two-port networks - khon kaen universityeestaff.kku.ac.th/~jamebond/182304/two port.pdf · two-port...

Two-port networks

Review of one ports

Various two-port descriptions

Terminated nonlinear two-ports

Impedance and admittance matrices of two-ports

Other two-port parameter matrices

The hybrid matrices

The transmission matrices

1-port 2-port 2-port 2-port 1-port

Thevenin’s Equivalent Circuit

Norton’s Equivalent Circuit

Nv

+

−

i

NOv

+

−

i

OCe

− +

0( ) ( ) ( , ) ( ) 0

t

OCv t e t h t i d tτ τ τ= + ≥∫For LTI network

0( ) ( ) ( ) ( ) 0

t

OCv t e t h t i d tτ τ τ= + − ≥∫In frequency domain

( ) ( ) ( ) ( )OCV s E s Z s I s= +

0

No independent sources

iD(t)

VD

vd(t)

+

-

vD(t)

+

-

t

va

VA

0t

vA=V

A+v

a

0

|va|p

Nonlinear

one port

0.750.5 0.55 0.6 0.65 0.7

0

1

1.5

2

2.5

3

3.5

0.5

vD(V)

iD(mA)

t

tBias pointQ

VD

ID

di

dv

vD (V)

iD (mA)

t

t

ID

0.699 0.6995 0.7 0.7005 0.7011.38

1.40

1.42

1.44

1.46

1.48

1.50

1.52

VD

For small dv

dv

di

TD nVVSD eII

/=

Td

TdTD

TdDTD

nVvD

nVvnVVS

nVvVS

nVvSD

eI

eeI

eIeIi

/

//

/)(/

=

=

== +

For DC bias

For DC bias + small signal

...!3!2

132

++++=xx

xex

From Taylor’s series expansion

++++= ...

!3!21

32xx

xIi DD

Where /d Tx v nV=

( ) dT

DDDD v

nV

IIxIi +=+≈ 1

For 1 or d Tx v nV=

d

dd

T

Dd

r

vv

nV

Ii =

=

di

T

D

nV

I

vD = V

D+ v

d

iD = I

D+ i

d

rd

id+

-

vd

+

-

Td

D

nVr

I

=

0.750.5 0.55 0.6 0.65 0.7

0

1

1.5

2

2.5

3

3.5

0.5

vD(V)

iD(mA)

t

tBias point

Q

VD

ID

Slope at Q point = 1

d

d

gr

=

iD

vD

V

10 kΩmA93.0

k10

V7.0V10=

Ω−

≈DI

V7.0≈DV

Ω=×

== 8.53mA93.0

mV252

D

Td

I

nVr

)100sin(µV5.53)100sin(mV01k10

ttr

rv

d

dd =

Ω+=

)100sin(V5.53V7.0 tvD µ+≈

Example If 10V 10mV sin(100 )V t= + find Dv

Two-port networks

LTI one ports

1V

1I

inZ inY

1

1

I

VZ in =

1

1in

IY

V=

Input impedance Input admittance

Fig. 1

Two-port networks

Example 1

Determine the input impedance of the circuit in Fig. 2

1Iβ

1I

inZ

Fig. 2

211

Z

VIII in

in +−== β 2(1 )in inV Z Iβ= + 2(1 )inZ Zβ= +

Example 2

Determine the output impedance of the circuit in Fig. 3

1Iβ

1I

outZ

outI

outV

Fig. 3

1 11

(1 ) outout

VI I I

Zβ β= − − = + 1

1

outout

out

V ZZ

I β= =

+

Two-port networks

Circuits can be considered by theirs terminal variables

Voltages and currents are terminal’s variables

Complex circuit can be analyzed more easily.

There are many kinds of two port parameters.

