Chapter 7

Response of First-order RL
and RC Circuits
7.1-2 The Natural Response of RL and RC
Circuits
7.3 The Step Response of RL and RC
Circuits
7.4 A General Solution for Step and Natural
Responses
7.5 Sequential Switching
7.6 Unbounded Response
1

Overview

Ch9-10 discuss steady-state response of

linear circuits to sinusoidal sources. The math
treatment is the same as the dc response
except for introducing phasors and
impedances in the algebraic equations.

From now on, we will discuss transient

response of linear circuits to step sources
(Ch7-8) and general time-varying sources
(Ch12-13). The math treatment involves with
differential equations and Laplace transform.
2

First-order circuit

A circuit that can be simplified to a Thvenin (or

Norton) equivalent connected to either a single
equivalent inductor or capacitor.

In Ch7, the source is either none (natural

response) or step source.
3

Key points

Why an RC or RL circuit is charged or

discharged as an exponential function of time?

Why the charging and discharging speed of an

RC or RL circuit is determined by RC or L/R?

What could happen when an energy-storing

element (C or L) is connected to a circuit with
dependent source?

Section 7.1, 7.2

The Natural Response of RL
and RC Circuits
1.

2.
3.

Differential equation & solution of a

discharging RL circuit
Time constant
Discharging RC circuit
5

What is natural response?

It describes the discharging of inductors or

capacitors via a circuit of no dependent source.

No external source is involved, thus termed as

natural response.

The effect will vanish as t . The interval

within which the natural response matters
depends on the element parameters.

Circuit model of a discharging RL circuit

Consider the following circuit model:

For t < 0, the inductor L is short and carries a

current Is, while R0 and R carry no current.

For t > 0, the inductor current decreases and the

energy is dissipated via R.
7

Ordinary differential equation (ODE), initial condition (IC)

For t > 0, the circuit reduces to:

IC depends on initial
energy of the inductor:

i (0 ) i (0 ) I 0 I s

By KVL, we got a first-order ODE for i(t):

d
L i(t ) Ri(t ) 0.
dt
where L, R are independent of both i, and t.
8

Solving the loop current

d
ODE : L i (t ) Ri (t ) 0, IC : i (0 ) I 0 I s ;
dt
di
R
L(di ) Ri (dt ) 0,
dt ,
i
L
i ( t ) di
R t
R t
i (t )
dt , ln i i ( 0 ) t 0 ,

i ( 0 ) i
L 0
L
i (t )
R
ln i (t ) ln i (0) ln
t,
I0
L
L
t
i (t ) I 0 e
, where
time constant
R
9

Time constant describes the discharging speed

The loop current i(t) will drop to e-1 ( 37%) of its

initial value I0 within one time constant . It will
be <0.01I0 after elapsing 5.

If i(t) is approximated by a linear function, it will

vanish in one time constant.

0.37 I 0
10

Solutions of the voltage, power, and energy

The voltage across R (or L) is:

0, for t 0 ,
abrupt change at t = 0.
v(t ) Ri(t )
t

RI 0 e
, for t 0 .

The instantaneous power dissipated in R is:

p (t ) i 2 (t ) R I 02 Re 2t , for t 0 .

The energy dissipated in the resistor R is:

t

w p(t )dt I R e 2t dt
0

2
0

w0 1 e 2t , w0 LI 02 2 , for t 0 .
initial energy stored in L

11

Example 7.2: Discharge of parallel inductors (1)

Q: Find i1(t), i2(t), i3(t), and the energies w1, w2

stored in L1, L2 in steady state (t ).

For t < 0: (1) L1, L2 are short, and (2) no current

flows through any of the 4 resistors,
i1 (0 ) 8 A, i2 (0 ) 4 A, i3 (0 ) 0,
w1 (0 ) (5 H)(8 A)2 2 160 J, w2 (0 ) (20)(4) 2 2 160 J.

12

Example 7.2: Solving the equivalent RL circuit (2)

For t > 0, switch is open, the initial energy stored

in the 2 inductors is dissipated via the 4 resistors.
The equivalent circuit becomes:
I0 = (8+4) = 12 A
Leq = (5H//20H) = 4 H

Req =
v(0-) = 0 8

The solutions to i(t), v(t) are:

t
2 t

i
(
t
)
I
e
12
e
A,
4

0.5 s,
Req 8
v(t ) Ri (t ) 96e 2t V.

