Transcript Two-Port Networks - ENCON

Two-Port Networks
Chapter 19
19.1 Introduction
• A two-port Network
is an electrical
network with two
separate ports for
input and output.
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19.2 Impedance Parameters
V1  z11I1  z12I 2
V2  z 21I1  z 22I 2
 V1   z11 z12   I1 
 I1 
 z  
V   z



 2   21 z 22  I 2 
I 2 
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V1
z11 
I1 I
,
2 0
V2
z 21 
I1 I
V1
z12 
I 2 I 0
1
, z 22
2 0
V2

I 2 I 0
1
z11 = Open-circuit input impedance
z12 = Open-circuit transfer impedance from port 1 to
port 2
z21 = Open-circuit transfer impedance from port 2 to
port 1
z22 = Open-circuit output impedance
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Fig 19.3
V1
V2
z11  , z 21 
I1
I1
V1
V2
z12  , z 22 
I2
I2
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19.4
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Fig 19.5
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Fig 19.6
1
V1  V2 , I1  nI 2
n
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Example 19.1
• Determine the z parameters for the circuit in Fig. 19.7.
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Example 19.1: PSpice
To find Z parameters
**Example 9.1
.Param Current_at_port2=1
**port2 = 0 : inject 0 A at port 2 ; 1st row
**port2 = 1 : inject 1 A at port 2 ; 2nd row
I1 0 1 DC {1-Current_at_port2}
*********************************************
R1 1 1x 20
R2 1x 0 40
R3 1x 2 30
*********************************************
I2 0 2 DC {Current_at_port2}
Vdummy 100 0 DC 1
.DC Vdummy 1 1 1
.STEP PARAM Current_at_port2 LIST 0 1
.PRINT DC V(1) V(2)
.END
ch19 Two-Port Networks
We need to edit here
for different circuits
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The Results
z11
z21
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3.000E+01
2.000E+01
z12
z22
2.000E+01
5.000E+01
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19.3 Admittance Parameters
I1  y11V1  y12V2
I 2  y 21V1  y 22V2
 I1   y11 y12   V1 
 V1 
 y  
I    y



 2   21 y 22  V2 
V2 
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I1
y11 
,
V1 V 0
2
I1
y12 
V2 V 0
1
I2
I2
y 21 
, y 22 
V1 V 0
V2 V 0
2
1
y11 = Short-circuit input admittance
y12 = Short-circuit transfer admittance from port 1 to
port 2
y21 = Short-circuit transfer admittance from port 2 to
port 1
y22 = Short-circuit output admittance
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Fig 19.13
V1
V2
y11  , y 21 
I1
I1
V1
V2
y12  , y 22 
I2
I2
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Example 19.3
• Obtain the y parameters for the  network shown in
Fig. 19.14.
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Example 19.3
1
y12   S  y 21
2
1
1
y11  y12   y11   y12  0.75 S
4
4
1
1
y 22  y12   y 22   y12  0.625 S
8
8
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Example 19.3
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Example 19.3
To find y parameters
**Example 19.3
.Param Voltage_at_port2=0
**port2 = 0 : apply 0 V at port 2 ; to find 1st row
**port2 = 1 : apply 1 V at port 2 ; to find 2nd row
V1 1x 0 DC {1-Voltage_at_port2}
Vport1 1x 1
****************************************************
R1 1 0 4
R2 1 2 2
R3 2 0 8
We just need to edit
here for different
networks.
****************************************************
Vport2 2x 2
V2 2x 0 DC {Voltage_at_port2}
Vdummy 100 0 DC 1
.DC Vdummy 1 1 1
.STEP PARAM Voltage_at_port2 LIST 0 1
.PRINT DC I(Vport1) I(Vport2)
.END
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The Results are
y11
y21
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7.500E-01
-5.000E-01
y12
y22
-5.000E-01
6.250E-01
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Example 19.4
• Determine the y parameters for the T network shown
in Fig. 19.17.
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Example 19.4
V1  Vo
Vo Vo  0
At node1,
 2I1 

8
2
4
V1  Vo
V1  Vo 3Vo
But I1 
, therefore, 0 

8
8
4
0  V1  Vo  6Vo  V1  5Vo
 5Vo  Vo
Hence, I1 
 0.75 Vo
8
I1  0.75Vo
and y11 

 0.15 S
V1
 5Vo
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Example 19.4
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Example 19.4
Vo  0
At node 2,
 2I1  I 2  0
4
or  I 2  0.25Vo  1.5Vo  1.25Vo
I 2 1.25Vo
Hence, y 21 

 0.25 S
V1  5Vo
Similarly, we get y12 and y 21 using Fig. 19.18(b). At node1,
0  Vo
Vo Vo  V2
 2I 1 

8
2
4
0  Vo
Vo Vo Vo  V2
But I1 
, therefore, 0   

8
8
2
4
or 0  Vo  4Vo  2Vo  2V2  V2  2.5Vo
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Example 19.4
I1  Vo /8
Hence, y12 