1V 2V

1I 2I

Fig. 4 A two port network

Common-Emitter (CE) Fixed-Bias Configuration

Removing DC effects of VCC and Capacitors

Small signal equivalent circuit

Hybrid equivalent model re equivalent model

Various two-port descriptions

( )g=i v 1 1 1 2

2 2 1 2

( , )

( , )

i g v v

i g v v

=

=or

( )r=v i 1 1 1 2

2 2 1 2

( , )

( , )

v r i i

v r i i

=

=or

Port currentPort voltage

1 1 1 2

2 2 1 2

( , )

( , )

v h i v

i h i v

=

=

Or hybrid

Two-port networks

The Y parameter

=

2

1

2221

1211

2

1

V

V

yy

yy

I

I

The admittance or Y parameter of a two port network is defined by

1 11 1 12 2

2 21 1 22 2

I y V y V

I y V y V

= +

= +

or in scalar form

The Y parameter

2 1

2 1

1 111 12

1 20 0

2 221 22

1 20 0

V V

V V

I Iy y

V V

I Iy y

V V

= =

= =

= =

= =

The Y parameters can found from

These parameters are call short-circuited admittance parameters

The Y parameter

Example 3

Determine the admittance parameters from the circuit in Fig 5.

1V 2V

1I 2I

10.5VFig 5.

1 1 1 2 1 2 1 2 1 2 2( ) ( )I Y V Y V V Y Y V Y V= + − = + −

2 1 3 2 2 2 1 2 1 2 3 20.5 ( ) (0.5 ) ( )I V Y V Y V V Y V Y Y V= + + − = − + +

1 2 21 1

2 2 32 20.5

Y Y YI V

Y Y YI V

+ − = − +

11 1 2 12 2

21 2 22 2 3

,

0.5 ,

y Y Y y Y

y Y y Y Y

= + = −

= − = +

The Y parameter

Example 4

Compute the y-parameter of the circuit in Fig.6

1V 2V

1I 2I1I

1VFig.6

1 1 1 1 1 1 1 2

1ˆ ˆ( ) 2 2I V V V V V V Va

= + − = − = −

2 1 1 1 1 1 22

1 1 1 2ˆ ˆ ˆ( )I I V V V V Va a a a

= − = − − + − = − +

1 1

2 22

12

1 2

I Va

I V

a a

−

= −

11 12

21 22 2

12,

1 2,

y ya

y ya a

= = −

= − =

Y parameter analysis of terminated two-port

1V 2V

1I 2I

LY

Fig. 9 Terminated two-port

=

2

1

2221

1211

2

1

V

V

yy

yy

I

I

Y-parameter equations

22 VYI L−=

11 12 11

21 22 20 L

y y VI

y y Y V

= +


From Crammer’s rules

21122211

122

2221

1211

22

121

1)(

)(0

yyYyy

IYy

Yyy

yy

Yy

yI

VL

L

L

L

−+

+=

+

+=

The input admittance Yin

12 2111

11 22( )in

L

y yY y

y y Y= −

+and

21 1 22 2

212 1

22

( )L

L

y V y Y V

yV V

y Y

= − +

= −+


12 211 11 1 12 2 11 1

22 L

y yI y V y V y V

y Y

= + = −

+

Gain: 2 21

1 22 L

V y

V y Y= −

+

1I

inY

sv

sR

11y 12 2y VLY22y21 1y V

1V 2V

2I

Fig 10 Terminated two-port Y-parameter model

Two-port networks

The Z parameter

1 11 12 1

2 21 22 2

V z z I

V z z I

=

The impedance or Z parameter of a two port network is defined by

1 11 1 12 2

2 21 1 22 2

V z I z I

V z I z I

= +

= +

or in scalar form

The Z parameter

2 1

2 1

1 111 12

1 20 0

2 221 22

1 20 0

I I

I I

V Vz z

I I

V Vz z

I I

= =

= =

= =

= =

The Z parameters can be found from

These parameters are call open circuit impedance parameters

The Z parameterExample 6Determine the impedance parameters from the circuit in Fig 11

Fig 11.

