Post on 26-Dec-2015
EMLAB
1
Chapter 7. First and second order transient circuits
EMLAB
2Contents
1. Introduction
2. First order circuits
3. Second order circuits
4. Application examples
EMLAB
31. Introduction
Camera flash discharge :
EMLAB
4Analysis
0)()(
R
t
dt
tdC CC
0)(1)(
tRCdt
tdC
C
RC
t
C eVt
0)(
EMLAB
52. First order transient circuits
Solution to 1st order differential equation :
f (t) = 0 → homogeneous equationf (t) ≠ 0 → inhomogeneous equation
xh(t) or xc(t) → homogeneous or complementary solutionxp(t) → inhomogeneous or particular solution
)()()(
tftaxdt
tdx
0)()(
taxdt
tdxc
c
)()()( txtxtx cp
EMLAB
6
a
AKtxp 1)(
Particular solution :
For polynomial functions f (t) , x(t) should also be a polynomial func-tion of the order of equal or lower degree.
cbtattxtttfex p 22 )(232)()
1)()() KtxAtfex p
AaKtaxdt
tdxAtftax
dt
tdxp
p 1)()(
)()()(
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7
Homogeneous solution :
CatCatc eKeKetx
22)(
00)()(
cc
cc ax
dt
dxtax
dt
tdx
dtaxddtax
dxax
dt
dxc
c
cc
c ][ln
Cattxc )(ln
atatcp eK
a
AeKKtxtxtx 221)()()(
The constant K2 can be found, if the value of x(t) is known at one instant of time.
Integrating both sides with respect to the arguments,
EMLAB
8General solution : 1st order differential equation
K1 : steady-state solution: x(t) → K1 as t→∞ when the second term becomes negligible.
τ : time constant
aeKKtx t 1
)( /21
EMLAB
9Simple RC circuit
)0( t
Sp VKKt )(
01
)( ttth e
RCeet
RC
1
RC
t
Shp eKVttt
2)()()(
0)()(
R
Vt
dt
tdC S
RC
Vt
RCdt
td S )(1)(
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10
S
S
VK
KVt
2
20)0(
RC
t
SS eVVt
)(
EMLAB
11Simple RL circuit
L
Vti
L
R
dt
tdiVtRi
dt
tdiL S
S )()(
)()(
tL
RS eK
R
Vti
2)(
0)0()0( 2 KR
Viti S
R
VK S 2
tL
RS
tL
RSS e
R
Ve
R
V
R
Vti 1)(
tL
R
SR eVtiRt 1)()(
EMLAB
12Example 7.1
Consider the circuit shown in Fig. 7.4a. Assuming that the switch has been in posi-tion 1 for a long time, at time t=0 the switch is moved to position 2. We wish to cal-culate the current i(t) for t > 0.
Initial condition :
][436
312)0( V
kk
kC
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13
0)()()(
21
R
t
dt
tdC
R
t
FCkRkR 100,3,6 21
0)(5)(
tdt
td
tAet )(Suppose
0)5(0)(5)(
tt eeAtdt
td
5
tt eAet 55 4)( ][4)0()0( VCC
][3
4)()( 5
2
mAeR
tti t
EMLAB
14Example 7.2
The switch in the network in Fig. 7.5a opens at t=0. Let us find the output volt-age vo(t) for t > 0.
][3
4)0()0( Aii LL
Initial condition :
Thevenin’s equivalent
EMLAB
15
0)(
)()( 311
dt
tdiLtiRRVS
HLRR 2,2,2 31
6)(2)(
tidt
tdi
teKti 223)(
][3
4)0()0( Aii LL 3
53
3
4
3
43)0( 22 KKi
teti 2
3
53)(
6)(2)(
tidt
tdi
to etit 2
3
106)(2)(
EMLAB
16Example 7.3Consider the circuit shown in Fig. 7.6a. The circuit is in steady state prior to time t=0,when the switch is closed. Let us calculate the current i(t) for t > 0.
Initial condition :
][2462
1236)0(
][32122012)0()46()0(
mAkkk
i
VikkC
EMLAB
17
06
)()(100
2
36)(
k
t
dt
td
k
t
)(t
180)(3
20)( t
dt
td
27)( tpt
h eKt 3
20
2)(
t
hp eKttt 3
20
227)()()(
5,3227)0( 22 KK
tet 3
20
527)(
][6
5
2
9
6
)()( 3
20
mAek
tti
t
EMLAB
18Problem-Solving Strategy
Step 1. We assume a solution for the variable x(t) of the form
Step 2. Assuming that the original circuit has reached steady state before a switch was thrown. Solve for the voltage across the capacitor, υc(0-) or the current through the Inductor iL(0-), prior to switch action.
