Post on 31-Mar-2015
V=V0sin(ωt)
V=V0cos(ωt)
V=V0sin(ωt+φ)
AC Circuit TheoryAmplitude
Phase
1
R
Apply AC current to a resistor…
I I=I0sin(ωt)
VR=IR=I0Rsin(ωt)
C
I
20
0
0C
tsinIC
1
tcosIC
1
dttsinIC1
CQ
V
…capacitor…
Voltage lags current by /2I
L 200L tsinLItcosLI
dtdI
LV
…and inductor
Voltage leads current by /22
Adding voltages
R1
R2
R3
I0sin(ωt)
V=I0R1sin(ωt)+I0R2sin(ωt)+I0R3sin(ωt)
=I0(R1+R2+R3)sin(ωt)
+
-
V(t)?
Kirchoff’s Voltage LawandOhm’s law
3
Adding voltages
R1
C1
I0sin(ωt)
tcosC
ItsinRIV 0
10
voltages out of phase…
V(t) =VR+VC
+
+
4
Complex numbers in AC circuit theory
jexpzzIm jzRez
Re(z)
Im(z)
θ j2=–1
Reason:
‘i’ is already taken
|z|cos(θ)
|z|s
in(θ
)
tj
tj
eVtsinVtV
eVtcosVtV
00
00
Im
Re
5
Complex Impedance
R 0
0C
L 0
j t
j t
j t
V IR I R e
1 IV Idt e
C j C
dIV L j I Le
dt
The ratio of the voltage across a component to the current through it when both are expressed in complex notation
RZR
C
1 jZ
j C C
LjZL
I=I0sin(ωt) I=I0ejωt
6
Complex ImpedanceReal part: resistance (R)
Imaginary part: reactance
C1
L
IphaseZphaseVphase
IZV
IZV
jωL
Cj1
R
Ohm’s law
7
Series / parallel impedances
Z1=R Z2=jωL
Impedances in series: ZTotal=Z1+Z2+Z3…
T nn
Z Z
Z1 Z2 Z3
Impedances in parallel
nT n
1 2 3
1 1
Z Z
1 1 1
Z Z Z
LR C
3
jZ
C
8
RC low pass filterR
1/(jwC)VIN VOUT
ZR
ZCCR
CIN
ZZZV
jeRC1
1V
RC1
RCj1V
RCj1RCj1
RCj11
V
RCj11
V
RV
ZZZV
V
2IN
2IN
IN
IN
Cj1
Cj1
INCR
CINOUT
RCtan 1
+ +
- -
9
RL high pass filterR
jwLVIN VOUTLjR
LjVV INOUT
je
1VV
2RL
RL
INOUT
LR
tan 1
+
- -
+
Here is a slight trick:Get comfortable withthe complex result. 10
Bode plot
0
1 R or
RC L
2
1VV
IN
OUT0
(-3dB)
11
DecibelsLogarithm of power ratio
IN
OUT2IN
2OUT
10 VV
log20VV
log10dB
VOUT/VIN dB
10 20
1 0
0.1 –20
0.01 –40
0.001 –60
12
Bode plot
13
Something Interesting
• Capacitor– At low frequencies (like DC) Open circuit
• Not a surprise, it’s got a gap!
– At high frequencies (“fast”) Short circuit!• Inductor
– At low frequencies Short circuit• Not a surprise, it’s just a wire really
– At high frequencies Open circuit!• Sometimes this can help you with your
intuition on the circuit’s behaviour. 14
EXPLAIN PHASORSGo to black Board and
15
LR CI0ejωt
LRC series circuit
IC
1LjR
IZZZ
VVVV
CLR
CLR
2
2
C1
LRZ
R
Ltan C
11
Purely resistive at LC
1R φ=0 Z=R
R
Ltantj2
20
C1
1
eC
1LRIV 16
L
LRC parallel circuit
R C
I0ejωt
R L C
VI I I I
Z
L1
CjR1
Z1
2 2
1Z
1 1j C
R L
1 1j C
R L 1 1
CR L
22
CL
1R1
1Z
C
L1
Rtan 1
17
Bandpass filterR
CVIN VOUTL
INLC
LCOUT V
ZRZ
V
CjLj
1Z1
LC
C1
jZL
1LC
IN
L1
L1
OUT V
Cj
R
Cj
V
2L12
IN
OUT
CR1
1VV
1VV
LC
1
0VV
0 0VV
IN
OUT0
IN
OUT
IN
OUT
+ +
18
Bandpass filter 2L
12IN
OUT
CR1
1VV
build a radio filter
f0=455kHz
Δf=20kHzLC
1f2 00
0
C=10nF
L=12.2μH
2
1
CR1
12
0L12
0
1CR 20L
12
0
R=389Ω
2 f
19
I(t)=I0ejωt
LCR series circuit – current driven
22 1
CLRZ
R
L C
111 tan
Re
Imtan
tjj eeZIIZV 0
L
R
C
I(t)
20
CLjRZ
CjLjRZ
1
1
Freq.Wild Stuff going on!
