Transcript No Slide Title
TWO-PORT NETWORKS
In many situations one is not interested in the internal organization of a network. A description relating input and output variables may be sufficient A two-port model is a description of a network that relates voltages and currents at two pairs of terminals
LEARNING GOALS
Study the basic types of two-port models Admittance parameters Impedance parameters Hybrid parameters Transmission parameters Understand how to convert one model into another
ADMITTANCE PARAMETERS
The network contains NO independent sources The admittance parameters describe the currents in terms of the voltages
y
21 determines the current flowing into port 2 when the port is short circuited and a voltage is applied to port 1
I
1
I
2
y
11
V
1
y
21
V
1
y
12
V
2
y
22
V
2 The first subindex identifies the output port. The second the input port.
The computation of the parameters follows directly from the definition
y
11
I V
1 1
V
2 0
y
12
I
1
V
2
V
1 0
y
21
I V
1 2
V
2 0
y
22
I
2
V
2
V
1 0
LEARNING EXAMPLE Find the admittance parameters for the network
I
1
I
2
y
11
V
1
y
21
V
1
y
12
V
2
y
22
V
2 Circuit used to determine
y
11 ,
y
21
I
2
I
1
I
2 ( 1 1 1 2 )
V
1 1 2
I
1
y
11
I
2 3 2 [
S
] 1 2
V
1
y
21 1 2 [
S
] Circuit used to determine
y
12 ,
y
22
I
2
I
1 1 2 3 1 3 2 3
I V
2 2
y
22 3 5 5 6
V
2 5 6 [
S
]
y
12 1 2 [
S
] Next we show one use of this model
An application of the admittance parameters Determine the current through the 4 Ohm resistor
I
1
I
2
y
11
V
1
y
21
V
1
y
12
V
2
y
22
V
2
I
1
I
1 3 2
V
1 1 2
V
2
I
2 1 2
V
1 2
A
,
V
2 5 6
V
2 4
I
2
I
2 1
V
4 2 The model plus the conditions at the ports are sufficient to determine the other variables.
2 0 3 2
V
1 1 2
V
1 1 2
V
2 5 6 1 4
V
2
V
1 13
V
2 6
V
2 8 11 [
V
]
I
2 2 11 [
A
]
LEARNING EXTENSION
I
1
V
1 Find the admittance (Y) parameters
I
2
V
2
I
1
I
2
y
11
V
1
y
21
V
1
y
12
V
2
y
22
V
2
I
2
V
1
I
1
I
1 1 ( 21 1 42 )
V
1 3 42
V
1
I
2 42 21 42
I
1
y
11 1 14 [
S
]
y
21 1 21 [
S
]
I
1
I
2
V
2
I
2 2 21 1 21
V
2
I
1 10 .
5 21 10 .
5
I
2
y
22 1 7 [
S
]
y
12 1 21 [
S
]
LEARNING EXTENSION 10
A I
1
V
1
I O I
2 Use the admittance (Y) parameters to find the current Io Conditions at I/O ports
I
1 10
A I
2
I o
1 5
V
2
I
2 5
V
2 Replace in model 10 1 14
V
1 1 21 ( 5
I o
)
I o
1 21
V
1 1 7 ( 5
I o
) Solve for variable of interest
I o
420 [ 98
I
1
I
2
y
11
V
1
y
21
V
1
y
12
V
2
y
22
V
2
A
]
y
11 1 14 [
S
]
y
21 1 21 [
S
]
y
22 1 7 [
S
]
y
12 1 21 [
S
]
IMPEDANCE PARAMETERS
The network contains NO independent sources
V
1
V
2
z
11
I
1
z
21
I
1
z
12
I
2
z
22
I
2 The ‘
z parameters’
can be derived in a manner similar to the Y parameters
