Transcript RC circuit – natural response

First Order Circuit
• Capacitors and inductors
• RC and RL circuits
RC and RL circuits (first order circuits)
Circuits containing no independent sources
•
Excitation from stored energy
•
‘source-free’ circuits
•
Natural response
Circuits containing independent sources
•
DC source (voltage or current source)
•
Sources are modeled by step functions
•
Step response
•
Forced response
Complete response = Natural response + forced response
RC circuit – natural response
Assume that capacitor is initially charged at t = 0
+
ic
C
iR
vc
 vc(0) = Vo
R

Objective of analysis: to find expression for vc(t) for t >0
i.e. to get the voltage response of the circuit
Taking KCL,
C
dv c v c

0
dt
R


t
dv c
dt

vc (0) v c
o RC
v c (t )  v c (0)e
dv c
v
 c
dt
RC
dv c
dt

vc
RC

vc (t )



1
t
RC
ln
OR
v c (t)
1

t
v c (0 )
RC
v c ( t )  Voe

1
t
RC
RC circuit – natural response
t / 
• Can be written as v c ( t )  v c (0)e
,  = RC  time constant
• This response is known as the natural response
vC(t)
 Voltage decays to zero exponentially
Vo
 At t=, vc(t) decays to 37.68% of its
initial value
 The smaller the time constant the
faster the decay
0.3768Vo
t
t=
v c (t )  v c (0)e

1
t
RC
RC circuit – natural response
The capacitor current is given by: iC  C
dv c
dt
iC  

And the current through the resistor is given by
iR 
Vo e
R
pR  v RiR 
2
t

R
The energy loss (as heat) in the resistor from 0 to t:

t
ER  pR dt 
0

t
o
Vo2e
2
R
t

t
  Vo2 2 t 
dt  
e 
 2 R
 0
t
2 
1 2 
ER  CVo 1  e  


2


t

v C (t)
Ve
 o
R
R
The power absorbed by the resistor can be calculated as:
Vo2e


t

RC circuit – natural response
As t  , ER 
1 2
CVo
2
As t  , energy initially stored in capacitor will be
dissipated in the resistor in the form of heat
t
2 
1 2 
ER  CVo 1  e  


2


RC circuit – natural response
PSpice simulation
1
+
ic
C
RC circuit
c1 1 0 1e-6 IC=100
r1 1 0 1000
.tran 7e-6 7e-3 0 7e-6 UIC
.probe
.end
iR
vc
R

0
100V
50V
0V
0s
1.0ms
2.0ms
3.0ms
4.0ms
V(1)
Time
5.0ms
6.0ms
7.0ms
RC circuit – natural response
PSpice simulation
RC circuit
.param c=1
c1 1 0 {c} IC=100
r1 1 0 1000
.step param c list 0.5e-6 1e-6 3e-6
.tran 7e-6 7e-3 0 7e-6
.probe
.end
1
+
ic
C
iR
vc
R

0
100V
c1 = 1e-6
c1 = 3e-6
50V
0V
0s
c1 = 0.5e-6
1.0ms
2.0ms
3.0ms
4.0ms
V(1)
Time
5.0ms
6.0ms
7.0ms
RL circuit – natural response
iL

vL
+
L
Assume initial magnetic energy stored in L at t = 0
+
vR

R
 iL(0) = Io
Objective of analysis: to find expression for iL(t) for t >0
i.e. to get the current response of the circuit
Taking KVL,
L
diL
 iLR  0
dt


t
diL
R

dt
i
L
iL ( 0 ) L
o
iL (t )  iL
diL
iR
 L
dt
L
diL
R
  dt
iL
L

iL ( t )

R
 t
(0)e L

OR
ln
iL ( t )
R
 t
iL (0)
L
iL ( t )  Io
R
 t
e L
RL circuit – natural response
• Can be written as
iL (t )  iL (0)e t / 
,  = L/R  time constant
• This response is known as the natural response
iL(t)
 Current exponentially decays to zero
Io
 At t=, iL(t) decays to 37.68% of its
initial value
 The smaller the time constant the
faster the decay
0.3768Io
t
t=
iL (t )  iL
R
 t
(0)e L
RL circuit – natural response
di
vL  L L
dt
The inductor voltage is given by:
 v L  Io Re
And the voltage across the resistor is given by
pR 
2
t

The energy loss (as heat) in the resistor from 0 to t:

t
ER  pR dt 
0

t
o
Io2 Re
2
t

t
  2 2 t 
dt   Io Re  
 2
 0
t
2 
1 2 
ER  LIo 1  e  

2 

t

v R  iL ( t )R  I Re
o
The power absorbed by the resistor can be calculated as:
v RiR  Io2 Re


t

RL circuit – natural response
As t  , ER 
1 2
LIo
2
As t  , energy initially stored in inductor will be
dissipated in the resistor in the form of heat
t
2 
1 2 
ER  LIo 1  e  

2 

RL circuit – natural response
PSpice simulation
1

vL
+
R
L
RL circuit
L1 0 1 1 IC=10
r1 1 0 1000
.tran 7e-6 7e-3 0 7e-6 UIC
.probe
.end
+
vR

0
10A
5A
0A
0s
1.0ms
2.0ms
3.0ms
4.0ms
I(L1)
Time
5.0ms
6.0ms
7.0ms
RL circuit – natural response
PSpice simulation
RL circuit
.param L=1H
L1 0 1 {L} IC=10
r1 1 0 1000
.step param L list 0.3 1 3
.tran 7e-6 7e-3 0 7e-6 UIC
.probe
.end
1

+
vR

vL
R
L
+
0
10A
L1 = 1H
L1 = 3H
5A
0A
0s
1.0ms
2.0ms
3.0ms
4.0ms
I(L1)
L1 = 0.5H
Time
5.0ms
6.0ms
7.0ms