Circuit Theory -EM II

Dr David OttawayBraggs 412

Ph 8313 5165david.ottaway@adelaide.edu.au

Circuit Theory Outline DC Circuits Resistors and Voltage

Sources AC Circuits- Resistors, Capacitors and

Inductors Frequency Response Transfer functions Time Domain Response

Active Circuits

DC Circuits and Networks Ohms Law and Resistors Power and Resistors Ideal voltage and current sources Kirchoffs Laws (Apply to AC circuits as well) Resistor and Voltage Source Networks Thevenins Equivalent Circuits Nortons Equivalent Circuits Voltage Dividers (Apply to AC circuits as well) Parallel and Series resistors

Resistors and Ohms Law

( ) ( )resistorV t RI t=

Symbol

Ohm's Law:

V is the voltageR is the Resistance (SI Unit Ohm ())I is the current

2 possibilities

Power Dissipation in an electrical component

2

2

( ) ( ) ( )( )( ) /

Power t I t V tI t RV t R

=

=

=

Apply Ohms Law

Can be derived from the fundamental definition of Voltage ie Voltage = Energy/Charge

Ideal Voltage Sources

An ideal voltage source provides a fixed voltage regardless of the resistance placed across it

A useful model that does not exist in reality

A large battery is an excellent approximation

Battery Ideal Voltage Source

The sum of the changes in potential around any closed path of a circuit must be zero

From Giancolli

Kirchoffs Loop Rule

Based on the law of conservation of energy

Example: Resistor Combinations -Series

Req = R1 + R2 + R3

Kirchhoff s Junction RuleAt any junction point, the sum of all currents

entering the junction must equal the sum of all current leaving the junctionThis is based on the law of conservation of charge

B

At Junction A

I = I1 + I2 Also at B

Note Some currents will need to be negative

Resistor Combinations - Parallel

eq 1 2 3

1 1 1 1R R R R

= + +

Resistor Network Analysis

R1

R3

R2R3

Vs1

Vs2

Use Kirchoffs Laws to derive 5 Equations in 5 unknowns

Thevenins Equivalent Circuits Any terminal networks of voltage

sources and resistors is equivalent to a single resistor in series with a single voltage source

Figure from Wikipedia

Calculating Vth and Rth The value of Vth is value of the voltage when no

current is drawn from the circuit The value of Rth is determined by:

Where Ishort is the current drawn from the terminals when the source is short circuited

Rth is also the resistance between the terminals when all voltage sources are replaced by zero resistance

thth

short

VRI

=

Thevenins Equivalent CircuitExample: Wheatstone Bridge

R1VR2

R3 R4

Vout

Nortons Theorem

Any network of batteries and resistors is equivalent to a perfect current source in parallel with a resistor

INO is the short circuit current, Ishort (See Thevenins) and RNO = RTh (See Thevenins)

Figure from Wikipedia

AC Circuits Capacitors and Inductors required Will consider continuous periodic signals and

transient response Periodic Signals

Best way to deal with is complex circuit analysis Consider the transfer functions of

RC Circuits RL Circuits LCR Circuits

Circuit Components - Inductor

( ) ( )inductor dI tV t L dt=

Symbol

V is the voltageL is the Inductance (SI Unit Henry (H))I is the current (SI Unit Ampere (A))

Circuit Components - Capacitor

( ) ( ) ( )0t

capacitor

I t dtQ tV t

C C= =

Symbol

V is the voltageC is the capacitance (SI Unit Farad (F))I is the current

Q is the charge stored on the plates

Current vs Voltage (Capacitor)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1

-0.5

0

0.5

1Current and Voltage for a 1uF Cap

time(s)

V

o

l

t

a

g

e

(

V

o

l

t

s

)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1

-0.5

0

0.5

1x 10-4

C

u

r

r

e

n

t

(

A

m

p

s

)

Voltage lags the currentVoltage

Current vs Voltage (Inductor)

