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Unformatted text preview: ELEC 101 Transient Circuits 1 2009/10 Spring Transient Circuit
Example
Study the behavior of a network as a function of
time after a sudden change in the network occurs
due to the switches opening and closing. Example switch opens at t = 0
iL t=0 2k You have to key in the pass code
within a certain time (say 30 seconds)
after entering your house. t The alarm will be ON after 30 seconds
if any intruder has triggered the
sensor of the alarm system. + VL(t) 5A
L 2k VL(t) 0 10 kV
a high voltage arc across the switch ELEC 101 1
a Transient Circuits 2 2009/10 Spring Switch Operations
t=0 t < 0, switch is opened
t = 0 and t > 0, switch is closed c
t=0 t=0 t < 0, switch is closed
b t=0 t = 0 and t > 0, switch is opened
t < 0, switch is closed
t = 0 and t > 0, switch is opened ELEC 101 2 Transient Circuits 3 Continuity of Capacitor Voltage VC Capacitor voltage VC is unchanged after switching
or VC before switching = VC just after switching
or VC is continuous after the switching Example Find VC(t) at t < 0, t = 0, and t > 0.
t=0 + VC(t) C
1mF 1m
5V Given C has no
initial charge 2009/10 Spring ELEC 101 a Transient Circuits 4 Before switching i.e. at t < 0
i. switch is opened
+
ii. C has no initial
VC(t)
charge Q Hence VC = Q/C = 0 c 2009/10 Spring t > 0 i.e. switch is closed
VC charges up to 5V
(will be discussed later) VC(t)
C
1mF 1m 5V 5V c
a VC ( t 0) 0V 0
b t VC is continuous at t = 0 (after switching)
b Just after switching i.e. at t = 0
d +
i. switch is closed VC(t) ii. Since using the
continuity theorem,
VC is unchanged
after the switching C
1mF VC (t 0 ) 0V
Hence VC just after switching
= VC before switching VC (t 0 ) VC (t 0) 0V 1m
5V At t = i.e. the switch closed
for a long time
i. circuit is at steady state
ii. C behaves like an open circuit VC ( ) 5V ELEC 101 3 Transient Circuits 5 C acts like an open circuit at steady state
At steady state, the capacitor is fully charged
(discharged), no current flows in the capacitor
and the capacitor behaves like an open circuit
(iC = 0). 4 Capacitor Current Arc
Capacitor current iC NOT continuous with
time.
The capacitor current can “jump” from one
value to another value.
Sometime it can cause a high current arc. 2009/10 Spring ELEC 101 Transient Circuits 6 2009/10 Spring VC(t)
Sometime we use
VC(0) to represent the capacitor voltage
right before t = 0, and t = 0+
t = 0 VC(0+) to represent the capacitor voltage
right after t = 0.
Therefore, VC(0+) = VC(0).
Continuity theorem of capacitor voltage 0
t= 0 t=0 t
t= 0+ iC(t)
t = 0+ On the other hand, iC(0+) ≠ iC(0).
If a switching occurs at t = 0, capacitor
current right after t = 0 may not
necessary equals to the capacitor current
right before t = 0. t = 0 t = 0 t 0
t = 0+
t=0 ELEC 101 Transient Circuits 7 2009/10 Spring Find iC(t) at t < 0, t = 0+, and t > 0. Example iC(t) t=0
+
iC(t) VC(t) 1m C
1mF + C has no
initial charge VC(t) C
1mF 1m
5V iC(t 0) 5kA 5V Hence iC is not continuous at t = 0.
