Download - AC IRCUITS
AC CIRCUITS
Ho Kyung Kim, Ph.D.
School of Mechanical Engineering
Pusan National University
Basic Experiment and Design of Electronics
Outline
• Capacitor & inductor
• RC circuits
• RLC circuits
• Storing energy in a electric field (capacitive energy storage)
– in the form of a charge separation when appropriately polarized by an electric field
– i.e., voltage
• Acting as an open circuit in DC
• Capacitance, C/V = F
Capacitor
dielectricsCVQ
)()( tCvtq
Taking a differentiation;
t
CC dttiC
tv ')'(1
)(
We can have;
Or
00
')'(1
)( VdttiC
tvt
tCC
where
0
')'(1
)( 00
t
CC dttiC
ttvV
initial voltage due to some charge stored at time t0
dt
tdvCti
dt
tdq CC
)()(
)(
• Energy storage in capacitors
t
C
t
C
t
C
t
C
dttiCCC
dttiC
dttiC
dttiC
tvtvtvtv
')'(111
')'(1
')'(1
')'(1
)()()()(
321
321
321
''
)'()'(')'()'(')'()( dt
dt
tdvCtvdttitvdttPtW C
CCCCC)(
2
1)( 2 tCvtW CC
Series circuit Parallel circuit
• Storing energy in a magnetic field
– current flowing
• Acting as a short circuit in DC
• Inductance, Vs/A = Wb/A = H
Inductor
dt
tdiLtv L
L
)()(
t
LL dttvL
ti ')'(1
)(
00
')'(1
)( IdttvL
tit
tLL
0
')'(1
)( 00
t
LL dttvL
ttiI
We can have;
Or
where
• Energy storage in inductors
Series circuit Parallel circuit
dt
tdiLLL
dt
tdiL
dt
tdiL
dt
tdiL
tvtvtvtv
)(
)()()(
)()()()(
321
321
321
)(
2
1)()()()()( 2 tLi
dt
d
dt
tdiLtitvtitP L
LLLLL
')'(
2
1
'')'()( 2 dttLi
dt
ddttPtW LLL )(
2
1)( 2 tLitW LL
• To describe the behavior of a voltage or a current during the transition between two distinct steady-state conditions
Transient analysis
v or i
t
steady-state
region
steady-state
region
transient
region
RC circuit
v(t) C
R
0')'(1
)()(
t
CCS dttiC
tRitv)()( titi CR
Applying KVL;
Differentiating;
0)(1)()(
tiCdt
tdiR
dt
tdvC
CS
dt
tdv
Rti
RCdt
tdi SC
C )(1)(
1)(
Another way;
dt
tdvCti
R
tvtvti C
CCS
R
)()(
)()()(
)(1
)(1)(
tvRC
tvRCdt
tdvSC
C
In general, any circuit containing a single energy storage element;
)()()(
001 tfbtxadt
tdxa
)()()(
0
0
0
1 tfa
btx
dt
tdx
a
a
)()()(
tfKtxdt
tdxS
first-order ordinary differential equation
first-order system equation
time constant = RC = V/I Q/V [time/charge] [charge] = [time]
or = L/R = Vs/I/(V/I) [time]
DC gain or static sensitivity
0')'(1)(
)()(
t
S dttiCdt
tdiLtRitv 0')'(
1)()()(
t
S dttiCdt
tdiLtRitv
RLC circuit
v(t) C
R LApplying KVL;
Differentiating;
0)(1)()()(
2
2
tiCdt
tidL
dt
tdiR
dt
tdvS
dt
tdvti
Cdt
tidL
dt
tdiR S )(
)(1)()(
2
2
Another way;
)()()()(
2
2
tvtvdt
tvdLC
dt
tdvRC SC
CC
dt
tdvCti C )(
)(
In general, any circuit containing two energy storage elements;
)()()()(
0012
2
2 tfbtxadt
tdxa
dt
txda
)()()()(
0
0
0
1
2
2
0
2 tfa
btx
dt
tdx
a
a
dt
txd
a
a
)()()(2)(1
2
2
2tfKtx
dt
tdx
wdt
txd
wS
nn
2
0
a
awn
20
1 1
2 aa
a
second-order ordinary differential equation
second-order system equation
where natural frequency
damping ratio
• Capacitor voltages and inductor currents cannot change instantaneously
DC steady-state condition
dt
tdvCti C
C
)()(
dt
tdiLtv L
L
)()(
– instantaneous changing iC(t) or vL(t)
– PC or PL (impossible!!!) since power is energy per time or v i
v or i
tt = 0
abrupt change
infinite slope
• DC steady state refers to circuits that have been connected to a DC (voltage or current) source for a very long time ( t )
– assuming that all voltages and current in the circuits have become constant
– all derivatives in the governing equations go to zero
)()()(
tfKtxdt
tdxS )()(
)(2)(12
2
2tfKtx
dt
tdx
wdt
txd
wS
nn
dt
tdvCti C
C
)()(
dt
tdiLtv L
L
)()(
iC(t) 0 as t vL(t) 0 as t
Steady-state capacitor current; Steady-state inductor voltage;
Open circuit at DC Short circuit at DC
Transient response of first-order circuit
)()()(
tfKtxdt
tdxS
VB
t = 0R
C vC(t)
Considering that f(t) is a forcing function, switched on at time t = 0;
FKtxdt
tdxS )(
)(
0)()(
txdt
tdxN
N
)()( tx
dt
tdx NN
/)( t
N etx
for t 0 I.C., x(t =0) = x(0)
solution = natural response + forced response= homogeneous solution + particular solution
natural response when t = 0
or
x(t)
t
e-1
= 0.368
FKtxdt
tdxSF
F )()(
)()( xFKtx SF
)()()()( // xeFKetxtxtx t
S
t
FN
)()0()0( xxtx )()0( xx
)1)(()0(
)()]()0([)(
/
/
t
t
exx
xexxtx
forced response for t 0
or
Then, we have the complete response
applying I.C.;
Therefore,
t
x(t)
x()
x(t)
t
x(0) + x()
x(0)
Transient response of second-order circuit
0)()()( tvtiRtv CSTT
)()( tvtv LC dt
tdiLtv L
C
)()(
0)()()()(
tidt
tdvC
R
tvtvL
C
T
CT0)()()( tititi LCS
vT
R
C vC(t) L vL(t)
iC(t) iL(t)iS(t)
Applying KVL;
Applying KCL;
)()()(
)(1
2
2
tidt
tidLC
dt
tdi
R
Ltv
RL
LL
T
T
T
)(1
)()()(
2
2
tvR
tidt
tdi
R
L
dt
tidLC T
T
LL
T
L
Then, we have;
or
dt
tdv
R
Ltv
dt
tdv
R
L
dt
tvdLC T
T
CC
T
C )()(
)()(2
2
Why not vC(t)?
