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# Capacitor i-v equation in action

Demonstrates the capacitor i-v equation by deriving the voltage on a capacitor driven by a current source. Written by Willy McAllister.
The capacitor is one of the ideal circuit elements. Let's put a capacitor to work to see the relationship between current and voltage. The two forms of the capacitors's i-v equation are:
i, equals, start text, C, end text, start fraction, d, v, divided by, d, t, end fraction v, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, start subscript, 0, end subscript, start superscript, T, end superscript, i, start text, d, end text, t, plus, v, start subscript, 0, end subscript
start text, C, end text is the capacitance, a physical property of the capacitor.
start text, C, end text is the scale factor for the relationship between i and d, v, slash, d, t.
start text, C, end text determines how much i gets generated for a given amount of d, v, slash, d, t.
v, start subscript, 0, end subscript is the initial voltage across the capacitor, at t, equals, 0.
In this article we'll work with the integral form of the capacitor equation. Our example circuit is a current source connected to a 1, mu, start text, F, end text capacitor.

## Voltage before, during, and after the current pulse

Suppose we apply a 2, start text, space, m, A, end text pulse of current to the 1, mu, start text, F, end text capacitor for 3 milliseconds. We'll assume the initial voltage across the capacitor is zero.
$i(t)=\begin{cases}2 \text{ mA}&; ~~0 \lt t \lt 3\,\text{ms} \\ 0&; ~~ \text{at other times}\end{cases}$
start text, C, end text, equals, 1, mu, start text, F, end text
v, start subscript, 0, end subscript, equals, 0
What is the capacitor voltage, v, left parenthesis, t, right parenthesis?
We use the integral form of the capacitor equation to solve for v, left parenthesis, t, right parenthesis in three separate chunks: before, during, and after the current pulse.

### Before the pulse

Before the current pulse left parenthesis, t, is less than, 0, right parenthesis, no current is flowing, so no charge accumulates on start text, C, end text. Therefore, v, start subscript, left parenthesis, t, is less than, 0, right parenthesis, end subscript, equals, 0. We didn't even have to use the equation.

### During the pulse

For any time T during the current pulse left parenthesis, 0, \lt, t, \lt, 3, start text, m, s, end text, right parenthesis, charge accumulates on start text, C, end text and the voltage rises. We can apply the capacitor equation to find out how v changes,
v, left parenthesis, T, right parenthesis, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, start subscript, 0, end subscript, start superscript, T, end superscript, i, start text, d, end text, t, plus, v, start subscript, 0, end subscript
Since i is constant during this time, we can take it outside the integral. We can also ignore v, start subscript, 0, end subscript, since it's zero.
v, left parenthesis, T, right parenthesis, equals, start fraction, i, divided by, start text, C, end text, end fraction, integral, start subscript, 0, end subscript, start superscript, T, end superscript, start text, d, end text, t
v, left parenthesis, T, right parenthesis, equals, start fraction, i, divided by, start text, C, end text, end fraction, t, vertical bar, start subscript, 0, end subscript, start superscript, T, end superscript
v, left parenthesis, T, right parenthesis, equals, start fraction, i, divided by, start text, C, end text, end fraction, T, start text, v, o, l, t, s, end text
This is the equation of a line with slope i, slash, start text, C, end text, valid any time during the current pulse. The slope is:
start fraction, i, divided by, start text, C, end text, end fraction, equals, start fraction, 2, times, 10, start superscript, minus, 3, end superscript, start text, A, end text, divided by, 1, times, 10, start superscript, minus, 6, end superscript, start text, F, end text, end fraction, equals, 2000, start text, v, o, l, t, s, slash, s, e, c, o, n, d, end text
By the end of the pulse, T, equals, 3, start text, m, s, end text, the voltage across the capacitor rises to:
v, start subscript, left parenthesis, T, equals, 3, start text, m, s, end text, right parenthesis, end subscript, equals, 2000, start text, v, o, l, t, s, slash, s, end text, dot, 0, point, 003, start text, s, end text, equals, 6, start text, v, o, l, t, s, end text

### After the pulse

After the pulse left parenthesis, 3, start text, m, s, end text, \lt, t, right parenthesis - The current falls to 0, so charge stops accumulating on the capacitor. Since no charge is moving, we should expect the voltage not to change. We can confirm this by applying the capacitor equation at starting time t, equals, 3, start text, m, s, end text, and starting voltage v, start subscript, 3, start text, m, s, end text, end subscript, equals, 6, start text, V, end text.
v, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, start subscript, 3, start text, m, s, end text, end subscript, start superscript, T, end superscript, 0, start text, d, end text, t, plus, 6, equals, 6, start text, v, o, l, t, s, end text
The current has stopped, so the charge stays put, and the capacitor voltage remains at 6, start text, V, end text.
Assembling the three chunks together gives us v, left parenthesis, t, right parenthesis,
Try it yourself. Adjust the size and duration of the current pulse (start color #1fab54, start text, g, r, e, e, n, end text, end color #1fab54 dot).
• How many ways can you achieve an ending voltage of 4, start text, V, end text?
• What happens to v, left parenthesis, t, right parenthesis if the current pulse goes negative?
This circuit configuration (current source driving a capacitor) has a nickname, it is called an integrator.