October 31, 2001
7.4 The step response of an RC circuit
t=0
Is
+
vc(t)
R
C
-
i
We will check the circuit response after the switch is closed.
Using KCL , we can write the following differential equation from the circuit :
C
dvc (t ) vc (t )
+
= Is
dt
R
(9.1)
To rearrange the equation
dvc (t )
v (t ) I
= −( c − s )
dt
RC C
(9.2)
dvc (t )
1
=−
(vc (t ) − RI s )
dt
RC
(9.3)
dvc (t )
1
=−
dt
(vc (t ) − RI s )
RC
Change the variables name and take the integral of both sides,
vC ( t )
∫
Vo
t
dy
1
= −∫
dτ
( y − RI s )
RC
0
We have
ln
vC (t ) − RI s
1
=−
t
V0 − RI s
RC
(9.4)
1
−
t
vC (t ) − RI s
= e RC
V0 − RI s
vC (t ) = RI s + (V0 − RI s )e
−
1
t
RC
for t ≥ 0
We know that V0 is the initial voltage in the capacitor and time constant
If V0 is the initial voltage in the capacitor is zero then,
(9.5)
τ = RC .
307-09-1
vC (t ) = RI s + (1 − e
1
− t
τ
) , for t ≥ 0
(9.6)
The current in the capacitor yields the differential equation
t
−

dvC (t )
d
RC
i (t ) = C
= C  RI S + (V0 − RI S )e 
dt
dt 

1 − RCt
)e ,
i (t ) = C (V0 − RI s ) (−
RC
V  −

i (t ) =  I s − 0  e RC , for t ≥ 0 +
R

t
(9.7)
Special case: if V0 the initial voltage in the capacitor is zero then,
i (t ) = I s e
−
t
RC
for
t ≥ 0+
(9.8)
Example:
20k
a
t=0
b
+
40V
0.25uF
Vo
8k
40k
i0
160k
75V
60k
-
Switch has been in position a for a long time.
Switch moves to position b at t=0.
v0 (t ) for t ≥ 0
b. Find i0 (t ) for t ≥ 0
a. Find
Switch has been in position a for a long time, The initial value of
V0 =
V0
60
40 = 30 V
60 + 20
We can find the Norton equivalent circuit for the right site of the circuit
307-09-2
8k
40k
+
Vo
160k
75V
40k
1.5mA
-
Voc =
160
( −75) = −60 V
(40 + 160)
RTh = 8 + 40 ||160 = 40k Ω
The Norton current source will be
IN =
VOC
−60
=
= −1.5 mA
RTh 40(103 )
+
40k
0.25uF
1.5mA
Vo
-
τ = RC = 0.25(10−6 )40(103 ) = 0.01
vC (t ) = RI s + (V0 − RI s )e
−
t
τ
, t≥0
vC (t ) = −40(103 )1.5(10−3 ) + (30 − (−40(103 )1.5(10−3 )) ) e−100t
vC (t ) = −60 + (30 + 60)e−100t = −60 + 90e −100t V for t ≥ 0
b.
V  −t 
30  −100t

,
i (t ) =  I s − 0  e RC =  −1.5(10−3 ) −
e
40(103 ) 
R


i (t ) = ( −2.25 ) e−100t mA for t ≥ 0 +
307-09-3