MAE140 Linear Circuits 23

Equivalent Circuits & Reduction (T&R Chap 2,3)

Circuit EquivalenceTwo circuits are equivalent if they have the same i-v

characteristics at a specified pair of terminals

Terminal = external connection to two nodes = port

Our aim is to simplify analysis replacing complicatedsubcircuits by simpler equivalent circuits

Example

Association of resistors

MAE140 Linear Circuits 24

Equivalent Circuits & Reduction (T&R Chap 2,3)

Resistors in Series

Resistors in Parallel

R2R1

A

B

REQ

A

B

=

R2

R1

A

B

REQ

A

B

=

MAE140 Linear Circuits 25

Example 2-11 (T&R, 5th ed, p. 34)

Consider the circuit

Compute equivalent cctsfrom AB and from CD R3

R2

R1

D

CA

B

32

321

3232

11||

RR

RR

RRRRR

CDEQ+

=!!"

#$$%

&+==

'

32

321321 ||

RR

RRRRRRR

ABEQ+

+=+=

MAE140 Linear Circuits 26

Example 2-10 (T&R, 5th ed, p. 31) Revisited Can you find iA?

All you need to know is

vA=30V

+

+

+++

- -

-

--30V

100Ω

200Ω300ΩvA v

3

v2

v1i2iA

MAE140 Linear Circuits 27

Example 2-10 (T&R, 5th ed, p. 31) Revisited Can you find iA?

All you need to know is

vA=30V

Answer

iA = -iR =-30/150 = -200 mA;

+

+

+++

- -

-

--30V

100Ω

200Ω300ΩvA v

3

v2

v1i2iA

150Ω++

- -30V vA

iA

=

iR

+

-vR

MAE140 Linear Circuits 28

Equivalent ccts

Equivalent ccts for resistive networks are familiarreductions of parallel and series connectionsThe equivalent cct depends on the port

From an external view the cct could be replaced byits equivalentThe internal cct variables are now unavailable

How would you compute them?

Could we substitute for the cct below?

R3R2

R1

Restof cct 2

Restof cct 1

MAE140 Linear Circuits 29

Equivalent sources

i-v relationships determine equivalence

+-vS

Restof cct

R

+

-

v

i

+ -vR

iSRestof cctR

+

-

v

i

iR

+-

Restof cct

R

+

-

v

i

+ -vR

iS R

Restof cctR

+

-

v

i

iR

vS / R

MAE140 Linear Circuits 30

Equivalent Sources (cntd)i

vSRestof cctR

+

-

v+- Svv =

Sii =iSRestof cct

R1

+

-

v+ -vR

i

i

vSRestof cct

+

-

v+-

iSRestof cct

+

-

v

i

MAE140 Linear Circuits 31

Equivalent Sources (cntd)i

Restof cct

+

-

v 21 iiiS +=

21 vvvS +=

i

vSRestof cctR

+

-

v+-

iSRestof cct

+

-

v

i

i1 i2

i

v1Restof cct

+

-

v+-

+-v2

MAE140 Linear Circuits 32

Equivalent Sources (cntd)i

Restof cct

+

-

v

21 iiiS ==

i

vSRestof cctR

+

-

v+-

iSRestof cct

+

-

v

i

i1

i2 i

v1

Restof cct

+

-

v

+- v2 -

+

21 ii =

Only works ifthen

21 vv =

21 vvvS ==

Only works ifthen

MAE140 Linear Circuits 33

Circuit Reduction

For ladder networks

Reduce complexity by successively replacing elements bytheir equivalents

What happens with three elements in series or in parallel?

…

a b c

c a bc b a

a c b

M

+ -i v

They are all equivalentWe can commute elements

MAE140 Linear Circuits 34

T&R, 5th Ed, Example 2-22 p 49

Find v

Reduce the left end

+_

10Ω20Ω

20Ω 20Ω 10Ω

10Ω

15V

_+ v

10Ω

20Ω 20Ω 10Ω

10Ω

0.75A

_+ v20Ω

10Ω

20Ω 20Ω 20Ω0.75A

_+ v20Ω

10Ω

10Ω 10Ω0.75A

_+ v

10Ω10Ω

10Ω

_+ v_+7.5V Vv 5.2=

MAE140 Linear Circuits 35

Voltage & Current Dividers

S

total

vR

Rv

11 =

321 RRRRtotal ++=

S

total

vR

Rv

33 =S

total

vR

Rv

22 =; ;

+

+

+

+

-

-

-

-

vS

v1

v2

v3R3

R2

R1

i

iS

i1 i2 i3

G1 G2 G3

+

-

v

321 GGGGtotal ++=

S

total

iG

Gi

11 = S

total

iG

Gi

22 = S

total

iG

Gi

33 =;;

i

iR

G1

=

MAE140 Linear Circuits 36

Voltage Dividers

Often we use a voltage dividerto provide an input voltage toa cct elementWhen would this work?

When the “rest of cct” doesnot draw much currentcompared to R2

Why is this?What is it asking of theequivalent of the rest of cct?

Note that this is a verycommon circuit used to“bias” a transistor to anoperating voltage

++-

-

vS voR2

R1

i

Restof cct

MAE140 Linear Circuits 37

Thévenin and Norton Equivalent Ccts

Thévenin’s TheoremIf the source cct in a two-terminal interface is linear, then

the interface signals v and i do not change when thesource cct is replaced by its Thévenin equivalentNote: nobody says the load must be linear!

Thévenin Equivalent CircuitvT is the open-cct voltage of sourceRT is evaluated from short-cct current

Sourcecct

Loadcct

i+_v

+_vT

RT Loadcct

+_vT

RT voc

+

_

isc

+_vT

RTocT vv =

sc

ocT

T

oc

T

Tsc

i

vR

R

v

R

vi

=

==

MAE140 Linear Circuits 38

Thévenin’s Theorem Proof

Linearity of the Source cct is the key – superpositionHook up a test current source to cct

itest yields voltage vtest

Part I, itest,1=0 but i and v sourcesin Source cct left ON then vtest,1=voc=vT

Part II, itest,2≠0 and sources left OFF in Source cct then vtest,2=-RTitest,2

By linearity of the Source cct vtest is the sum of these partsfor any choice of itest

This establishes the i-v relationship for any load cct

Sourcecct

i

_vtest itest

MAE140 Linear Circuits 39

Thévenin and Norton Equivalent Ccts

Norton’s TheoremIf the source cct in a two-terminal interface is linear, then

the interface signals v and i do not change when thesource cct is replaced by its Norton equivalent

Norton Equivalent CircuitiN is the short-cct currentRT is evaluated from open-cct voltage

Sourcecct

Loadcct

i+_v

iN RN

Loadcct

voc

+

_

isc

T

T

N

T

scN

R

v

R

v

ii

=

=

=

T

sc

ocN

NscNNoc

R

i

vR

RiRiv

=

=

==

iN RN

iN RN

MAE140 Linear Circuits 40

Example 3-16 p.110 T&R, 5th ed

Find the Thévenin and Norton equivalent ccts of

Find the voltage, current and power if load is 50Ω

Answer: vT=-30V; iN=-417mA; RN=RT=72ΩV=-12.3V; i=-246mA; p=3.03W

+ -

Load

60Ω

30Ω40Ω20Ω

20Ω 50V

+

-

i

v