Step 3. Recall that voltage across a capacitor and the current flowing through an in-ductor cannot change in zero time after the switch is changed. Solve for the initial value of the variable x(0+) using υc(0+) = υc(0-), iL(0+) = iL(0-).
/21)( teKKtx
Step 4. Assuming that steady state has been reached after the switches are thrown,draw the equivalent circuit, valid for t >5τ by replacing the capacitor by an open cir-cuit or the inductor by a short circuit.
)()(5
xtxt
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19
Step 5. Since the time constant for all voltages and currents in the circuit will be the same, it can be obtained by reducing the entire circuit to a simple series circuitcontaining a voltage source, resistor, and a storage element (i.e., capacitor or in-ductor) by forming a simple Thévenin equivalent circuit at the terminals of the storage element. This Thévenin equivalent circuit is obtained by looking into the circuit from the terminals of the storage element.
ThTh R
LCR ,
Step 6. Using the results of steps 3, 4, and 5, we can evaluate the constants.
)()0(),(
)(,)0(
21
121
xxKxK
KxKKx
/)]()0([)()( texxxtx
EMLAB
20Example 7.4
The circuit shown in Fig. 7.8a is assumed to have been in a steady-state condition prior to switch closure at t=0. We wish to calculate the voltage v(t) for t>0.
/21)( teKKt
][3
8
31
61
31
3636
4
24)0( AiL
Initial condition (before the switch action)
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21
Right after the switch action : (t = 0+)
Applying Kirchhoff current law to node υ1,
][3
20)0(0
12
)0(8
6
)0(
4
24)0(1
111 V
][3
52)0(24)0( 1 V
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22
Steady state solution :
][24)( V ][2123
12
121
61
41
1
ThR
][22
4s
R
L
Th
][3
2024
3
52)()0(],[24)(
)(,)0(
21
121
VKVK
KKK
][3
2024)]()0([)()( 2/ Veet
tt
EMLAB
23Pulse response
)0(1
)0(0)(
t
ttu
EMLAB
24Unit step function
)(1
)(0)(
0
00 tt
ttttu
)]()([)( TttuAt
EMLAB
25Example 7.6Let us determine the expression for the voltage υo(t). Since the source is zero for all negative time, the initial conditions for the network are zero
)]3.0()([9)( ttut
0)0()0( tt CC
0)0(84
8)0(
t
kk
kt Co
4)(846
8)(
t
kkk
kto
)()1(4
)()0()()(
4.0
1
tue
etttt
t
oooo
kkkkRTh 4)84(||6
)3.0()1(4)( 4.0
3.0
2
tuet
t
o
)3.0()1(4)()1(4
)()()(
4.0
3.0
4.0
21
tuetue
ttttt
oo
9
0.3
EMLAB
267.3 Second order transient circuit
Integrating both sides,
)()()(1
00
tidt
dCtidxx
LR SL
t
t
)()()(
10
0
tdt
diLtdxxi
CiR SC
t
t
dt
di
CLCdt
d
RCdt
d S112
2
dt
d
LLC
i
dt
di
L
R
dt
id S12
2
EMLAB
27Solution to 2nd order differential equation :
Homogeneous differential equation :
20201 ,2 aa
Trial solution :
Characteristic equation :
ζ : damping ratioω0 : natural frequency
2
)()()a
AtxAtfex p
02 2002
2
ccc x
dt
dx
dt
xd
0212
2
ccc xa
dt
dxa
dt
xd
)(212
2
tfxadt
dxa
dt
xd
)()()( txtxtx cp
stc Ketx )(
0)2(2 200
2200
2 stststst KessKesKeKes
02 200
2 ss
Standard form :
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28
Characteristic equation :
Homogeneous solution :
Initial conditiontsts
cp eKeKa
Atxtxtx 21
212
)()()(
212
)0( KKa
Ax
22110
)(KsKs
dt
tdx
t
02 200
2 ss
12002,1 s
tstsc eKeKtx 21
21)(
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29
(1) |ζ|>1 : over damped
(2) |ζ|=1 : critically damped
(3) |ζ|<1 : under-damped
tsts eKeKA
tx 21212
0
)(
2120
)0( KKA
x 2211)0( KsKs
dt
dx
tAtAeA
tx ddt
sincos)( 212
0
0
120
)0( KA
x 210)0( KK
dt
dxd
tetKKA
tx 0212
0
)(
12
0
)0( KA
x 210)0( KK
dt
dx
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30
t
t
ttt
eBtBtx
BtBtxe
txetxee
txtdt
dxt
dt
xd
0
0
00
0
)()(
)(
0])([0])([1
0)()(2)(
21
21
2002
2
Critically damped: ζ=1
Derivation for the critically damped case :
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31
)(ts
Typical unit step responses with ζ changed
(1) |ζ|>1 : over damped
(2) |ζ|=1 : critically damped
(3) |ζ|<1 : under-damped
02 2002
2
ccc x
dt
dx
dt
xd
02
2
LC
i
dt
di
L
R
dt
id
L
CR
LC 2,
10
(ζ is proportional to R.)