0)()()( tvtiRtv CSTT 0)()()( tititi LCSfrom and
0)()()()(
tidt
tdvC
R
tvtvL
C
T
CT
we can have
There is no unique method to arrive at the final equation!
)()]()([)( tvtitiRtv CLCTT
differentiating the above equation and using
① asymptotically tending to final value of 1
– due to KS = 1 or x(t) = f(t) when steady state (all derivatives are zero)
② oscillating with a approximate period of 6 s
– due to wn = 1 or T = 2p/wn 6.28 s
③ decaying as time goes
– due to
– when = 1, no longer overshoot like that of first-order system
)()()(2)(1
2
2
2tfKtx
dt
tdx
wdt
txd
wS
nn
t
No
rma
lize
d A
mp
litu
de
1
T = 2p/wn
Let's consider the response of second-order systemto switched unit input with KS = 1, wn = 1, = 0.2
)()()(2)(1
2
2
2tfKtx
dt
tdx
wdt
txd
wS
nn
solution = natural response + forced response= homogeneous solution + particular solution
(1) Natural response
0)()(2)(1
2
2
2
tx
dt
tdx
wdt
txd
wN
N
n
N
n
st
N etx )(
012
2
2
sww
s
nn
tsts
N eetx 21
21)(
14)2(2
1 222 nnnnn wwwwws
021 2
2
stst
n
st
n
eesw
esw
or
natural response
characteristic polynomial or equation
characteristic roots s1 and s2
① overdamped solution ( > 1; real and distinct roots)
1
1
1 2
nn ww
1
1
2 2
nn ww
where
21
2221
/
2
/
1
)1(
2
)1(
121)(
tt
twwtwwtsts
N
ee
eeeetx nnnn
1 = 2 = 1, wn = 1, = 1.5
② critically damped solution ( = 1; real and repeated roots)
where
1 = 2 = 1, wn = 1, = 1
/
2
/
1
212121)(
tt
twtwtsts
N
tee
teeteetx nn
nw
1
③ underdamped solution ( < 1; complex conjugate roots)
where1 = 2 = 1, wn = 1, = 1.5
tjwwtjwwtsts
Nnnnn eeeetx
)1(
2
)1(
121
2221)(
assuming that 1 = 2 = ;
twetwe
eeetx
d
tw
n
tw
tjwtjwtw
N
nn
nnn
cos2)1cos(2
)(
2
)1()1( 22
21 nd ww damped natural frequency
– wd wn as 0
– time constant for exponential decay, = 1/ wn
– (oscillation decays quickly) as (more damping)
(2) Forced response
)()()(2)(1
2
2
2tfKtx
dt
tdx
wdt
txd
wS
nn
FKtx SF )(
FKxtx SF )()(
switched DC sources (or excitations) constant excitation all derivatives are zero
forced response (DC steady-state solution)
(3) Complete response
① overdamped case
② critically damped case
③ underdamped case
)(
)()()(
)1(
2
)1(
1
22
xee
txtxtx
tjwwtjww
FN
nnnn
)(
)()()(
)1(
2
)1(
1
22
xee
txtxtx
twwtww
FN
nnnn
)(
)()()(
21
xee
txtxtx
twtw
FN
nn
• unknown constants 1 and 2
– from two I.C.'s
)0()0( xtx
)0(')(
0
xdt
tdx
t
Laplace-transformed models of circuit elements
R
L
C
Transfer function
• Called “Network function”
• Providing the system engineer with a great deal of knowledge about the systems operation, since dynamic properties of transfer functions are governed by the system poles
• In deriving a TF, all IC’s are set equal to zero
From the Laplace transform with all the IC’s = 0,
outputor response
network or transfer function
input orforcing function
- Zeros = roots of numerator- Poles = roots of denominator
Pole-zero plot
Recall the natural responses:
Second-order network:
XX
X
X
X
Example
vi(t) C
R L
vo(t)