So the formula for charging a capacitor is:

$$v_c(t) = V_s(1 – exp^{(-t/\tau)})$$

Where \$V_s\$ is the charge voltage and \$v_c(t)\$ the voltage over the capacitor.

If I want to derive this formula from 'scratch', as in when I use Q = CV to find the current, how would I go about doing that?

Same with the formula for discharge:

$$V_c(t) = V_s \cdot e^{(-t/\tau)}$$

## Best Answer

^{simulate this circuit – Schematic created using CircuitLab}It's a pretty straightforward process. There are three steps:

Let's go through this. Instead of using an actual step function, I'm going to use a DC input and assume the capacitor starts out discharged. First, you write a KVL equation:

$$V_i = v_R + v_C$$

In circuit analysis, we like to use current instead of charge. So instead of \$Q = CV\$, we use \$i = C \frac {dV}{dt}\$. The resistor and capacitor share the same current, so:

$$i_R = i_C = C \frac {dv_C}{dt}$$

You can put this into the KVL equation:

$$v_R = Ri_R = Ri_C = RC \frac {dv_C}{dt}$$ $$V_i = RC \frac {dv_C}{dt} + v_C$$

This is a first-order linear differential equation. Using, some algebra, you can rearrange it into a solvable form:

$$RC \frac {dv_C}{dt} = V_i - v_c$$ $$\frac {dv_C} {V_i - v_C} = \frac {dt} {RC}$$

Integrating both sides gives:

$$-\ln (V_i - v_C) = \frac t {RC} + C_0$$

You can get rid of the \$\ln\$ by moving the negative sign and making both sides a power of \$e\$:

$$V_i - v_C = e^{-t/RC + C_0} = e^{-t/RC}e^{C_0}$$

\$C_0\$ is a constant of integration, so \$e^{C_0}\$ is also a constant. Let's rename it to \$C_1\$ for convenience:

$$V_i - v_C = C_1e^{-t/RC}$$

The differential equation is solved, but there's still an unknown (\$C_1\$). You can find its value if you know the initial condition of the circuit. In this case, I said that the capacitor started out discharged (\$v_C = 0\$ at \$t = 0\$), so let's use that:

$$V_i - 0 = C_1e^{-0/RC} = C_1 \cdot 1$$ $$C_1 = V_i$$

Now you can find the fully-solved equation:

$$V_i - v_C = V_ie^{-t/RC}$$ $$v_C = V_i - V_ie^{-t/RC}$$ $$v_C = V_i(1 - e^{-t/RC})$$

Is this correct? At \$t=0\$, you have:

$$v_C = V_i(1 - 1) = 0$$

and at \$t=\infty\$, you have:

$$v_C = V_i(1 - 0) = V_i$$

So the capacitor starts out discharged, ends up fully-charged, and in between there's an exponential decay. That's correct!

For the discharge, \$V_i = 0\$ and the initial condition is that the capacitor is charged to a nonzero value, which I'll call \$V_0\$. You can use these to solve for \$C_1\$ again:

$$0 - V_0 = C_1e^{-0/RC}$$ $$C_1 = -V_0$$ $$-v_C = -V_0e^{-t/RC}$$ $$v_C = V_0e^{-t/RC}$$