EMLAB
32Example 7.7
( R=2, C=1/5F, L=5 H)
0)0()(1
0 dt
dCidxx
LR L
t
01
2
2
LCdt
d
RCdt
d
Let us assume that the initial conditions on the storage elements are iL(0) = -1 A and υC(0) = 4 V. Let us find the node voltage v(t) and the inductor current.
05.22
2
dt
d
dt
d
stKet )(
015.22 ss
5.022
5.15.2
2
45.25.2 2
ors
Applying Kirchhoff current law to node υ(t),
Trial solution :
Characteristic equation :
EMLAB
33
tt eKeKt 5.02
21)( General solution :
Initial condition for υ(0)
4)0( 21 KK
Initial conditions for 0tdt
d
210
5.02)(
KKdt
t
t
are used to determine unknown coefficients.
0)( 0 dt
dCti
R L
Initial conditions can be extracted from KCL on node υ
5)(1
0 tiCRCdt
dL
2,2 21 KK
EMLAB
34
tt eet 5.02 22)(
EMLAB
35Example 7.8
The series RLC circuit shown in Fig. 7.18 has the following parameters: C=0.04 F, L=1 H, R=6 , iL(0) = 4 A, and υC(0) = -4 V. Let us determine the expression for both the current and the capacitor voltage.
0)0()(1
0 dt
diLdxxi
CiR C
t
02
2
LC
i
dt
di
L
R
dt
id 02562
2
idt
di
dt
id
02562 ss
jj
s 432
86
2
10066 2
stKeti )(
Applying Kirchhoff voltage law to the loop,
Trial solution :
Characteristic equation :
EMLAB
36
)4sin4cos()( 213 tKtKeti t
Initial condition for i(0) : 4)0( 1 Ki
Initial conditions for 0tdt
di
014640)()(1
00
dt
di
dt
diLtdxxi
CiR C
t
t
20dt
di
2043)4cos44sin4()4sin4cos(3 21213
213
0
KKtKtKetKtKedt
di tt
t
2,4 21 KK
)4sin24cos4()( 3 tteti t
EMLAB
37Example 7.9Determine the current i(t) and the voltage υ(t) in the figure.
0)()(1
00
dt
diLtdxxi
CiR C
t
t
AiV
HLFCRR
LC 2
1)0(,1)0(
2,8
1,8,10 21
Using KVL : 0)(1 tiRdt
diL C
2
)(Rdt
dCti
Using KCL :
01
2
211
22
2
LCR
RR
dt
d
L
R
CRdt
d096
2
2
dt
d
dt
d
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38
0962 ss
stKet )(
3s
Trial solution :
Characteristic equation :
tetKKt 321 )()(
Initial condition for υ(0)
11)0()0( KC
Initial conditions for 0tdt
d
tt eKeKdt
d 31
32 3
5.0)0(20
iRdt
dC
t
3185.0
)0()0(
20
CRC
i
dt
d
t
33 120
KKdt
d
t
62 K
tett 3)61()( tetRdt
dCti 3
2 2
3
2
1)(
EMLAB
39Example 7.11
Let us determine the output voltage υ(t) for t>0.
