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 Thévenin (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?
4
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.
6
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 τ  , for t  0 .

The energy dissipated in the resistor R is:
t
t
w   p(t )dt   I R  e 2t  τ dt 
0


2
0
0
 w0 1  e 2t τ  , 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

  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


4

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 ) 

 t  

R
V0 R e
, for t  0 .

The instantaneous power dissipated in R is:
v 2 (t ) V02 2t τ 
p(t ) 

e
, for t  0 .
R
R

The energy dissipated in the resistor R is:
w
t
0


2
CV
0
p (t )dt   w0 1  e 2t τ  , 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
 t  
, v(t )  V0 e
 t  
.
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:
G
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 Thévenin 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 Thévenin 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