There
is only one operational amplifier in the student edition PSpice library,
but it is the very popular 741
type.The PSpice part number is uA741.The
input pins are 3 (+) and 2 (-), the output pin is 6, and pins 4 and 7 must
be connected to power supplies, typically –15 and +15 volts as shown in
Figure 1.Pins 1 and 5 can be ignored
for this experiment.If you want
to arrange the symbol so that the inverting input is uppermost, you can
do so by ctrl-R, ctrl-R, ctrl-F (Rotate, Rotate,
Flip.)
Figure
1
1.Inverting
amplifier
One of the most common op amp circuits is the simple inverting amplifier.Build the circuit of Figure 2, choosing values such that: (1) R1, the input impedance to the amplifier, is as large as possible, (2) the feedback resistor Rf is no larger than 270K, and (3) the output is as large as possible without distortion.The power supplies to the op amp must be +15 and –15 volts.The input comes from a sinusoidal voltage source Vsin of frequency 1 KHz, amplitude 1 volt, and having a Thevenin equivalent resistance Rth of 600 Ohms.
Figure 2
Unlike
idealized op amps, the 741 op amp does not allow the output voltage to
reach the “rails” of +15 and –15 volts.What
is the largest output amplitude you can obtain, without distorting the
sinusoid?(Hint: overdrive the amplifier
to discover the largest amplitude, even with distortion, that you can obtain
at the output.Measure it with the
cursor.)Calculate the voltage gain
required to obtain that amplitude from the 1-volt input specified, then
calculate the circuit values to obtain that gain.When
your circuit works, get a hard copy of the PSpice plot of the output.
2.Non-inverting
amplifier
In
the circuit of Figure 3, the 100mV, 50Kohm source shown represents a small,
high-impedance input which is to be amplified to 10 Volts. With high-impedance
sources such as this, the non-inverting op amp circuit of Figure 3 is used
because of its high input impedance.Assuming
that the op amp is ideal and using 100K for R4, calculate the
value of R5 to obtain the 10 volt output.Build
the circuit and determine the output voltage.You
will find that it differs somewhat from the theoretical result.Adjust
the value of R5 to get an output of 10 volts with this realistic
op amp.
Figure 3
3.Integrator
The circuit in Figure 4 is the standard op amp integrator circuit.Ideally, the output is –(1/RC) times the integral of Vin(t).In practice, a large resistor is usually placed across the capacitor to avoid drifting into saturation, but we will omit that large resistor here.Determine what the output should be if the input Vin(t) is as shown in Figure 5, assuming the capacitor starts off uncharged and that the RC product is equal to 1.In PSpice, the initial voltage condition on the capacitor is done by selecting the capacitor and using Edit/Attributes to set IC to 0.
The voltage source shown is the VPWL part (PieceWise Linear Voltage source), and to create the input voltage shown, click on it to turn it red, then use Edit/Attributes to provide values for the Time-Voltage pairs which define the signal: T1=0, V1=0;T2=1, V2=2;T3=2, V3=2;T4=2.001, V4= -1, and so on.Note that the abrupt drop near t = 2 must be given a nonzero time interval, since no transitions can occur in zero time (either in PSpice or in reality).
Some practical issues arise in this circuit.Build
the circuit, using an RC product equal to 1, and run the simulation to
verify your calculation.Try using
R = 1 and C = 1 and observe the output.Does
circuit integrate well with these values?This
combination is impractical because C = 1 is impossibly large.Try
using R = 1012 and C = 10-12.Does
the circuit integrate well with these values?This
combination is impractical because the resistive value is impossibly large
(far larger than the input resistance of the op amp.)Find
what you consider to be a practical combination of R and C (with the RC
product equal to 1) for your circuit, use it in your simulation, and get
a hard copy of a plot showing the input and output.

Figure 4
Figure 5: Vin(t) to the integrator circuit of Figure 4
4.Low
Pass Filter
The circuit of Figure 6 is a low pass filter designed
to “pass” low frequencies and attenuate high frequencies.An
ideal low pass filter would have an output which was a perfect replica
of the input if the frequency of the input sinusoid was below the “cutoff
frequency”, and an output of zero if the input frequency was above the
cutoff, though realistic filters do not have such an ideal cutoff.To
test the behavior of this active filter, use the VSIN sinusoidal source,
and use Edit/Attributes to set its DC offset voltage VOFF to 0, its amplitude
VAMPL to 1 volt, and its frequency FREQ to various values to see how the
filter responds at various frequencies.Simulate
the circuit using Analysis/Setup/Transient.Use
a Final Time of 6 ms and a Step Ceiling of 6 us, to get smooth graphs.Explore
frequencies from 250 Hz to 3 KHz and at each frequency you try (by modifying
the FREQ attribute of the source, then simulating the circuit) measure
the size of the output sinusoid.Use
Matlab to make a graph of this “frequency response.”The
“cutoff frequency” for a low pass filter is by convention the frequency
at which the output amplitude is 0.707 times the input amplitude.By
experimenting, discover the cutoff frequency for this active filter.
except
of course you would use your own data for the vectors frequency
and magnitude.That is the
bare minimum but you could add red circles at the data points by substituting
plot(frequency,
magnitude, frequency, magnitude, ’ro’);grid
or
you could use
semilogx(frequency,
magnitude);grid
to
plot the frequency on a semilogarithmic axis, and so on.Type
help
plot for more plotting information.
Figure
6
Extra
credit portion:
You may recall that diodes can be used to “clip” waveforms, but they do
a rather mediocre job of it.The
circuit below can be used as an “ideal clipper”.Simulate
it and get a graph of the input and output on the same axes.For
the voltage source use the VSIN part, at a low frequency (100 Hz
or lower).
Figure 7