown at the input blocks DC (infinite reactance) and easily passes AC (low reactance). This is a common isolation technique used at the input and output of AC circuits.
Questions
A. If the input signal decreases, what happens to the collector voltage? _____
B. If you apply a sine wave to the input, what waveform would you expect at the collector? _____
Answers
A. The collector voltage, VC, increases.
B. A sine wave, but inverted as shown in Figure 8.4.
Figure 8.4
9 Figure 8.4 shows the input and output sine waves for an amplifier circuit.
The input voltage Vin is applied to the base. (Strictly speaking, it is applied across the base-emitter diode.) The voltage variations at the collector are centered on the DC bias point VC, and they will be larger than variations in the input voltage. Therefore, the output sine wave is larger than the input sine wave (that is, amplified).
This amplified output signal at the collector can be used to drive a load (such as a speaker).
To distinguish these AC variations in output from the DC bias level, you indicate the AC output voltage by Vout. In most cases, Vout is a peak-to-peak value.
Questions
A. What is meant by VC? _____
B. What is meant by Vout? _____
Answers
A. Collector DC voltage, or the bias point
B. AC output voltage
The ratio of the output voltage to the input voltage is called the voltage gain of the amplifier.
To calculate the voltage gain of an amplifier, you can measure Vin and Vout with an oscilloscope. Measure peak-to-peak voltages for this calculation.
; 10 For the circuit shown in Figure 8.4 you can calculate the voltage gain using the following formula:
In this equation:
RL is the load resistance. In this circuit, the collector resistor, RC, is the load resistance.
Rin is the input resistance of the transistor. You can find Rin (often called hie) on the data or specification sheets from the manufacturer. In most transistors, input resistance is approximately 1 kΩ to 2 kΩ.
You can find Vout by combining these two voltage gain equations:
Solving this for Vout results in the following equation. Here, the values of Rin = 1 kΩ, Vin = 1 mV, RC = 1 kΩ, and β = 100 were used to perform this sample calculation.
Questions
A. Calculate Vout if Rin = 2 kΩ, Vin = 1 mV, RC = 1 kΩ, and β = 100. _____
B. Find the voltage gain in both cases. _____
Answers
A. Vout = 50 mV
B. AV = 100 and AV = 50
This simple amplifier can provide voltage gains of up to approximately 500. But it does have several faults that limit its practical usefulness.
Because of variations in β between transistors, VC changes if the transistor is changed. To compensate for this, you must adjust RB.
Rin or hie varies greatly from transistor to transistor. This variation, combined with variations in β, means that you cannot guarantee the gain from one transistor amplifier to another.
Both Rin and β change greatly with temperature; hence the gain is temperature-dependent. For example, a simple amplifier circuit like that discussed in this problem was designed to work in the desert in July. It would fail completely in Alaska in the winter. If the amplifier worked perfectly in the lab, it probably would not work outdoors on either a hot or cold day.
Note An amplifier whose gain and DC level bias point change as described in this problem is said to be unstable. For reliable operation, an amplifier should be as stable as possible. In later problems, you see how to design a stable amplifier.
A Stable Amplifier
11 You can overcome the instability of the transistor amplifier discussed in the first ten problems of this chapter by adding two resistors to the circuit. Figure 8.5 shows an amplifier circuit to which resistors RE and R2 have been added. R2, along with R1 (labeled RB in the previous circuits), ensures the stability of the DC bias point.
Figure 8.5
By adding the emitter resistor RE, you ensure the stability of the AC gain.
The labels in Figure 8.6 identify the DC currents and voltages present in the circuit. These parameters are used in the next several problems.
Figure 8.6
Question
In designing an amplifier circuit and choosing the resistor values, there are two goals. What are they? _____
Answer
A stable DC bias point, and a stable AC gain
12 Look at the gain first. The gain formula for the circuit shown in Figure 8.6 is as follows:
This is a slight variation on the formula shown in problem 10. (The complex mathematical justification for this is not important right here.) Here, the AC gain is not affected by transistor β and transistor input resistance, so the AC gain will be constant, regardless of variations in these parameters.
Questions
Use the circuit shown in Figure 8.6 with RC = 10 kΩ and RE = 1 kΩ to answer the following questions:
A. What is the AC voltage gain for a transistor if its β = 100? _____
B. What is the gain if β = 500? _____
Answers
A. 10
B. 10
13 This problem provides a couple of examples that can help you understand how to calculate voltage gain and the resulting output voltage.
Questions
A. Calculate the voltage gain (AV) of the amplifier circuit shown in Figure 8.6 if RC = 10 kΩ and RE = 1 kΩ. Then, use AV to calculate the output voltage if the input signal is 2 mVpp. _____
B. Calculate the voltage gain if RC = 1 kΩ and RE = 250 ohms. Then, use AV to calculate the output voltage if the input signal is 1 Vpp. _____
Answers
A.
