The purpose of this experiment is to illustrate the “golden rules” of negative feedback for a variety of circuits. These concepts permit you to create and understand a vast number of practical circuits using only two simple rules. The op-amps used in this experiment are fully compensated, i.e. the open-loop phase shift is less than 135
◦ to reduce the danger of oscillation. However, for future applications you should remember that a compensated op-amp can oscillate if additional phase shift is introduced accidentally in the feedback loop.
The ideal op amp has the following characteristics:
1. The open loop gain ( A ), the gain when there is no feedback in the op amp circuit, is infinite.
2. The input impedance is infinite, implying that no current flows between the inverting and non-inverting terminals of the op amp.
3. There is no output impedance, so the output voltage is independent of the output current.
4. The op amp has infinite bandwidth, so the frequency response is flat.
5. The op amp has no input offset voltage, that is when the inverting and non-inverting inputs are shorted together, the output voltage is zero.
6. Since the inputs draw no current, and have no input offset voltage, v
+ v
− .
≈
7. The output can change infinitely fast, that is the output is not slew rate limited.
The most important of these rules in terms of circuit analysis are that the inputs draw no current and the inverting ( v
−
) and non-inverting ( v
+
) input voltages are approximately the same. So, the key rules to analyzing op amps are as follows:
1.
v
−
≈ v
+
1

2.
i
− = i
+ = 0
A simple example of these rules can be examined using the inverting op amp as shown in Fig. 1. To begin the analysis, consider the voltage at the noninverting (+) input of the op amp. From the characteristics of the ideal op amp, it is known that no current flows in to or out of the inputs. Because of this, no current flows through resistor
R
3 , so there is no voltage drop across this resistor. The voltage v
+
Since the other side of
R
3 must then be equal to the voltage before the resistor.
is connected to ground, v
+ = 0.
Further, since v
+
≈ v
−
, v
− input, it can be shown that
= 0. Now, analyzing the node at the non-inverting v o
− v
−
R
2
+ v
R o
2 v i
− v
−
R
+
1 v
R i
1
= 0
= 0 v o
=
−
R
2
R
1 v i
In practical op amps, the rate at which the output voltage can change is limited.
The maximum rate of change of the op amp output voltage is called the slew rate. The slew rate for the 741 op amp is typically 0.7
V
µ s
. The maximum peak to peak output voltage obtainable is given by
V max,pp =
Δ V
Δ t
πf
(1)
Op amps also have finite bandwidth owing to capacitance within the op amp’s circuitry. This is characterized by the gain bandwidth product. The bandwidth
( B ), closed loop gain ( G
0
), and unity gain frequency ( f
T
) obey the relationship
G
0
B
= f
T
(2)
2

The closed loop gain is the expected gain with feedback. The inverting op amp, for example, has a closed loop gain of
−
Hz to f bc
For an op amp with feedback,
R
2
R
1
. The bandwidth is distance from 0
, the point at which the closed loop gain decreased by a factor of √
2
.
G
=
G
1 +
0 jf f b c
(3)
Figure 1: High frequency performance circuit
In this section you will compare the performance of a general purpose 741 opamp to a relatively high speed 318 op-amp in the non-inverting amplifier circuit shown below. Always use bypass capacitors, and turn off the power supplies before components are changed.
1. Calculate the mid frequency closed-loop gain. For both op-amps, calculate the closed loop break frequency and the maximum amplitude signal that can be obtained at f bc without slew rate distortion. Use the op amp properties given in Table 1. Include these calculations in your laboratory report.
2. Use the slew rate to calculate the rise and fall time of an output pulse whose amplitude is 10V. Draw a sketch of
V o for a 10 kHz input square wave. Include these calculations in your laboratory report.
3