a t < 0 switch is opened for a long time
C = open circuit c + iC(t) VC(t) C
1mF i C (t 0) i C (t 0 ) 0A
b t = 0 switch is closed
i. C has no initial charge
ii. Since VC (at t = 0+) = VC (at t < 0)
= VC(0) = 0V iC(0) 5V V (0 ) 5V 0V
C 5kA
1m
1m 1m t>0 i.e. switch is closed VC charges up to 5V 5V
d t iC() Since VC() = 5V 5VV () 5V5V
C 0A
1m
1m
+
iC(t) i C () 0 A VC(t) C
1mF 1m
5V ELEC 101 Transient Circuits 8 2009/10 Spring VC(t)
5V
b t d VC is continuous after switching
(at t = 0, VC = 0 and t = 0+, VC = 0) t>0 c t=0
t<0 a iC(t) t 0 t=0 +
iC(t) VC(t) high current arc C
1mF 1m
5V 5kA
c t>0
d
a t<0 t
t 0
b t=0 iC is not continuous after switching
(at t = 0, iC = 0 and t = 0+, iC = 5 kA) ELEC 101 5 Transient Circuits 9 Continuity of Inductor Current iL Inductor current iL is unchanged after switching
or iL before switching = iL just after switching
or iL is continuous after the switching 6 L acts like a short circuit at steady state
At steady state, a constant current flows through
the inductor. The voltage across the inductor is
zero and the inductor behaves like an short
circuit (VL = 0). 2009/10 Spring ELEC 101 Transient Circuits 10 Example Find iL(t) at t < 0, t = 0+, and t > 0.
t=0
circuit is at steady
state for t < 0
iL(t)
1A and switch is
L
1k 1k
opened at t = 0 a t<0 c 2009/10 Spring t>0 d t switch is closed
iL(t) circuit is at steady state
L acts like a short circuit iL(t)
1A L switch is opened 1A L
1k 1k i L () 0A 1k 1k
i. Since the circuit is at steady state,
ii. L behaves like a short circuit Hence the current transient iL(t)
iL(t) i L (t 0) 1A
b t=0 a 1A switch is opened c iL is discharged to 0A t<0
iL(t) 1A L
1k 1k i. Since iL is unchanged after the switching
Hence iL (t = 0+) = iL (t < 0) = 1A d
0
b
t
iL is continuous
at t = 0 (after switching) i L ( t ) 0 A ELEC 101 7 Transient Circuits 11 Inductor voltage Arc
Inductor voltage VL is NOT continuous with time.
The inductor voltage can “jump” from one value
to another value.
Sometime it can cause a high voltage arc. Example Find VL(t) at t < 0, t = 0+, and t > 0 .
t=0 +
iL(t) VL(t) L 1A
1k 1k 2009/10 Spring Example:
Fluorescent
Lamp
Starter ELEC 101 a Transient Circuits 12 Since iL(t < 0) = iL (t = 0+) = iL(0) = 1A t < 0 switch is closed for a
long time VL(0) iL(0) 1k L = short circuit Hence VL is not continuous at t = 0 since
1A +
VL(0+) 1k 1k VL (t 0) 0V t>0 d t switch is opened
circuit is at steady state
L acts like a short circuit Since iL() = 0A V () iL() 2 k02 k 0 V
L t = 0 switch is opened +
VL(t) Since iL(0+) = iL(<0) = 1A +
VL(t) L 1k iL(0+) = 1A VL (t 0 ) VL (t 0)
c b L 1A1k1kV iL(<0) = 1A
+
VL(t) L 2009/10 Spring 1A L
1k 1k VL(t 0 ) 1kV a
0 1A
1k 1k VL(t)
t
c d b
iL is NOT continuous 1 kV high voltage arc ELEC 101 8 Transient Circuits 13 iL(t) equation 2009/10 Spring To find time constant after switching,
kill V and I sources (short V source
and open I source) Step 2 This equation will be given in exam I L (t ) I L ( ) [I L (0 ) I L ( )] e t / is time constant = L/R after switching means t > 0, switch is open iL(t) 1A current
source is killed
(replaced by
open circuit) L 1mH 1k 1k
Example Find iL(t) for t = 0+ and t > 0. t=0
iL(t)
1mH 1A L
1k 1k Step 1
At t = 0+ L 1 mH 1 s
R 1 k Find iL(), and substitute
into the iL(t) equation
iL(0+), i L (0 ) i L ( t 0) 1 A At t = switch is opened, L is short i L ( ) 0 A Step 3 Substitute IL(), IL(0), into IL(t)
equation i L (t ) i L ( ) [i L (0 ) i L ( )] e t / 0 A [1 A 0 A]e t / 1s 1 e t / 1s A iL(t)
1A
0 t ELEC 101 Transient Circuits 14 Example Find i(t) for t > 0. Circuit is at steady state for t < 0.
t=0 8V a L
2H t<0 VO(t) Neon
lamp 100 4k Neon lamp: Need a high voltage to ignite. i(t) Lamp ignites when VO > 300V.