 0.05 S
V2 2.5Vo
Vo  V2
At node 2,
 2I1  I 2  0
4
1
2Vo
or  I 2  0.25Vo  (2.5)Vo 
 0.625Vo
4
8
I 2 0.625Vo
Thus, y 22 

 0.25 S
V2
2.5Vo
Notice that y12  y 21 in this case, since the network
is not reciprocal.
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PSpice
To find y parameters
**Example 19.3
.Param Voltage_at_port2=0
**port2 = 0 : apply 0 V at port 2 ; to find 1st row
**port2 = 1 : apply 1 V at port 2 ; to find 2nd row
V1 1xx 0 DC {1-Voltage_at_port2}
Vport1 1xx 1
****************************************************
R1 1 1x 8
R2 1x 0 2
R3 1x 2 4
F1 1x 2 Vport1 2
****************************************************
Vport2 2xx 2
V2 2xx 0 DC {Voltage_at_port2}
Vdummy 100 0 DC 1
.DC Vdummy 1 1 1
.STEP PARAM Voltage_at_port2 LIST 0 1
.PRINT DC I(Vport1) I(Vport2)
.END
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The Results are
y11
1.500E-01
y12
y21
-2.500E-01
y22
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-5.000E-02
2.500E-01
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19.4 Hybrid Parameters
V1  h11I1  h12 V2
I 2  h 21I1  h 22 V2
V1  h11 h12   I1 
 I1 
 h 
 I   h



 2   21 h 22  V2 
V2 
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V1
V1
h11 
, h12 
I1 V 0
V2 I 0
2
1
I2
I2
h 21 
, h 22 
I1 V 0
V2 I 0
2
1
h11 = Short-circuit input impedance
h12 = Open-circuit reverse voltage gain
h21 = Short-circuit forward current gain
h22 = Open-circuit output admittance
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Fig 19.20
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Example 19.5
• Find the hybrid parameters for the two-port network
of . 19.22.
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Example 19.5
From Fig. 19.23(a),
V1  I1 (2  3 6)  4I1
V1
 4
Hence, h11 
I1
2
6
I1  I1
 I2 
3
63
2
I2
Hence, h12   
3
I1
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Example 19.5
From Fig. 19.23(b),
2
6
V2  V2
V1 
3
63
V1 2

Hence, h12 
V2 3
Also, V2  (3  6)I 2  9I 2
Thus, h 22
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I2 1
 S

V2 9
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Fig 19.31
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V2
V2
a
, b
V1 I 0
I1 V 0
1
I2
c
,
V1 I 0
1
1
2
d
I1 V 0
1
a = Open-circuit voltage gain
b = Negative short-circuit transfer impedance
c = Open-circuit transfer admittance
d = Negative short-circuit current gain
AD  BC  1, ad  bc  1
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Example 19.8
• Find the transmission parameters for the two-port
network in Fig. 19.32
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Example 19.8
From Fig. 19.33(a),
V1  (10  20)I1  30I1 and V2  20I1  3I1  17I1
Thus
V1 30I1
I1
I1
A

 1.765, C 

 0.0588 S
V2 17I1
V2 17I1
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Example 19.8
From Fig. 19.33(b),
V1  Va Va

 I2  0
10
20
But Va  3I1 and I1  (V1  Va ) / 10,
 Va  3I1 , V1  13I1
3I1
17
 I1 
 I 2  0  I1  I 2
20
20
Therefore,
I1 20
V1
 13I1
D 
 1.176, B  

 15.29 
I 2 17
V2 (17 / 20)I1
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19.6 Relationships Between
Parameters
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Example 19.10
• Find [z] and [g] of a two-port network if
 10 1.5 
[T]  

2
S
4


• Solution:
If A  10, B  1.5, C  2, D  4, the determinant of the
matrix is
T  AD  BC  40  3  37
From Table 19.1,
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Example 19.10
T 37
A 10
 18.5

z11    5, z12 
2
C
C 2
D 4
1 1
z 21    0.5, z 22    2
C 2
C 2
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T
C 2
   3.7
g11    0.2, g12  
10
A
A 10
B 1.5
1 1
 0.15
g 21    0.1, g 22  
A 10
A 10
0.2 S  3.7 
 5 18.5
, [g ]  
Thus, [z ]  



0.15
1
.
0
2
5
.
0




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19.7 Interconnection of Networks
[z ]  [z a ]  [z b ]
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Example 19.12
• Evaluate V2/V1 in the circuit in Fig. 19.42.
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Example 19.12
This may be regarded as two - ports in series.
For N b ,
z12b  z 21b  10  z11b  z 22b
Thus,
12 8  10 10  22 18 
[z ]  [z a ]  [z b ]  





8
20
10
10
18
30

 
 

But
V1  z11I1  z12I 2  22I1  18I 2
V2  z 32I1  z 22I 2  18I1  30I 2
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Example 19.12
Also, at the input port V1  Vs  5I1
V2
and at the output port V2  20I 2  I 2  
20
18
 Vs  5I1  22I1  V2  Vs  27 I1  0.9V2
20
30
2 .5
 V2  18I1  V2  I1 
V2
20
18
2 .5
 Vs  27 
V2  0.9V2  2.85V2
18
V2
1
And also,

 0.3509
Vs 2.85
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Example 19.13
• Find the y parameters of the two-port in Fig. 19.44.
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Example 19.14
• Find the transmission parameters for the circuit in Fig.
19.46.
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