Voltage leads the current

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1

-0.5

0

0.5

1Current and Voltage for a 1uH Inductor

time(s)

V

o

l

t

a

g

e

(

V

o

l

t

s

)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-1

-0.5

0

0.5

1x 104

C

u

r

r

e

n

t

(

A

m

p

s

)

Voltage

Complex Analysis

( )2

( )2

1 1( )2 2

1 1( )2 2

1 1

i t i t

i t i t

i ti t i tddt

i ti t i t

Cos t e e

Sin t e e

e i e e

e dt e ei

pi

pi

+

= +

=

= =

= =

Electronic Systems as Linear Systems

if = costhen = cos +

Vin(t) TF() Vout(t)

Voltage Amplitude Voltage Phase

Phase and amplitude are real numbers

Amplitude and Phase

0 2 4 6 8 10 12 14 16 18 20-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time (s)

V

o

l

t

a

g

e

(

v

)

VinVout

What is the transfer function, amplitude and relative phase for the above ???

Amplitude and Phase

0 2 4 6 8 10 12 14 16 18 20-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time (s)

V

o

l

t

a

g

e

(

v

)

VinVout

AinAout

T T

12

2.22 2 2.36

out

in

ATFA

Trad

T pi pi

=

= = =

Linear SystemsRemember: cos =

+

So: = cos =

+

and = cos +

= 2

+

= 2

+

=

2 +

2

Linear Systems is a complex number

=

=

, = cos + !"( )

Summary of Complex Impedance

1CZ i C

=

Inductor

Capacitor

Resistor

LZ i L=

RZ R=

Transfer Function of RC Circuit

10-2 10-1 100 101 10210-2

10-1

100Transfer Function of RC circuit

Angular frequency w/w0

A

m

p

l

i

t

u

d

e

10-2 10-1 100 101 102-100

-80

-60

-40

-20

0

Angular frequency w/w0

P

h

a

s

e

(

d

e

g

r

e

e

s

)

Transfer Function of RL Circuit

10-2 10-1 100 101 10210-2

10-1

100Transfer Function of RL circuit

Angular frequency w/w0

A

m

p

l

i

t

u

d

e

10-2 10-1 100 101 1020

50

100

Angular frequency w/w0

P

h

a

s

e

(

d

e

g

r

e

e

s

)

Transfer Function of LCR Circuit

10-2 10-1 100 101 10210-3

10-2

10-1

100Transfer Function of LCR circuit

Angular frequency w/w0

A

m

p

l

i

t

u

d

e

10-2 10-1 100 101 102-100

-50

0

50

100

Angular frequency w/w0

P

h

a

s

e

(

d

e

g

r

e

e

s

)

B=0.1B = 1B=10

Transfer Function of LCR Notch

10-2 10-1 100 101 10210-4

10-2

100Transfer Function of LCR circuit

Angular frequency w/w0

A

m

p

l

i

t

u

d

e

10-2 10-1 100 101 102-100

-50

0

50

100

Angular frequency w/w0

P

h

a

s

e

(

d

e

g

r

e

e

s

)

B=0.1B = 1B=10

Summary of Circuits Considered

( )( )

out

in

VRV i L

=

Vin() Vout()

Vin() Vout()

Vin() Vout()

RC Circuit

RL Circuit

RLC Circuit

( )( )

11

out

in

VV i RC

=

+

( )( ) ( )21

out

in

LiV RLV LC i R

=

+

Transient Response Will consider two circuits with step function

inputs RC Filter RLC

The time domain response is trivial if Laplace Transforms are used Not taught anymore in the Science part of maths(Following slides summarize this, not examinable and unfortunately cannot be used in exam)

Transient Approach1. Use Kirchoffs Loop Law to develop initial

equation, use V& = ' (, V) =*+,()

+ and V- = .

