18 November 2016
is that when there is negative feedback, they work very
hard to keep the + and - inputs at the same voltage. In the
case of this circuit, the + input is at circuit ground potential
(let’s say zero volts relative to the power supplies), so the
op-amp will work to keep the - input at that same potential.
As we increase the input voltage, the current flowing
through the capacitor will be the time derivative (time rate
of change) of that voltage.
That would normally cause the - input to rise, but
the op-amp wants it to go back down so that it stays the
same as the + input. In order to do that, it must match
the current flowing through the capacitor with the current
flowing through the feedback resistor by decreasing the
output voltage in proportion to the capacitor’s current.
This follows Kirchhoff’s current law if we assume that
the op-amp’s input current is negligible. So, the op-amp’s
output voltage will be proportional to the capacitor’s
current, and so it will be the negative of the derivative of
the input voltage. The constants of proportionality will be
the capacitor’s capacitance and the resistor’s resistance. It
will follow the relationship in Figure 4, written to equate
the two currents.
To do integration instead of differentiation, we reverse
the roles of the resistor and capacitor as you can see in
Figure 5. In that case, the voltage at the output of the op-amp must make the current through the capacitor match
the current flowing through the input resistor. The only
way to do that is to essentially invert the voltage-current
relationship of the capacitor. This gives us an integral
instead of a differential.
The integration is considered over some time interval
(see Figure 6 for the relationship). Note that there is a
dual (like a mirror image) of this circuit that can be made
with an inductor. Inductors have a similar time derivative
relationship to a capacitor, with the roles of voltage and
To observe this on an oscilloscope, all you need to
do is provide an input waveform. Many of the available
microcontrollers (like an Arduino) have a PWM, or pulse
width modulated output that can be used with a filter to
produce sinusoidal AC waveforms.
That’s not ideal for this application, though, since you
can’t observe other shapes and how they’re changed by
Some Arduinos also have a DAC (
digital-to-analog converter) output, so that arbitrary waveforms
can be reproduced with a table of samples. There
is an Arduino tutorial that describes a waveform
generator at https://www.arduino.cc/en/Tutorial/
It uses a Due, which is unfortunately no longer
available. There are several newer boards that will work,
though. Look at the table on their website that can be
found at https://www.arduino.cc/en/Products/Compare.
There is a column labeled “Analog In/Out.” Choose
one that has at least one output. Then, you can choose
values for R and C that will give you an observable output
using the equations provided. As for op-amps, the figures
use a TL072, but any common device you might have in
the parts drawer should work. Just give it a ±15V supply if
that’s what the device wants. NV
n FIGURE 6. Relationship for integrator circuit.
QUESTIONS and ANSWERS
n FIGURE 5. Active integrator circuit.
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n FIGURE 3. Active differentiator circuit.
n FIGURE 4. Relationship for differentiator circuit.