NOT a Very Good Idea
L
R
C
V(t)=V0ejωt
LCR series circuit – voltage driven
22
C1
LRZ
R
Ltan C
11
j0 eZV
ZV
I
V(t)
maximum RV
I
0I 0
0I
00
2
2
0
C1
LR
VI
21
R/L=106
R/L=105
22
23
Q value No longer on CP2 syllabus (moved to CP3)
peak energy stored
Q 2energy dissipated per cycle
High Q value – narrow resonance
24
FWHM
Q0 Frequency at resonance
Diff. in Freq. to FWHM
Not a
lway
s he
lpfu
l
Q valueFor what we did last time.
High Q value – narrow resonance 25
L
CR
L
CRQ
FWHM
22
1
200
Dw
LC circuit – power dissipation
Power is only dissipated in the resistor
proof – power dissipated inductor and capacitor – none
T
0dttVtI
T1
P
2
T
tcosXIV
tcosLIV
tcosIC
1V
tsinItI
0X
0L
0C
0
Lor C
1X
reactance
0tsinXIT2
1dttcostsinXI
T1
P/2T
0
220
T
0 0C,L
L
R
CI(t)
26
LRC Power dissipation
L
R
C
V(t) Z=Z0ejφ
I
V(t)=
C1
LjRZ
22
0 C1
LRZ
R
Ltan C
1
tsinZV
tI
tsinVtV
0
0
0
tsintsinZV
tItVtP0
20
Instantaneous power 27
Power dissipation
tsintsinZV
tItVtP0
20
sintcoscostsintsinIV 00
dtsintcostsincostsinIVT1
PT
0
200
0 dtcost2cos1TIV
PT
0 2100
T
0
00 cos2t
TIV
cos2IV 00 cosIV RMSRMS
cosφ = power factor 28
POWER IN COMPLEX CIRCUIT ANALYSIS
On black board
But then go ahead and flip to next slide now anyway.
29
Power factor
RMSRMSIVP
power apparentpower average
cos
Resistive load Z=R cosφ=1
Reactive load Z=X cosφ=0
30
Bridge circuits
V(t) V
Z1 Z3
Z2Z4
To determine an unknown impedance
Bridge balanced when VAVB=0
VA VB
2 4
1 2 3 4
Z ZV t V t 0
Z Z Z Z
2 4
1 2 3 4
2 3 4 4 1 2
2 3 4 1
Z ZZ Z Z Z
Z Z Z Z Z Z
Z Z Z Z 31
CP2 September 2003
32
33
34
35
BACKUP SLIDESPhasors and stuff
36
R
PhasorsI I=I0sin(ωt)
VR=IR=I0Rsin(ωt)
C
I
20
0
0C
tsinIC
1
tcosIC
1
dttsinIC1
CQ
V
I
L 200L tsinLItcosLIdtdI
LV
37
tsinRIV 0R
tcosC
IV 0
C
VT
Phasors
ωt V0sin(ωt)
V0cos(ωt)
R CI0sin(ωt) tcos
CI
tsinRIV 00T
38
RC phasorsR C
I0sin(ωt) tcosC
ItsinRIV 0
0T
RI0
CI0
VT=V0sin(ωt+φ)
CR1
tan
C1
RIV2
200
I0sin(ωt)
V0sin(ωt+φ)
CR1
tan 1 39
Phasors – mathematics
tcosC
ItsinRIV 0
0T
VT=V0sin(ωt+φ)
CR1
tan
C1
RIV2
200
Asin(ωt) + Bcos(ωt) = Rsin(ωt + φ)
Rsin(ωt + φ) = Rsin(ωt)cos(φ) + Rcos(ωt)sin(φ)
A=Rcos(φ) B=Rsin(φ)
A2 + B2 = R2
B/A = tan(φ)40
RL phasorsR L
I0sin(ωt)
tcosLItsinRIV 00T
RI0
0LIVT=V0sin(ωt+φ)
RL
tan
LRIV 2200
I0sin(ωt)V0sin(ωt+φ)
RL
tan 141
RL filterR
LVIN VOUT
2RL
RL
22IN
OUT
1
LR
1L
VV
0VV
0
1VV
IN
OUT
IN
OUT
high pass filter
RI0
0OUT LIV 22
00 LRIV
42
Phasors: RC filterR
CVIN VOUTRI0
CI
V 0OUT
22
00IN C1
RIVV
2
2
C12
IN
OUT
RC1
1
R
1C
1VV
1VV
0
0VV
IN
OUT
IN
OUT
low pass filter43
CP2 June 2003
44
45
46