z
11
V
1
I
1
I
2 0
z
21
V
2
I
1
I
2 0
z
12
V
1
I
2
I
1 0
z
22
V
2
I
2
I
1 0
LEARNING EXAMPLE Find the Z parameters
V
1
V
2
z
11
I
1
z
21
I
1
z
12
I
2
z
22
I
2
z
11
V
1
I
1
I
2 0
z
21
V
2
I
1
I
2 0 Write the loop equations
V
1
V
2 2
j I
2 1
I
2
j
4 (
I
1
j
4 (
I
2
I
2 )
I
1 ) rearranging
V
1
V
2 ( 2
j
4
I j
4 )
I
1 1
j
2
I
2
j
4
I
2
z
12
z
11 2
z
21
j
4
j
4
V
1
I
2
I
1 0
z
12
z
22
j
4
j
2
z
22
V
2
I
2
I
1 0
LEARNING EXAMPLE Use the Z parameters to find the current through the 4 Ohm resistor
V
1
V
2
z
11
I
1
z
21
I
1
z
12
I
2
z
22
I
2 Output port constraint
V
2 4
I
2 Input port constraint
V
1 12 0 ( 1 )
I
1
V
1
V
2 ( 2
j
4
I j
4 )
I
1 1
j
2
I
2
j
4
I
2 0
j
4
I
1 ( 4 12 ( 3
j
4 )
I
1
j
2 )
j
4
I
2
I
2 ( 3
j
4 )
j
4 48
j
( 16 ( 4
j
2 )( 3
j
4 ))
I
2
I
2 1 .
61 137 .
73
V
1 LEARNING EXTENSION
I
1
V
1
V
2 12
I
1 3
I
2 6 (
I
6 (
I
1 1
I
2
I
2 ) )
z
11 18 ,
z
21 6 ,
I
2
V
2
z
12 6
z
22 9 Find the Z parameters. Find the current on a 4 Ohm load with a 24V input source
V
1
I
1 24
V
V
2
I
2 4
V
1
V
2 18
I
1 6
I
1 6
I
9
I
2 2 output port constraint :
V
2 input port constraint :
V
1 4
I
2 24 [
V
] 24 0 6 18
I
1
I
1 6
I
2 13
I
2 24 ( 39 6 )
I
2 ( 3 )
I
2 24 33 [
A
]
HYBRID PARAMETERS
The network contains NO independent sources
V
1
I
2
h
11
I
1
h
21
I
1
h
12
V
2
h
22
V
2
h
11
h
12
V
1
I
1
V
1
V
2
V
2 0
I
1 0
h h
21 22
I
2
I
1
I
2
V
2
V
2 0
I
1 0
h
11 short circuit input impedance
h
12 open circuit reverse voltage gain
h
21 short circuit forward current gain
h
22 open circuit output admittance These parameters are very common in modeling transistors
LEARNING EXAMPLE Find the hybrid parameters for this circuit Non-inverting amplifier
V
1
I
2
h
11
I
1
h
21
I
1
h
12
V
2
h
22
V
2 Equivalent linear circuit
I R
2
I DS V
1 (
R i
R
1 ||
R
2 )
I
1
h
11
R i
R R
1 1
R
2
R
2
I
2
I R
2
I DS
R
1
R
1
R
2
I
1
AR i I
1
R o h
21
AR i R o
R
1
R
1
R
2
V
1
R
1
R
1
R
2
V
2
h
12
R
1
R
1
R
2
V i
0
h
22
I
2
R o
|| (
V
2
R
1
R o R o
(
R R
1 1
R R
2 2 )
R
2 )
LEARNING EXTENSION Find the hybrid parameters for the network
V
1
I
1
I
2
V
2
I
1
I
2
V
1
V
2
V
1 ( 12 ( 6 || 3 ))
I
1
h
11 14
I
2 6 3 6
I
1
h
21 2 3 0
V
1
I
2
h
11
I
1
h
21
I
1
h
12
V
2
h
22
V
2
V
1
I
1 0
I
2
V
2
V
1 3 6 6
V
2
h
12 2 3
I
2
V
2 9
h
22 1 9 [
S
]
LEARNING EXTENSION
I
1
I
2
V
1
V
2
h
11 14 ,
h
21 2 3
h
12 2 3
h
22 1 9 [
S
] Determine the input impedance of the two-port 4
V
1
I
2
h
11
I
1
h
21
I
1
h
12
V
2
h
22
V
2
R in
V
1
I
1 output port constraint :
V
2 4
I
2
V
1
I
2
h
11
I
1
h
21
I
1
h
12 ( 4
I
2 )
h
22 ( 4
I
2 )
I
2 1
h
21 4
h
22
I
1
V
1
h
11 1 4
h
12
h
21 4
h
22
I
1
R in
14 4 ( 2 / 3 )( 2 / 3 ) 1 4 ( 1 / 9 ) 14 16 13 15 .
