Step Response
• Circuit’s behavior to the sudden application of a
DC voltage or current.
• Energy is being stored in the inductor or
capacitor
ECE 201 Circuit Theory I
1
Step Response of an RL Circuit
J1
Key = Space
R
1kOhm
i(t)
Vs
12 V
+
v(t)
L
1mH
-
The switch is closed at t = 0.
Determine the current i(t) and the voltage v(t)
for t>0.
Allow for the possibility of an initial current I0.
ECE 201 Circuit Theory I
2
Step Response of a RL Circuit
J1
Key = Space
R
1kOhm
i(t)
Vs
12 V
+
v(t)
L
1mH
-
• Write KVL around the loop.
di
Vs  iR  L
dt
ECE 201 Circuit Theory I
3
di
dt
di iR  Vs
R V 

  i  s 
dt
L
L R 
R V 
di    i  s  dt
L R 
di
R
  dt
V
L
i s
R
t
t
dx
R
t Vs   L 0 dy
0 x 
R
V 
i (t )   s 
 R  Rt
ln
V
L
I0  s
R
V 
i (t )   s 
R
 R   e  L t
V
I0  s
R
Vs  iR  L
V 
V
i (t )  s   I 0  s
R 
R
R
  L t
e

ECE 201 Circuit Theory I
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Look at the result
• Consider the case when there is no
initial current in the inductor.
Vs 
Vs 
i (t )    I 0   e
R 
R
R
 t
L
R
Vs Vs  L t
i (t )   e
R R
R


 t 
Vs
i (t )  1  e  L  

R 

ECE 201 Circuit Theory I
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Plot the current i(t)
ECE 201 Circuit Theory I
6
Determine the voltage v(t)
Vs 
Vs 
i (t )    I 0   e
R 
R
di
v(t )  L
dt
R
  t
L
Vs 
 R 
v(t )  L     I 0   e
R
 L 
v(t )  Vs  I 0 R  e
R
 t
L
R
 t
L
ECE 201 Circuit Theory I
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Plot the voltage v(t)
ECE 201 Circuit Theory I
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Example 7.5
• The switch has been in position “a” for a long
time. At time t = 0, the switch moves from
position a to position b. The switch is a “makebefore- break” type: that is, the connection at
position b is established before the connection
at position a is broken, so there is no interruption
of the current through the inductor.
R1
b
2 Ohm
a
J1
V
24 V
+
i(t)
v(t)
Key = Space
R2
8 Ohm
I
8A
L
200mH
-
ECE 201 Circuit Theory I
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Determine the initial value of the current
R1
b
2 Ohm
a
J1
V
24 V
Key = Space
I0 = - 8A
R2
8 Ohm
I
8A
L
200mH
The 200mH inductor looks like a short circuit and
carries an initial current of 8A upwards.
Therefore, I0 = -8A.
ECE 201 Circuit Theory I
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Determine the final value of the current and
the time constant
R1
b
2 Ohm
a
J1
V
24 V
Key = Space
i(∞)
R2
8 Ohm
I
8A
L
200mH
• The final value of the current is (24V/2Ω) = 12A
• The time constant is (L/R) = (200mH/2Ω) =
100ms
ECE 201 Circuit Theory I
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R1
b
2 Ohm
a
J1
V
24 V
R2
8 Ohm
Key = Space
i(t)
I
8A
L
200mH
R
Vs 
Vs   L t
i (t )    I 0   e
R 
R
 2 
24 
24   0.2 t
i (t ) 
  8   e
2 
2 
i (t )  12  20e 10t A
ECE 201 Circuit Theory I
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Complete Solution for the Current and
Voltage
• The current in an inductor cannot change
instantaneously
– Let t = (0-) be the time just before the switch is
opened, and t = (0+) be the time just after the
switch is opened
– Then i(0-) = i(0+) =I0, and
i (t )  I 0 e
R
-  t
L
,t  0
  RL  
v(t )  i (t ) R  I 0 R e  t  , t  0 


ECE 201 Circuit Theory I
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Determine the initial voltage across the
inductor just after the switch moves to “b”
R1
2 Ohm
V
24 V
b
a
J1
Key = Space
R2
8 Ohm
I
8A
L
200mH
di
v(t )  L
dt
v(t )  0.2  200e 10t 
v(t )  40e 10tV
v(0 )  40V
ECE 201 Circuit Theory I
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Does this result make sense?
R1
b
2 Ohm
-
vR(0+)
V
24 V
a
J1
+
i(0+)
Key = Space
+
R2
8 Ohm
I
8A
L
vL(0+) 200mH
-
vL  i (0  )(2)  24  0
vL  i (0  )(2)  24
Voltage @ R1 adds to the battery voltage
vL  (8)(2)  24  16  24
vL  40V
ECE 201 Circuit Theory I
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Determine the time it takes for the inductor
voltage to reach 24 Volts
v(t )  40e 10t  24
24
10 t
e 
40
 24 
10t  ln  
 40 
 1   24 
t     ln    0.1(0.5108)
 10   40 
t  0.05108s  51.08ms
ECE 201 Circuit Theory I
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Plot i(t) and v(t)
ECE 201 Circuit Theory I
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ECE 201 Circuit Theory I
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Oscilloscope Results
v(t )  40e10t
i(t )  12  20e10t
ECE 201 Circuit Theory I
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