AC CIRCUIT

22
ESE211 Lecture 3 Phasor Diagram of an RC Circuit V(t)=V m sin(ωt) R V i (t) C V o (t) Current is a reference in series circuit KVL: V m = V R + V C V R V m I m ϕ V C

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BASIC AC CIRCUIT explanation

Transcript of AC CIRCUIT

ESE211 Lecture 3

Phasor Diagram of an RC Circuit

V(t)=Vm sin(ωt) R

Vi(t) C Vo(t)

• Current is a reference in series circuit

KVL: Vm = VR + VC

VR

Vm

Im ϕ

VC

ESE211 Lecture 3

Phasor Diagram of an RL Circuit

V(t)=Vmsin(ωt) VR (t) R

Vi(t) L Vo(t)

KVL: Vm = VR + VL

Vm

ϕ

VL

Im VR

ESE211 Lecture 3

Phasor Diagram of a Series RLC Circuit

VC VL

C L

Vi R VR

KVL: Vm = VC + VL + VR

VL

Vm

VL-C

VR Im

VC

• Voltages across capacitor and inductor compensate each other

ESE211 Lecture 4

Integrating Circuit

VR Vi(t) R Vo(t) C

V V Vi R= + C or V V Vi R= +2 2

C

• Assume high frequency 1/ωC << R then VR >> VC V VR i≈

V CI dt

RCV dtO i

t

i

t

= ≈∫ ∫1 1

0 0

Vi

V V dtO i∝ ∫

• Accurate integrating can be obtained at high

frequency that leads to a low output signal

ESE211 Lecture 4

Differentiating Circuit

VC Vi(t) C Vo(t) R

V V Vi R= + C or V V Vi R= +2 2

C

• Assume low frequency 1/ωC >> R then VR << VC VR

Vi V V C i≈ VC

I dt I Cddt

Vi i

t

i i≈ ⇒ =∫1

0

V V I R RCddt

VO R i i= = ≈

V

ddt

VO i∝

• An accurate result can be achived at low frequency.

ESE211 Lecture 4

Transient Processes

in Passive Circuits

RC Circuit without a Source

• The circuit response is due only to the energy stored in the

capacitor

t=0

C R

• The capacitor C is precharged to the voltage V0 Applying KCL:

C dV/dt +V/R = 0 • This is the 1st order differential equation

ESE211 Lecture 4

Solution of the Differential Equation

dV/dt + V/RC = 0

dV/V = -dt/RC

ln V = -t/RC +a

V = A e -t/RC , A=e+a • e is the base of the natural logarithms, e = 2.718…

• Continuity requires the initial condition:

At t = 0 V(0) = V0

V(t) = V0 e -t/RC

• The response governed by the elements themselves

without external force is called a natural response

ESE211 Lecture 4

The Time Constant V(t)

V0

0.368 V0 0.05 V0

τ time 3τ

• The time constant τ = RC is the rate at which the natural

response decays to zero

• Time τ is required to decay by a factor of 1/e=0.368 • After the time period of 3τ the transient process is considered to

be completed

ESE211 Lecture 4

Determination of the Time Constant

V0 V/e τ t

V1

V

1) From the solution of the equation

By definition: as the time when the signal decreases by 1/e

V(t+τ)/V(t) = 1/e

The time t can be chosen arbitrarily

2) From the slope of the line at t=0 Differentiating the solution

dV/dt = -{V0/τ } e –t/τ

V1(t) = -{V0/τ }t +V0 V1(τ) = 0

ESE211 Lecture 4

Transient Response of an Integrating RC Circuits

VR

Vi VC

KVL: Vi = VR + VC

Vi

t

VO =VC

VR t

t

ESE211 Lecture 4

Transient Response of a Differentiating RC Circuits

VC

Vi VR

KVL: Vi = VC + VR

Vi

t

VO = VR

t

VC

ESE211 Lecture 5

Transfer Function Vi(t) Vo(t) Vi(t) = Vimcos(ωt) Vo(t) = Vomcos(ωt+ϕ)

Vom(ω) ϕ(ω) Vo(t) =Re { Vo e jωt}, Vo = Vome jϕ

i

o

VVT )()( ω

ω =

Amplitude response |T(ω)| = Vom(ω)/Vim

Phase response ∠T(ω) = ϕ(ω)

ESE211 Lecture 5

The Transfer Function of

an Integrating RC Circuit R Vi(t) C Vo(t)

RCjCj

R

CjV

VTi

o

ωω

ωωω

+=

+==

11

1

1)()(

ωτω

jT

+=

11)(

τ = RC

ESE211 Lecture 5

-3 -2 -1 0 1 2 30.0

0.5

1.0

1000ωο100ωο

10ωοωο0.1ωο

0.01ωο0.001ωο

0.707

1000ωο100ωο10ωοωο0.1ωο

0.01ωο0.001ωο

|Vo/V

in|

Frequency (ω)

