AC Electric Circuits

Passive Integrator and Differentiator Circuits

25 questions By Tony R. Kuphaldt

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Question 1 of 25

∫f(x) dx Calculus alert!

Calculus is a branch of mathematics that originated with scientific questions concerning rates of change. The easiest rates of change for most people to understand are those dealing with time. For example, a student watching their savings account dwindle over time as they pay for tuition and other expenses is very concerned with rates of change (dollars per year being spent).

In calculus, we have a special word to describe rates of change: derivative. One of the notations used to express a derivative (rate of change) appears as a fraction. For example, if the variable S represents the amount of money in the student’s savings account and t represents time, the rate of change of dollars over time would be written like this:

dS
dt

The following set of figures puts actual numbers to this hypothetical scenario:

Date: November 20

Saving account balance (S) = $12,527.33

Rate of spending ([dS/dt]) = -5,749.01 per year

List some of the equations you have seen in your study of electronics containing derivatives, and explain how rate of change relates to the real-life phenomena described by those equations.

Reveal answer

Voltage and current for a capacitor:

i = C dv
dt

Voltage and current for an inductor:

v = L di
dt

Electromagnetic induction:

v = N d φ
dt

I leave it to you to describe how the rate-of-change over time of one variable relates to the other variables in each of the scenarios described by these equations.

Follow-up question: why is the derivative quantity in the student’s savings account example expressed as a negative number? What would a positive [dS/dt] represent in real life?

Challenge question: describe actual circuits you could build to demonstrate each of these equations, so that others could see what it means for one variable’s rate-of-change over time to affect another variable.

Notes:
The purpose of this question is to introduce the concept of the derivative to students in ways that are familiar to them. Hopefully the opening scenario of a dwindling savings account is something they can relate to!

A very important aspect of this question is the discussion it will engender between you and your students regarding the relationship between rates of change in the three equations given in the answer. It is very important to your students’ comprehension of this concept to be able to verbally describe how the derivative works in each of these formulae. You may want to have them phrase their responses in realistic terms, as if they were describing how to set up an illustrative experiment for a classroom demonstration.
Question 2 of 25

How does the rate of charge flow (current) into and out of a capacitor relate to the amount of voltage across its terminals? How does the rate of water flow into and out of a vessel relate to the amount of water stored in that vessel?

Reveal answer

Rather than simply give you an answer here, I’ll let you figure this out for yourself. Think very carefully about the water-in-a-vessel analogy when answering this question! Fill a glass with water, if necessary, to gain an intuitive understanding of these quantities.

Notes:
The existence of such an appropriate analogy for capacitor action makes an explanation unnecessary, even if the concept takes a bit of thought to comprehend at first. It is important that students clearly distinguish the quantities of current, voltage, and charge in a capacitor circuit just as they clearly distinguish the quantities of liquid height, flow rate, and liquid volume in a hydraulic system.
Question 3 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.

Suppose we had an oscilloscope capable of directly measuring current, or at least a current-to-voltage converter that we could attach to one of the probe inputs to allow direct measurement of current on one channel. With such an instrument set-up, we could directly plot capacitor voltage and capacitor current together on the same display:

For each of the following voltage waveforms (channel B), plot the corresponding capacitor current waveform (channel A) as it would appear on the oscilloscope screen:

Note: the amplitude of your current plots is arbitrary. What I’m interested in here is the shape of each current waveform!

Reveal answer

Follow-up question: what electronic device could perform the function of a “current-to-voltage converter” so we could use an oscilloscope to measure capacitor current? Be as specific as you can in your answer.

Notes:
Here, I ask students to relate the instantaneous rate-of-change of the voltage waveform to the instantaneous amplitude of the current waveform. Just a conceptual exercise in derivatives.