Chapter 1

TWO PORT CIRCUITS

 Course Outcomes
2

 Ability to explain and analyze special

types of circuit such as two port
networks.
 Introduction
3

 What is a port?
 It is a pair of terminals through which a current may enter or
leave a network.

 One port or two terminal circuit

 Two Port Circuits/Networks
4

• It is an electrical network with two separate ports for

input and output.

• No independent sources.

I1 I2
+ +
Input Output
V1
_ Port The Network Port
V2
_
 Network/Circuit Equations
5

 The parameters are defined in terms of open and short circuit

conditions of the two ports.
 Impedance Z parameters

V1 = z11I1 + z12I2 I1 = y11V1 + y12V2

V2 = z21I1 + z22I2 I2 = y21V1 + y22V2

 Admittance Y parameters
V1 = h11I1 + h12V2
 Hybrid H parameters
I2 = h21I1 + h22V2

V1 = AV2 - BI2
 Transmission A, B, C, D parameters
I1 = CV2 - DI2
  z11 z12 
Z parameters z
 21 z22 
6

V z11 is the impedance seen looking into port 1

z = 1
11 I
1 I =0 when port 2 is open.
2

V z12 is a transfer impedance. It is the ratio of the

z = 1 voltage at port 1 to the current at port 2 when
12 I I =0
2 1 port 1 is open.

V z21 is a transfer impedance. It is the ratio of the

z = 2
21 I I =0 voltage at port 2 to the current at port 1 when
1 2
port 2 is open.
V
z = 2 z22 is the impedance seen looking into port 2
22 I I =0
2 1 when port 1 is open.
 Impedance parameters (Z in ohm)
7

 Assume no independent source in the network

V1 = z11 I 1 + z12 I 2 V1   z11 z12   I1   I1 

V  =  z z   I  = z  I 
V2 = z21 I 1 + z22 I 2  2   21 22   2   2
Impedance parameters
8

V1 V2
z11 = and z 21 =
I1 I 2 =0
I1 I 2 =0

z11 = Open-circuit input impedance

z21 = Open-circuit transfer impedance
from port 1 to port 2

V1 V2
z12 = and z 22 =
I2 I1 =0
I2 I1 = 0

z12 = Open-circuit transfer impedance

from port 2 to port 1
z22 = Open-circuit output impedance
 9

• When z11 = z22, the two-port network is said to be

symmetrical.
• When the two-port network is linear and has no dependent
sources, the transfer impedances are equal (z12 = z21), and
the two-port is said to be reciprocal.
 Example 1a
10

Given the following circuit.

I1 I2
8 10 

+ +

V1 20  20  V2

_ _

Find the Z parameters for the above network.

Answer 1a
11
The Z parameter equations can be expressed in
matrix form as follows.

V1   z11 z12   I 1 

V  =  z   
 2   21 z 22   I 2 

12
 Example 1b
13

 Determine the Z-parameters of the following circuit.

Answer 1b
14
 Exercise
15

 Find z parameter for the below circuit:

I1 5 I2

+ +
V1 20 15 V2
− −
Solution
16
 Exercise
17

 Determine the Z-parameters of the following circuits;

 Admittance parameters (y)

Assume no independent source in the network

I1 = y11V1 + y12 V2 I1   y11 y12  V1  V1 

I 2 = y 21V1 + y 22 V2 I  =  y y  V  = y V 
 2   21 22   2   2

where the y terms are called the admittance parameters, or simply y

parameters, and they have units of Siemens.
18
  y11 y12 
y parameters y
 21 y22 
19

I y11 is the admittance seen looking into port 1

y = 1
11 V
1 V =0 when port 2 is shorted.
2

I y12 is a transfer admittance. It is the ratio of the

y = 1 current at port 1 to the voltage at port 2 when
12 V V =0
2 1 port 1 is shorted.