Leq

13

Example 7.2: Solving the inductor currents (3)

The two inductor currents i1(t), i2(t) can be

calculated by v(t):

96e 2t V

1 t
1 t
i1 (t ) i1 (0) v(t )dt 8 96e 2t dt 1.6 9.6e 2t A.
L1 0
5 0
1
i2 (t ) i2 (0)
L2

1 t
2t
2t

v
(
t
)
d
t

96
e
d
t

1
.
6

2
.
4
e
A.
0

20 0
t

14

Example 7.2: Solutions in steady state (4)

Since

i1 (t ) 1.6 9.6e 2t 1.6 A,

i2 (t ) 1.6 2.4e 2t 1.6 A,

the two inductors form a closed current loop!

The energies stored in the two inductors are:

1
w1 (t ) (5 H)(1.6 A) 2 6.4 J,
2
1
w2 (t ) (20)(1.6) 2 25.6 J,
2

which is ~10% of the initial energy in total.

15

Example 7.2: Solving the resistor current (5)

By current division, i3(t) = 0.6i4(t), while i4(t) can

be calculated by v(t):
i4

v(t )
96e 2t
i4 (t )

9.6e 2t A.
4 (15 // 10)
10
3
i3 (t ) i4 (t ) 5.76e 2t A.
5
16

Circuit model of a discharging RC circuit

Consider the following circuit model:

For t < 0, C is open and biased by a voltage Vg,

while R1 and R carry no current.

For t > 0, the capacitor voltage decreases and the

energy is dissipated via R.
17

ODE & IC

For t > 0, the circuit reduces to:

IC depends on initial
energy of the capacitor:

v (0 ) v (0 ) V0 Vg

By KCL, we got a first-order ODE for v(t):

d
v(t )
C v(t )
0.
dt
R
where C, R are independent of both v, and t.
18

Solving the parallel voltage

d
v(t )
ODE : C v(t )
0,
dt
R

IC : v(0 ) V0 Vg ;

v(t ) V0 e t , where RC time constant

Reducing R (loss)
and parasitic C is
critical for highspeed circuits.

19

Solutions of the current, power, and energy

The loop current is:

v(t ) 0, for t 0 ,
abrupt change at t = 0.
i (t )

R
V0 R e
, for t 0 .

The instantaneous power dissipated in R is:

v 2 (t ) V02 2t
p(t )

e
, for t 0 .
R
R

The energy dissipated in the resistor R is:

w

2
CV
0
p (t )dt w0 1 e 2t , w0
, for t 0 .
2

initial energy stored in C

20

Procedures to get natural response of RL, RC circuits

1.

Find the equivalent circuit.

2.

Find the initial conditions: initial current I0

through the equivalent inductor, or initial voltage
V0 across the equivalent capacitor.

3.

Find the time constant of the circuit by the

values of the equivalent R, L, C:

L R , or RC ;
4.

Directly write down the solutions:

i (t ) I 0 e

, v(t ) V0 e

.
21

Section 7.3
The Step Response of
RL and RC Circuits
1.
2.

Charging an RC circuit
Charging an RL circuit

22

What is step response?

The response of a circuit to the sudden

application of a constant voltage or current
source, describing the charging behavior of the
circuit.

Step (charging) response and natural

(discharging) response show how the signal in a
digital circuit switches between Low and High
with time.

23

ODE and IC of a charging RC circuit

IC depends on
initial energy of
the capacitor:

v (0 ) v (0 ) V0

Derive the governing ODE by KCL:

v (t )
d
dv
1
Is
C v (t ),
v V f ,
R
dt
dt

where RC , V f I s R are the time constant

and final (steady-state) parallel voltage.

24

Solving the parallel voltage, branch currents

1
1
dv
dv
v V f ,
dt ,

dt
v Vf

v(t )

V0

v (t ) V f
1 t
t v (t ) V f
dv
e t ,
dt , ln
,
0
V0 V f
V0 V f
v I s R

v (t ) V f V0 V f e t .

The charging and discharging processes have

the same speed (same time constant = RC).

The branch currents through C and R are:

V0 t
d
v(t )

iC (t ) C v(t ) I s e , iR (t )
, for t 0.
dt
R
R

25

Example 7.6 (1)

Q: Find vo(t), io(t) for t 0.

