Transcript Slide 1

Two-Port Network
1
TWO PORT NETWORKS
Two port networks are a useful tool for describing idealized
components. The basic device schematics are simple, but
each set of parameters views the device differently.
The basic device is seen below, along with the various
parameter sets,
I1
I2
+
v1
-
+
s-domain
circuit
v2
-
2
z-parameters
V1  z11I1  z12 I 2
V2  z21I1  z22 I 2
y-parameters
I1  y11V1  y12V2
I 2  y21V1  y22V2
a-parameters
V1  a11V2  a12 I 2
I1  a21V2  a22 I 2
b-parameters
V2  b11V1  b12 I1
I 2  b21V1  b22 I1
3
h-parameters
V1  h11I1  h12V2
I 2  h21I1  h22V2
g-parameters
I1  g11V1  g12 I 2
V2  g 21V1  g 22 I 2
PARAMETER VALUES
Obviously some of the parameters are impedance, while
others are admittance. They can be easily determined by
setting other parameters to zero, and measuring relevant
voltages/currents.
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z-Parameters (impedance)
The values are as below,
5
The equivalent circuit for the z-parameters is shown below,
I1
+
I2
z22
z11
+
V1
-
+
z21I1
z11I2
-
+
V2
-
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EXAMPLE
Find the z parameters for the circuit shown
SOLUTION
V1
z11 
I1

2020

 10 
I 2 0
40
7
V2
z 21 
I1
V2
z 22 
I2
I 2 0
0.75V1

 7.5 
V1 / 10

1525

 9.375 
I1  0
40
When port 1 is open, I1 is zero and the voltage V1 is
V2
20   0.8 V2
V1 
5  20
V2
I2 
9.375
8
V1
z12 
I2
I1  0
0.8 V2

 7.5 
V2 / 9.375
The following measurements pertain to a two-port circuit
operating in the sinusoidal steady state. With port 2 open, a
voltage equal to 150 cos 4000t V is applied to port 1. The
current into port 1 is 25 cos (4000t - 45°) A, and the port 2
voltage is 100 cos (4000t + 15°) V. With port 2 short-circuited,
a voltage equal to 30 cos 4000t V is applied to port 1. The
current into port 1 is 1.5 cos (4000t + 30°) A, and the current
into port 2 is 0.25 cos (4000t + 150°) A. Find the a parameters
that can describe the sinusoidal steady-state behavior of the
circuit.
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SOLUTION
V1  1500 V ;
0
I1  25  45 A;
0
V2  10015 V ; I 2  0 A.
0
V1
a11 
V2
I1
a21 
V2
I 2 0
15000
0


1
.
5


15
100150
25  45
0

 0.25  60 S
0
10015
0
I 2 0
V1  300 V ;
0
I1  1.530 A;
0
V2  0 V ; I 2  0.25150 A.
0
10
V1
 300
0
a12 


120

30

0
I 2 V 0 0.25150
0
2
I1
a21 
I2
 1.530
0


6

60
0.251500
0
V2  0
Drill Exercise
The following measurements were made on a two-port resistive circuit.
With 10 mV applied to port 2 and port 1 open, the current into port 2 is 025
µA, and the voltage across port 1 is 5 µV. With port 2 short-circuited and
50 mV applied to port 1, the current into port 1 is 50 µA, and the current
into port 2 is 2 mA. Find the h parameters of the network.
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RELATIONSHIPS AMONG THE TWO-PORT PARAMETERS
I1  y11V1  y12V2 ,
I 2  y21V1  y22V2 ;
 I1 y12 
I y 
y22
y12
2
22 

V1 

I1 
I2
y
 y11 y12  y
y

 21 y22 
 y11 I1 
y

I2 
y21
y11
21

V2 

I1 
I2
y
y
y
12
y22
z11 
;
y
y21
z21  
;
y
y12
z12  
;
y
y11
z22  
.
y
To find the z parameters as functions of the a parameters,
V1  z11 I1  z12 I 2 ,
V2  z21 I1  z22 I 2 ;
V1  a11V2  a12 I 2
I1  a21V2  a22 I 2
1
a22
V2 
I1 
I2
a21
a21
13
 a11a22

a11
V1 
I1  
 a12  I 2
a21
 a21

a11
z11 
;
a21
a
z12 
.
a21
1
z 21 
;
a21
a22
z 22 
a21
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EXAMPLE
1.
2.
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1.
V1
h11 
I1 V
24x103

 1.2 k
6
20x10
I2
h21 
I1
103

 50
6
20x10
2 0
V2  0
The parameters h12 and h22 cannot be obtained directly from the open circuit test.
V1
a11 
V2
I1
a21 
V2
I 2 0
I 2 0
10x103

 0.25x103 ;
 40
10x106

 0.25x106 S ;
 40
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V1
a12  
I2
I1
a22  
I2
V2  0
V2  0
24x103

