TWO-PORT NETWORKS

In many situations one is not interested in the internal organization of anetwork. A description relating input and output variables may be sufficient

A two-port model is a description of a network that relates voltages and currentsat two pairs of terminals

Admittance parametersImpedance parametersHybrid parametersTransmission parameters

LEARNING GOALS

Study the basic types of two-port models

Understand how to convert one model into another

ADMITTANCE PARAMETERS

The admittance parameters describe the currents in terms of the voltages

2221212

2121111

VyVyIVyVyI

The first subindex identifies

the output port. The secondthe input port.

1 port to applied is voltage a and circuited-short is port

the when 2 port into flowingcurrent the determines 21y

The computation of the parameters follows directly from the definition

02

112

01

111

12

VV V

IyVIy

02

222

01

221

12

VV V

IyVIy

The network contains NO independent sources

LEARNING EXAMPLEFind the admittance parameters for the network

2221212

2121111

VyVyIVyVyI

][23)

211( 1111 SyVI

2111, yy determine to used Circuit

2I

1212 21

211 VIII ][

21

21 Sy

2212 , yy determine to used Circuit][

65

31

21

2222 SyVI

][21

6553

323

12221 SyVII

Next we show one use of this model

An application of the admittance parameters

2221212

2121111

VyVyIVyVyI

212

211

65

21

21

23

VVI

VVI

221 4,2 IVAI

The model plus the conditions at theports are sufficient to determine theother variables.

21

21

41

65

210

21

232

VV

VV

22 41VI

][112

][1186

13

2

2

21

AI

VV

VV

Determine the current through the4 Ohm resistor

LEARNING EXTENSIONFind the admittance (Y) parameters

1I

1V

2I

2221212

2121111

VyVyIVyVyI

1I

1V

2I

2V

111 423)

421

211( VVI

12 422142 II

][211

][141

21

11

Sy

Sy

1I 2I

2V

21

22

5.10215.10211

212

II

VI

][211

][71

12

22

Sy

Sy

LEARNING EXTENSION

5A10

OI

Use the admittance (Y) parameters to find the current Io

Conditions at I/O portsAI 101

2

22 51

II

VI

o

1I

1V2I

2V 2221212

2121111

VyVyIVyVyI

][211

][141

21

11

Sy

Sy

][211

][71

12

22

Sy

Sy

Replace in model

)5(71

211

)5(211

14110

1

1

oo

o

IVI

IV

Solve for variable of interest ][98420 AIo

IMPEDANCE PARAMETERS

The network contains NO independent sources

2221212

2121111

IzIzVIzIzV

The ‘z parameters’ can be derived in a manner similar to the Y parameters

01

221

01

111

22

II I

VzIVz

02

222

02

112

11

II I

VzIVz

LEARNING EXAMPLE Find the Z parameters

Write the loop equations

)(42)(42

1222

2111

IIjIjVIIjIV

24442

2221

1211

jzjzjzjz

212

211

244)42(

IjIjVIjIjV

rearranging

01

221

01

111

22

II I

VzIVz

02

222

02

112

11

II I

VzIVz

2221212

2121111

IzIzVIzIzV

LEARNING EXAMPLEUse the Z parameters to find the current through the 4 Ohmresistor

2221212

2121111

IzIzVIzIzV

Output port constraint

22 4IV

Input port constraint

11 )1(012 IV

212

211

244)42(

IjIjVIjIjV

21 )24(40 IjIj

21 4)43(12 IjIj

)43( j4j

2))43)(24(16(48 Ijjj 73.13761.12I

LEARNING EXTENSIONFind the Z parameters. Find the current on a 4 Ohm load with a 24V input source

