RC Differentiator: What is Physically Happening?

[dsdsuster]

Hi guys,

I have a quick question I think. Let's consider an RC differentiator driven by a 0-5V square wave which is initially at 0V. When the input voltage suddenly makes the transition to 5V one side of the capacitor will jump to 5V. I've read that the other side (side with the resisitor) will follow, the explanation being that the capacitor cannot acquire a voltage difference instantly.

Can anyone explain what is physically happening in this case? My thoughts are that since electric field lines flow out of positive charge, and both sides of the capacitor suddenly jump from 0 to 5V, positive charge is simultaneously rushing in on both sides. Not sure why this would happen.

Thank You

 

[rude man]

Hi guys,

I have a quick question I think. Let's consider an RC differentiator driven by a 0-5V square wave which is initially at 0V. When the input voltage suddenly makes the transition to 5V one side of the capacitor will jump to 5V. I've read that the other side (side with the resisitor) will follow, the explanation being that the capacitor cannot acquire a voltage difference instantly.

Can anyone explain what is physically happening in this case? My thoughts are that since electric field lines flow out of positive charge, and both sides of the capacitor suddenly jump from 0 to 5V, positive charge is simultaneously rushing in on both sides. Not sure why this would happen.

Thank You

The input is applying +5V to both sides of the capacitor. There is an electric field between either side of the capacitor and ground, but that is not associated with charge in the the capacitor. It's associated simply with the source voltage.

You are right, the output follows the input instantaneously if the step voltage is infinitely fast (zero rise time). And it's because the capacitor needs time to change its voltage and the current is limited by the output resistor.

 

[dsdsuster]

Thanks rudeman,

Can you clarify what you mean by this?

The input is applying +5V to both sides of the capacitor.

I understand that this is the ultimate effect, but the way I'm thinking of things right now is that the plus side of the battery goes from 0-5V and the minus goes from 0-0V.

In the short time frame when both terminals rise to 5V, I'm curious about what physically, in terms of charge and electric fields, will cause the negative terminal of the capacitor to change voltage since only apparently the positive side is receiving charge.

 

[rude man]

There is no E field in the capacitor at t = 0+ since the capacitor has not had time to charge at all. So the integral of the E field times any distance between the two capacitor nodes = 0 at t = 0+. But that integral is also the potential difference between the two plates. So both plates must be at 5V.

The capacitor voltage does not change at t = 0+. It's zero.

Replace your C with an R and your R with an L. This is also a 'differentiator' (I prefer to call it a high-pass network. Neither circuit is actually a pure differentiator). Again, when the voltage step is applied, both sides of the resistor go to +5V instantly even though there is at that instant no current through the resistor.

 

[CWatters]

What Rudeman said

It takes time to charge a capacitor so when a step change in voltage is applied to the circuit there is no corresponging step change in the voltage drop across the capacitor, that stays at whatever it was before, in this case 0V.

The resistor behaves differently. There is no limit as to how fast the voltage drop across the resistor can change (at least for the purposes of understanding this problem).

So the step change appears across the resistor.