1 2 1 2 1 2

10 10 104 ( ) (4 )V I I I I I

s s s= + + = + +

2 2 1 2 1 2

10 10 103 ( ) (3 )V I I I I I

s s s= + + = + +

11 12

21 22

10 4 10

10 3 10

s

z z s sZ

z z s

s s

+

= = +

1V

1I+ -24I

0.1F

3

+ +

- -2V

2I

In frequency domain

The Y parameter

Example 7

Compute the z-parameter of the circuit in Fig.12

Fig.12

1 1 1 1 3V R I R I= −

2 3 2 3 3V R I R I= +

1V

1I R2

+ +

- -2V

2I

R3R13I

1 1 3 2 1 2 3 30 ( )R I R I R R R I= − + + + +

313 1 2

1 2 3 1 2 3

RRI I I

R R R R R R= −

+ + + +

The Z parameter2

1 311 1 1 2

1 2 3 1 2 3

1 2 3 1 31 2

1 2 3 1 2 3

( )

( )

R RRV R I I

R R R R R R

R R R R RI I

R R R R R R

= − ++ + + +

+= +

+ + + +2

1 3 32 1 3 2

1 2 3 1 2 3

1 3 3 1 21 2

1 2 3 1 2 3

( )

( )

R R RV I R I

R R R R R R

R R R R RI I

R R R R R R

= + −+ + + +

+= +

+ + + +

++

+

++

++++

+

=

321

213

321

31

321

31

321

321

2121

1211

)()

))(

RRR

RRR

RRR

RR

RRR

RR

RRR

RRR

zz

zz

Z parameter analysis of terminated two-port

1V 2V

1I 2I

LZ

Fig. 14 Terminated two-port

1 11 12 1

2 21 22 2

V z z I

V z z I

=

Z-parameter equations

2 2LV Z I= −

11 12 11

21 22 20 L

z z IV

z z Z I

= +


From Crammer’s rules

1 12

22 22 11

11 12 11 22 12 21

21 22

0 ( )

( )

L L

L

L

V z

z Z z Z VI

z z z z Z z z

z z Z

+ += =

+ −

+The input impedance Zin

12 2111

22in

L

z zZ z

z Z= −

+and

21 1 22 2

212 1

22

( )L

L

z I z Z I

zI I

z Z

= − +

= −+


12 211 11 1 12 2 11 1

22 L

z zV z I z I z I

z Z

= + = −

+

Gain: 2 1 2 21 21

1 22 22

in L L

s s in s L in L in s

ZV V V Z z Z z

V V V Z Z z Z Z z Z Z Z= = =

+ + + +i i i

1I

inZ

sv

sR

11z

12 2z I

LZ

22z

21 1z I

1V 2V

2I

Fig 15 Terminated two-port Z-parameter model

Z parameter analysis of terminated two-portExample 9

The circuit in Fig 16 is a two-stage transistor amplifier. The Z-parameters for each stage are

6

2 6

1.0262 10 6,790.8Z

1.0258 10 6,793.5

×=

× 1 6

350 2.667Z

10 6,667

=

−

Determine a) The input impedance and 2inZ inZb) The overall voltage gain

c) Check the matching of the load and output impedance

sV k1V2V

1I2I

outI

outV

inZ 2inZFig 16


Solution12 21

2 1122

66 6790.8 1.0258 10

1.0262 106793.5 16

3,159

in

L

z zZ z

z Z= −

+

× ⋅= ⋅ −

+= Ω

21

22 22

616(1.0258 10 )

(16 6793.5)3,159

0.7629

out L

L in

V Z z

z Z ZV=

+

⋅=

+

=


12 2111

22 1

62.667 10

3506667 1224.7

687.9

in

L

z zZ z

z Z= −

+

×= +

+= Ω

2 1 21

1 22

61224.7 10

1224.7 6667 75 687.9

203.4

L

L s ins

V Z z

Z z Z ZV=

+ +

− = + + = −

Ω=== 7.12243159//2000//2 21 inL ZkZ

2 1 2 21 21

1 22 22

in L L

s s in s L in L in s

ZV V V Z z Z z

V V V Z Z z Z Z z Z Z Z= = =

+ + + +i i i

0.902 225.6


The overall voltage gain

VV

V

V

V

V

V

VA

s

out

s

outVS

/2.155

)4.203(7629.0

2

2

−=

−×=

==

Out put impedance

02

2

=

=sV

outI

VZ

The detail is left to the student to show that

12 2122

11out

s

z zZ z

R z= −

+


Therefore the load is closely matched to the output impedance

12 211 22

11

62.667 106667

0.5 350

14.276 k

out

s

z zZ z

R z= −

+

×= +

+= Ω

2 1 // 2 1.7542 ks outR Z k= = Ω6

6

6790.8 1.0258 106793.5

1754.24 1.0262 10

16.93

outZ⋅ ×

= −+ ×

= Ω

The h-parameter (Hybrid parameter)

