What happens to the capacitor in RC circuit when square wave flips to negative

[markusz]

Hi experts,

Sorry for the (maybe) confusing title. Let me show you the circuit:

As you can see this is a simple RC circuit, the time constant is big so the capacitor never fully charges. What I'm trying to understand is what happens when square wave flips from +5V to -5V? It seems to accelerate the discharging.

The formulas I evaluated with differential equation are only for charging and discharging. AFAIK, as long as the RC keeps the same, charging from 0 to say 63% of Vs should take the same time of discharging from 63% of Vs to 63% * 27* of Vs, that is, around 1RC. So it really catches me off when the oscilloscope shows that it charges to some 3.5+V in half of the period but discharges to 0 in much shorter time.

Now that I re-think the problem, actually, if I simply discharge the capacitor with NO VOLTAGE applied at all, technically it should take forever to discharge it to 0. So the negative voltage must be PULLING it to zero, and then charge it from the other side.

I guess I answered my own question, but the next question is: How do I calculate how much time it takes a capacitor that has a Vc charge to drop to 0 under a negative voltage -Vs?
 

**Edit**
I thought about the second question for a bit and figured I should be able to calculate by using this model:
Say that the capacitor is already been charged to 63% * Vs, and then we apply -Vs to it. So this is actually discharging + charging from the other side. My reasoning: If it's just discharging, it won't drop to 0 forever (in ideal world); If it's just charging from the other side, "charging" 63% * Vs to 0 (i.e. a negative charge) should take exactly the same time, and it doesn't match the chart. So it must be discharging AND charging from the other side.

[IanB]

The voltage on the top plate of the capacitor (between R and C) will follow the source voltage with a first order lag having a time constant determined by R and C. If R and C do not change, the time constant will be fixed.

If we let V_C be the capacitor voltage, then the rate of change of V_C at any instant will depend on the difference between V_C and the source voltage V_S, and on the time constant (on the values of R and C).

[markusz]

Ahhh! So this is where I got wrong, I don't know Vs is actually (Vs-0), and now Vs = -Vs, the difference is -2Vs. And basically the building up of opposite charges is a lot faster due to that.

[Capernicus]

if you have a small time constant it counts as as ac resistor.   except it stays cool and doesnt heat up.   

[Terry Bites]

The circuit is measuring the integral of the applied voltage.


You can also visualise what's happening in the frequency domain. A squarewave is the sum of odd numbered sinewave harmonics. The RC circuit attenuates the higher harmonics and alters their phase relationships to create a triangle wave.

[TimFox]

Very basic explanation:  you cannot change the voltage across a capacitor instantaneously, nor can you change the current through an inductor instantaneously.
Past that, use the mathematical expressions.

[Terry Bites]

This is beginners forum.