Transient response of RC and RL circuits
ENGR 40M lecture notes — July 26, 2017
Chuan-Zheng Lee, Stanford University
Resistor–capacitor (RC) and resistor–inductor (RL) circuits are the two types of first-order circuits: circuits
either one capacitor or one inductor. In many applications, these circuits respond to a sudden change in an
input: for example, a switch opening or closing, or a digital input switching from low to high. Just after the
change, the capacitor or inductor takes some time to charge or discharge, and eventually settles on its new
steady state. We call the response of a circuit immediately after a sudden change the transient response, in
contrast to the steady state.
A first example
Consider the following circuit, whose voltage source provides vin (t) = 0 for t < 0, and vin (t) = 10 V for t ≥ 0.
R
+
+ C vout
vin (t)
−
−
A few observations, using steady state analysis. Just before the step in vin from 0 V to 10 V at t = 0,
vout (0− ) = 0 V. Since vout is across a capacitor, vout just after the step must be the same: vout (0+ ) = 0 V.
Long after the step, if we wait long enough the circuit will reach steady state, then vout (∞) = 10 V.
What happens in between? Using Kirchoff’s current law applied at the top-right node, we could write
10 V − vout dvout
=C .
R dt
Solving this differential equation, then applying the initial conditions we found above, would yield (in volts)
t
vout (t) = 10 − 10e− RC .
v vin (t)
10
vout (t)
t
τ = RC 2τ 3τ 4τ
What’s happening? Immediately after the step, the current flowing through the resistor—and hence the
capacitor (by KCL)—is i(0+ ) = 10RV . Since i = C dvdtout , this current causes vout to start rising. This, in
turn, reduces the current through the resistor (and capacitor), 10 V−v R
out
. Thus, the rate of change of vout
decreases as vout increases. The voltage vout (t) technically never reaches steady state, but after about 3RC,
it’s very close.
Transient response equation
It turns out that all first-order circuits respond to a sudden change in input with some sort of exponential
decay, similar to the above. Therefore, we don’t solve differential equations every time we see a capacitor or
an inductor, and we won’t ask you to solve any.
Instead, we use the following shortcut: In any first-order circuit, if there is a sudden change at t = 0, the
transient response for a voltage is given by
v(t) = v(∞) + [v(0+ ) − v(∞)]e−t/τ ,
where v(∞) is the (new) steady-state voltage; v(0+ ) is the voltage just after time t = 0; τ is the time
constant, given by τ = RC for a capacitor or τ = L/R for an inductor, and in both cases R is the resistance
seen by the capacitor or inductor.
The transient response for a current is the same, with i(·) instead of v(·):
i(t) = i(∞) + [i(0+ ) − i(∞)]e−t/τ .
What do we mean by the “resistance seen by the capacitor/inductor”? Informally, it means the resistance
you would think the rest of the circuit had, if you were the capacitor/inductor. More precisely, you find it
using these steps:
1. Zero out all sources (i.e. short all voltage sources, open all current sources)
2. Remove the capacitor or inductor
3. Find the resistance of the resistor network whose terminals are where the capacitor/inductor was
About the time constant
The time constant τ (the Greek letter tau) has units of seconds (verify, for both RC and R/L), and it
governs the “speed” of the transient response. Circuits with higher τ take longer to get close to the new
steady state. Circuits with short τ settle on their new steady state very quickly.
More precisely, every time constant τ , the circuit gets 1 − e−1 ≈ 63% of its way closer to its new steady
state. Memorizing this fact can help you draw graphs involving exponential decays quickly.
After 3τ , the circuit will have gotten 1 − e−3 ≈ 95% of the way, and after 5τ , more than 99%. So, after a
few time constants, for practical purposes, the circuit has reached steady state. Thus, the time constant is
itself a good rough guide to “how long” the transient response will take.
Of course, mathematically, the steady state is actually an asymptote: it never truly reaches steady state.
But, unlike mathematicians, engineers don’t sweat over such inconsequential details.
v(t)
v(0+ )
˜63% of ˜95% of
[v(0+ )−v(∞)] [v(0+ )−v(∞)]
v(∞)
t
τ 2τ 3τ 4τ 5τ
2
Example 1
Example 2