H-parameter is the combination of Z and Y parameter defined by

1 11 12 1

2 21 22 2

V h h I

I h h V

=

1 11 1 12 2

2 21 1 22 2

V h I h V

I h I h V

= +

= +

or in scalar form

H-parameter is commonly used in transistor modeling.

The h-parameter

2

2

1

1

1 12 2111 11

1 11 220

2 21 2121

1 11 220

2 12 2122 22

2 11 220

1 12 1212

2 11 220

1

1

V

V

I

I

V z zh z

I y z

I y zh

I y z

I y yh y

V y z

V y zh

V y z

=

=

=

=

= = = −

= = = −

= = − =

= = − =

The h parameters can found from

The h-parameter

1I

inZ

sv

sR11h

12 2h V

LZ

22h

21 1h I

1V 2V

2I

Fig 17 Hybrid parameter model

The h-parameter

Example 10

Determine the h-parameter of the two-port circuit shown in Fig. 18

1V 2V

2I1I

2V

Fig. 18

21ˆ1

Va

V = 1 2I aI= −

2 2 2 1 2ˆ R

V V RI I Va

= − = +

1 1 22

1RV I V

aa= +

2 2 2 22

1 2

ˆ ˆ

10

V V V VI

R R R

I Va

−= = − +

= − +

−=

2

12

2

1

01

1

V

I

a

aa

R

I

V

The h-parameter

Example 10

Find the h-parameter of the circuit in Fig. 19 assuming L1=L2=M=1H

1V 2V

2I1I

2V

2L1L

1I

Fig. 19

In frequency domain

2111ˆ sMIIsLV +=

111ˆ VII −=

1 1 2 1 1(1 )sL V sMI sL I+ − =

The h-parameter

2 2 2 1 2 2 1 1ˆ ˆ ( )V sL I sMI sL I sM I V= + = + −

2 2 2ˆV V I= +

2 2 2 1 1(1 ) ( )V sL I sM I V= + + −

1 2 2 1 2(1 )sMV sL I sMI V− + = −

In matrix form

−=

+−

−+

2

11

2

1

2

1

1

0

)1(

1

V

I

sM

sL

I

V

sLsM

sMsL

1

1 1 11

2 2 2

1 0

(1 ) 1

V sL sM IsL

I sM sL VsM

−+ −

= − + −

The h-parameter

1

1 1

2 2

1 0

(1 ) 1

V Is s s

I Vs s s

−+ −

= − + −

With L1=L2=M=1 H

+−+=

2

1

112

1

V

I

ss

ss

s

The inverse hybrid parameter

(g- parameter)g-parameter is defined by

1 11 12 1

2 21 22 2

I g g V

V g g I

=

1 11 1 12 2

2 21 1 22 2

I g V g I

V g V g I

= +

= +

or in scalar form

g-parameter is an alternative form of hybrid representation.

