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FIRST ORDER TRANSIENT CIRCUITS
IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTS
CANNOT CHANGE INSTANTANEOUSLY.
EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES A
TRANSIENT BEHAVIOR
FIRST ORDER CIRCUITS
Circuits that contain a single energy storing elements.
Either a capacitor or an inductor
ANALYSIS OF LINEAR CIRCUITS WITH INDUCTORS AND/OR CAPACITORS
THE CONVENTIONAL ANALYSIS USING MATHEMATICAL MODELS REQUIRES THE DETERMINATION
OF (A SET OF) EQUATIONS THAT REPRESENT THE CIRCUIT.
ONCE THE MODEL IS OBTAINED ANALYSIS REQUIRES THE SOLUTION OF THE EQUATIONS FOR
THE CASES REQUIRED.
FOR EXAMPLE IN NODE OR LOOP ANALYSIS OF RESISTIVE CIRCUITS ONE REPRESENTS THE
CIRCUIT BY A SET OF ALGEBRAIC EQUATIONS
THE MODEL
WHEN THERE ARE INDUCTORS OR CAPACITORS THE MODELS BECOME LINEAR ORDINARY
DIFFERENTIAL EQUATIONS (ODEs). HENCE, IN GENERAL, ONE NEEDS ALL THOSE TOOLS
IN ORDER TO BE ABLE TO ANALYZE CIRCUITS WITH ENERGY STORING ELEMENTS.
A METHOD BASED ON THEVENIN WILL BE DEVELOPED TO DERIVE MATHEMATICAL MODELS
FOR ANY ARBITRARY LINEAR CIRCUIT WITH ONE ENERGY STORING ELEMENT.
THE GENERAL APPROACH CAN BE SIMPLIFIED IN SOME SPECIAL CASES WHEN THE FORM
OF THE SOLUTION CAN BE KNOWN BEFOREHAND.
THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOME
PARAMETERS.
TWO SUCH CASES WILL BE DISCUSSED IN DETAIL FOR THE CASE OF CONSTANT SOURCES.
ONE THAT ASSUMES THE AVAILABILITY OF THE DIFFERENTIAL EQUATION AND A SECOND
THAT IS ENTIRELY BASED ON ELEMENTARY CIRCUIT ANALYSIS… BUT IT IS NORMALLY LONGER
WE WILL ALSO DISCUSS THE PERFORMANCE OF LINEAR CIRCUITS TO OTHER SIMPLE INPUTS
AN INTRODUCTION
INDUCTORS AND CAPACITORS CAN STORE ENERGY. UNDER SUITABLE CONDITIONS THIS ENERGY
CAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERS
OF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT
With the switch on the left the capacitor receives
charge from the battery.
Switch to the right
and the capacitor
discharges through
the lamp
GENERAL RESPONSE: FIRST ORDER CIRCUITS
Including the initial conditions
the model for the capacitor
t
t
x
t
t
1


voltage or the inductor current e x (t )  e  x (t ) 
 e fTH ( x )dx * / e 
0
will be shown to be of the form
t 
0
0
dx
(t )  ax (t )  f (t ); x (0)  x0
dt
dx
  x  fTH ; x (0)  x0
dt
x(t )  e

t t 0

x(t0 ) 
1

t
e

tx

fTH ( x)dx
t0
THIS EXPRESSION ALLOWS THE COMPUTATION
OF THE RESPONSE FOR ANY FORCING FUNCTION.
WE WILL CONCENTRATE IN THE SPECIAL CASE
WHEN THE RIGHT HAND SIDE IS CONSTANT
Solving the differential equation
using integrating factors, one
tries to convert the LHS into an
exact derivative
 is called the " time constant."
t
dx
1
  x  fTH /* e 
dt

t
t
t
dx 1 
1
e
 e x  e fTH
dt 

t

t0
d   1 
 e x   e fTH
 
dt 

t
t
it will be shown to provide significan t
informatio n on the reaction speed of the
circuit
The initial time, to , is arbitrary. The
general expression can be used to
study sequential switchings .
FIRST ORDER CIRCUITS WITH
CONSTANT SOURCES
dx
  x  fTH ; x (0)  x0
dt
x (t )  e

t t0
t
1
x (t0 )   e

t

tx

fTH ( x )dx
0
x(t )  e
x(t )  e


t t 0

t t 0

x(t0 ) 
t
x(t )  e
e

tx



x(t0 ) 
fTH

t
e
x
0
t
t



 
x(t0 )  fTH e  e  e 



t
0
x(t )  fTH  x(t0 )  fTH e

t t 0

t  t0
If the RHS is constant
t t0
t
 
e  e 

 t
fTH


The form of the solution is
tx

dx
t0
x(t )  K1  K 2e

t t 0

; t  t0
TIME
CONSTANT
TRANSIENT

t
x


Any variable in the circuit is of
the form
e e 
x(t )  e

t t0

x(t0 ) 
fTH

e

t t
x


 e dx
t0
y(t )  K1  K2e

t t 0

; t  t0
Only the values of the constants
K_1, K_2 will change
EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF
THE TIME CONSTANT
Tangent reaches x-axis in one time constant
Drops 0.632 of initial
value in one time constant
With less than 2% error
transient is zero
beyond this point
A QUALITATIVE VIEW:
THE SMALLER THE THE TIME
CONSTANT THE FASTER THE
TRANSIENT DISAPPEARS
THE TIME CONSTANT
The following example illustrates
the physical meaning of time
constant
Charging a capacitor
vC  v S RS a
KCL@a :
RS
dv v  v S
+
C c C
0

dt
RS
vS
C v
c

_ The model
dv b
C C
dt
dvC
RTH C
 vC  vTH
dt
Assume
v S  V S , v C ( 0)  0
  RTH C
t

