# Passive integrator & differentiator : Question 3

#### sub3

Joined Sep 11, 2010
3
Hi ,
I am new to this forum and in the short duration that i have surfed through this website i have found it to be very helpful.

In Question 3 in the Passive integrator and differentiator workbook the problem statement is forcing a voltage across the capacitor and asking for the current waveforms.

This is somewhat incorrect because if we assume ideal voltage source (represented by the function generator) and ideal capacitor then there is conflict where these two connect.

Would it be appropriate to treat the input as current waveforms and plot current waveforms as output ?

#### Ghar

Joined Mar 8, 2010
655
The inconsistency shows up only if you attach a voltage source that has a different value than the voltage on the capacitor without series impedance. This creates impulses of current which take on an infinite value.
This isn't much of an inconsistency... In a real circuit there is always some impedance making true impulses not exist and the behaviour will be very similar regardless.
Impulses are just as real as ideal voltage sources, it's not really a problem.

A smoothly varying voltage applied to a capacitor will result in a current that's the derivative of that voltage (without impulses), since the voltage difference between the source and the capacitor is infinitesimal.
A voltage like that square wave results in periodic impulses of current because of the finite voltage difference in zero time.

In the real circuit there is of course impedance, function generators are actually 50 ohm outputs, and you can't ever have a voltage with zero rise/fall time.
For frequencies where 1/sC >> 50 ohms the situation is essentially the same as an ideal voltage source connected across the capacitor.

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#### sub3

Joined Sep 11, 2010
3
Yes Ghar. I agree with all you have said.

But my question is bit academic.

Let's assume that the voltage/signal generator is ideal voltage source (no source impedance) and all the cables are perfect. There is no initial charge on the capacitor.

Also lets assume that the capacitor is perfect.

An ideal voltage source will tend to impose the voltage on the capacitor instantaneously. However, the capacitor cannot let that happen (assume the other side of capacitor is a ideal ground).

The capacitor voltage should slowly rise. But because the source is ideal that cannot happen.

This is what is was trying to say that there is a conflict. So to explain the i=c dv/dt i thought it would be easier if we did not use a ideal voltage source and instead use a ideal current source.

But i guess, if the current has a triangular of pulse shape we really have di/dt so i guess that does not help explain the operation of a capacitor.

Is this why almost all explanations use a voltage source.

Is it possible to frame and explain the problem using a ideal current source ?

#### Ghar

Joined Mar 8, 2010
655
The academic explanation is that for a change in voltage on a capacitor in zero time you require an impulse of current with infinite amplitude but finite energy.

If you apply a sinusoid which starts at zero to an uncharged capacitor there is no impulse and you merely get i = C*dV/dt which is another sinusoid. There is nothing 'weird' with it.

You are correct, most academic explanations avoid all this and just work with a current source. There isn't really a problem though, they just don't like impulses and it is easier to work with a current source because of it.

Similarly they excite inductors with voltage sources to avoid the same problem. But again, it's not really a problem. Personally I think that approach of avoiding impulses obfuscates the devices, making them seem more mysterious than they are. Circuit theory still works when applying a voltage source to a capacitor, you just need to work with impulses.

In reality you can very easily put a voltage source on a capacitor, you just need to be aware that it might demand large current spikes which is exactly what circuit theory would say.

Basically all of this confusion comes from the fact that instructors prefer integrals because you can easily integrate discontinuous functions.

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• sub3