Introduction to
Op-Amps
Swarthmore College
Engineering 72: Electronic Circuit Application
Lab # 1
September 14, 2004
Heather Jones & David Luong
Professor Erik Cheever
Procedure
The procedures can be found at the following web address:
http://www.swarthmore.edu/NatSci/echeeve1/Class/e72/E72L2/Lab2(OpAmp).html
Analysis
1. Op-Amp Circuits
a.
a. Inverting Amplifier
Schematic 1
The internal resistances between the inputs (v+ and v-) and ground are infinite. Since v+is
v
grounded, v - is also grounded. As a result, the current into the node at v – is simply in
R2
v
by Ohm’s Law. The current across the R1 resistor is also in . By voltage division,
R2
v
vout   in R1 .
R2
So the gain of the circuit is
vout
R
 1 .
vin
R2
Figure 1—Inverting Amplifier with R1 = 10kOhms
Figure 2—Inverting Amplifier with R1 = 30kOhm
b.
Non-Inverting Amplifier
Schematic 2
The node between R1 and R2 is vin since v+ = v-. Using a voltage divider,
R1
vin  vout
R1  R2
Rewriting,
vout
R
 1 1 .
vin
R2
A Multisim produced schematic and simulation are shown below.
VCC
12V
XFG1
7
1
5
U1
3
6
R1
2
1.00kOhm_1%
LF411CH
4
V1
12 V
R2
10.0kOhm_1%
XSC1
G
T
B
Schematic 3
Figure 3—Multisim Simulation
A
c. Ideal Integrator
Schematic 4
Assume that R1 has no effect on the circuit. Since this is essentially an inverting
amplifier,
vout
Z
 c ,
vin
R2
where Z c is the impedance of the capacitor. It is equal to
1
vout
sC
1
 1 
.
vin
R2
sC1 R2
1
. Rewriting,
sC1
Figure 4 –Ideal Integrator with R1 having infinite resistance
Figure 5–Ideal Integrator with R1 = 1MOhm
2. The resistor labeled R1 in the above diagram allows the integrator to deal with a DC
signal. When this resistor is removed, the gain of the circuit is Vo(s)/Vi(s) = -1/(RsC),
which goes to infinity as s goes to zero, so any DC signal will cause the amplifier to
saturate.
3. Our calculated peak-to-peak value was 500 mV. Referring to the vout/vin relationship
of the integrator circuit, we have values for the capacitor and resistor. To determine vout,
we place vin on the right side of the equation, and replace it with its Laplace equivalent.
Knowing that 1/s^2 is merely –ω^2, vout becomes 1/( ω^2*C1*R2). The time domain
equivalent is vout(t) = 1/(C1*R2)*t. Noting that the input frequency is 1 kHz, it takes
half that or .5 kHz to reach a peak. Substituting that and values for C1 and R2, vout(t) =
500 mV. Our measured value was 660 mV. Below is a Multisim simulation of the
integrator circuit.
Figure 6—Multisim Simulation
4. We measured a slew rate of 11.69 V/s for the 411 op-amp, which is close to the rate
of 10 V/s listed in the manufacturer’s specifications.
5. An amplifier with a gain of one can be used as a buffer to transfer a voltage between
circuits where a direct connection would have allowed current to flow and changed the
voltage.
6. In the comparator circuit, if the voltage at the inverting input terminal is greater than
the voltage at the non-inverting input terminal, there will be a closed switch inside the
comparator, connecting the output to ground. In this case, the resistor R3, which is
connected between Vcc and the output, prevents a short circuit between Vcc and ground.
When the voltage at the non-inverting input terminal is greater than that at the inverting
terminal, the switch is opened, and now no current can flow through R3, and the voltage
at the output is pulled up to Vcc.
7. Schmitt Trigger
Schematic 5
a. The circuit looks at the inputs at v+ and v- and compares them. The circuit
creates two different thresholds, and checks the number of times the input
signal crosses the thresholds. Depending on the threshold that is crossed
(either high or low), the output changes states—either high or low.
Specifically, when the input crosses the high threshold the output is low.
When it crosses the low threshold, the output is high.
b. When v- is greater than v+, the switch in the circuit, which is connected to
ground and the output, is closed. That makes vout = 0 volts, and by voltage
R1 k 
division, v  5
. [R1 is in parallel with R3]. For R1 = R2 =
R1 R3  R2
R3 =10 k , v =5/3 volts. When v- is less than v+, the switch is opened.
R1
By voltage division, v  5
=2.5 volts. Also vout is also 2.5 volts
R1  R2
because no current can flow through R3 as there is no place for it to go.
c. From part b, the lower threshold output voltage is 0 volts, and the high is
2.5 volts.
d. See the printout below.
Figure 7—Oscilloscope Output
8. For the comparator and Schmitt trigger circuits, our output could range from 0-5V,
with threshold voltages at 2.5V or 1.67V. To get an input that was centered near this
threshold band, and would thus cross through it periodically, we had to add in a DC offset
in the function generator. To see this offset on the oscilloscope, we had to turn on DC
coupling (AC coupling would have subtracted out the signal mean and shown the signal
centered around 0V).
9. Relaxation Oscillator
Schematic 6
Figure 8-Oscilloscope Output
When the output is high, that means that v+ is greater than v- and the switch is open.
This charges up the capacitor and thus increases v- until it becomes greater than v+. This
causes the switch to close, pulling the output to a lower voltage than v-, which in turn
causes the capacitor to discharge around the feedback loop until v+ is greater than vonce again. This condition causes the switch to open, and the process repeats, making the
circuit oscillate. The shape of the output is trivial—it is either high or low depending on
the relative magnitudes of the inverting and non-inverting inputs. V+ and v- increase and
decrease over time due to the oscillatory nature of the circuit as evidenced by the step
triangle waves. The measured frequency of oscillation is 115 kHz, and our calculated
one is 1/(R3*C) = 83 kHz.
Figure 9-Multisim Simulation