23 Verification
R in
12 6 || 7 12 42 13
TRANSMISSION PARAMETERS
ABCD parameters
A
V V
2 1
I
2 0
B
V
1
I
2
V
2 0 The network contains NO independent sources
V
1
I
1
AV
2
CV
2
BI DI
2 2
C
I
1
V
2
I
2 0
D
I I
2 1
V
2 0 A B C open circuit voltage ratio negative short open D negative short circuit transfer impedance circuit transfer admittance circuit current ratio
LEARNING EXAMPLE Determine the transmission parameters
V
1
I
1
AV
2
CV
2
BI DI
2 2
A
V V
2 1
I
2 0
C
I
1
V
2
I
2 0
B
V
1
I
2
V
2 0
D
I I
2 1
V
2 0 when
I
2 0
V
2 1 1
j
1
j
V
1
A
1
j
V
2 1
j
I
1
I V
2 1
j
I
2 when
V
2 1 1
j
1
j
I
1 0 1 1
j
I
1
D
1
j
V
1 1 ( 1 || 1
j
)
I
1 2 1
j
j
( 1
j
)
I
2
B
2
j
LEARNING EXTENSION
V
1
I
1
I
3
I
2 Determine the transmission parameters
V
2
V
1
I
1
AV
2
CV
2
BI DI
2 2
A
V V
2 1
I
2 0
C
I
1
V
2
I
2 0
B
V
1
I
2
V
2 0
D
I I
2 1
V
2 0 when
I
2 0
V
2 10 .
10 .
5 5 21
V
1
A
3
I
3 42 42 21
I
1
V
2 10 .
5
C
I V
2 1 1 7 [
S
] When
V
2 0
I
2 42 42 21
I
1
D
3 2
V
1 ( 42 || 21 )
I
1 14
I
1 14 ( 3 2
I
2 )
B
21
PARAMETER CONVERSIONS
If all parameters exist, they can be related by conventional algebraic manipulations.
As an example consider the relationship between Z and Y parameters
V
1
V
2
z
11
I
1
z
21
I
1
z
12
I
2
z
22
I
2
V
1
V
2
z
11
z
21
z
12
z
22
I I
2 1
I
1
I
2
z
11
z
21
z
12
z
22 1
V V
2 1
y
11
y
21
y
12
y
22
V V
2 1
y
11
y
21
y
12
y
22
z
11
z
21
z
12
z
22 1 1
Z
z
22
z
21
z z
12 11 with
Z
z
11
z
22
z
21
z
12 In the following conversion table, the symbol stands for the determinan t of the correspond ing matrix
Z
z
11
z
21
z
12
z
22 ,
Y
y
11
y
21
y
12
y
22 ,
H
h
11
h
21
h
12
h
22 ,
T
A C B D
INTERCONNECTION OF TWO-PORTS
Interconnections permit the description of complex systems in terms of simpler components or subsystems The basic interconnections to be considered are:
parallel, series and cascade
PARALLEL: Voltages are the same.
Current of interconnection is the sum of currents The rules used to derive models for interconnection assume that each subsystem behaves in the same manner before and after the interconnection SERIES: Currents are the same.
