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Two-Port Networks
Introduction
 In general, a network may have n ports. The current entering one
terminal leaves through the other terminal so that the net current
entering the port equals zero.
 Thus, a two-port network has two terminal pairs acting as access
points. The current entering one terminal of a pair leaves the other
terminal in the pair.
 Three-terminal devices such as transistors can be configured into
two-port networks.
 Knowing the parameters of a two-port network enables us to treat
it as a “black box” when embedded within a larger network.
1
Introduction
 To characterize a two-port network requires that we relate the
terminal quantities V1, V2, I1, and I2, out of which two are
independent.
 The various terms that relate these voltages and currents are called
parameters. Our goal is to study six sets of these parameters.
 As with op amps, we are only interested in the terminal behavior
of the circuits.
Reminder: Only two of the four
variables (V1, V2, I1, and I2) are
independent.
2
Impedance Parameters
V1 = z11 I1 + z12 I 2
V2 = z21 I1 + z22 I 2
The z terms are called the impedance parameters, or simply
z parameters, and have units of ohms.
The values of the parameters can be evaluated by setting
I1 = 0 (input port open-circuited) or
I2 = 0 (output port open-circuited).
Therefore, the z parameters are also called the open-circuit
impedance parameters.
3
Impedance Parameters
Note: The two equations associated with a set of parameters
determine how the individual parameters are obtained.
V1 = z11 I1 + z12 I 2
V2 = z21 I1 + z22 I 2
V1
z11 =
I1
V2
z21 =
I1
I 2 =0
V1
z12 =
I2
I 2 =0
V2
z22 =
I2
[ ]
[ ]
[ ]
I1 =0
[ ]
I1 =0
Sometimes z11 and z22 are called driving-point impedances, while z21
and z12 are called transfer impedances.
A driving-point impedance is the input impedance of a two-terminal
(one-port) device. Thus, z11 is the input driving-point impedance with
the output port open-circuited, while z22 is the output driving-point
impedance with the input port open-circuited.
4
Impedance Parameters
 Determination of the z parameters 
 When z11 = z22, the two-port network is
said to be symmetric. This implies that the
network has mirror-like symmetry about
some center line; that is, a line can be
found that divides the network into two
similar halves.
 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. This
means that if the points of excitation and
response are interchanged, the transfer
impedances remain the same.
5
Impedance Parameters
 Any two-port that is made entirely of resistors, capacitors, and
inductors must be reciprocal.
 A reciprocal network can be replaced by the T-equivalent circuit.
 If the network is not reciprocal, a more general equivalent network
is obtained.
Reciprocal network (z12 = z21)
Non-reciprocal network
Note that both equivalent circuits satisfy the z-parameter equations.
V1 = z11 I1 + z12 I 2 ,
V2 = z21 I1 + z22 I 2
6
Impedance Parameters
 Some two-port networks parameters might not exist. As an
example, consider the ideal transformer. It is impossible to express
the voltages in terms of the currents.
An ideal transformer has no z parameters.
However, it does have hybrid parameters (later).
7
Example z-1 — Method 1
Determine the z parameters for the following circuit.
V1
z11 =
I1
V2
z21 =
I1
V1
z12 =
I2
V2
z22 =
I2
= 12
I1 =0
= 4 + 12 = 16
I1 =0
= 1 + 12 = 13
I 2 =0
= 12
I 2 =0
é13  12 ù
ú
[ z ] = êê
ú
ë12  16 û
Non-symmetric, reciprocal
8
Example z-1 — Method 2
z12 = 12  = z21
z11 - z12 = 1 

