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