R of the lamp is very large. Circuit is at steady state.
L acts like a short circuit. i(<0) 8V 2009/10 Spring 8V VO(<0)
100 4k
2H i(t 0) i(<0) 8V 0.08 A
100 Neon lamp is OFF as
VO(t<0) = 8 V < 300V ELEC 101 b Transient Circuits 15 at t = 0+ switch is connected to 4 k i(t 0 ) i(t 0) 0.08 A VO (0 ) 8V i(0) 4 k 2009/10 Spring i( t ) i() [i(0) i()] e t 2mA [80mA 2mA] e t / 0.5 ms 8V 0.08A 4k 312 V
312V
8V
8V 4k
i(0+)
2H c i(t)
Magnitude of
VO > 300V.
Hence neon
lamp ignites. 100
320V
i(0+) at t = L acts like a short circuit. i ( ) d VO(0+) 8V 2mA
4100 at t > 0
100 4k
2H L
R
2H 0.5 ms
4100 80 mA 2 mA
0
t=0 t ELEC 101 9 Transient Circuits 16 VC(t) equation Step 2 This equation will be given in exam VC (t) VC () [VC (0 ) VC ()] et / 2009/10 Spring To find time constant after switching,
kill V and I sources (short V source
and open I source) after switching means t > 0 , switch is closed is time constant = CR
C
1mF 1m kill V source Find VC(t) for t = 0+ and t > 0. Example CR 1 mF 1 m 1 s t=0
+ VC(t) C
1mF 1m
5V Step 3 Substitute VC(), VC(0), into
VC (t) equation VC (t ) VC ( ) [VC (0 ) VC ( )] e t / Step 1 C has no initial charge At t = 0+, VC (0 ) VC ( t 0) 0V 5V [0V 5V ]e t / 1s 5 (1 e t / 1s )V VC(t)
5V At t = , switch is closed, C is open VC () 5V 0 t ELEC 101 Transient Circuits 17 11 Proof for continuity of iL and VC 10 + iC VC iL +
VL CVC ( t ) 2
EC 2 Li ( t )
EL L
2 2 V VC(0) = A E(t)
tangent
at t = 0
0 VC (t) A e t VC (t 5 τ) VC () Energy EC and EL must be continuous
with time (unchanged after switching) a 2009/10 Spring t VC() 0.368 A
VC(5) 0.007 A
VC() 0 A Hence VC and iL must be continuous with time. 99.3%
b Another short proof iC C
If VC dVC ( t )
dt VL L di L (t )
dt (or iL ) is not continuous, infinite iC (or VL ) is required. Hence dVC/dt and
diL/dt must be finite, or VC and iL must be
continuous with time. ELEC 101 12 Transient Circuits 18 13 Proof of VC(t) equation VT Proof of iL(t) equation I(t) R
t=0 C 2009/10 Spring I(t) t=0
+
V(t) at t 0 IT R L +
V(t) at t 0 dI
V (IT I) R L
dt
dt
dI
dI L / R IT I
I IT VT V
dV
I
C
R
dt
dt
dV
dV CR VT V
V VT V(t) VT A e I(t) IT A etR / L t / CR t 0
t t 0 V() VT 0 V(t) V() [V(0) VT ] e t / CR I(0) IT A t V(0) VT A I() IT 0 I(t) I() [I(0) IT ] etR / L ELEC 101 14 Transient Circuits 19 15 C in parallel i 2009/10 Spring C in series i i i C1 V C1 C2 V Ceff Ceff C1 C2 V V V1 V2 Proof V 1
1
( ) idt
C1 C2 dV
dV
dV
dV C2 (C1 C2 ) Ceff
dt
dt
dt
dt Example C1 = 3F and C2 = 6F.
Find Ceff .
3F
9V 6F idt +18C V2 = 3V 6V
9V Ceff = 3F + 6F = 9F 1
Ceff C1 = 3F and C2 = 6F.