z11 = 1  + z12 = 13 
z22 - z12 = 4 

z22 = 4  + z12 = 16 
é13  12 ù
ú
[ z ] = êê
ú
ë12  16 û
9
Example z-2
Find I1 and I2 in the following circuit.
V1 = 40 I1 + j 20I 2 = 100V0
V2 = j 30 I1 + 50I 2 = -10I 2  I1 = j 2I 2
100V0 = j80 I 2 + j 20I 2
100V0
 I2 =
= 1A- 90
j100
I1 = j 2I 2 = 2A0
10
Example z-3
Calculate I1 and I2 in the following two-port.
z11 I1 + z12 I 2 = V1 = 2V30- 2I1
z21 I1 + z22 I 2 = V2 = 0
2V30 = 8I1 - j 4I 2
0 = - j 4I1 + 8I 2  I 2 = j 0.5 I1
2V30 = 8I1 + 2I1
Answer:
I1 = 20030 mA
I 2 = 100120 mA
11
Example z-4
Determine the s domain expressions for the z parameters.
V1
z11 =
I1
V2
z21 =
I1
V1
z12 =
I2
V2
z22 =
I2
= 10
I1 =0
= (0.2 s + 10) 
I1 =0
I2
2
10 s + 2

= + 10 =
s
s
=0
= 10
I 2 =0
é10 s + 2
ê
[ z ]= ê s
ê
êë 10
ù
10 ú
ú
ú
0.2 s + 10úû
Non-symmetric, reciprocal
12
Example z-5
Find the z parameters for the following circuit.
z11 =
V1
I1
= 20  (5 + 15)  = 10
I 2 =0
When I 2 = 0, then V2 =
V2
z21 =
I1
z22
V2
=
I2
V1
I2
0.75V1
=
= 7.5 
V1 10
= 15  (5 + 20)  = 9.375
I1 =0
When I1 = 0, then V1 =
z12 =
I 2 =0
15
V1 = 0.75V1
15 + 5
=
I1 =0
20
V2 = 0.8V2
20 + 5
0.8V2
= 7.5 
V2 9.375
é 10 
7.5  ù
ú
[ z ] = êê
ú
ë 7.5  9.375 û
Non-symmetric, reciprocal
13
Admittance Parameters
 The y terms are known as the
admittance parameters (or, simply, y
parameters) and have units of Siemens.
I1 = y11V1 + y12V2
I 2 = y21V1 + y22V2
 Since the y parameters are obtained by
short-circuiting the input or output port,
they are also called the short-circuit
admittance parameters.
I1
I1
y11 =
y12 =
[S]
[S]
V1 V =0
V2 V =0
2
I2
y21 =
[S]
V1 V =0
2
1
I2
y22 =
V2
[S]
V1 =0
14
Admittance Parameters
 For a two-port network that is linear and has no dependent sources,
the transfer admittances are equal (y12 = y21).
 A reciprocal network (y12 = y21) can be modeled by a equivalent circuit.
 If the network is not reciprocal, a more general equivalent network
is obtained.
Reciprocal network (y12 = y21)
Non-reciprocal network
Note: Impedance and admittance parameters are collectively referred
to as immittance parameters.
15
Example y-1 — Method 1
Obtain the y parameters for the  network.
I1
1
1
=
+
= 0.25S
y11 =
V1 V =0 20 5
2
I2
1
== -0.2S
y21 =
5
V1 V =0
2
I1
y12 =
V2
V1 =0
I2
y22 =
V2
1
1
=
+
= 0.267S
15 5
V =0
=-
1
1
= -0.2S
5
é 0.25S -0.2S ù
ú
[ y ] = êê
ú
ë-0.2S 0.267Sû
Non-symmetric, reciprocal
16
Example y-1 — Method 2
1
= -0.2S = y21
y12 = 5
1
y11 + y12 =
20
1
y22 + y12 =
15


1
- y12 = 0.25S
y11 =
20
1
- y12 = 0.267S
y22 =
15
é 0.25S -0.2S ù
ú
[ y ] = êê
ú
ë-0.2S 0.267Sû
17
Example y-2
Obtain the y parameters for the following T network.
I1
I1
1
=
= 0.2273S
y11 =
V1 V =0 (2 + 4  6)
2
I2

