Physics 322, Lab 7: Op-Amps
Name:_______________
Key concepts
Follower, non-inverting amplifier, inverting amplifier, voltage summer, current-to-voltage conversion
Text Material
Diefenderfer, Sections 9-1 to 9-7
Equipment
Resistor substitution box,
function generator,
digital oscilloscope
Components needed
(1) 741 operational amplifier
(1) 100 kΩ resistor
assorted other resistors depending on student design
Some background
In Lab 6 we studied the single transistor amplifier. Although useful for simple switching and power
amplifier circuits the single transistor amplifier suffers from several disadvantages. The biggest of these is that the
gain is rather small and somewhat temperature dependent. Thus the common-emitter amplifier studied in Lab 6 is
not a good choice for an application that requires long term stability. Fortunately many years have gone by since the
invention of the transistor and during that time electrical engineers have perfected the “Operational Amplifier” (or
Op-amp for short). This is an integrated circuit device consisting of many transistor amplifiers that are cascaded
together to make an amplifier with enormous gain (106 or more!) and exceptional linearity. Furthermore, use of the
op-amp in circuits with negative feedback leads to extremely stable operation (gain is independent of temperature,
specific device characteristics, and other factors.). In today’s lab we will study 6 different op-amp circuits:
Part A: Follower
Part B: inverting amplifier
Part C: non-inverting amplifier
Part D: difference amplifier
Part E voltage summer
Part F: current-to-voltage conversion
Op-Amp Golden Rules
Op-amp circuits are easy to understand as long as you remember the four rules for an ideal op-amp as stated on
page 185 of the text:
1.)The circuitry of an op-amp operated in a closed negative feedback loop will adjust its output in any way it can in
order to make the inverting (-) and non-inverting (+) terminals of the device equal in voltage.
2.)The inputs draw no current. In other words, the input impedance is infinite.
3.)The voltage gain, (Vout/Vin), is infinite.
4.)The output impedance is zero.
Part A: Follower
A “follower” circuit is a buffer circuit (Vin=Vout) for which the
output impedance is much smaller than the input impedance. Thus it
is a circuit that provides current gain. In Lab 6 we constructed a
follower using a single transistor. We now examine a follower
constructed from an op-amp as in the schematic diagram shown here.
As you see in the diagram the op-amp requires a bipolar power
supply (both +15VDC and -15VDC with respect to ground).
Question 1:
(a)Construct the op-amp follower shown in the schematic diagram
using the pin-out given here.
(b)Drive the follower with a 100mVpp sine wave of 10kHz. Is
Vin=Vout as it should be? Verify that the amplitude and phase of the
two signals is the same.
Vin=__________________, Vout=____________________
Phase of Vout relative to Vin =____________________
Beware: Integrated circuits (IC) are
low current devices and are easily
damaged by wiring mistakes. Always
double-check your circuit before
turning on power supplies and signal
sources. If an IC gets hot chances are
you have connected power to the chip
incorrectly. Turn off the power
immediately and check your circuit!
(c)Vary the input frequency. At what upper frequency does the output drop from being equal to the input? Is there a
lower frequency limit where this also happens?
Upper frequency limit =_____________________________
Lower frequency limit = ____________________________
(d)Does the follower amplifier work for DC input signals? Feed in 1 or 2 VDC and see what happens.
Vin =______ Vout =________
Now slowly increase Vin until it exceeds the power supply voltage Vs. What
happens?
Question 2: The input impedance of the op-amp follower is very large. Convince yourself of this as follows: First
charge up a 1 uF capacitor to 5 volts and measure the voltage across the capacitor directly with a DMM. From the
rate of voltage decay you can estimate the input resistance of the DMM (It should be roughly 10 MOhms.). Now
charge the capacitor up again and measure its voltage using the follower circuit as a buffer. (That is, connect the
capacitor to the follower input and the DMM to the follower output.) Use the rate of decay of the voltage to
estimate the input resistance of the follower circuit.
Vinitial
DMM
Follower+DMM
Vfinal
∆t
RinputC
Rinput
Part B: Inverting amplifier
The op-amp can also be used to make voltage amplifiers
which have much better stability and higher gain than
the single transistor common-emitter amplifier from Lab
11. The circuit at the right is called an inverting
amplifier because Vin and Vout are of opposite
polarities for DC signals (and 180 degrees out of phase
for AC signals). It can be shown (see text) that the gain
of this amplifier is approximately:
R
Vout
Gain 
 2
Vin
R1
Question 3:
(a)Construct the inverting amplifier circuit shown above with R1=100kΩ and drive it with a 10 mVpp sine wave at
1000 Hz. Display both Vin and Vout simultaneously on your scope. What is the phase difference between the two
signals?
Phase shift = _________________________
(b)Measure the gain as a function of R1 and compare it with the theoretical gain of the circuit using the table below.
Does theory match experiment for all gains?
