EXPERIMENT 8. PART A: OP-AMP A-C Signal Amplifier
The op amp makes an excellent small signal amplifier. In this experiment, you will use a
741 to amplify small audio frequency signals. You will also see how the gain of the circuit
affects the frequency response. The op amp cannot deliver much power to a load. Here you
will see how to interface complementary-symmetry transistors to an op amp to get more
output power.
EQUIPMENT
dual-power supplies ± 15,
oscilloscope,
DMM (bench top)
Function generator
(1) 741 op amp
(1) TIP 31 or equivalent NPN power transistor, (1) TIP 32 or equivalent PNP power
transistor
 (1) 10-micro.F, 25-WVDC capacitor
 (1) 5-KΩ pot; (1) 10-KΩ pot
 (1) 100-KΩ, 1/2-W ± 5% resistor; (2) 10-KΩ, 1/2-W ± 5% resistors;
 (3) 1-KΩ, 1/2-W ± 5% resistors; (1) 100-Ω, 1/2-W ± 5% resistors
 (2) 4.7KΩ, 1/2-W ± 5% resistors; (1) 9.1KΩ, 1/2-W ± 5% resistor






Fig. E8-2
PROCEDURE
1. Review the data sheet for a 741 OP-Amp and add the appropriate pin numbers to
the circuit drawings in this Lab. Adjust VCC to +15.0 V and VEE to -15.0V. Build the
above circuit and before connecting voltage divider (that has two 4.7KΩ resistors and a
5KΩ pot) set the negative input to ground and adjust the NULL pot for 0.00 volts on
the output. A suggested component placement is shown in the pictures below. Adjust
the 10KΩ pot for 0.00 V on the output using the DMM.
NOTE: You should be able to make V0 go positive and negative a volt or so with
respect to ground as you vary the 10KΩ pot (If you cannot, recheck your wiring).
Adjust Vin to the following voltages using the 5KΩ pot and record the VO in the table
below.
Remove the ground from the negative input and connect the voltage divider as shown
above (that has two 4.7KΩ resistors and a 5KΩ pot). Adjust the voltage divider for the
DC input voltages below and record expected voltages and measured voltages.
The above amplifier circuit is (a) non-inverting (b) inverting amplifier with a gain of
________.
DC Voltage in (three
digit value)
0.00 V
0.20 V
1.00 V
-1.00 V
-0.20 V
Calculated VO (three
digit value)
0.00 V
Without input.
Measured VO (three digit
value)
With input DC voltages.
2. Ensure RN is adjusted so that V0 = 0VDC with zero inputs.
3. Modify the circuit to match the drawing below, a suggested component placement is
shown in the picture below the circuit drawing.
4. Connect the signal generator to the input, and adjust it for vin = 0.1 Vp-p at f = 1
KHz. Note: you may need to add a 20 dB pad as shown to reach 0.1 Vp-p at 1kHz on
the input. Measure ac voltage out.
vo = _________
Calculate the a-c gain of the circuit.
Av = vo / vin =
Fig. E8-2A
Table E8-2
f (Hz)
Vo
R F = 100 KΩ
1
2
3
4
5
6
7
100
500
1K
5K
10K
50K
100K
R F = 10 KΩ
8
9
10
11
12
13
220
Electronic Troubleshooting
100
1K
10K
50 K
100K
500K
Gain
5. Now, adjust the generator frequency to each of the values listed in Table E8-2,
lines 8 through 13, and record the measured output voltage and the gain. Ensure that
vin remains at the same peak-peak value.
6. Next, ensure vin = 0.1 Vp-p at f = 1 KHz and change RF so as to make the
voltage gain equal to 100 = 1 KHz.
7. Repeat your measurements for each frequency listed on lines 1 through 7.
Fig. E8-2B
8. Shut off the power, change RF back to 10K ohms, and add the
complementary-symmetry stage to your op amp circuit as shown above and in
the picture below.
9. Connect a d-c voltmeter to the top of RL as shown, and apply power. Adjust RN
down while watching the d-c voltage across RL. It should go positive and negative
slightly if your circuit is working properly. Adjust the null control so that the d-c
voltage across RL = 0. RL in this circuit represents the speaker in an audio amplifier.
Since no blocking capacitor is used between the amplifier output and the speaker,
you can appreciate the need to null the output to zero so as to keep dc out of the
speaker.
10. Apply a 1 KHz signal of 0.2 VP-P to the input. Now observe the signal across the
load with an oscilloscope. Is it distorted? (a)Yes. (b)No. Why?