I y21 is a transfer impedance. It is the ratio of the

y = 2
21 V V =0 current at port 2 to the voltage at port 1 when
1 2
port 2 is shorted.
I
y = 2 y22 is the admittance seen looking into port 2
22 V V =0
2 1 when port 1 is shorted.
Admittance parameters
20

I1 I2
y11 = and y 21 =
V1 V2 =0
V1 V2 =0

y11 = Short-circuit input admittance

y21 = Short-circuit transfer admittance
from port 1 to port 2

I1 I2
y12 = and y 22 =
V2 V1 =0
V2 V1 =0

y12 = Short-circuit transfer admittance

from port 2 to port 1
y22 = Short-circuit output admittance
 Example 2
21

 Determine the y-parameters of the following circuit.

Answer 2
22
 Example 3
23

 Determine the y-parameters of the following circuit.

Answer 3
24
  h11 h12 
Hybrid parameters h
 21 22 
h 

Assume no independent source in the network

V1 = h11I1 + h12 V2 V1   h11 h12  I1  I1 

I  = h    = h   
I 2 = h 21I1 + h 22 V2  2   21 h 22  V2  V2 

where the h terms are called the impedance parameters, or simply h

parameters, and each parameter has different units, refer above.
25
 26

Assume no independent source in the network

V h11= short-circuit input V1 h12 = open-circuit reverse

h11 = 1 h12 =
I1 impedance () V2 voltage-gain
V2 = 0 I1 = 0

I I2
h 21 = 2 h2 1= short-circuit h 22 = h22 = open-circuit output
I1 forward current gain V2 I1 = 0 admittance (S)
V2 = 0
 Example 4
27

 Determine the h-parameters of the following circuit.

Answer 4
28
 Find the hybrid parameters for the network
I1 I2
+ +
V1 = h11 I1 + h12V2
V1 V2
I 2 = h21 I1 + h22V2
− −

29
 A B
Transmission parameters C D 
 
30

Assume no independent source in

the network

V1 = AV2 − BI2 V1  A B  V2  V2 

I1 = CV2 − DI 2 I  =  C D − I  = T − I 
1    2  2

where the T terms are called the transmission parameters, or simply T or ABCD
parameters, and each parameter has different units.
 ABCD Parameters
31
V1
 open-circuit voltage ratio = A=
V2 I 2 =0

 negative short-circuit transfer impedance ()

V1
 = B=−
I 2 V =0
2

I1
 open-circuit transfer admittance (S)= C=
V2 I 2 =0

I1
 negative short-circuit current ratio = D=−
I 2 V =0
2
 Example 5
32

 Determine the ABCD parameters of the following

circuit.
Answer 5
33
 Cramer’s Rule
34

 Use to solve the simultaneous equation

 Eg,
 I1   y11 y12  V1 
I  =  y   
 2   21 y22  V2 
 Determinant,  = y11 y22 − y12 y21
 RELATIONSHIPS BETWEEN
PARAMETERS

I1 y12
I2 y22 y22 I1 y12 I 2
V1 = = − = z11 I1 + z12 I 2
y11 y12 y y
y21 y22

y11 I1
y21 I2 y11 I 2 y21 I1
V2 = = − = z 22 I1 + z 21 I 2
y11 y12 y y
y21 y22
35
Compare these two above equations:
36

y22 y12
z11 = z12 = −
y y
y21 y11
z 21 = − z 22 =
y y
Two Port Parameter Conversions
37
To go from one set of parameters to another, locate the set of parameters
you are in, move along the vertical until you are in the row that contains
the parameters you want to convert to – then compare element for element

H
z11 =
h22

38
 Example 7
39

 Find [z] and [g] of a two-port network if  10 1.5  

[T] =  
 2 S 4 
Solution 7:
Answer 7
40
 ANALYSIS OF TERMINATED TWO-PORT
CIRCUITS
41

I1 I2
Zg
+ Two-port model +
Vg of a
V1 V2 ZL
network
− −
6 Characteristics of Terminated Two-port Circuit
42

