by Earl Boysen
1. VB = _____
2. VE = _____
3. IC = _____
4. VR = _____
5. VC = _____
6. AV = _____
Answers
1.
2. VE = 1.2 2 0.7 = 0.5 volt
3.
4. VR = 10 kΩ × 0.5 mA = 5 volts
5. VC = 10 volts 2 5 volts = 5 volts (This is the bias point.)
6. With the capacitor:
7. Without the capacitor:
29 You can determine the lowest frequency the amplifier will satisfactorily pass by following these simple steps:
1. Determine the value of RE.
2. Calculate the frequency at which XC = RE/10. Use the capacitor reactance formula. (This is one of those “rules of thumb” that can be mathematically justified and gives reasonably accurate results in practice.)
Questions
For the circuit shown in Figure 8.28, find the following.
A. RE = _____
B. f = _____
Answers
A. RE = 1 kΩ (given in the circuit diagram)
B. So, you set XC = 100 ohms, and use this formula:
So, the following is the result:
30 For the circuit shown in Figure 8.29, follow the steps given in problems 28 and 29 to answer the following questions.
Figure 8.29
Questions
1. VB = _____
2. VE = _____
3. IC = _____
4. VR = _____
5. VC = _____
6. With capacitor:
AV = _____
Without capacitor:
AV = _____
7. Low frequency check:
f = _____
Answers
Your answers should be close to these.
1. 1.18 volts
2. 0.48 volts
3. 1 mA
4. 4.7 volts
5. 5.3 volts (bias point)
6. With capacitor: 376
Without capacitor: 10
7. 57 Hz (approximately)
The JFET as an Amplifier
31 Chapter 3 discussed the JFET in problems 28231, and Chapter 4 discussed the JFET in problems 37241. You may want to review these problems before answering the questions in this problem. Figure 8.30 shows a typical biasing circuit for a JFET.
Figure 8.30
Questions
A. What type of JFET is depicted in the circuit? _____
B. What value of VGS would you need to turn the JFET completely ON? _____
C. What drain current flows when the JFET is completely ON? _____
D. What value of VGS would you need to turn the JFET completely OFF? _____
E. When a JFET is alternately turned completely ON and OFF in a circuit, what type of component are you using the JFET as? _____
Answers
A. N-channel JFET.
B. VGS = 0 V to turn the JFET completely ON.
C. Drain saturation current (IDSS).
D. VGS should be a negative voltage for the N-channel JFET to turn it completely OFF. The voltage must be larger than or equal to the cutoff voltage.
E. The JFET is being used as a switch.
32 You can use a JFET to amplify AC signals by biasing the JFET with a gate to source voltage about halfway between the ON and OFF states. You can find the drain current that flows in a JFET biased to a particular VGS by using the following equation for the transfer curve:
In this equation, IDSS is the value of the drain saturation current, and VGS(off) is the gate to source voltage at cutoff. Both of these are indicated on the transfer curve shown in Figure 8.31.
Figure 8.31
For the transfer curve shown in Figure 8.31, IDSS = 12 mA and VGS(off) = −4 volts. Setting the bias voltage at VGS = −2 volts returns the following value for the drain current:
Questions
Calculate the drain current for the following:
A. VGS = 21.5 _____
B. VGS = 20.5 volts _____
Answers
A. 4.7 mA
B. 9.2 mA
Note Data sheets give a wide range of possible IDSS and VGS(off) ) values for a given JFET. You may need to resort to actually measuring these with the method shown in Project 4-2.
33 For the circuit shown in Figure 8.30, you choose the value of the drain to source voltage, VDS, and then calculate the value of the load resistor, RD, by using the following equation:
For this problem, use ID = 3 mA, and a drain supply voltage (VDD) of 24 volts. Calculate the value of RD that results in the specified value of VDS; this is also the DC output voltage of the amplifier.
Question
Calculate the value of RD that will result in VDS = 10 volts. _____
Answers
34 The circuit shown in Figure 8.32 (which is referred to as a JFET common source amplifier) applies a 0.5 Vpp sine wave to the gate of the JFET and produces an amplified sine wave output from the drain.
Figure 8.32
The input sine wave is added to the −2 volt bias applied to the gate of the JFET. Therefore, VGS varies from −1.75 to −2.25 volts.
Question
Using the formula in problem 32, calculate ID for the maximum and minimum values of VGS.
Answer
For VGS = 21.75 volts, ID = 3.8 mA
For VGS = 22.25 volts, ID = 2.3 mA
35 As the drain current changes, VRD (the voltage drop across resistor RD) also changes.
Question
For the circuit shown in Figure 8.32, calculate the values of VRD for the maximum and minimum values of ID you calculated in problem 34. _____
Answer
This corresponds to a 7 Vpp sine wave.
36 As the voltage drop across RD changes, the output voltage also changes.
Question
For the circuit shown in Figure 8.32, calculate the values of Vout for the maximum and minimum values of VRD you calculated in problem 35. _____
Answer
Therefore, the output signal is a 7 Vpp sine wave.
