Fall 2000
EE3301 Experiment 1
NONINVERTING OPERATIONAL AMPLIFIER CIRCUITS
Relevant sections of textbook:
Chapter 2 Operational Amplifiers
2.4 Noninverting Amplifier
2.4 Voltage-Follower (Buffer) Amplifier
1 Introduction
A unity gain (“voltage follower”) amplifier circuit has a direct connection between its output and its
negative input terminal. The resistance, RL , represents the effect of a circuit or component (for example,
an earphone or a transducer) that might be attached to the amplifier circuit. It is known as the load
resistance.
−
+
Vin
−
+
+
Vout
RL
−
Test
Figure 1: Unity gain voltage follower
Since for an ideal model the input difference voltage, v+ − v− , is zero, the output voltage, vout , is equal to
vin .
vout = vin
(1)
To the extent that the input current is zero, no signal power is supplied by the voltage source vin . The
amplifier input behaves as a nearly infinite resistance, a resistance that is generally too large to measure
with conventional techniques. For values of land resistance, RL , that do not result in excessively large
values of load current (vout /RL ), the output voltage is equal to vin and independent of RL . This
corresponds to the amplifier circuit having , for its output, a Thévenin equivalent circuit with a zero
resistance. While for an actual amplifier circuit the Thévenin resistance is necessarily finite, it is
generally very small. To achieve an output voltage that is greater than its input voltage, a resistor voltage
divider circuit may be used for the feedback connection between the output of the amplifier and its
negative input terminal.
The voltage v− may be expressed in terms of the output voltage:
v− =
R1 vout
R1 + R2
(2)
Assuming that the ideal operational amplifier model applies, the input difference voltage, v+ − v− , is zero.
Av =
R2
vout
= 1+
vin
R1
1
(3)
R1
R2
−
+
+
+
Vout
RL
Vin
−
−
Figure 2: Non-inverting amplifier
Except for a zero value of feedback resistance, R2 , or an infinite value of R1 , the voltage gain is greater
than one. Voltage gains as great as 100 may readily be achieved with operational amplifier circuits. As
for the unity gain circuit, the output voltage tends to be independent of the load resistance for reasonable
values of load resistance.
2 Experimental Procedure
The following circuit will be used for the initial parts of the experimental procedure. A “dual” power
−
7
+
2
6
10 k Ω
+
Vin
+
4
3
RL
2.2 k Ω
+
VCC = +15V
−
Vout
−
−
+
VEE = +15V
( note polarity!)
−
Test
Figure 3: Non-inverting amplifier experimental setup
supply, that is, one with output voltages of both +15 V and -15 V is desirable. The integrated circuit
operational amplifier, an LF 356, is fabricated within a standard 8-pin package (Fig. 4). An adjustable
input voltage, vin , is obtained from the 10 kΩ potentiometer of the circuit. The 1 MΩ resistor in series
with the input of the amplifier (pin 3 of the op amp) serves to limit the input current. An excessively large
current could occur for vin too large.
1. Assemble the circuit shown in Fig. 3 on an experimenter board, being careful to correctly wire the
2
8 7 6 5
1 2 3 4
Figure 4: Top view of “dual-inline-pin (DIP) integrated circuit package
integrated circuit. There should be no connection to pins 1, 5, and 8, which will not be used in this
experiment.
2. After turning on the supply, use a multimeter to indicate the output voltage, vout . If the circuit is
operating properly, this voltage should change as the potentiometer setting is changed. A range for
vout of at least -12 V to +12 V should occur. Obtain data to plot vout as a function of vin , using
approximately 2 V steps for vin . The multimeter probe may be moved to obtain readings of both vin
and vout . The voltage vin should be determined at the potentiometer connection not at the
operational amplifier input terminal. Be sure to obtain data for the maximum excursion in vin , that
is, for -15 V and +15 V. The input voltage, vin , is determined not only by the potentiometer setting,
but also by the input resistance of the amplifier (the resistance between pin 3 of the op amp and the
ground or common connection of the circuit). As a result of the 1 MΩ resistor, the input voltage of
the op amp will also depend on the very small input current of the amplifier.
3. With the multimeter connected so as to indicate vout , adjust the potentiometer to result in a voltage
of approximately 5 V for vout . Observe the effect of placing a short circuit across the 1 MΩ resistor
of the circuit. If vout changes, record its new value. If the amplifier input current is truly zero, the
voltage across the 1 MΩresistor will be zero and vout will not be affected by the short circuit.
4. A determination of the dependence of the output voltage on the load resistance, RL , is desired. With
the short circuit removed, adjust the potentiometer for an output voltage, vout , of approximately 5
V. Record this voltage. Remove the 2.2 kΩ load resistor and again record the voltage (RL = ∞).
Determine vout for resistance values of 100 kΩ, 10 kΩ, 1 kΩ, and 470 Ω for RL . Be careful not to
disturb the potentiometer setting while changing resistors.
5. By altering the feedback path of the circuit (the connection from the output of the op amp to its
minus input), a voltage gain may be achieved. Assemble the resistance network as shown in Fig. 5
for this part of the circuit (the potentiometer and the power supply connection should not be
altered). The original 2.2 kΩ resistance should be used for the load resistance. Determine, using
the multimeter, the dependence of vout on vin for this circuit. Approximately 1 V increments are
appropriate for vin . If the circuit is working properly, vout should be twice vin , except for an input
voltage which results in an excessively large value of vout . Replace R2 with a 100 kΩ resistor and
repeat the measurements. Since a relatively small value of vin will result in an output voltage which
is limited by the amplifier, quite small increments for vin will need to be used for this measurement.