V1

V2


I2
46
1
=⋅
= -0.0909S
y21 =
4  6 + 2 6
V1 V =0
2
I1
y12 =
V2
I2
y22 =
V2
=V1 =0
24
1
⋅
= -0.0909S
2  4 + 6 2
1
=
= 0.1364S
(6 + 2  4)
V1 =0
é 0.2273S -0.0909Sù
ú
[ y ] = êê
ú
ë-0.0909S 0.1364S û
18
Example y-3
Obtain the s domain expressions of the y parameters.
I1
1 1 + s 2 LC
y11 =
= sC + =
V1 V =0
sL
sL
2
I2
1
y21 =
=V1 V =0
sL
2
I1
y12 =
V2
1
=sL
V =0
I2
y22 =
V2
1 1 + s 2 LC
= sC + =
sL
sL
V =0
1
1
é1 + s 2 LC
ê
ê sL
[ y]= ê
ê
ê -1
êë
sL
1 ùú
sL úú
1 + s 2 LC úú
sL úû
Symmetric, reciprocal
19
Example y-4
Determine the y parameters for the following two-port.
Short circuit Port 2
V1 - Vo
Vo
Vo - 0V
= 2I1 +
+
I1 =
8
2
4
V1 - Vo
Vo
+3
0=
8
4
0 = V1 - Vo + 6Vo
 V1 = -5 Vo
20
Example y-4 cont’d
-5Vo - Vo
= -0.75S ⋅ Vo
8
I1 -0.75S ⋅ Vo
y11 =
=
= 0.15S
V1
-5 Vo
I1 =
I2
Vo - 0V
+ 2I1 + I 2 = 0
4
-I 2 = 0.25 S ⋅ Vo -1.5S ⋅ Vo = -1.25S ⋅ Vo
I 2 1.25S ⋅ Vo
y21 =
=
= -0.25S
V1
-5 Vo
21
Example y-4 cont’d
Short circuit Port 1
Vo
Vo - V2
0V - Vo
I1 =
= 2I1 +
+
8
2
4
Vo
Vo
Vo - V2
0=+
+
8 2
4
0 = -Vo + 4Vo + 2Vo - 2V2  V2 = 2.5 Vo
22
Example y-4 cont’d
- Vo 8
I1
y12 =
=
= -0.05S
V2
2.5 Vo
Vo - V2
+ 2I1 + I 2 = 0
4
Vo 2.5Vo 2Vo
-I 2 =
= -0.625S ⋅ Vo
4
4
8
0.625S ⋅ Vo
I2
=
= 0.25S
y22 =
V2
2.5 Vo
é 0.15S -0.05Sù
ú
[ y ] = êê
ú
ë-0.25S 0.25S û
23
Hybrid Parameters
 The z and y parameters of a two-port network do not always exist.
So there is a need for developing other sets of parameters.
 This third set of parameters is based on making V1 and I2 the
dependent variables.
V = h I +h V
1
11 1
12 2
I 2 = h21 I1 + h22V2
 The h terms are known as the hybrid parameters (or, simply,
h parameters) because they are a hybrid combination of ratios.
 The values are:
V1
h11 =
[ ]
I1 V =0
V1
h12 =
V2
I2
h21 =
I1
I2
h22 =
V2
2
V2 =0
I1 =0
[S]
I1 =0
24
Hybrid Parameters
 To be specific, h11 = short-circuit input impedance
h12 = open-circuit reverse voltage gain
h21 = short-circuit forward current gain
h22 = open-circuit output admittance
 For reciprocal networks, h12 = −h21.
V1 = h11 I1 + h12V2
I 2 = h21 I1 + h22V2
The h-parameter equivalent network of
a two-port network
25
Example h-1: Ideal Transformer
1
V1 = V2 , I1 = -nI 2
n
V1 = h11I1 + h12V2
I 2 = h21I1 + h22V2
é
ê 0
ê
[h] = ê
ê 1
êêë n
1ù
ú
nú
ú
ú
0ú
úû
26
Example h-2
Find the hybrid parameters for the following two-port network.
V1
V1