37 Table 8.1 shows the results of the calculations made in problems 34–36 including the DC bias point.
Table 8.1 Calculation Results
Question
What are some characteristics of the AC output signal? _____
Answer
The output signal is a 7 Vpp sine wave with the same frequency as the input sine wave. As the input voltage on VGS increases (toward 0 volts), the output decreases. As the input voltage decreases (becomes more negative), the output voltage increases. This means that the output is 180 degrees out of phase with the input.
38 You can calculate the AC voltage gain for the amplifier discussed in problems 34–37 by using the following formula:
The negative sign in this formula indicates that the output signal is 180 degrees out of phase from the input signal.
Question
Calculate the AC voltage gain for the amplifier discussed in problems 34–37. _____
Answer
39 You can also calculate the AC voltage gain by using the following formula:
In this equation, gm is the transconductance and is a property of the JFET. It is also called the forward transfer admittance. A typical value for gm is usually provided for JFETs in the data sheet from the manufacturers. You can also use the data in Table 8.1 to calculate gm using the following formula:
In this equation, Δ indicates the change or variation in VGS and the corresponding drain current. The unit for transconductance is mhos.
Questions
A. Using the data from Table 8.1, what is the value of gm for the JFET used in the amplifier? _____
B. What is the corresponding AC voltage gain? _____
Answers
A.
B. Av = 2(0.003)(4670) = 214, the same result you found in problem 38
40 Design a JFET common source amplifier using a JFET with IDSS = 14.8 mA and VGS(off) = 23.2 volts. The input signal is 40 mVpp. The drain supply is 24 volts.
&nb
sp; Questions
A. Determine the value of VGS that will bias the JFET at a voltage near the middle of the transfer curve. _____
B. Calculate the drain current when VGS is at the value determined in step A, using the formula in problem 32. _____
C. Choose a value of VDS and calculate the value of RD using the formula in problem 33. _____
D. Calculate the maximum and minimum values of VGS that result from the input signal, and the corresponding values of drain current using the procedure in problem 34. _____
E. Calculate the maximum and minimum values of Vout that result from the input signal using the procedures in problems 35 and 36. _____
F. Calculate the gain of the amplifier. _____
Answers
A. VGS = 21.6 volts
B. ID = 3.7 mA
C. For VDS = 10 volts,
D. VGS will vary from 21.58 to 21.62 volts. Use the formula to calculate values of drain current. ID will vary from 3.79 to 3.61 mA.
E. VRD will vary from 14.3 to 13.6 volts. Therefore, Vout will vary from 9.7 to 10.4 volts.
F.
41 Use the results of problem 40, question D, to answer the following question.
Questions
Calculate the transconductance of the JFET and the AC voltage gain using the formulas discussed in problem 39. _____
Answers
This is close to the value you found in problem 40, question F.
42 Figure 8.33 shows a JFET amplifier circuit that uses one power supply, rather than separate power supplies for the drain and gate used in the amplifier discussed in problems 34–41.
Figure 8.33
The DC voltage level of the gate is zero because the gate is tied to ground through RG. Therefore, the voltage drop across RS becomes the gate to source voltage. To design the circuit, you must find values for both RS and RD. Use the same bias point for this problem as you used for the amplifier discussed in problems 34–41: VGS = 22 volts and ID = 3 mA. Follow these steps:
1. Calculate RS, using the following formula, recognizing that VRS = VGS:
2. Calculate RD using the following formula, using VDS = 10 volts, the same value you used for the amplifier discussed in problems 34–41:
3. Calculate XCS using the following formula:
Then, calculate CS using the following formula:
4. Calculate the peak-to-peak output voltage using the procedures shown in problems 34–36.
5. Calculate the AC voltage gain using this formula:
Note Choose the value of CS so that its reactance is less than 10 percent of RS at the lowest frequency you need to amplify. The DC load for the JFET is RD plus RS. The AC load is RD only because CS bypasses the AC signal around RS, which keeps the DC operating point stable. The use of CS reduces the gain slightly because you now use a smaller RD to calculate the AC voltage swings at the output.
Questions
A. What is the value of RS? _____
B. What is the value of RD? _____
C. What is the value of CS? Assume f = 1 kHz. _____
D. Calculate the peak-to-peak Vout for Vin = 0.5 Vpp. _____
E. What is the voltage gain? _____
Answers
A.
B.
C. XCS = 66.7 ohms, CS = 2.4 μF
D. The AC drain current will still vary from 3.8 to 2.3 mA, as in problem 37. The voltage across RD is now 6 Vpp because RD is 4 kΩ. The output voltage is also 6 Vpp.
E. The gain is 12.
The Operational Amplifier
43 The operational amplifier (op-amp) in use today is actually an integrated circuit (IC). This means that the device has numerous transistors and other components constructed on a small silicon chip. These IC op-amps are much smaller and, therefore, more practical than an amplifier with equivalent performance that is made with discrete components.