,()

-/

0

2. Differential to covert to first or second order differential equation

3. Solve

Summary of Transient Solutions

= 0 1 2 exp 2 (67

Vin(t) Vout(t)RC Circuit

Vin(t)RLC CircuitVout(t)

= 08

= 08

= 0 1 2(*0

exp 2( 2*7 9:; 0

Transient Response Using Laplace Transforms

(Not examinable and techniques cannot be used in exam/test)

Transient Response of R, L and C Circuits

Start by looking at the Differential Equations

Easier way is to use the S plane version of complex analysis

Utilize the power of Laplace Transforms Deriving equation in s space and transforming

them back to the time domain

Summary of S plane impedance

1CZ i C

=

RZ R=

LZ sL=Inductor

Capacitor

Resistor

LZ i L=

1CZ

sC=

RZ R=

S plane

Laplace Transformations

0

22

2

11

1

0

( ) { ( )} ( )

{ ( ) ( )} { ( )} { ( )}( ){ } { ( )} (0)

( ) (0){ } { ( )} (0)

( ) (0){ } { ( )} (0) .....

1{ ( ) } { ( )}

st

n nn n

n n

t

F s L f t e f t dt

L af t bg t aL f t bL g tdf tL sL f t f

dtdf t dfL s L f t sf

dt dtdf t dfL s L f t s f

dt dt

L f t dt L f ts

= =

+ = +

=

=

=

=

Definition of Laplace Transform

Linearity

Relationship for 1st derivative

Relationship for 2ndderivative

Relationship for nthderivative

Relationship for integral

Common Laplace Transforms Pairs

( )

( )

2 2

2 2

2 21

2 2

12 2

2 2

( ) { ( )} ( ) ( ) { ( )} ( )1 sin( )

1/ ( ) cos( )

1/ / ( 1)! sin( )1 ( tan( / ))1 1

sin( )( 1)! ( )( )1 c( ) ( )

n n

at

n at atn

at at

F s L f t f t F s L f t f tt t

s

ss u t t

s

ss t n t

s

e as a

t e e tn s as a

a s ae e

s s a s a

= =

+

+

+ + +

+

=

+

+ ++

+

+ + +

( )( ) ( )( )

os( )

1 1 1( ) ( )( ) ( )at bt at bt

t

se e ae be

s a s b a b s a s b b a

+ + + +

End of Transient Response Using Laplace Transforms

Operational Amplifiers Electronic Circuits made up

of transistors that electrically amplify signals

Op Amps are amongst the most utilized and versatile of all electronic components

Cost between a few cents and a few $100 dollars

Operational Amps Both Voltage and Power Gain

Transformers can increase the voltage of oscillating signals but cannot increase power Not DC and low frequency signals

Op Amps are active hence can amplify both voltage and power

Crucial for many applications such as stereo systems

CD playerLight signal->Voltage signal ->Amplified -> Sound Signal

Operational Amplifiers

V+ Non inverting inputV

-

Inverting inputVout OutputG Open loop gainVout = (V+- V-)G

V+

V-

Vout+

-

Ideal Op Amp has infinite open loop gain, infinite input impedance and zero output impedance

Operational Amplifiers -The RealityParameter Ideal Typical Values

Open Loop Gain (G) Infinite < 106-107

Input Impedance (Rin)

Infinite Typically 106-1012Ohm

Output Impedance (Rout)

Zero 40 2000 Ohms

V+

V-

Vout

+

-

+

Vs = (V+- V-)GVs

Rout

Rin

Two Golden Rules of Op AmpsRule 1 The output of an Op Amp provides

whatever voltage is necessary to make the two input voltages equal

Rule 2 The input draws no current

Apply when negative feedback is applied

Op Amp Circuit Inverting Amplifier

Vout+

-

Vin

Rb

Ra

Op Amp Circuit Non-Inverting Amplifier

Vout+

-

Vin

Rb

Ra

Op Amp Circuit Integrator

Vout+

-

VinR

C

Op Amp Circuit Differentiator

Vout+

-

Vin

R

C

Op Amp Circuit Summing Amplifier

Vout+

-

V1

Rb

RaV2

Ra