HLFCRR 2,4
1,2,10 21
24)(1 tiRdt
diL C
2
)(Rdt
dCti
LCLCR
RR
dt
d
L
R
CRdt
d 241
2
211
22
2
48127
2
2
dt
d
dt
d
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40
481272
2
dt
d
dt
d
Particular solution : Atp )(
48121272
2
Adt
d
dt
dp
pp
4A
Homogeneous solution :
sth Ket )(
4301272 orsss
tthp eKeKttt 4
23
14)()()(
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41
Initial conditions :
][212)0()0(21
2 VRR
RCC
][112
)0()0(21
ARR
ii LL
0)0()0(
12020
CRC
i
dt
d
Rdt
dC
tt
043,24)0( 210
21
KKdt
dKK
t
6,8 21 KK
tt eet 43 684)(
Before the switch action:
Using KCL :
EMLAB
42Application example 7.13Consider the high-voltage pulse generator circuit shown in Fig. 7.26. This circuit is capable of producing high-voltage pulses from a small dc voltage. Let’s see if this circuit can produce an output voltage peak of 500 V every 2 ms, that is,500 times per second.
1
0
1)(
T
indtVL
tiSwitch in position 1 :L
TVi in
p1
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43
Switch in position 2 :
0)()(
0)()(
tiL
R
dt
tditRi
dt
tdiL
)(
0
1
)(Tt
L
R
eiti
)(
0
1
)()(Tt
L
R
KeRtit
500)( 110 K
L
RTVT in 500
10
10053
11
KT
L
RTVin
][5],[11 AImsT p
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44
Consider the simple RL circuit, shown in the figure, which forms the basis for essen-tially every dc power supply in the world. The switch opens at t=0. Vs and R have been chosen simply to create a 1-A current in the inductor prior to switching. Let us find the peak voltage across the inductor and across the switch.
Application example 7.15
Before the switch action :
L
Vti
L
R
dt
tdiVtiRR
dt
tdiL S
SSL )()(
)()()(
][1)0( AR
Vi
S
S
After the switch action :
tL
R
SS
tL
RS e
RRVKe
R
Vti
11)(
LR
SR
),( SLSL RRRRR
S
S
LSS
LS
LSSS
S
SS
R
V
RRR
RV
RRRVK
R
VK
R
Vi
)(
11)0(
)(
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45
Inductor voltage :
tL
R
S
LSL e
R
RV
dt
diLt 1)(
Switch voltage :
t
L
R
SS
LL eV
R
RtiRt )()(switch
1
S
L
R
R0t
)(switch t
SS
L VR
R
(inductive kick)
LS RR
L
• At the peak inductor voltage is negative infinity! This voltage level is caused by the attempt to disrupt the inductor current instantaneously.
• The peak switch voltage jumps up to positive infinity.• This phenomenon is called inductive kick, and it is the nemesis of power supply
designers.
EMLAB
46Circuit design to avoid inductive kick
Before the switch action :
t
CSCLS dxxiC
VtiRRdt
tdiL
0)(
1,)()(
)(
R
Vi S )0(
After the switch action :
02
2
LC
i
dt
di
L
R
dt
id
iRout
SR
LR
C
S
t
LS VdxxiC
tiRRdt
tdiL 0 )(
1)()(
)(
EMLAB
47
01
2 2200
2 LC
sL
Rsss
]/[10,1 60 srad Set arbitrarily
][199],[10 SL RRRnFC
020 2,
1 L
R
LC ][2001010010122
],[10101010010
11
660
86122
0
LR
nFL
C
tetKKti610
21 )()(
][1)0( 1 AKi
SLSt
ViRRdt
diL
)0()(0
Initial conditions :
44
0
10199)12001(10)0()(1
iRRVLdt
diLSS
t
tt etKKeKdt
di 66 10621
102 )10)((
16
20
10 KKdt
di
4442 10991019910100 K
EMLAB
48
tetti6105 )109.91()(
tLout ettiRt
6105 )109.91(199)()(
][199)0( Vout
EMLAB
49Application example 7.16
][24 V
• DC voltage sources or power supplies are fed from AC outlets on walls.
• The ac waveform is converted to a quasi-dc voltage by an inexpensive ac–dc con-verter whose output contains remnants of the ac input and is unregulated.
• A higher quality dc output is created by a switching dc–dc converter. Of the sev-eral versions of dc–dc converters.
• We will focus on a topology called the boost converter.