You can purchase op-amps in different case configurations. Some of these configurations are the Transistor Outline (TO) metal package, the flat pack, and the dual in-line pin (DIP) package. You can also find two op-amps (dual) or four op-amps (quad) in a single IC.
Their size, low cost, and wide range of applications have made op-amps so common today that they are thought of as a circuit device or component in and of themselves, even though a typical op-amp may contain 20 or more transistors in its design. The characteristics of op-amps closely resemble those of an ideal amplifier. Following are these characteristics:
High input impedance (does not require input current)
High gain (used for amplifying small signal levels)
Low output impedance (not affected by the load)
Questions
A. What are the advantages of using op-amps? _____
B. Why are op-amps manufactured using IC techniques? _____
Answers
A. Small size, low cost, wide range of applications, high input impedance, high gain, and low output impedance.
B. Because of the large numbers of transistors and components that are required in the design of an op-amp, they must be constructed on a single, small silicon chip using IC manufacturing techniques to be of a reasonable size.
44 Figure 8.34 shows the schematic symbol for an op-amp.
Figure 8.34
An input at the inverting input results in an output that is 180 degrees out of phase with the input. An input at the noninverting input results in an output that is in phase with the input. Both positive and negative voltage supplies are required, and the data sheet will specify their values for the particular op-amp you use. Datasheets usually contain circuit diagrams showing how you should connect external components to the op-amp for specific applications. These circuit diagrams (showing how a particular op-amp can be used for various applications) can be useful to the designer or the hobbyist.
Questions
A. How many terminals does the op-amp require, and what are their functions? _____
B. How is the output related to the input when the input is connected to the inverting input? _____
Answers
A. Five—two input terminals, one output terminal, two power supply terminals.
B. The output is 180 degrees out of phase with the input.
45 Figure 8.35 shows a basic op-amp circuit. The input signal is connected to an inverting input, as indicated by the negative sign. Therefore, the output signal will be 180 degrees out of phase with the input.
Figure 8.35
You can find the AC voltage gain for the circuit using the following equation:
Resistor RF is called a feedback resistor because it forms a feedback path from the output to the input. Many op-amp circuits use a feedback loop. Because the op-amp has such a high gain, it is easy to saturate it (at maximum gain) with small voltage differences between the two input terminals. The feedback loop allows the operation of the op-amp at lower gains, allowing a wider range of input voltages. When designing a circuit, you can choose the value of the feedback resistor to achieve a specific voltage gain. The role of the capacitors in the diagram is to block DC voltages.
Questions
A. Calculate the value of RF that would give the amplifier an AC voltage gain of 120. _____
B. Calculate AC Vout if AC Vin is 5 mVrms. _____
Answers
A.
B.
The output signal is inverted with respect to the input signal.
46 Use the op-amp circuit shown in Figure 8.35 to build an amplifier with an output voltage of 12 Vpp, an AC voltage gain of 50, and with Rin = 6.8 kΩ.
Questions
A. Calculate the value of RF. _____
B. Calculate the value of Vin required to produce the output voltage specified earlier. _____
Answers
A.
B.
Project 8.2: The Operational Amplifier
Objective
The objective of this project is to demonstrate how AC voltage gain changes when you use feedback resistors of different values in an op-amp circuit.
/> General Instructions
After the circuit is set up, you measure Vout for each value of RF, and find AV, using the ratio Vout/Vin. You also determine a calculated AV using the ratio RF/Rin in each case to determine how close the calculated AV is to the measured AV.
Parts List
You need the following equipment and supplies:
One 0.1 μF capacitor.
Two 10 kΩ, 0.25-watt resistors.
One 51 kΩ, 0.25-watt resistor.
One 100 kΩ, 0.25-watt resistor.
One 150 kΩ, 0.25-watt resistor.
One 220 kΩ, 0.25-watt resistor.
One 270 kΩ, 0.25-watt resistor.
One 330 kΩ, 0.25-watt resistor.
One 380 kΩ, 0.25-watt resistor.
Two terminal blocks.
Two 6-volt battery packs (4 AA batteries each).
One function generator.
One oscilloscope.
One breadboard.
One OPA134 operational amplifier. This op-amp comes in a few different packages; get the 8-pin dual in-line (DIP) version. Figure 8.36 shows the pinout diagram for the OPA134. When you insert the op-amp into the breadboard, try not to bend any of the leads. The leads on dual in-line packages are fragile and will break off if you bend them more than once or twice.
Figure 8.36
Step-by-Step Instructions
Set up the circuit shown in Figure 8.37 using the 51 k resistor for RF. Figure 8.38 shows the battery connections. 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, look at the photos of the completed circuit in the “Expected Results” section. One unusual aspect of this circuit you may want to look for in the photos is how the 2V bus of one 6-volt battery pack should be connected to the +V bus of the other 6-volt battery pack.