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Capacitor i-v equations


5m read
·Nov 11, 2024

We're going to talk about the equations that describe how a capacitor works, and then I'll give you an example of how these equations work.

So, the basic equation of a capacitor says that the charge Q on a capacitor is equal to the capacitance value times the voltage across the capacitor. Okay, so here's our capacitor over here. Let's say we have a voltage on it of plus or minus V, and it has a, we say it has a capacitance value of C. That's a property of this device here.

C is equal to, just looking at the equation over there, C is equal to the ratio of the charge stored in the capacitor divided by the voltage on the capacitor. So what we mean by stored charge is if a current flows into this capacitor, it can leave some excess charge on the top. I'll just mark that with plus signs, and there'll be a corresponding set of minus charges on the other plate of the capacitor.

This collection of excess charge will be Q+ and this down here will be Q-, and they're going to be the same value. What we say here is when the capacitor is in this state, we say it's storing this much charge, and we'll just name one of these numbers here. They're going to be the same with opposite signs. So that's what it means for a capacitor to store charge.

What I want to do now is develop some sort of expression that relates the current through a capacitor to the voltage. We want to develop an IV characteristic. So this will correspond to sort of like Ohm's law for a capacitor—what relates the current to the voltage.

The way I'm going to do that is to excise this equation by causing some changes, and in particular, we'll change the voltage on this capacitor and we'll see what happens over here. So when we say we're going to change a voltage, that means we're going to create something—a condition of dV/dt, the change in voltage per change in time.

I can do that by taking the derivative of both sides of this equation here. I've already done it for this side, and over here what I'll have is dQ/dt. I took the derivative of both sides just to be sure I treated both sides of the equation the same.

Let's look at this little expression right here. This is kind of interesting; this is the change of charge with change of time, and that's equal to—that's what we mean by current. That is current, and the symbol for current is I. So dQ/dt is current, essentially by definition we give it the symbol I, and that's going to be equal to C(dV/dt).

This is an important equation; that's basically the IV relationship between current and voltage in a capacitor. What it tells us is that the current is actually proportional to, and the proportionality constant is C. The current is proportional to the rate of change of voltage, not to the voltage itself, but to the rate of change of voltage.

All right, now what I want to do is find an expression that expresses V in terms of I. So here we have I in terms of dV/dt. Let's figure out if we can express V in terms of some expression containing I. The way I do that is I need to eliminate this derivative here.

I'm going to do that by taking the integral of this side of the equation, and at the same time, I'll take the integral of the other side of the equation to keep everything equal. So what that looks like is the integral of I with respect to time is equal to the integral of C(dV/dt) with respect to time dt.

On this side, I have basically—I do something like this and I have the integral of dV. So this looks like an anti-derivative; this is an integral acting like an anti-derivative. What function has a derivative of dV? That would be just plain V.

So I can rewrite this side of the equation: the constant C comes out of the expression and we end up with V on this side, just plain V. That equals the integral of I dt. So we're partway through—we're developing what's going to be called an integral form of the capacitor IV equation.

What I need to look at next is what are the bounds on this integral? The bounds on this integral are basically minus time equals minus infinity to time equals some time T, which is sort of like the time now, and that equals capacitance times voltage.

Let me take the C over on the other side, and actually, I'm going to move V over here onto the left. Then I can write this as 1/C, this is the normal looking version of this equation. I integrate from minus infinity to time T; T, time—big T is time right now.

What this says is that the voltage on a capacitor has something to do with the summation or the integral of the current over its entire life, all the way back to T equals minus infinity. This is not so convenient. What we're going to do instead is we're going to pick a time, and we'll call it, we'll pick a time called T equals z, and we'll say that the voltage on the capacitor was equal to U, let's say V_KN with some value.

Then what we'll do is we're going to change the limit on our integral here from minus infinity to time T equals 0. Then we'll use the integral from, instead 0 to the time we're interested in. So that equation looks like this; we're just going to change the limits on the integral and we have the integral now.

But we have to actually account for all the time before T equals z, and what we do there is we just basically add V_KN, whatever V_KN is—that's the starting point at time equals z. The integral takes us from time 0 until time now; this is the integral form of the capacitor equation.

I want to actually make one more little change. This is the current at V as a function of big T. What we really want to write here is we want to write V of little t. Just stylistically, this is what we like this equation to look like.

So I want my limits on my integral to be 0 to little t, and now I need to sort of make up a new replacement for this T that's inside here. I can call it something else; I can call it tau. This is basically just a little fake variable dtau plus V_KN.

And this is now—we finally have it. This is the integral form of the capacitor equation. We have the other form of the equation that goes with this, which was I equals big C(dV/dt). So there are the two forms of the capacitor equation.

Now, I want to do an example with this one here just to see how it works when we have a capacitor circuit.

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