Find Ceff and V2 . Ceff 3F // 6F 2F
Example Ceff V C2 Proof i C1 1
1
1 Ceff C1 C2 18C
+18C
18C +18C
2F
18C V2 = 3V ELEC 101 16 17 L in parallel 2009/10 Spring L in series
i i i
V Transient Circuits 20 i is equivalent to
V L1 L2 is equivalent to
L1 V 1
1
1 L eff L1 L 2
Proof 1
1 ) Vdt
L1 L 2 i 1
L eff V2 Vdt V V1 V2 (L1 L2 ) Example
Example 3H V Leff Leff L1 L2 Proof i( V1 L2 Leff di
di Leff
dt
dt Find Leff . Find Leff . 6H Leff = 3H//6H = 2H 3H
Leff 6H Leff = 3H + 6H = 9H Leff ELEC 101 Example Transient Circuits 21 Find iO(t) for t 0.
Circuit is stable for t < 0. Kill 12V
(replace by a short) = CR
= 200F(3k //3k) 0.3s 3k V 3k iO
C
200F 12V 6k t=0 3 Find VC(t) first for t 0.
Then use VC(t) to find iO(t). use VC (t) VC () [VC (0 ) VC ()] e
Find VC(0)
for t < 0 t / 3k V 3k iO
C 12V 2 12V 4
6k VC () 12V for t 0 3k V 3k
C
200F 5 Use VC(t) to find iO(t) for t 0
6k 3k
6V
6k Substitute VC(0+), VC() and into VC(t) equation for t 0 9k t/ 9V VC (0 ) VC (0 ) 9V VC(t) VC() [VC(0 ) VC()]e
12k 6 [9 6]et/0.3s V 6 3et/0.3s V
iO(<0)=12V/12k = 1mA Find time constant 12V 200F Find VC () VC() 6V VC(0) 9V 0.3s VC ( 0) 12V 3k V 3k
C 3k V 3k
C 1 2009/10 Spring iO(t) V (t)
C 2 1et/0.3s mA
3k t VC(t) 6 3e0.3s V
9V
6V 0 0.3s iO(t) 2 1e t
t
0.3s mA 3mA
2mA
1mA 0 0.3s t ELEC 101 Transient Circuits 22 Example Charge and discharge of flash bulb Close S1 to charge and close S2 to discharge.
Find iC(t) and VC(t) .
camera
t=0
flash
3k V (t) S2
S1
C
15V 1 100F iC(t) 3 a b At t < 0, circuit is in steady state.
Hence C acts like an open circuit.
Therefore VC(<0) = 0V.
3k V (t) S2
C
100F At t = , i.e. S1 is closed for a long time. VC VC () 15V
c Find time constant CR 100 F 3 k 0.3 s Charge 15V At t = 0, S1 is closed. VC (0) VC ( 0) 0V d S1 2009/10 Spring iC(t) Substitute into equation and find the
voltage across C in the charging process VC (t) V() [V(0 ) V()] et / CR 15V [0 15V] et / 300ms 15V(1 et / 300ms ) ELEC 101 2 Transient Circuits 23 2009/10 Spring Discharge
After S1 is closed for a long time, VC = 15V.
S1 is then opened. At t = 0, S2 is closed.
3k V (t) S2
C S1
15V 100F iC(t) VC(t) is continuous
1 3 2 charge VC(t) 15 (1 e t / 300ms )V 15V 15 et / 0.3msV a VC (0 ) VC ( 0) 15V b VC ( ) 0 V c CR 100 F 3 0.3 ms d VC ( t ) V() [V(0 ) V()] e t / CR 0 [15V 0] e t / 0.3ms 15V e t / 0.3ms discharge tangent
0 300ms S1 close
S2 open 0 0.3ms S2 close
S1 open ELEC 101 Transient Circuits 24 2009/10 Spring Example Find iC(t) from VC(t).
1 S1 Charge
15V 3k V (t) S2
C
iC(t) is NOT continuous iC(t) 100F 1 iC (0 ) 5 mA i (t) iC() 0A
15V VC(t)
iC(t) 5et 300ms mA
3k C
15V 5mA
3 k Discharge
15V S1 discharge 5 et / 300msmA 0.3ms 0 300ms
2 2 charge 0 5 e t / 0.3ms A 3k V (t) S2
C
100F iC (0 ) 5 A iC () 0A V (t)
iC(t) C 5et 0.3ms A
3 iC(t) 3 15V 5A
3 High current
arc (5A) lights
up flash bulb ...
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This note was uploaded on 01/28/2011 for the course ELEC 101 taught by Professor Chan during the Fall '09 term at HKUST.
 Fall '09
 CHAN

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