B.
Although the amplifier circuit shown in Figure 8.6 produces stable values of voltage gain, it does not produce high values of voltage gain. For various reasons, this circuit is limited to voltage gains of 50 or less. Later, this chapter discusses an amplifier circuit that can produce higher values of voltage gain.
14 Before you continue, look at the current relationships in the amplifier circuit shown in Figure 8.6 and an approximation that is often made. You can calculate the current flowing through the emitter resistor with the following equation:
In other words, the emitter current is the sum of the base and the collector currents.
IC is much larger than IB. You can, therefore, assume that the emitter current is equal to the collector current.
Question
Calculate VC, VE, and AV for the circuit shown in Figure 8.7 with VS = 10 volts, IC = 1 mA, RC = 1 kΩ, and RE = 100 ohms. _____
Figure 8.7
Answer
15 For this problem, use the circuit shown in Figure 8.7 with VS = 10 volts, IC = 1 mA, RC = 2 kΩ, and RE = 1 kΩ.
Question
Calculate VC, VE, and AV. _____
Answers
16 For this problem, use the circuit shown in Figure 8.7 with VS = 10 volts and IC = 1 mA.
Questions
Find VC, VE, and AV for the following values of RC and RE:
A. RC = 5 kΩ, RE = 1 kΩ _____
B. RC = 4.7 kΩ, RE = 220 ohms _____
Answers
A.
B.
Biasing
17 In this problem, you see the steps used to calculate the resistor values needed to bias the amplifier circuit shown in Figure 8.8.
Figure 8.8
You can determine values for R1, R2, and RE that bias the circuit to a specified DC output voltage and a specified AC voltage gain by using the following steps.
Read the following procedure and the relevant formulas first, and then you will work through an example.
1. Find RE by using the following:
2. Find VE by using the following:
3. Find VB by using the following:
4. Find IC by using the following:
5. Find IB by using the following:
6. Find I2 where I2 is 10IB. (Refer to the circuit shown in Figure 8.6.) This is a convenient rule of thumb that is a crucial step in providing stability to the DC bias point.
7. Find R2 by using the following:
8. Find R1 by using the following:
9. Steps 7 and 8 might produce nonstandard values for the resistors, so choose the nearest standard values.
10. Use the voltage divider formula to see if the standard values you chose in step 9 result in a voltage level close to VB found in step 3. (“Close” means within 10 percent of the ideal.)
This procedure produces an amplifier that works, and results in a DC output voltage and AC gain that are close to those specified at the beginning of the problem.
Questions
Find the values of the parameters specified in each of the following questions for the circuit shown in Figure 8.9 if AV = 10, VC = 5 volts, RC = 1 kΩ, β = 100, and VS = 10 volts.
Figure 8.9
Work through steps 1–10, referring to the steps in this problem for formulas as necessary.
1. Find RE.
2. VE = _____
3. VB = _____
4. IC = _____
5. IB = _____
6. I2 = _____
7. R2 = _____
8. R1 = _____
9. Choose the standard resistance values that are closest to the calculated values for R1 and R2.
R1 = _____
R2 = _____
10. Using the standard resistance values for R1 and R2, find VB.
VB = _____
Answers
You should have found values close to the following:
1
. 100 ohms
2. 0.5 volt
3. 1.2 volts
4. 5 mA
5. 0.05 mA
6. 0.5 mA
7. 2.4 kΩ
8. 16 kΩ
9. 2.4 kΩ and 16 kΩ are standard values. (They are 5 percent values.) Alternative acceptable values would be 2.2 kΩ and 15 kΩ.
10. With 2.4 kΩ and 16 kΩ, VB = 1.3 volts. With 2.2 kΩ and 15 kΩ, VB = 1.28 volts. Either value of VB is within 10 percent of the 1.2 volts calculated for VB in step 3.
Figure 8.10 shows an amplifier circuit using the values you calculated in this problem for R1, R2, and RE.
Figure 8.10
18 Follow the steps in problem 17 to answer the following questions.
Questions
Find the values of the parameters specified in each question for the circuit shown in Figure 8.9 if AV = 15, VC = 6 volts, β = 100, RC = 3.3 kΩ, and VS = 10 volts.
1. RE = _____
2. VE = _____
3. VB = _____
4. IC = _____
5. IB = _____
6. I2 = _____
7. R2 = _____
8. R1 = _____
9. Choose the standard resistance values that are closest to the calculated values for R1 and R2.
R1 = _____
R2 = _____
10. Using the standard resistance values for R1 and R2, find VB.