1I

1V

2I

2V

)(63)(612

2122

2111

IIIVIIIV

9,66,18

2221

1211

zzzz

4

V24

212

211

96618IIVIIV

1I

1V

2I

2V

22 4IV :constraint port output][241 VV :constraint port input

21

21

136061824IIII

)3(

2)639(24 I

][3324

2 AI

HYBRID PARAMETERS

The network contains NO independent sources

2221212

2121111

VhIhIVhIhV

admittance output circuit-opengain current forward circuit-short

gain voltage reverse circuit-openimpedance input circuit-short

22

21

12

11

hhhh

These parameters are very common in modeling transistors

01

221

01

111

22

VV I

IhIVh

02

222

02

112

11

II V

IhVVh

LEARNING EXAMPLE Find the hybrid parameters for this circuit

Non-inverting amplifier Equivalent linear circuit2221212

2121111

VhIhIVhIhV

21

21111211 )||(

RRRRRhIRRRV ii

2RI

DSI

o

iDSR R

IARIRR

RIII 11

21

122

21

121 RR

RRARho

i

21

1122

21

11 RR

RhVRR

RV

)(||0

21

22 RRR

VIVo

i

)( 21

2122 RRR

RRRho

o

LEARNING EXTENSIONFind the hybrid parameters for the network

1I

1V

2I

2V2221212

2121111

VhIhIVhIhV

1I

1V

2I

02 V

14))3||6(12( 1111 hIV

32

636

2112

hII

1V

2I

2V01 I

32

636

1221

hVV

][91

9 222

2 ShVI

LEARNING EXTENSION

4

Determine the input impedance of the two-port

2221212

2121111

VhIhIVhIhV

1I

1V

2I

2V 22 4IV :constraint port output

1

1

IVRin

)4()4(

2221212

2121111

IhIhIIhIhV

122

212 41

Ih

hI

122

2112111 41

4 IhhhhV

][91

32

32,14

2221

1211

Shh

hh

23.15131614

)9/1(41)3/2)(3/2(414inR

Verification

1342127||612inR

TRANSMISSION PARAMETERS

The network contains NO independent sources

221

221

DICVIBIAVV

02

1

02

1

22

II V

ICVVA

02

1

02

1

22

VV I

IDIVB

ratio current circuit-short negative Dadmittance transfer circuit-open C

impedance transfer circuit-short negative Bratio voltage circuit openA

ABCD parameters

LEARNING EXAMPLEDetermine the transmission parameters

221

221

DICVIBIAVV

02

1

02

1

22

II V

ICVVA

02

1

02

1

22

VV I

IDIVB

jAV

j

jV

111

1

12

jVII

jV

2

112

1

02 I when02 V when

112 11

11

1

Ij

I

j

jI

jD 1

211 )1(12)1||1(1 Ij

jjI

jV

jB 2

LEARNING EXTENSIONDetermine the transmission parameters

221

221

DICVIBIAVV

02

1

02

1

22

II V

ICVVA

02

1

02

1

22

VV I

IDIVB

1I

1V

2I

2V

02 I when

3215.10

5.1012

AVV

3I

][71

5.10214242

2

1213 S

VICVII

02 V When

23

214242

12

DII

)23(1414)21||42( 2111 IIIV

21B

PARAMETER CONVERSIONS

If all parameters exist, they can be related by conventional algebraic manipulations.As an example consider the relationship between Z and Y parameters

2

11

2221

1211

2

1

2

1

2221

1211

2

1

2221212

2121111

VV

zzzz

II

II

zzzz

VV

IzIzVIzIzV

2

1

2221

1211

VV

yyyy

1

2221

1211

2221

1211

zzzz

yyyy

1121

12221zzzz

Z

12212211 zzzzZ with

matrix ingcorrespondthe of tdeterminan the for stands symbol the table, conversion following the In

DCBA

hhhh

yyyy

zzzz

THYZ ,,,2221

1211

2221

1211

2221

1211

INTERCONNECTION OF TWO-PORTSInterconnections permit the description of complex systems in terms of simplercomponents or subsystems

The basic interconnections to be considered are: parallel, series and cascade

PARALLEL: Voltages are the same.Current of interconnectionis the sum of currents

CASCADE:Output of first subsystemacts as input for thesecond

The rules used to derive modelsfor interconnection assume thateach subsystem behaves in thesame manner before and afterthe interconnection

SERIES: Currents are the same.Voltage of interconnection is the sumof voltages

aaaba

aaa

a

aa

a

aa VYI

yyyy

YVV

VII

I

2221

1211

2

1

2

1 ,,bbb VYI manner similar a In

baba

baba

VVVVVVIIIIII

222111

222111

,,

:sconstraint ctionInterconne

ba

ba

VVVIII

ba YYY

VYYVYVYI babbaa )(

Parallel Interconnection: Description Using Y Parameters

YVIVV

yyyy

II

2

1

2221

1211

2

1

ndescriptioctionInterconne

Series interconnection using Z parameters

bbbaaa IZVIZV ,subsystem each of nDescriptio

ba

ba

VVVIII

sconstraint ctionInterconneIZZIZIZV baba )(

ba ZZZ

SERIES: Currents are the same.Voltage of interconnection is the sumof voltages

Cascade connection using transmission parameters

ba

ba

baba

IIIIVVVV

VVII

2211

2211

1212

:sconstraint ctionInterconne

a

a

aa

aa

a

a

IV

DCBA

IV

2

2

1

1

b

b

bb

bb

b

b

IV

DCBA

IV

2

2

1

1

221

221

DICVIBIAVV

2

2

1

1

IV

DCBA

IV

2

2

1

1

IV

DCBA

DCBA

IV

bb

bb

aa

aa

Matrix multiplication does not commute.Order of the interconnection is important

CASCADE:Output of first subsystemacts as input for thesecond

LEARNING EXAMPLEFind the Y parameters for the network

112

2j1I

1V

2I

2V

1I

1V

2I

2V

12

121 2II

IjVV

21,

21

21

1211 jySjj

y aa

SjySjy aa 21;

