# Sketching RC exponentials

Voltages and currents often have an exponential shape. We look at some properties of exponential curves and learn tricks to rapidly sketch accurate waveforms.

## What we are trying to show…

Suppose you have an $RC$ circuit with known $R$, $C$, and $V_o$, and a blank voltage vs time plot with labeled $t$ and $v$ axes… Can you sketch the $v(t)$ curve right off the top of your head?

You know what a falling exponential curve looks like in general. They all have a similar droopy shape. The only remaining interesting thing is to know how fast the the curve falls—how droopy is it? That’s the only missing piece of info you need if you want to sketch it.

The video goes through a sequence of pleasingly clever math tricks and observations that makes it easy to quickly draw a reasonably correct exponential curve. There is no profound or significant wisdom here, it is just a handy trick to amaze your friends.

We know the $RC$ natural response. It is an exponential,

$v(t) = V_o e^{-t/RC}$

We make some clever observations about this equation and make those observations into a handy manual technique.

Observation 1. We know the solution for the $RC$ natural response is $e^{(something)t}$. What if we take the derivative of the solution? That would give us an expression for the slope of the natural response. Taking the derivative of $e^{(something)t}$ brings the “$something$” down out in front of the exponential to become part of the leading coefficient. The slope expression is,

$\dfrac{dv}{dt} = -\dfrac{1}{RC} V_o e^{-t/RC}$

Observation 2. is based on a property of an exponent. Anything raised to the $0$ power $= 1$. If we set $t = 0$ then $e^{t/RC}$ becomes $e^0$. That makes the exponential term $e^0 = 1$. We are left with whatever coefficient was sitting in front of the exponential. If we use this idea on the expression for the slope, we get,

slope at $t = 0$ is $-\dfrac{V_o}{RC}$

The slope at $t = 0$ involves the value of the two components and the starting voltage value. That should make intuitive sense. Now we know the slope of the exponential curve at $t=0$. So what? More cleverness…

Observation 3. We take the slope we just figured out for the exponential curve and use it to draw a straight line. We are counting on the fact that we are pretty good at quickly drawing a straight line with a known slope. To draw the line we need two points. One of them is where the exponential passes through the voltage axis (where $t = 0$) and $v = V_o$. Put a dot at $[0, V_o]$.

The line going through $t = 0$ has a slope of $-V_o/RC$. How steep is that? A tiny bit of geometry tells you the line passes through the horizontal axis at $t = RC$. That is also an easy point to locate. Multiply ohms times farads to get seconds (the time constant) and put a dot at $[RC, 0]$.

Draw a line between the two points. It will have the same slope as the natural response at $t = 0$. Now the plot has a straight line , but no exponential… yet. As you can see in the video, if you hold the plot out at arm’s length, the straight line is a rough approximation to the exponential curve. But we can do better.

Observation 4. Go back to the original $v(t)$ natural response equation and plug in the time constant, $t = RC$. Compute the voltage after one time constant has passed,

$v(t=RC) = V_o e^{-RC/RC} = V_o e^{-1}$

What voltage does this represent? $e^{-1}$ is just a number. If you compute $e^{-1}$ you get roughly $0.37$. That means $v(t=RC) = V_o \times 0.37$, or saying it another way, when time advances to $t = RC$ the voltage has fallen to $37\%$ of $V_o$. Put a dot on the plot at this point, $[RC, 37\%$ of $V_o]$. This voltage is down by $2/3$ from its original value. You know the natural response curve goes through this point.

Observation 5. A third point on the natural response curve is at $t = reallylongtime$. After a really long time the voltage falls to $0$. That’s what happens to every natural response.

Sketch: Now you have three points on the plot that you know the exponential goes through. It intersects $V_o$ at $t=0$ and the slope of the exponential is exactly the slope of the line. That gives you an accurate initial steepness. Then the curve goes through the point $[RC, 37\%$ of $V_o]$, and finally you know it fades away to $0$ after a long time. This is enough to sketch a reasonably accurate version of the exponential curve.

Pretty good trick, eh? I think it’s pretty brilliant.

Created by Willy McAllister.