A capacitor integrates current

So now I have my two capacitor equations; the two forms of the equation. One is I in terms of V, and the other is V in terms of I. We're going to basically look at this equation here and do a little exercise with it to see how it works.

I'm going to draw a little circuit here, and it's going to have a current source and a capacitor. The value of the capacitor is one microfarad, and the value of our current source, we'll call it Big I, and it's going to actually look like a pulse like that. It will go from zero to 3 milliamps and then back to zero over here.

The amount of time it takes, that's going to be this time here, is going to be 3 milliseconds. The question I want to answer is what is V of T right here? We're going to use this integral equation to figure that out.

So over here is where we're going to put our answer. This is going to be T, and this will be I. Here's I, and our plot right down here will be T, and here's V of T. We'll plot I this way on here in time. It's zero, and then it goes up to some value, and then it goes over. So it's a pulse of current, and we said that that was zero. This is 3 milliamps, and this is 3 milliseconds.

Now what we want to do is we want to find V of t. One of the things we have to do, we have to make an assumption about V KN here, and we're going to assume for our problem here that V KN equals z volts. What that means is there's no charge; there's zero charge stored on this capacitor when we start the experiment.

So now let's look at three different time periods. Let's look at the period before the pulse, during the pulse, and after the pulse. So we'll break the problem into three parts. Part one is before, and we'll do that by just looking up here. Now we decided that V KN was zero in the before state, so we put a little zero there. We decided that I is zero, so that means that the term inside the integral is zero.

What that means is 0 + 0 is equal to V of T. So before the pulse, the capacitor equation tells us the voltage is zero. Okay, now let's go during the pulse. Let's do the second period of time. Let's do during the pulse.

So now we have to be a little more careful. We have V equals 1 / C integral from now time equals z to time equals sometime T. What's I of, what's I during the pulse? Well it's sitting right here. I is a constant, so we write in 3 milliamps.

Don't forget the starting voltage. What's our starting voltage? Well, we look right here and the starting voltage is 0o plus 0. So now we can solve this: V = 1 / C time 3 milliamps comes out times the integral from 0 to T of DT, and that equals, let's plug in C this time.

3 milliamps divided by 1 microfarad times what does this evaluate to? The integral from 0 to T of DT is just T. Let's do a little bit of arithmetic here to reduce this. Milliamps is 10^-3, and microfarads is 10^-6.

So let me move this up; we'll keep our plots on there. What we end up with is V equals 3 * 10^-3 * 10^-6. So that's 3 * 10^3 or 3,000 * time. And what's that? That's the equation of a line, and it has a slope of 3,000.

What's the units here? This is 3,000 volts per second. So I'll go over here and we'll sketch this in. This is going to be a line that's a straight ramp with a constant slope like that. It's going to be happening all during this pulse.

We can ask, what's this value right here? What's that voltage right there? Okay, let's work that out. T is 3 milliseconds, so let's plug in three milliseconds right here. So V at 3 milliseconds equals 3,000 * 3 milliseconds. That equals 3 * 3 is 9, and 3,000 * this is an exponent of -3, so it's going to be 9 volts.

So this value right here, right there, is nine volts. So that says the voltage on our capacitor during the pulse, during the current pulse, rises in a straight line up to 9 volts. We got that from this integral that we did. The capacitor is integrating the current, adding up the current; it's integrating this pulse to get an ever-rising voltage.

Okay, so now we've solved the capacitor equation during the pulse. Let's now go back to what happens after the pulse. We'll do that over in the corner over here. So now let's solve what happens after the current pulse. What happens to this voltage from here on? Does it go down? Does it go up? Does it go straight sideways?

Let's find out and let's use our capacitor equation to do this. So now what we're doing is we're going to define, we're going to do a new integral and we're going to start at time equals 3 milliseconds. What is our voltage at 3 milliseconds? What's V not?

So V in this case equals 9 volts. That's for this period of time after 3 milliseconds, and we'll use our equation again. V is 1 / C times the integral of I of t DT plus V KN.

Alright, so let's see. It goes from time — it goes from time 3 milliseconds to time T. So time T now is out here somewhere on the time scale, and we know the voltage right here; we filled it in. There's the voltage right there, we're going to fill that in, and we're going to work out what's the voltage out here after 3 milliseconds.

So to solve this integral, we look and see, well what's I of t? What's I of t during this time? Let's look at our chart; the pulse is at zero. Oh look, so this entire term right here is zero, and what is V KN? V KN we decided was right there; V KN is 9 volts.

That's where we started from, and so the answer here after the pulse goes away is V equals 9 volts. I can sketch that in up here, and basically it just goes straight sideways at the new voltage.

So that's how we solve a capacitor problem—a really simple one—and it happens that we were integrating a current pulse, and what we got was a voltage ramp.