Voltage of interconnection is the sum of voltages CASCADE: Output of first subsystem acts as input for the second
Parallel Interconnection: Description Using Y Parameters Interconne ction descriptio n
I I
2 1
I
YV
y
11
y
21
y
12
y
22
V V
1 2
I a
I
1
a I
2
a
,
V a
V V
1
a
2
a
,
Y a
y
11
a y
21
a
Interconne ction constraint s :
y
12
y
22
b a
I a I
1
V
1
I
1
a
V
1
a
I
1
b
V
1
b
, ,
I
2
V
2
I
2
a
V
2
a
I
2
b
V
2
b
V I
Y a V a
I a
V a
I b
V b
In a similar manner
I b
Y b V b
I
Y a V a
Y b V b
(
Y a
Y b
)
V Y
Y a
Y b
Series interconnection using Z parameters SERIES: Currents are the same.
Voltage of interconnection is the sum of voltages Descriptio n of each subsystem
V a
Z a I a
,
V b
Z b I b
Interconne ction constraint s
I a V
I b
V a
I
V b
V
Z a I
Z b I
(
Z a
Z b
)
I Z
Z a
Z b
Cascade connection using transmission parameters CASCADE: Output of first subsystem acts as input for the second
V
1
a I
1
a
Interconne ction constraint s :
I
2
a V I
1 1
V
1
a I
I
1
b
1
a V I
2 2
V
2
a V
2
b I
2
b
V
1
b
A C a a B a D a
V
I
2
a
2
a
V
1
b I
1
b
A C b b B b D b
V
I
2
b
2
b
V
1
I
1
A C a a V
1
I
1
AV
2
CV
2
BI DI
2 2
V
1
I
1
A C B D
V
I
2 2 Matrix multiplication does not commute.
Order of the interconnection is important
B D a a
A C b b B D b b
V
2
I
2
LEARNING EXAMPLE Find the Y parameters for the network
V
1
V
1
I
1
I
1
V
1
I
2
V
2
I
1
j
2
I
1 1
j
2
I
2
V
2
y
11
a
2 1
j
j
1 2
S
,
y
21
a
j
1 2
S
;
y
12
a
j
1 2
y
22
a
1
j
2
S Y
3 5 1 5
j j
1 2 1 2 2 1
I
2
V
2
V
1
V
2 2
I
1
I
1
I
2 3
I
2
Y b
2 1 1 3 1 1 5 3 1 2 1 1 5 2 5
j
1 2
j
1 2 [
S
]
Find the Y parameters for the network using a direct approach
I
1
V
1
V x I
2
V
2
V x
1
V x
1
V
1
V x
V
2 2 0
V x
2
V
1
V
2 5
I
1
I
2
V
1
V
2 1
V x
2
V x
V
1
j V
2 2
V
2
V
1
j
2 Replace V x and rearrange
LEARNING EXAMPLE Find the Z parameters of the network Network A Use direct method, or given the Y parameters transform to Z … or decompose the network in a series connection of simpler networks
Z a
2 3 3 2 2 2 2
j j j
2 3 2 2 4 3 2
j j j
Z b
1 1 1 1
Z
Z a
Z b
5 3 5 3 4 2 2 2
j j j j
5 3 5 3 2 2 6 2
j j j j
Network B
LEARNING EXAMPLE Find the transmission parameters By splitting the 2-Ohm resistor, the network can be viewed as the cascade connection of two identical networks
A C B D
1
j
j
2 1
j j
1
j
j
A C B D
( 1
j
( 1
j
) 2
j
) ( 2 ( 1
j
)
j
j
)(
j
) 2 1
A C j j
B D
1
j
j
2 1
j j
( 1
j
)( 2
j
( 2
j
) ( 2
j
) ( 1
j
)( 1
j
) 2
j
)
A C B D
1 2
j
4
j
2 2 2 2 4 6
j
1 4
j
2 2 2 2
LEARNING by APPLICATION Given the demand at the receiving end, determine the conditions on the sending end
V
1
I
1
AV
2
CV
2
BI DI
2 2 Transmission parameters are best suited for this application In the next slide we show how to determine the transmission parameters for the line. Here we assume them known and use them for analysis |
V P
2 |
V
3
L
|
V L
300
kV
||
I L
| (line
pf
I
2
P
cos 1 3
V L pf
Conditions at the | voltage)
pf I L
| |
I
2 |
V
2
P
3
V L pf
|
V L
sending end
V
1
I
1
AV
2
CV
2
BI DI
2 2 | 0
P
received
P
sent
P total Q total f
3 |
V line
||
I line
3 |
V line
||
I line
Power factor angle
I line
| | cos sin
f f V line V
1
pf
sending
P
sending
I
1
Determining the transmission parameters for the line
V
1
I
1
AV
2
CV
2
BI DI
2 2
I
2 0
A
V V
2 1
R
Z L
Z C Z C
0 .