 Input impedance (Zin=V1/I1) or admittance

(Yin=I1/V1)
 Output current, I2
 Thevenin voltage and impedance (ZTh, VTh) with
respect to port 2
 Current gain I2/I1
 Voltage gain V2/V1
 Voltage gain V2/Vg
 6 characteristics in term of z parameters

 4 parameter equations that describe the circuit:

V1 = z11 I1 + z12 I 2 …1

V2 = z 21 I1 + z 22 I 2 …2

V1 = Vg − I1Z g …3

V2 = − I 2 Z L …4

43
 1st characteristic (input impedance)

In Eq(2) we replace V2 with –I2ZL and solve for I2

− z21I1
I2 =
Z L + z22
…..(5)

Then substitute this Eq into eq(1), Zin=V1/I1

z12 z21
Z in = z11 −
z22 + Z L
44
 2nd characteristic (output current, I2)

We solve Eq.(1) for I1 after replacing V1 with the right-

hand side of Eq.(3) the result is :

Vg − z12 I 2
I1 =
z11 + Z g
Then using Eq(5)
− z21Vg
I2 =
( z11 + Z g )( z22 + Z L ) − z12 z21
45
3rd characteristic (Thevenin voltage @ impedance)

V1
V2 I 2 =0
= z21I1 = z21
z11
But V1=Vg-I1Zg , and then I1= Vg/(Zg+Z11)

z21
V2 = VTh = Vg
I 2 =0
Z g + z11

46
 Impedance Thevenin

WHEN Vg is replacing by a short circuit (Vg=0), Eq(3)

reduces to:
V1=-I1 Zg
Substituting the above eq into Eq(1) gives:
− z12 I 2
I1 = ……(6)
z11 + Z g
Now use Eq(6) to replace Eq(2) with the result that:

V2 z12 z 21
= Z Th = z 22 −
I2 Vg = 0
z11 + Z g

47
 4th characteristic (current gain)
48

From equation (5), the current gain is given by:

I2 − z21
=
I1 Z L + z22
5th characteristic (voltage gain V2/V1)

By replacing I2 in Eq.(2) with its value from Eq.(4);

thus:
 − V2 
V2 = z21I1 + z22   …..7
 ZL 

 − V2 
z11 I1 = V1 − z12  
 ZL 
V1 z12V2
I1 = + …..8
z11 z11Z L
49
 50

We now replace I1 in Eq.(7) with Eq.(8) and solve the

resulting expression for V2/V1:

V2 z 21Z L
=
V1 z11Z L + z11 z 22 − z12 z 21
z 21Z L
=
z11Z L + z
 6th characteristic (voltage gain V2/Vg)

To derive the voltage ratio V2/Vg, we first combine Eqs(1)

(3)(4) to find I1 as a function of V2 and Vg :

z12V2 Vg
I1 = + ….(9)

Z L ( z11 + Z g ) z11 + Z g
We now use Eq.(9) and Eq(4) in conjunction with Eq.(2) to
derive an expression involving only V2 and Vg; that is:

z21 z12V2 z21Vg z22

V2 = + − V2
Z L ( z11 + Z g ) z11 + Z g Z L
51
 52

Finally, we can manipulate to get the desired voltage

ratio:

V2 z21Z L
=
Vg ( z11 + Z g )( z22 + Z L ) − z12 z21
 Example 6
53

 The ABCD parameters of the two-port network below are

 4 20Ω 
0.1S 2 

 The output port is connected to a variable load for maximum
power transfer. Find RL and the maximum power transferred.
I1 I2

V1 V2

Answer: VTH = 10V V; RL = 8; Pm = 3.125W.

 END of CHAPTER 1

Never put off till tomorrow what you can do today.

Never trouble another for what you can do yourself.

Nothing is troublesome that we do willingly.

Thomas Jefferson

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