capacitorchargeformula-derivation
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:
$$V_c(t) = V_s \cdot e^{(-t/\tau)}$$
My contribution is to point out a circuit that suits your title: "A path for capacitor's charging, and another for discharging it". It is a solution commonly used to drive a N-channel mosfet/IGBT in the configuration high-side (load grounded). This avoids the use of P-channel mosfet, typically showing higher RDSon. But otherwise imposes a complication when designing N-channel gate drivers (except using a gate voltage larger than the drain voltage - a rare situation). The scheme shown below uses a capacitor as a "floating" gate voltage for the Mosfet. This solution is already integrated in most commercial gate drivers (pick a specific datasheet and note the presence of the diode and capacitor around the component). Motor control and power conversion are typical applications.
Note: When I started to answer, the author had not yet supplemented the question with the last figure (I took some time drawing my figures). The web page no mantain updated, when editing. I write this so the reader does not think I simply repeated that idea. It happens on forums.
The more general form of the first equation is
$$v_C(t) = v_C(0)\cdot e^{\frac{-t}{RC}}$$
This must be so since, for \$t=0\$
$$e^{\frac{-0}{RC}} = 1$$
So, assuming the capacitor has charged up to the source voltage before you disconnect the source at \$t=0\$, we have
$$v_C(0) = V_S$$
and then
$$t(v_C) = - RC \ln \frac{v_C}{V_S} = RC \ln \frac{V_S}{v_C} $$
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:
$$0 - V_0 = C_1e^{-0/RC}$$ $$C_1 = -V_0$$ $$-v_C = -V_0e^{-t/RC}$$ $$v_C = V_0e^{-t/RC}$$