2

2

1

1

1 12 21 2211 11

1 11 220

2 21 21 2121

1 11 220

2 12 21 1122 22

2 11 220

1 12 12 1212

2 11 220

11 22 12 21

1

1

I

I

V

V

I y y hg y

V z y h

V z y hg

V z y h

V z z hg z

I z y h

I z y hg

I z y h

h h h h h

=

=

=

=

= = = − =∆

= = = − = −∆

= = − = =∆

= = − = = −∆

∆ = −where

The g parameters can found from

Inverse hybrid parameter model

Conversion of Two-port parameters

=I YV

∴ =V ZYV

Two port parameters can be converted to any form as follows

From

=

2

1

2221

1211

2

1

V

V

yy

yy

I

I

And 1 11 12 1

2 21 22 2

V z z I

V z z I

=

1−=Z Y and 1−=Y Z

=V ZI

11 22 12 21

11 22 12 21

Z z z z z

Y y y y y

∆ = −

∆ = −

22 12

11 12

21 22 21 11

z z

y y Z Z

y y z z

Z Z

− ∆ ∆= − ∆ ∆

22 12

11 12

21 22 21 11

y y

z z Y Y

z z y y

Y Y

− ∆ ∆= − ∆ ∆

where

Conversion of Two-port parametersFrom y to h

=

2

1

2221

1211

2

1

V

V

yy

yy

I

I

11 1 1 12 2

21 1 2 22 2

y V I y V

y V I y V

− = − +

− + =

11 1 12 1

21 2 22 2

0 1

1 0

y V y I

y I y V

− − = −

1

1 11 12 1

2 21 22 2

0 1

1 0

V y y I

I y y V

−− −

= −


1 12 1

2 21 11 22 12 21 211

11V y I

I y y y y y Vy

− = −

Hence

12

11 1111 12

21 22 21 12 2122

11 11

1 y

y yh h

h h y y yy

y y

− = −


It can be shown that for the terminated two-port with h-parameter the following equations can be derived

122

212 I

Yh

hV

L+−=

L

inYh

hhh

I

VZ

+−==

22

211211

1

1

2 12 2122

2 11out

s

V h hZ h

I h Z= = −

+

2 21

1 22( )L in

V h

V h Y Z= −

+

and2 1 2 21

1 22

1

( )VS

s s L in s

V V V hA

V V V h Y Z Z= = = −

+ +

Transmission parameter

The t-parameter or transmission parameters are used in power systemand it is called ABCD parameter. The transmission parameter is defined by

1 11 12 2

1 21 22 2

V t t V

I t t I

= −

This means that the power flows into the input port and flow out to theload from the output port.

t-parameter can be calculated from

2 2

2 2

1 111 12

2 20 0

1 121 22

2 20 0

I V

I V

V Vt t

V I

I It t

V I

= =

= =

= = −

= = −

Open or short circuit at

the output port

1 2

1 2

V VA B

I IC D

= −

or


Example 11

1V 2V

2I1I

2V Fig 20

Determine the t-parameter of the circuit shown in Fig 20.

)(1ˆ1

2221 RIVa

Va

V −==

21 aII −=

−

=

∴

2

21

1

1

0 I

V

aI

VaR

a


One of the most importance characteristics of the two-port circuit witht-parameter is to determine the overall cascade parameter.

1V2V 3V 4V

1I 2I− 3I 4I−

−=

2

21

1

1T

I

V

I

V3 4

23 4

TV V

I I

= −

3232 , IIVV =−=

1 41 2

1 4

T TV V

I I

= −

Therefore

2 1

2 1

V VA B

I IC D

′ ′ = ′ ′ −

Inverse Transmission parameter

1 1

1 1

2 2

1 10 0

2 2

1 10 0

I V

I V

V VA B

V I

I IC D

V I

= =

= =

′ ′= = −

′ ′= = −

Interconnection of two-port network

Two port networks can be connected in series parallel or cascaded

Series and parallel of two-port have 4 configurations

Series input-series output (Z-parameter)

Series input-parallel output (h-parameter)

Parallel in put-series output (g or h-1-parameter)

Parallel input-parallel output (Y-parameter)

With proper choice of parameters the combined parameters can be added together.

Interconnection of two-port network

Z1

Z2

Z=Z1+Z2

V1

+ + +

+ +

-

- -

- -

V11

V12

V21

V22

Y1

Y2

Y=Y1+Y2

V1V2

+ +

- -

Example Bridge-T network

N1 // N2

For network N2

For network N1

[ ] 41

0 1

ZT

=

Y-parameters of the bridge-t network are

two-port networks - khon kaen universityeestaff.kku.ac.th/~jamebond/182304/two port.pdf · two-port...

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