RC transients

The idea

A capacitor can’t change its voltage instantly; it has to fill with charge through whatever resistance is in the way. So when you switch on an RC circuit, the voltage climbs along a curve instead of jumping. In a robot this happens every time the power supply charges the capacitor bank (DC link) in front of a motor drive through the cable resistance.

The time constant

$\tau =RC\text{(seconds)}$

Time	1τ	2τ	3τ	4τ	5τ
Charged to	63.2%	86.5%	95.0%	98.2%	99.3%

After 5τ, call it fully charged (or discharged).

Deriving the curves

This is your first engineering ODE (ordinary differential equation): its answer is a function of time, not a number.

1. Component laws: $u_R=iR$ and $i=C\frac{d{u}_C}{dt}$ (current is the rate of change of the capacitor’s charge).
2. KVL round the loop while charging from $V$: $V=u_R+u_C=RC\frac{d{u}_C}{dt}+u_C$.
3. Separate the variables and integrate from $u_C=0$ at $t=0$:

$u_C(t)=V\left(1-e^{-t/\tau }\right)i(t)=\frac{V}{R}{e}^{-t/\tau }$

Discharging from $V_0$ (no source, so $RC\frac{d{u}_C}{dt}+u_C=0$):

$u_C(t)=V_0{e}^{-t/\tau }i(t)=-\frac{V_0}{R}{e}^{-t/\tau }$

The minus sign means the current now flows the other way.

Energy

• A charged capacitor holds $E=\frac{1}{2}C{V}^2$.
• Charging it through a resistor always wastes exactly as much energy as it stores, whatever R is. Half the energy from the source becomes heat.

Capacitors combine the opposite way to resistors

• Parallel: $C=C_1+C_2$, same voltage rating. Two 500 µF, 24 V capacitors → 1000 µF, 24 V.
• Series: $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$, voltage ratings add. Two 500 µF, 24 V in series → 250 µF, 48 V.

Worked example

A 12 V supply charges a 100 µF capacitor through 47 kΩ. Find τ, the voltage after 4.7 s and after 10 s, the time to reach 9 V, and the starting current.

Answer

$\tau =47000\times 100\times 10^{-6}=4.7$ s At 4.7 s (1τ): $12(1-e^{-1})=7.59$ V At 10 s: $12(1-e^{-10/4.7})=12(1-0.119)=10.57$ V To 9 V: $9=12(1-e^{-t/4.7})$ → $e^{-t/4.7}=0.25$ → $t=4.7\ln 4=6.52$ s $i(0)=12/47000=0.255$ mA. Stored when full: $\frac{1}{2}\times 100\times 10^{-6}\times 12^2=7.2$ mJ.

Common mistakes

• Using µF without the $10^{-6}$.
• Using the charging equation for a discharge.
• Taking a log of the wrong side: isolate $e^{-t/\tau }$ first, then take ln of both sides.

Check yourself

1. A 10 Ω, 220 µF circuit. How long until it's fully charged?

$\tau =2.2$ ms, so about $5\tau =11$ ms.

2. Why is it said a capacitor "looks like a short circuit" at switch-on?

At $t=0$ it has no voltage across it, so the current is the full $V/R$, as if the capacitor were a wire.