21

2221

212

211

32

IIVIIV

2113

51

3112 1

bY

][

21

52

21

51

21

51

21

53

Sjj

jjY

1I

1V

2I

2V

xV

520

2112121 VVVVVVVV

xxxx

22

21

1222

2111

jVVVVI

jVVVVI

x

x

Find the Y parameters for the network using a direct approach

rearrange and V Replace x

LEARNING EXAMPLEFind the Z parameters of the networkNetwork A

Network BUse direct method,or given the Y parameters transform to Z… or decompose the network in a seriesconnection of simpler networks

jj

j

jjj

Za

2342

232

232

2322

1111

bZ

jj

jj

jj

jj

ZZZ ba

2365

2325

2325

2345

LEARNING EXAMPLEFind the transmission parameters

jjjj

DCBA

121

By splitting the 2-Ohm resistor,the network can be viewed as thecascade connection of two identicalnetworks

jjjj

jjjj

DCBA

121

121

2

2

)1()2())(1()1()1)(2()2)(1()2()1(

jjjjjjjjjjjjjj

DCBA

22

22

24122264241

jjjj

DCBA

sentreceivedPP

pfIVP

VVkVVV

LL

LL

||||3

0||300|| 22 voltage) (line

LEARNING by APPLICATIONGiven the demand at the receiving end, determinethe conditions on the sending end

221

221

DICVIBIAVV

Transmission parametersare best suited for thisapplication

In the next slide we show how to determine the transmission parameters for theline. Here we assume them known and use them for analysis

pfVPIIL

L 3|||| 2 flinelinetotal

flinelinetotal

IVQ

IVP

sin||||3

cos||||3

lineV

lineI

Power factor anglef

pfpfV

PIL

12 cos

3

221

221

DICVIBIAVV

end sending

the at Conditions

sendingsending PpfIV

1

1

Determining the transmission parameters for the line221

221

DICVIBIAVV

2

12 0

VVAI

27.09590.0C

CL

ZZZR

2

12 0

VICI

TI CCL

CT Z

VIZZR

ZI 212

SZ

ZZRCC

CL 13.9010.97522

2

12 0

IVBV

84.8400.100)( 21 LL ZRBIZRV

2

12 0

IIDV

27.09590.0

C

CL

ZZZR

LEARNING EXAMPLEDetermine the effect of the load on the voltage gain kRkRRMRA oi 49,1,500,1,000,20 21

5011

2 RRgain Ideal

2221212

2121111

VhIhIVhIhV

Using the hybrid parameters

21 VI for solving and gEliminatin

21122211

2111212 hhhh

IhVhV

22 IRV L :port output at Constraint

111

21122211

212

2

V

Rhhhhh

hV

V

L

for Solving

LL RRhhhhh

hG 247.11

88.4911

21122211

21

Effect of load resistance

Hybrid parameters are computed in next slide

Computing the hybrid parameters for non-inverting amplifier (repeat earlier example)

Non-inverting amplifier Equivalent linear circuit2221212

2121111

VhIhIVhIhV

21

21111211 )||(

RRRRRhIRRRV ii

2RI

DSI

o

iDSR R

IARIRR

RIII 11

21

122

21

121 RR

RRARho

i

21

1122

21

11 RR

RhVRR

RV

)(||0

21

22 RRR

VIVo

i

)( 21

2122 RRR

RRRho

o

LEARNING BY DESIGNGain required = 10,000 on a load of 1kOhm

5001

000,20

o

i

RMR

Aparameters Amplifier

000,1011

2 RRgain Ideal

kRMR

1999.9

1

2

For the final solution we will need to cascadeamplifiers. Hence the transmission parameterswill prove very useful

2121

22

21

11

21

22112112

21

1h

DhhC

hhB

hhhhh

hA H

table conversion the From

221

221

DICVIBIAVV

22 IRV L

LL RRBAV

VI

7.1661

66671

1

2

1

geliminatin

mShh

hMh

0.1,100.4

100.1,001.1

227

21

41211

Analysis of solution:-Even with infinite load the maximum gainis only 6,667Likely causes:-R2 is higher than input resistance Ri-Desired gain is comparable to the maximumgain, A, of the Op-Amp

Proposed solution:-Cascade two stages, each with ideal gain of100. This also lowers R2 to 99kOhm

Analysis of proposed solution

Since the two stages will be cascaded, the transmission parameters of the proposed solution will be

baba

baba

DBBABCBAAA

2

2

2

2

1

1

IV

DCBA

DCBA

IV

DCBA

DCBA

IV

bb

bb

aa

aa

bb

bb

aa

aa

LRBAV

V

1

1

2 Effect of load resistance. G=10,000

Two-Ports

Identical stages