R1
Vin (mV pp)
Vout (mV pp)
Observed Gain
(Vout/Vin)
Theoretical Gain
(-R2/R1)
100kΩ
10 kΩ
1 kΩ
100 Ω
(c)Set R1=10kΩ, verify that the circuit is still working and then use the scope to measure the voltage between the
inverting and non-inverting input of the op-amp.
V+- (peak to peak) = __________________________
Is Golden Rule #1 being obeyed? Is the signal between the inputs comparable in size to the source signal or is it
much smaller?
Question 4:
Design a circuit with gain of -0.1. Draw the schematic here, then build it and verify that the gain is close to your
predicted value.
Predicted gain=_______________________
Actual gain = Vout/Vin =
/
=
Part C: Non-Inverting amplifier
The inverting amplifier suffers from having an input
impedance equal to the input resistor R1. The value of
this resistance is usually fairly small so as to make the
gain of the amplifier large. Thus the inverting amplifier
ends up being a significant load to a signal source. The
non-inverting amplifier shown at the right does not have
this problem. Its input impedance is the very large input
impedance of the op-amp’s FET front end. It can be
shown (see text) that the gain of this amplifier is
approximately:
R
Vout
Gain 
 1 2
Vin
R1
Question 5:
(a)Construct the non-inverting amplifier circuit shown above with R1=100kΩ and drive it with a 10 mVpp sine
wave at 1000 Hz. Display both Vin and Vout simultaneously on your scope. What is the phase difference between
the two signals?
Phase shift = _________________________
(b)Measure the gain as a function of R1 and compare it with the theoretical gain of the circuit using the table below.
Does theory match experiment for all gains?
R1
Vin (mV pp)
Vout (mV pp)
Observed Gain
(Vout/Vin)
Theoretical Gain
(1+R2/R1)
100kΩ
10 kΩ
1 kΩ
100 Ω
(c)Set R1=10kΩ, verify that the circuit is still working and then use the DMM to measure the voltage between the
inverting and non-inverting input of the op-amp.
V+- (rms) = __________________________
Is Golden Rule #1 being obeyed? Is the signal between the inputs comparable in size to the source signal or is it
much smaller?
Part D: Difference amplifier
In all of the amplifiers studied above one of the input
terminals is ground. What happens if you need to
amplify a voltage but neither of the input connections
are at ground? What you need is a difference amplifier
like that shown in the schematic diagram to the right.
This amplifier amplifies the difference between the two
inputs, V1 and V2. If R1=R2 and R3=RF then the gain
of such an amplifier is given by:
V
R
Gain  out  F
V2  V1 R1
Question 6:
Construct the difference amplifier circuit shown above with RF=100kΩ and other resistors chosen so that the
nominal gain is 10. Try feeding in different values of DC voltages V1 and V2 to see if the gain really is 10. One
easy way of obtaining your two voltages is to wire up a voltage divider between the two terminals of a DC power
supply. Use a potentiometer as one of the resistors in the divider so you can vary the output voltages.(Be sure to use
fairly large enough resistances in the divider so that the resting current isn’t too large, say 50 mA or so.) You can
use a DMM to measure V2-V1 directly. Sketch your circuit here including component values. Try at least 4
different V2-V1 values. Record your data in the table below.
V2-V1
Vout
Observed gain
Vout/(V2-V1)
Theoretical Gain
RF/R1
Part E: Voltage summer
Op-amps can also be used to perform mathematical
operations on analog signals. For example the circuit
shown here is an adder circuit. Vout is given by:
V V V
Vout  ( 1  2  3 ) RF
R1 R2 R3
Thus the output is proportional to the sum of the three
currents passing through the three input resistors. Or, if
R1=R2=R3 the output is simply proportional to the sum
of the three input voltages.
Question 7:
Construct the difference amplifier circuit shown above using 10kΩ for all resistances. Use the variable bipolar DC
voltage supply and the fixed 5 VDC power supply as voltage inputs. (Make sure the grounds of these supplies are
tied to the ground of the op-amp power supply.) Try a few different values for V1, V2, V3, and see if the output is
what you expect. Record your data below.
V1
V2
V3
Observed output
Expected output
-(V1+V2+V3)
Part F: Current to voltage conversion
Say you have a photodiode circuit that outputs a current to
ground that depends on the light intensity hitting the diode.
You need some way of converting the diode’s output current
to a voltage so you can measure it with a voltmeter. You
don’t want to simply use a resistor since this would
introduce a voltage drop in the circuit which would, in turn,
affect the diode current. The circuit shown on the right will
produce an output voltage which is proportional to the
current flowing into the input terminal without altering your
diode circuit in any significant way (no voltage drops!).
Vout  Iin RF
Question 8:
Construct the current to voltage converter shown above starting with RF= 1kΩ. Construct a simple current source
using a DC power supply and variable resistor (potentiometer). Place a DMM ammeter in the circuit to measure the
current into the inverting input of the op-amp. Try several different values of RF and input current and check to see
if Vout  Iin RF . Record your data below.
Iinput
RF
Observed Vout
Theoretical Vout = IinRF