-3 -2 -1 0 1 2 310-3

10-2

10-1

100 slope-20 dB/dec-3 dB

Frequency (ω)

|Vo/V

in|

-60

-40

-20

0

V 0 [dB]

2211)(

τωω

+=T

Amplitude Response of

the Integrating RC Circuit

• The magnitude of the transfer function is

• The Integrating RC circuit is a LOW-PASS filter

ESE211 Lecture 5

-3 -2 -1 0 1 2 3

-90

-45

0

1000 ω ο 100 ω ο 10 ω ο ω ο 0.1 ω ο 0.01 ω ο 0.001 ω ο

V o

Frequency ( ω )

[ ][ ])(Re

)(Imarctan)(ωω

ωTTT =∠

)arctan()1)(1(

1)( ωτωτωτ

ωτω −=

−+−

∠=∠jj

jT

Phase Response of

the Integrating RC Circuit

The angle of the transfer function is

ESE211 Lecture 5

The Transfer Function of

a Differentiating RC Circuit C Vi(t) R Vo(t)

)()()()()(

ωωωω

ωRCi

o

XIRI

VVT

⋅⋅

==

RCjRCj

CjR

RX

RTRC ω

ω

ωω

ω+

=+

==11)(

)(

2222 1)(

1)1()(

τωωτωτ

τωωτωτ

ω+

+=

+−

=jjjT

τ = RC

ESE211 Lecture 5

-3 -2 -1 0 1 2 30.0

0.5

1.0

1000ωο100ωο

10ωοωο0.1ωο

0.01ωο0.001ωο

V0=0.707 V

in

1000ωο100ωο

10ωοωο

0.1ωο0.01ωο

0.001ωο

|V

o/Vin|

Frequency ( ω)

-3 -2 -1 0 1 2 310-3

10-2

10-1

100

slope-20 dB/dec

or-6 dB/oct

-3 dB point at ω0

Frequency ( ω)

|V

o/Vin|

-60

-40

-20

0

V 0 [dB]

222211

)()(τω

ωττω

ωτωτω

+=

++

=jT

Amplitude response of the Differentiating

RC Circuit

• Amplitude response of the RC circuit is the magnitude of T(ω)

• The Differentiating RC circuit is a HIGH-PASS filter

ESE211 Lecture 5

-3 -2 -1 0 1 2 3

0

45

90

1000ωο100ωο10ωοωο0.1ωο0.01ωο

0.001ωο

V 0

Frequency ( ω)

++

∠=∠ 221)()(

τωωτωτ

ωjT

ωτω

1arctan)( =∠T

Phase response of the Differentiating

RC Circuit

• Phase response of the RC circuit is the phase angle of T(ω)

ESE211 Lecture 5

Frequency Response of a Series RLC Circuit

VC VL

C L

Vi R VR

KVL: Vi = VC + VL + VR

VL

VL-C

VR Im

VC

Vi

• Voltages across capacitor and inductor compensate each other

ESE211 Lecture 5

-3 -2 -1 0 1 2 30.0

0.5

1.0

1000ωο100ωο10ωοωο0.1ωο0.01ωο0.001ωο

V0=0.707 Vin

1000ωο100ωο10ωοωο0.1ωο0.01ωο0.001ωο

|Vo/V

in|

Frequency ( ω)

-3 -2 -1 0 1 2 30.0

0.5

1.0

1.5

|VL/Vin||VC/Vin|

Frequency (ω)

|VC/

V in|,

|VL/V

in|

22

0

11|/|

−+

=++

=

CLR

R

RCj

Lj

RVV IN

ωωω

ω

Amplitude Response of a Series RLC Circuit

ESE211 Lecture 5

-3 -2 -1 0 1 2 3

-180

-90

0

1000 ωο100ωο10ωοωο0.1ωο0.01ωο0.001ωο

V o

Frequency (ω)

-3 -2 -1 0 1 2 310-6

10-4

10-2

100

slope-40 dB/dec

|Vo/V

in|

-120

-80

-40

0

V 0 [dB]

Amplitude and Phase Response of

the Ladder RC Circuit R R Vin C C Vo

1000ωο100ωο10ωοωο0.1ωο0.01ωο0.001ωο Frequency (ω)

ESE211 Lecture 5

Amplitude and Phase Response of a Wien-Bridge Circuit

R C

BandWidth

-3 -2 -1 0 1 2 30.0

0.33

V0=0.707 Vmax

1000ωο100ωο10ωοωο0.1ωο0.01ωο0.001ωο

|V/V

|

Frequency ( ω)

0.245

o

in

Vin R C Vo Cj

RCjR

CjR

ZZT

TOTAL

OUT

ωω

ωω

11

1

)( 1

1

++

+

+

== −

Frequency (

-3 -2 -1 0 1 2 3 -100

0.001 ω ο

-50

0

50

100

V o

1000 ω ο 100 ω ο 0.01 ω ο 0.1 ω ο 10 ω ο ω ο ω )