9590 0 .
27
I T V
2
V
1 0 (
R
B Z
V
1
I
2
L
)
I
2
B
R
Z L
100 .
00 84 .
84
I T I
2
R
Z Z C L
2
Z C
0
C
I V
2 1
I
1
C
V
2
Z C R
Z L Z C
2 2
Z C
975 .
10 90 .
13
S V
2 0
D
I I
2 1
R
Z L
Z C Z C
0 .
9590 0 .
27
LEARNING EXAMPLE Determine the effect of the load on the voltage gain
A
20 , 000 ,
R i
1
M
,
R o
500 ,
R
1 1
k
,
R
2 49
k
Hybrid parameters are computed in next slide Ideal gain 1
R
2 50
R
1
G
h
21
h
11
h
22
h
12
h
21
h
11
R L
49 .
88 1 1 .
247
R L
Using the hybrid parameters
V
1
h
11
I
1
h
12
V
2
I
2
h
21
I
1
h
22
V
2 Eliminatin g
I
1 and solving for
V
2
V
2
h h
21
V
1 11
h
22
h h
12 11
I h
21 2 Constraint at output Solving for
V
2
V
2
h
11
h
22
h h
12 21
h
21 port
h
11
R L V
1 :
V
2
R L I
2 Effect of load resistance
Computing the hybrid parameters for non-inverting amplifier (repeat earlier example) Non-inverting amplifier
V
1
I
2
h
11
I
1
h
21
I
1
h
12
V
2
h
22
V
2 Equivalent linear circuit
I R
2
I DS V
1 (
R i
R
1 ||
R
2 )
I
1
h
11
R i
R R
1 1
R
2
R
2
I
2
I R
2
I DS
R
1
R
1
R
2
I
1
AR i I
1
R o h
21
AR i R o
R
1
R
1
R
2
V
1
R
1
R
1
R
2
V
2
h
12
R
1
R
1
R
2
V i
0
h
22
I
2
R o
|| (
V
2
R
1
R o R o
(
R R
1 1
R R
2 2 )
R
2 )
LEARNING BY DESIGN Gain required = 10,000 on a load of 1kOhm From the conversion table
A
H h
21
h
12
h
21
h
21
h
11
h
22
B
h
11
h
21
C
h
22
h
21
D
1
h
21
V
1
AV
2
BI
2
R L I
2 Amplifier
A
20 , 000 parameters Ideal gain 1
R R i o
1
M
500
h
11 1 .
001
M
,
h
12
R
2 10 , 000
R
1
R
2 9 .
999 1 .
0 10
R
1 4 1
k
M
h
21 4 .
0 10 7 ,
h
22 1 .
0
mS
For the final solution we will need to cascade amplifiers. Hence the transmission parameters will prove very useful Analysis of solution: -Even with infinite load the maximum gain is only 6,667 Likely causes: -R2 is higher than input resistance Ri -Desired gain is comparable to the maximum gain, A, of the Op-Amp
V
2
V
1 1
A
B R L
6667 1 166 .
7
R L
Proposed solution: -Cascade two stages, each with ideal gain of 100. This also lowers R2 to 99kOhm
Analysis of proposed solution Since the two stages will be cascaded, the transmission parameters of the proposed solution will be
A
B
V I
1 1
A C a a A a A b A a B b
B a C b
B a D b
V
2
V
1
B D a a
A C b b
A
1
B R L
Identical stages
B D b b
V
2
I
2
A C a a B D a a
A C b b
B D b b
V
2
I
2 Effect of load resistance. G=10,000 Two-Ports