AC Electric Circuits

Passive Integrator and Differentiator Circuits

25 questions By Tony R. Kuphaldt

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Question 4 of 25

∫f(x) dx Calculus alert!

According to the “Ohm’s Law” formula for a capacitor, capacitor current is proportional to the time-derivative of capacitor voltage:

i = C dv
dt

Another way of saying this is to state that the capacitors differentiate voltage with respect to time, and express this time-derivative of voltage as a current.

We may build a simple circuit to produce an output voltage proportional to the current through a capacitor, like this:

The resistor is called a shunt because it is designed to produce a voltage proportional to current, for the purpose of a parallel (“shunt”)-connected voltmeter or oscilloscope to measure that current. Ideally, the shunt resistor is there only to help us measure current, and not to impede current through the capacitor. In other words, its value in ohms should be very small compared to the reactance of the capacitor (R_shunt << X

C).

Suppose that we connect AC voltage sources with the following wave-shapes to the input of this passive differentiator circuit. Sketch the ideal (time-derivative) output waveform shape on each oscilloscope screen, as well as the shape of the actual circuit’s output voltage (which will be non-ideal, of course):

Note: the amplitude of your plots is arbitrary. What I’m interested in here is the shape of the ideal and actual output voltage waveforms!

Hint: I strongly recommend building this circuit and testing it with triangle, sine, and square-wave input voltage signals to obtain the corresponding actual output voltage wave-shapes!

Reveal answer

Follow-up question: given that R_shunt << X

C in order that the resistance does not impede the capacitor current to any significant extent, what does this suggest about the necessary time-constant (τ) of a passive differentiator circuit? In other words, what values of R and C would work best in such a circuit to produce an output waveform that is as close to ideal as possible?

Notes:
This question really is best answered by experimentation. I recommend having a signal generator and oscilloscope on-hand in the classroom to demonstrate the operation of this passive differentiator circuit. Challenge students with setting up the equipment and operating it!
Question 5 of 25

Generally speaking, how many “time constants” worth of time does it take for the voltage and current to “settle” into their final values in an RC or LR circuit, from the time the switch is closed?

Reveal answer

If you said, “five time constants’ worth” (5 τ), you might not be thinking deeply enough! In actuality, the voltage and current in such a circuit never finally reach stable values, because their approach is asymptotic.

However, after 5 time constants’ worth of time, the variables in an RC or LR circuit will have settled to within 0.6% of their final values, which is good enough for most people to call “final.”

Notes:
The stock answer of “5 time constants” as the amount of time elapsed between the transient event and the “final” settling of voltage and current values is widespread, but largely misunderstood. I’ve encountered more than a few graduates of electronics programs who actually believe there is something special about the number 5, as though everything grinds to a halt at exactly 5 time constants worth of time after the switch closes.

In reality, the rule of thumb of “5 time constants” as a settling time in RC and LR circuits is an approximation only. Somewhere I recall reading an old textbook that specified ten time constants as the time required for all the variables to reach their final values. Another old book declared seven time constants. I think we’re getting impatient as the years roll on!
Question 6 of 25

Suppose a fellow electronics technician approaches you with a design problem. He needs a simple circuit that outputs brief pulses of voltage every time a switch is actuated, so that a computer receives a single pulse signal every time the switch is actuated, rather than a continuous “on” signal for as long as the switch is actuated:

The technician suggests you build a passive differentiator circuit for his application. You have never heard of this circuit before, but you probably know where you can research to find out what it is! He tells you it is perfectly okay if the circuit generates negative voltage pulses when the switch is de-actuated: all he cares about is a single positive voltage pulse to the computer each time the switch actuates. Also, the pulse needs to be very short: no longer than 2 milliseconds in duration.

Given this information, draw a schematic diagram for a practical passive differentiator circuit within the dotted lines, complete with component values.

Reveal answer

Did you really think I would give you the component values, too? I can’t make it too easy for you!

Challenge question: An alternative design to the differentiator circuit shown above is this:

This circuit would certainly work to create brief pulses of voltage to the computer input, but it would also likely destroy the computer’s input circuitry after a few switch actuations! Explain why.

Notes:
The behavior of a differentiator circuit may be confusing to students with exposure to calculus, because the output of such a circuit is not strictly related to the rate of change of the input voltage over time. However, if the time constant of the circuit is short in comparison to the period of the input signal, the result is close enough for many applications.