h11 =
h12 =
[ ]
I1 V =0
V2
2
I2
h21 =
I1
I2
h22 =
V2
V2 =0
I1 =0
[S]
I1 =0
V1
h11 =
= 2 + 3  6 = 4
I1 V =0
2
h21 =
I2
I1
=V2 =0
6
2
=(3 + 6)
3
27
Example h-2 cont’d
h12 =
V1
V2
I2
h22 =
V2
é
ê4 
ê
[h] = ê
ê 2
êêë 3
2ù
ú
3ú
ú
1 ú
Sú
9 úû
6
2
=
=
(3 + 6) 3
I =0
1
1
1
=
= S
(3 + 6) 9
I =0
1
Note that h12 = −h21
since the network is reciprocal
28
Example h-3
Determine the h parameters for the following circuit.
V1
= 20  5 = 4
h11 =
I1 V =0
2
é 4
0.8 ù
ú
[ h ] = êê
ú
ë-0.8 1.067Sû
h12 = −h21
I2
h21 =
I1
20
== -0.8
(20 + 5)
V =0
V1
h12 =
V2
20
=
= 0.8
(5 + 20)
I =0
I2
h22 =
V2
1
=
= 1.067S
15  (5 + 20) 
I =0
2
1
1
29
Example h-4
Find the impedance at the input port of the following circuit.
I 2 =  V2 Z L  V2 50k
(1) V1 =h11 I1  h12V2
(2) I 2 =h21 I1  h22V2  V2 Z L
h12 h21Z L
(1) V1 =h11 I1 
I1
1  h22 Z L
 h21Z L
I1
 V2 
1  h22 Z L
h12 h21Z L
 1.667k
Z in  V1 I1  h11 
1  h22 Z L
30
Inverse Hybrid Parameters
 A set of parameters closely related to the h parameters are the
g parameters or inverse hybrid parameters. These are used to
describe the terminal currents and voltages as
I1 = g11V1 + g12 I 2
V2 = g 21V1 + g 22 I 2
The g-parameter model of a two-port
network
 For reciprocal networks, g12 = −g21.
31
Inverse Hybrid Parameters
 The values of the g parameters are determined as
I1 = g11V1 + g12 I 2
V2 = g 21V1 + g 22 I 2
I1
g11 =
V1
V2
g 21 =
V1
[S]
I 2 =0
I 2 =0
I1
g12 =
I2
V1 =0
V2
g 22 =
I2
[ ]
V1 =0
 Thus, the inverse hybrid parameters are specifically called
g11 = open-circuit input admittance
g12 = short-circuit reverse current gain
g21 = open-circuit forward voltage gain
g22 = short-circuit output impedance
32
Example g-1
Find the g parameters as functions of s for the following circuit.
I1
I2

V1

V2


I
g11 = 1
V1
V2
g 21 =
V1
I2
I1 = g11V1 + g12 I 2
V2 = g 21V1 + g 22 I 2
1
[S]
=
s +1
=0
I1
g12 =
I2
1
=s +1
V =0
1
=
s +1
=0
V2
g 22 =
I2
1 1⋅ s
s 2 + s +1
[]
= +
=
s 1+ s
s ( s + 1)
V =0
I2
1
1
Note that g12 = −g21 since the network is reciprocal
33
Example g-2
Find the g parameters from measured data:
measured (port 2 open): V1  50 mV, I1  5μA
V2  200 mV
BLACK
BOX
measured (port 1 short): V2  10 mV, I1  2μA
I 2  0.5μA
I1
g11 =
V1
V2
g 21 =
V1
=
I 2 =0
I 2 =0
5μA
= 0.1mS
50 mV
200
=
=4
50
I1
g12 =
I2
g 22 =
V2
I2
V1 =0
2
== -4
0.5
=
V1 =0
10 mV
= 20 k
0.5μA
g12 = −g21
34