For t < 0, the switch is connected to Terminal 1

for long, the capacitor is an open circuit:
60 k
vo (0 ) ( 40 V)
30 V, io (0 ) 0.
( 20 60) k

26

Example 7.6 (2)

At t 0+, the charging circuit with two terminals

2 and G can be reduced to a Norton equivalent:

vo (0 ) vo (0 )

V0 30 V

v0
I s 1.5 mA
27

Example 7.6 (3)

The time constant and the final capacitor

voltage of the charging circuit are:
RC (40 k)(0.25 F) 10 ms,
V f I s R (1.5 mA)(40 k) 60 V.

vo (t ) V f V0 V f e t 60 90e 100t V.
io (t ) I s V0 R e t 2.25e 100t mA.

At the time of switching, the capacitor voltage

is continuous: vo (0 ) 30 V v0 (0 ) ( 60 90) V ,
while the current io jumps from 0 to -2.25 mA.

28

Charging an RL circuit
IC depends on
initial energy of
the inductor:

i (0 ) i (0 ) I 0
d
Vs Ri (t ) L i (t ), i (t ) I f I 0 I f e t ,
dt
Vs
L
where , I f .
R
R

The charging and discharging processes have

the same speed (same time constant =L/R).

29

Section 7.5
Sequential Switching

30

What is and how to solve sequential switching?

Sequential switching means switching occurs n (2)

times.

It is analyzed by dividing the process into n+1 time

intervals. Each of them corresponds to a specific
circuit.

Initial conditions of a particular interval are determined

from the solution of the preceding interval. Inductive
currents and capacitive voltages are particularly
important for they cannot change abruptly.

Laplace transform (Ch12) can solve it easily.

31

Example 7.12: Charging and discharging a capacitor (1)

Q:

v (t) = ? for t 0.
t =0
t =15 ms

IC: V0 v (0 ) 0

0 t 15 ms, the 400-V source charges the

capacitor via the 100-k resistor, Vf = 400 V,
= RC =10 ms, and the capacitor voltage is:

For

v1 (t ) V f (V0 V f )e t 400 400e 100t V.

32

Example 7.12 (2)

For t > 15 ms, the capacitor is disconnected from

the 400-V source and is discharged via the 50k resistor, V0 = v1(15 ms) = 310.75 V, Vf = 0,
= RC = 5 ms. The capacitor voltage is:

v2 (t ) V0e

( t t0 )

310.75e

200 ( t 0.015 )

V.

33

Section 7.6
Unbounded Response

34

Definition and reason of unbounded response

An unbounded response means the voltages or

currents increase with time without limit.

It occurs when the Thvenin resistance is

negative (RTh < 0), which is possible when the
first-order circuit contains dependent sources.

35

Example 7.13: (1)

Q: vo(t) = ? for t 0.

For t > 0, the capacitor seems to discharge

(not really, to be discussed) via a circuit with a
current-controlled current source, which can be
represented by a Thvenin equivalent.

36

Example 7.13 (2)

Since there is no independent source, VTh = 0,

while RTh can be determined by the test source
method:

vT
vT
vT
iT

7
,
10 k 20 k
20 k
RTh vT iT -5 k 0.
37

Example 7.13 (3)

For t 0, the equivalent circuit and governing

differential equation become:

V0 vo (0 ) 10 V, RC 25 ms 0,

vo (t ) V0e t 10e 40t V. ..grow without limit.

38

Why the voltage is unbounded?

10-k, 20-k resistors are in parallel,
i10k = 2i, the capacitor is actually charged (not
discharged) by a current of 4i!

Since

effect will increase vo, which will in turn

increase the charging current (i = vo/20 k) and
vo itself. The positive feedback makes vo soaring.

Charging

6i

2i
39

Lesson for circuit designers & device fabrication

engineers

Undesired interconnection between a capacitor

and a sub-circuit with dependent source (e.g.
transistor) could be catastrophic!

40

Key points

Why an RC or RL circuit is charged or

discharged as an exponential function of time?

Why the charging and discharging speed of an

RC or RL circuit is determined by RC or L/R?

What could happen when an energy-storing

element (C or L) is connected to a circuit with
dependent source?

41