Op Amp
318
Slew Rate f
T
70 V/
μ s 15 MHz
741 0.5
μ s 1.2 MHz
Table 1: Op amp properties
3. Build the circuit of Fig. 1. Use
R
1
= 1
.
1 kΩ,
R
2
= 10 kΩ, and
R
3
= 1 kΩ.
4. Check the circuit you just built. When there is no input, you should have
V
0
≤
0
.
1 V. Apply a 1 kHz sine wave and check that the gain conforms to theory.
5. Use a 741 op-amp and a 1 V, square wave input signal. It is good practice to observe the input and output signals simultaneously.
Measure f bc .
Be sure that the signal amplitude does not exceed the slew-limit in the frequency range you are using.
6. After you measure f bc
, increase the size of the signal and observe the amplitude at which slew-rate distortion becomes evident. Next, use a 10 kHz square wave and adjust the signal amplitude so that is approximately
10 V. Measure the positive and negative slew rate.
7. Repeat the previous steps using a 318 op amp. You should observe a large increase in the gain at the closed-loop break-frequency f bc
, which indicates that the phase shift of the 318 exceeds 90 degrees in this frequency range.
You also see that the 318 has better high frequency gain than the 741.
The slew rate of the 318 is so large that you probably cannot observe slew-rate distortion in this circuit. The output rise and fall for a square wave are much faster for the 318 than for the 741. Observe this fact, but it is not necessary to measure the slew rate or sketch. You should also notice a damped overshoot on the output pulse, which is caused by the excess phase shift.
During this portion of the experiment, you will build a ramp generator. This circuit utilizes resistors and capacitors to provide a ramped voltage which is reset during each period of the input square wave. When the square wave voltage v i is negative, the electrons that flow through R capacitor
C f linearly as
C f
R
1 by the op amp.
D shorting out
C f charges. When which rapidly discharges
C f v i
1
2 are pulled in to the feedback prevents current from flowing through during this part of the input voltage’s cycle. The output voltage v goes positive, a large current flows through
. When v o reaches -0.6 V,
D
2 is forward biased,
, and the output voltage stays at a constant -0.6 V.
o
R
1 increases
4

Figure 2: Ramp Generator
1. Build the circuit in Fig. 2. Use
C f
R
1
= 1 kΩ,
= 100 pF. Use 1N914 diodes for
D
1 and
R
D
2
2
.
= 51 kΩ,
R
3
= 1 kΩ, and
2. Apply a 50 kHz, 10 Vpp square wave to your circuit using the signal generator. Sketch the output voltage. Record the slope of the ramp and its amplitude.
3. For your lab report, derive an equation for the output voltage of the ramp generator. Using this equation, predict the slope and amplitude of the ramp. Compare the values calculated to your measured values.
The summing circuit will combine two signals and the output voltage will be the sum of the signals with unity gain, assuming that all resistors are of the same value.
1. Build the circuit in Fig. 3. For all resistors, use R = 1 kΩ. Do not connect v
2 initially. Use a 10 kHz, 5 Vpp sine wave for v
1
.
2. Observe v o and v
1
. What is the gain?
3. Attach the TTL/CMOS output of the signal generator to the point marked v
2
.
4. Observe v
2 on the oscilloscope.
5. Observe and sketch v o
.
5

Figure 3: Summing Circuit
A capacitor is included in the feedback loop of the following inverting amplifier.
The parallel 100 kΩ resistor is needed to set the quiescent conditions. It does not affect the signal when X c
R f
. The gain is given by v o v i
=
−
R f
R
1
(4)
1 +
1
R f
X c
2
≈ −
X c
R
1
And the phase by tan φ = −
X c
R f
(5)
1. Calculate the gain and expected phase shift of this circuit for a frequency of 1.6 kHz and 16 kHz if
R
1 = 100 kΩ, and
C f = 0
.
01
μ
F.
= 1 kΩ,
R
2 = 1 kΩ,
R f
2. Build the circuit in Fig. 4. Use R
1
C f
= 0 .
01 μ F.
= 1 kΩ, R
2
= 1 kΩ, R f
= 100 kΩ, and
3. Measure the gain and phase shift for the circuit. Use a 1 Vpp 1.6 kHz sine wave and a 1 Vpp 16 kHz sine wave.
This circuit yields a gain of
6

Figure 4: Reactance in the feedback loop v v o i
=
R
2
R
1
( v
2
− v
1 ) (6) that is, it amplifies the difference between input voltages v
1 and v
2
. For the circuit to work properly, the resistance in the feedback loop must be exactly equal to the series combination of
R
4 and
R
5 .
1. Build the circuit in Fig. 5. Use
R
1
R
4 = 5
.
1 kΩ, and for
R
5
= 1 kΩ, use a 10 kΩ trimpot.
R
2 = 10 kΩ,
R
3 = 1 kΩ,
2. Connect points A and B. Apply a 1 Vpp, 1 kHz sine wave to both points.
3. While observing v o
, adjust the trimpot until v o which will usually be no smaller than 10 mV.
is as small as possible,
4. Remove the connection between points A and B.
5. Connect point B to ground, and apply a 1 Vpp, 1 kHz sine wave to point
A. You should see that the output is the input voltage amplified 10 times.
•
Op Amps: (1) 741, (1) 318
•
Capacitors: (1) 0.01
μ
F, (1) 100 pF
•
Resistors: (4) 1 kΩ, (1) 1.1 kΩ, (1) 5.1 kΩ, (1) 10 kΩ, (1) 51 kΩ, (1) 100 kΩ
•
Diodes: (4) 1N914
7

Figure 5: Feedback difference amplifier
•
Trimpots: (1) 10 kΩ
8