 24 ;
3
10
20x106
3



20
x
10
.
3
10
a
h11 
;
a22
a21
h22 
.
a22
6
a  a11a22  a12 a21  10
 106
5
h12 

5
x
10
3
 20x10
 0.25x106
h22 
 12.5 S
3
 20x10
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Analysis Of The Terminated Two Port Circuit
I2
I1
Zg
+
V1
Vg
-
+
Two-port model
of network
V2
ZL
-
Six characteristics of the terminated two-port circuit define its terminal behavior:
1. The input impedance Zin = V1/I1, or the admittance Yin =I1/V1;
2. The output current I2 ;
3. The Thevenin voltage and impedance (VTh , ZTh ) with respect to port 2 ;
4. The current gain I2 / I1 ;
5. The voltage gain V2 / V1 and 6. the voltage gain V2 / Vg.
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V1  z11 I1  z12 I 2 ;
V2  z 21 I1  z 22 I 2 ;
V1  Vg  I1Z g ;
V2   I 2 z L .
 z21 I1
I2 
Z L  z22
z12 z21
1). Z in  V1 / I1  Z in  z11 
Z L  z22
I1 
Vg  z12 I 2
z11  Z g
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2). I 2 
z
11
 z21Vg
 Z g z22  Z L   z12 z21
The Thevenin voltage with respect to port 2
V1
3a).V2 I 0  VTh  z21 I1  z21
2
z11
Vg
V1  Vg  I1Z g  I1 
Z g  z11
V2
I 2 0
z21
 VTh 
Vg
Z g  z11
Thevenin or output impedance is ratio V2 / I2 when Vg is replace
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by short cicuit.
V2  z21 I1  z22 I 2 ;
 z12 I 2
I1 
;
z11  Z g
V2
3b).
I2
Vg  0
z12 z21
 ZTh  z22 
z11  Z g
The current gain
I2
 z21
4). 
I1 Z L  z22
The voltage gain V2 / V1
  V2 

V2  z21 I1  z22 
 ZL 
21
  V2 

z11 I1  V1  z12 
 ZL 
V1 z12V2
I1 

z11 z11Z L
V2
z21Z L
z21Z L
5). 

V1 z11Z L  z11 z22  z12 z21 z11Z Lz
The voltage ratio V2 / Vg
Vg
z12V2
I1 

Z L z11  Z g  z11  Z g
z 21Z g
z 21 z12V2
z 22
V2 


V2
Z L z11  Z g  z11  Z g Z L
22
V2
z21Z L
6). 
Vg z11  Z g z22  Z L   z12 z21
23
y12 y21Z L
Yin  y11 
1  y22 Z L
y parameters
I2 
y21Vg
1  y22 Z L  y11Z g  yZ g Z L
VTh 
 y21Vg
y22  yZ g
Z Th 
1  y11Z g
y22  yZ g
I2
y21

I1 y11  yZ L
V2
 y21Z L

V1 1  y22 Z L
V2
y21Z L

Vg y12 y21Z g Z L  1  y11Z g 1  y22 Z L 
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a parameters
a11Z L  a12
Z in 
a21Z L  a22
I2 
 Vg
a11Z L  a12  a21Z g Z L  a22 Z g
VTh 
Vg
a11  a21Z g
Z Th 
a12  a22 Z g
a11  a21Z g
I2
1

I1 a21Z L  a22
V2
ZL

V1 a11Z L  a12
V2
ZL

Vg a11  a21Z g Z L  a12  a22 Z g
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b parameters
b22 Z L  b12
Z in 
b21 Z L  b11
I2 
 V g b
b11 Z g  b21Z g Z L  b22 Z L  b12
VTh 
V g b
b22  b21 Z g
Z Th 
b11Z g  b12
b21Z g  b22
I2
 b

I1 b11  b21Z L
V2
bZL

V1 b12  b22 Z L
V2
bZL

Vg b12  b11 Z g  b22 Z L  b21 Z g Z L
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Example
I1
I2
500Ω
Zg
V1
50000
+
+
[b]
-
V2
ZL
5kΩ
-
The two port circuit shown above is described in term of its b parameters, the values of
which are
b11= -20, b12= -3000Ω, b21= -2 mS, and b22 = -0.2.
a. Find the phasor voltage V2.
b. Find the average power delivered to the 5 kΩ load.
c. Find the average power delivered to the input port.
d. Find the load impedance for maximum average power transfer.
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e. To find he maximum average power delivered to the load in (d)
Solution
a.

b   20 0.2   3000  2x10
3
  4  6  2

V2
 25000
10


3
Vg  3000  20500  0.25000  2 x10 5005000 19


 10 
V2   500  263.1600 V
 19 
b. The average power delivered to the 5000 Ω load is
263.162
P2 
 6.93W
25000
c. To find the maximum average power delivered to the input port

 0.25000  3000 400
Z in 

 133.33 
3
 2 x10 5000  20
3
28
500
I1 
 789 .47 mA
500  133 .33
0.789472
133.33  41.55W
P1 
2
d. The load impedance for maximum power transfer

 20500  3000 13,000
ZTh 

 10,833.33 
3
 2 x10 500  0.2 1.2
*
Z L  ZTh
 10,833.33 
e. The maximum average power delivered to ZL
V2
 0.8333 V2  0.8333500  416.67 V
Vg
1 416.672
P2 m ax 
 8.01W
2 10,833.33
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I2
I1
Zg
Vg
+
V1
+
Two port model
of a network
-
V2
ZL
-
The b parameters of the two-port network are :
b11 =2000/3, b12 =2/3 MΩ, b21 =1/15 S, and b22 =-100/3. The network is driven by a
sinusoidal current source having a maximum amplitude of 100 μA and an internal
impedance of 1000+j0 Ω. It is terminated in a resistive load of 10 kΩ.
a. Calculate the average power delivered to the load resistor.
b. Calculated the load resistance for maximum average power.
c. Calculate the maximum average power delivered to the resistor in (b).
30