11. Adjust the NULL control up and down while watching the output. Explain
what you observe.
12. Shut off the power, and reconnect the right end of RF to the top of RL, as shown
above. Turn on the power again and observe the output. What do you notice?
PART A QUIZ
1. The NULL control varies the (a)gain. (b)a-c signal level. (c)d-c
level at the output.
2. The a-c gain of Fig. E8-1A is equal to —RF/R1. (a)True. (b)False.
3. Using a gain of 100, the 741 can amplify frequencies up to
approximately
Hz.
4. Using a gain of 10, the 741 can amplify frequencies up to
approximately
Hz.
5. The complementary—symmetry transistors allow us to deliver more
power to a low-resistance load. This is because they (a) increase the
voltage gain. (b) act like emitter followers, and thus prevent loading on
the op amp.
6.
222
In step 10, severe crossover distortion was evident. This was
Electronic Troubleshooting
Operational Amplifiers 221
because (a) both transistors were biased at cutoff. (b) two power supplies
were used. (c)no blocking capacitor was used.
7. As the NULL pot was varied in Step 11, the crossover point
shifted. This is because (a) the gain changed. (b) as one transistor
became more forward biased, the other became more back biased.
8. In Step 12, the crossover distortion (a) increased. (b) decreased.
(c) remained the same. This is because (d)a slight forward bias was
applied to both transistors simultaneously. (e)the gain of the op amp is
very high before either transistor starts conducting, so the op amp makes
up for the base-emitter drop by going an additional 0.7 V positive or
negative.
What other
performance?
9.
simple
change
could
be
made
to
improve
10. When you are troubleshooting an amplifier like the one in Fig. E82B, you find that the speaker is burned open and that the NPN power
transistor is shorted. Before replacing the bad transistor, it is a good
idea to (a)try driving the speaker with just the PNP transistor. (b)check
the d-c level out of the op amp. (c)put a blocking capacitor between
the speaker and the output transistors.
11. The problem in the previous question could have been caused by
(a) RN becoming open or disconnected from VEE. (b) RE resistance
too low. (c) excessive crossover distortion.
PART B: OP-AMPS In The Differential Mode
The circuit configuration that uses an IC op amp to its fullest capability is the
differential mode amplifier. In this experiment, you will see the op amp's ability to amplify
a small desired signal, while rejecting much larger common-mode d-c and a-c voltages.
PROCEDURE
1. Build the circuit of Fig. E8-3A. Connect a clip lead between points I and II,
shorting them together.
2. Adjust Rx so that V1 = V2 = 1 VDC. Measure V0. Vo should be quite small.
Vo =
3.
Change Rx so that V1 = V2 = 3 VDC. Measure
Vo. Vo =
Do you notice any significant change from
Step 2? (a)Yes. (b)No.
Why?
4. Now remove the clip lead from between points I and II. Measure Vo.
Vo =
Measure V,.
V, =
Measure V2.
V2
NOTE Measure V1 and V2 carefully, because there
is only a small difference between them. You may
want to connect a voltmeter directly across Ry to
make this measurement.
E8-3 Circuits for experiment PART B
The following material is a continuation of 4 above. The difference voltage V1 —
V2 is amplified by the circuit to give V0. Determine the voltage gain.
Av= Vo / (V, — V 2) =
"Compare this to the theoretical gain.
5. Connect a signal generator from ground to points I and II as shown in Fig. E83B. Adjust the generator amplitude to 1 Vp-p. Note that both points I and II
receive the full generator voltage, because the two coupling capacitors both
connect to the generator output terminal.
The a-c signal from the generator represents a common-mode noise voltage
Vcm. Measure the a-c signal at the output of the amplifier using an oscilloscope.
(This voltage should be quite small.)
Vcmo =
6. What is the common-mode gain of this circuit?
Acm= Vcmo /Vcm =
Now determine the common-mode rejection ratio.
CMRR = Av/Acm =
Part B QUIZ
1. In step 2, shorting together points I and II caused the same 2-VDC
signal to be applied to both the inverting and non-inverting inputs. This
signal is called a --- (a) common-mode signal. (b) differential signal.
2. In step 3, the closed-loop gain of the amplifier was(a)0,(b)10, (c)infinite,
but the d-c output voltage was (d)near zero. (e)ten times the common-mode
input signal.
3. Steps 5 and 6 showed that --(a)a-c signals cannot be amplified by a
difference amp. (b) common-mode a-c signals are rejected by a differential
amp.