The amplifier will also function with a varying (time-dependent) input voltage. For an input voltage
that does not vary too rapidly, the output voltage will be a “magnified” version of the input voltage.
6. The dynamic behavior of the circuit for a varying input voltage will be ascertained for the 100 kΩ
feedback resistance, R2 . Remove the potentiometer circuit and connect the output of a laboratory
function generator directly to the input of the amplifier circuit, as shown in Fig. 6. Set the generator
for a frequency of 1 kHz (sinusoidal) and use a dual-trace oscilloscope display to simultaneously
observe vin and vout . A time base of .2 ms/cm and internal triggering are convenient for this
observation. A vertical sensitivity of 5 V/cm is desirable for the input used to observe vout and 1
3
R1
R2
10 k Ω
10 k Ω
2
−
6
3
+
+
RL
+
Vout
2.2 k Ω
Vin
−
−
Figure 5: Non-inverting circuit with gain, for section 2.5
2
Function
Generator
+
3
−
+
6
Vin
−
Figure 6: Dynamic behavior test
V/cm for vin (DC coupling for both inputs). If different graticule lines are used for the zero voltage
levels of the two inputs, the resultant display will be easier to interpret. Observe the effect of
changing both the amplitude of the function generator’s output and its offset voltage. For an
excessive amplitude or offset voltage , a limiting of the output voltage will occur. Verify that the
output voltage is indeed a magnified version of the input voltage when limiting does not occur.
Using the oscilloscope display, determine the ratio of the peak-to-peak value of vin and vout for a
function generator setting that does not result in limiting.
7. Determine the effect of changing the waveform selection of the function generator (triangular and
square) on the performance of the circuit. Observe also the effect of changing the frequency of the
function generator. The oscilloscope time base will need to be simultaneously adjusted as the
frequency is changed to maintain a convenient display. Record (qualitatively) the effect of
increasing the frequency of the function generator to its upper limit for each of the wave forms it is
capable of producing.
3 Conclusion
Data was obtained in section 2.2 of the experimental procedure for the dependence of vout on vin (unity
gain circuit). In section 2.5, a resistance feedback network was added to the amplifier circuit to increase
its voltage gain. Data was obtained for two different feedback resistances. Prepare a single graph of vout
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plotted as a function of vin for the three amplifier circuits tested. For values of |vout | less than 10 V, vout
should have been linearly dependent upon vin . The slope of the vout versus vin characteristic is the
experimentally determined voltage gain. Compare, by means of an appropriate table, the experimentally
determined voltage gains with those theoretically predicted using the idealized model for the operational
amplifier.
The effect of a large resistance in series with the input of the amplifier was determined in section 2.3 of
the experimental procedure. If it is assumed that the input of the amplifier circuit behaves as a very large
resistance, Ri , the data of this measurement may, in principle, be used to evaluate Ri . The potentiometer
may be assumed to produce a voltage that is independent of the input current of the amplifier. Hence, the
potentiometer behaves as an ideal voltage source (its Thévenin equivalent resistance may be assumed
zero).
1 MΩ
+
Pot
−
+
+
Vin
Ri
+
V’
=
’
out Vin
−
−
−
Figure 7: Equivalent circuit
The output voltages were determined for the circuit:
vout = vin 1M Ω shorted out
v0out = v0in 1 M Ω resistor in circuit
The ratio of the two measured output voltages depends on Ri :
Ri vin
Ri + 1MΩ
Ri
=
Ri + 1MΩ
v0in =
0
vout
vout
(4)
In principle, the voltage v0out should have been slightly smaller than vout . In practice, however, the
0 was greater
difference may have been smaller than that which the multimeter would indicate. If vout - vout
0
than zero, use this value to determine Ri . If the value of vout - vout was zero, assume that this voltage
could be as great as the least significant digit of the meter display. (For a 3-1/2 digit display, this is .01 V
0 small, the following approximation is convenient:
for a 20 V meter scale). For vout - vout
v0out
1
=
≈ 1 − 1MΩ/Ri
vout
1 + 1MΩ/Ri
0
vout − vout
≈ 1MΩ/Ri
vout
(5)
If the “least significant digit” is used, the value of Ri so obtained is a lower limit for Ri . For the
instruments used, it may be concluded that the equivalent input resistance is no smaller than this value.
The output terminal of the amplifier circuit may be represented by a Thévenin equivalent circuit. The
load resistance, RL , was connected to the right-hand side of the equivalent circuit. What were the values
of vout determined for an infinite load resistance and for a load resistance of 1 kΩ (section 2.4)? If
different values of vout were obtained, use the measured values to calculate a value for RT . If the
measured values of vout were identical, assume, as in the previous section, that they may have differed by
5
RT
+
+
VT
−
+
Vout
−
RL
−
Figure 8: Equivalent circuit
as much as a single digit of the indicating meter. For this case, the Thévenin equivalent resistance is no
greater than that calculated.
In sections 2.6 and 2.7 (the experimental procedure), the behavior of the amplifier circuit for a varying
input voltage was observed. Comment on the behavior of the circuit for a 1 kHz input signal. Did the
amplifier’s behavior differ from that suggested by the static measurements of the previous sections? What
was the effect of increasing the frequency of the input signal? How did the amplifier’s behavior deviate
from that observed for a non-varying input voltage? Summarize, using a few well-thought-out
paragraphs, what you observed in these two experimental sections.
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