2
3
4
5

t
e 
0.368 With less than 1%
0.135 error the transient
0.0498 is negligible after
0.0183 five time constants
0.0067
The solution can be shown to be
vC (t )  VS  VS e

t

transient
For practical purposes the
capacitor is charged when the
transient is negligible
  RTH C
CIRCUITS WITH ONE ENERGY STORING ELEMENT
THE DIFFERENTIAL EQUATION APPROACH
CONDITIONS
1. THE CIRCUIT HAS ONLY CONSTANT INDEPENDENT SOURCES
2. THE DIFFERENTIAL EQUATION FOR THE VARIABLE OF INTEREST
IS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS;
e.g., KCL, KVL. . . OR THEVENIN
3. THE INITIAL CONDITION FOR THE DIFFERENTIAL EQUATION
IS KNOWN, OR CAN BE OBTAINED USING STEADY STATE ANALYSIS
FACT: WHEN ALL INDEPENDENT SOURCES ARE CONSTANT
FOR ANY VARIABLE, y( t ), IN THE CIRCUIT THE
SOLUTION IS OF THE FORM
y( t )  K 1  K 2 e

( t  tO )

, t  tO
SOLUTION STRATEGY: USE THE DIFFERENTIAL EQUATION AND THE
INITIAL CONDITIONS TO FIND THE PARAMETERS K1, K 2 ,
If the diff eq for y is known
in the form
Use the initial condition to get
one more equation
y(0)  K1  K 2
dy
a1  a0 y  f We can use this
dt
info to find
the unknowns
y (0  )  y 0
Use the diff eq to find two
more equations by replacing
the form of solution into the
differential equation

t
t
dy
K 
y(t )  K1  K 2e  , t  0 
 2e 
dt

 K2 
a1  
e 
 




  a 0  K1  K 2 e 




f
a0 K1  f  K1 
a0
t
t
t

 f



 a1

  0    a1


a
K
e

0 2
a0
 

K 2  y(0)  K1
SHORTCUT: WRITE DIFFERENTIAL EQ.
IN NORMALIZED FORM WITH COEFFICIENT
OF VARIABLE = 1.
a1
dy
a dy
f
 a0 y  f  1
y
dt
a0 dt
a0

K1
FIND i (t ), t  0
EXAMPLE
x ( t )  K1  K 2 e

i (t )
VS  v R  v L  Ri (t )  L
t

,t  0
K1  x (); K1  K 2  x (0)
 vR 
KVL


vL
t
i(t )  K1  K 2e  , t  0
MODEL. USE KVL FOR t  0

di
(t )
dt
INITIAL CONDITION
t  0  i (0 )  0

i ( 0  )  0
inductor  i (0)  i (0 )
L di
VS   L
(t )  i (t ) 
R
R dt
R
STEP 2 STEADY STATE i ()  K  VS
1
R
STEP 1
t


STEP 3 INITIAL CONDITION

L 
VS 
ANS : i (t )  1  e R 
i (0)  K1  K 2
R



FIND vO (t ), t  0

t
vC (t )  K1  K 2 e , t  0

K1  vC (); K1  K 2  vC (0)
R1
C
R2
DETERMINE vc (t )
MODEL FOR t  0. USE KCL
dv
vC
dv
2
1
C C (t ) 
 0  ( R1  R2 )C C (t )  vc  0 vO (t ) 
vC (t )  vC (t )
dt
R1  R2
dt
24
3
STEP 1   ( R1  R2 )C  (6 103 )(100 106 F )  0.6 s
STEP 2

t
vC (t )  K1  K 2e  , t  0 K1  0
t
8  0.6
vO (t )  e [V ], t  0
3
INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0
STEP 3
vC (0)  8  K1  K 2  K 2  8[V ]

6
vC (0)  (12)V
9

vC (t )  8e

t
0.6 [V ], t
0
USING THEVENIN TO OBTAIN MODELS
Obtain the voltage across the capacitor
or the current through the inductor
Circuit
with
resistances
and
sources
a
Inductor
or
Capacitor
b
RTH a
Thevenin
VTH

b
Representation of an arbitrary
circuit with one storage element
RTH a
KCL@ node a
Inductor
or
Capacitor

RTH a
Use KVL
vR  vL  vTH
i

i

0

v

c
R

R
iR C

VTH 
L
VTH
vc
vR  RTH iL
dvC
vL


ic  C
_
iL
diL
dt

vL  L
b
b
v v
dt
Case 1.2
Case 1.1
iR  C TH
Current through inductor
Voltage across capacitor
RTH
diL
L
 RTH iL  vTH
dvC vC  vTH
dt
C

0
dt
RTH
 L  diL
vTH
dvC



i

RTH C
 vC  vTH
 R  dt L R  iSC
dt
TH
 TH 
ic
+