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50
The boost converter with switch set-tings for time intervals (a) ton and (b) toff.
)(1
)( 000 ononin
t
in tiIL
tVIdtV
Lti
on State #1 :
State #2 :
0)(
)()(1
)()(1
)(
ItitL
VV
tidtVVL
tidttL
ti
onoffoin
on
tt
t oinon
tt
t L
offon
on
offon
on
(I0 is the initial current at the be-ginning of each switching cycle)
00 IIL
tVt
L
VV oninoff
oin
Assuming that Vo is nearly constant,
EMLAB
51
0)(0
oninoffoinonin
offoin tVtVV
L
tVt
L
VVin
off
onoff Vt
ttV
0
)(1
)( 000 ononin
t
in tiIL
tVIdtV
Lti
on )()(1
)( on
tt
t oin tidtVVL
tioffon
on
EMLAB
52
inin
onoff
onin
off
onoff VD
V
ttt
Vt
ttV
1
1
1
10
onoff
on
tt
tD cycle)Duty (
EMLAB
53
)(2
)(1
21)( onon ttsttsino eKeKVt
Rdt
dCtiVt
dt
diL oo
ino
)(,)(
LC
Vt
LCdt
d
RCdt
d ino
oo )(11
2
2
State #2 :
)(1
)(1
onoontt
o tRC
tiCdt
d
on
State #1 :
0)()(
R
t
dt
tdC oo
0)(1)(
tRCdt
tdo
o RC
t
CC et )0()(
21)( KKVt inono
RC
t
oono
on
et )0()(
ont offt0
2211 sKsKdt
d
ontt
o
)(to
EMLAB
54Application example 7.20 : automobile ignition system
• The voltage source represents the standard 12-V battery.• The inductor is the ignition coil. The inductor’s internal resistance is
modeled by the resistor.• The switch is the keyed ignition switch.
Optimum starter operation requires an over-damped response for iL(t) that reaches at least 1 [A] within 100 ms after switching and remains above 1 [A] for between 1~and 1.5 s. let us find a value for the capacitor that will produce such a current waveform.
Requirement:
EMLAB
55
100 ms
1~1.5s
02
2
LC
i
dt
di
L
R
dt
id
0)0()(1
0 iR
dt
diLdxxi
C C
t
Characteristic equation :
0)())((2 21212
21200
2 ssssssssssss
stKeti )( 012
LCs
L
Rs
The values of s1 and s2 should be set to satisfy the requirements.
Requirement:
EMLAB
56
2121
1, ss
LCss
L
R
Comparing the two expressions, we see that
1221 0)0( KKKKiL
0)0()0(0
tC dt
diLRi
)()( 2112211 ssKsKsKtdt
diL
60125)0()0(1
0
C
t
RiLdt
di
60)( 121 ssK
2021 L
Rss
Derived constraints:
Derived constraints:
tstsL eKeKti 21
21)(
EMLAB
57
(1) Let us arbitrarily choose s1=3 and s2=17, which satisfies the condition.
29.460
121
ssK ][98
1
21
mFsLs
C ][)(29.4)( 173 Aeeti ttL
Design :
0.5s
Fail
(2) Let us arbitrarily choose s1=1 and s2=19, which satisfies the condition.
33.360
121
ssK ][263
1
21
mFsLs
C
][)(33.3)( 19 Aeeti ttL
1.5s
Success
EMLAB
58Application example 7.21 : Defibrillator
A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated quivering of the heart muscle fibers—by delivering an electric shock to the heart. Its key feature, as shown in the figure, is its voltage waveform. A simplified circuit dia-gram that is capable of producing the Lown waveform. Let us find the necessary values for the inductor and capacitor.
• The Lown waveform is oscillatory. → Under-damped (ζ <1).• The waveform starts from zero voltage.
LCteKtx dd
t 1,1,sin)( 0
201
0
Lown waveform
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59
• The period = 10 [ms]. 2002
Td
12250
3000
)4/3(
)4/( 2/4/3
1
4/1 0
0
0
T
T
T
o
o eeK
eK
T
T
250
49712ln2
0 T
• From the ratio of voltage peaks
012
LCs
L
Rs
02 200
2 ss
][3.504972
49722 0 mHR
LL
R
20
2220
20 1801)497()200()( dd
Characteristic eq. for a series RLC circuit.
• Natural frequency :
][311120
20
FL
CLC
T
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