Date post: | 13-Mar-2016 |
Category: |
Documents |
Upload: | callum-rodriquez |
View: | 57 times |
Download: | 1 times |
TWO-PORT NETWORKS
In many situations one is not interested in the internal organization of anetwork. A description relating input and output variables may be sufficient
A two-port model is a description of a network that relates voltages and currentsat two pairs of terminals
Admittance parametersImpedance parametersHybrid parametersTransmission parameters
LEARNING GOALS
Study the basic types of two-port models
Understand how to convert one model into another
ADMITTANCE PARAMETERS
The admittance parameters describe the currents in terms of the voltages
2221212
2121111
VyVyIVyVyI
The first subindex identifies
the output port. The secondthe input port.
1 port to applied is voltage a and circuited-short is port
the when 2 port into flowingcurrent the determines 21y
The computation of the parameters follows directly from the definition
02
112
01
111
12
VV V
IyVIy
02
222
01
221
12
VV V
IyVIy
The network contains NO independent sources
LEARNING EXAMPLEFind the admittance parameters for the network
2221212
2121111
VyVyIVyVyI
][23)
211( 1111 SyVI
2111, yy determine to used Circuit
2I
1212 21
211 VIII ][
21
21 Sy
2212 , yy determine to used Circuit][
65
31
21
2222 SyVI
][21
6553
323
12221 SyVII
Next we show one use of this model
An application of the admittance parameters
2221212
2121111
VyVyIVyVyI
212
211
65
21
21
23
VVI
VVI
221 4,2 IVAI
The model plus the conditions at theports are sufficient to determine theother variables.
21
21
41
65
210
21
232
VV
VV
22 41VI
][112
][1186
13
2
2
21
AI
VV
VV
Determine the current through the4 Ohm resistor
LEARNING EXTENSIONFind the admittance (Y) parameters
1I
1V
2I
2221212
2121111
VyVyIVyVyI
1I
1V
2I
2V
111 423)
421
211( VVI
12 422142 II
][211
][141
21
11
Sy
Sy
1I 2I
2V
21
22
5.10215.10211
212
II
VI
][211
][71
12
22
Sy
Sy
LEARNING EXTENSION
5A10
OI
Use the admittance (Y) parameters to find the current Io
Conditions at I/O portsAI 101
2
22 51
II
VI
o
1I
1V2I
2V 2221212
2121111
VyVyIVyVyI
][211
][141
21
11
Sy
Sy
][211
][71
12
22
Sy
Sy
Replace in model
)5(71
211
)5(211
14110
1
1
oo
o
IVI
IV
Solve for variable of interest ][98420 AIo
IMPEDANCE PARAMETERS
The network contains NO independent sources
2221212
2121111
IzIzVIzIzV
The ‘z parameters’ can be derived in a manner similar to the Y parameters
01
221
01
111
22
II I
VzIVz
02
222
02
112
11
II I
VzIVz
LEARNING EXAMPLE Find the Z parameters
Write the loop equations
)(42)(42
1222
2111
IIjIjVIIjIV
24442
2221
1211
jzjzjzjz
212
211
244)42(
IjIjVIjIjV
rearranging
01
221
01
111
22
II I
VzIVz
02
222
02
112
11
II I
VzIVz
2221212
2121111
IzIzVIzIzV
LEARNING EXAMPLEUse the Z parameters to find the current through the 4 Ohmresistor
2221212
2121111
IzIzVIzIzV
Output port constraint
22 4IV
Input port constraint
11 )1(012 IV
212
211
244)42(
IjIjVIjIjV
21 )24(40 IjIj
21 4)43(12 IjIj
)43( j4j
2))43)(24(16(48 Ijjj 73.13761.12I
LEARNING EXTENSIONFind the Z parameters. Find the current on a 4 Ohm load with a 24V input source
1I
1V
2I
2V
)(63)(612
2122
2111
IIIVIIIV
9,66,18
2221
1211
zzzz
4
V24
212
211
96618IIVIIV
1I
1V
2I
2V
22 4IV :constraint port output][241 VV :constraint port input
21
21
136061824IIII
)3(
2)639(24 I
][3324
2 AI
HYBRID PARAMETERS
The network contains NO independent sources
2221212
2121111
VhIhIVhIhV
admittance output circuit-opengain current forward circuit-short
gain voltage reverse circuit-openimpedance input circuit-short
22
21
12
11
hhhh
These parameters are very common in modeling transistors
01
221
01
111
22
VV I
IhIVh
02
222
02
112
11
II V
IhVVh
LEARNING EXAMPLE Find the hybrid parameters for this circuit
Non-inverting amplifier Equivalent linear circuit2221212
2121111
VhIhIVhIhV
21
21111211 )||(
RRRRRhIRRRV ii
2RI
DSI
o
iDSR R
IARIRR
RIII 11
21
122
21
121 RR
RRARho
i
21
1122
21
11 RR
RhVRR
RV
)(||0
21
22 RRR
VIVo
i
)( 21
2122 RRR
RRRho
o
LEARNING EXTENSIONFind the hybrid parameters for the network
1I
1V
2I
2V2221212
2121111
VhIhIVhIhV
1I
1V
2I
02 V
14))3||6(12( 1111 hIV
32
636
2112
hII
1V
2I
2V01 I
32
636
1221
hVV
][91
9 222
2 ShVI
LEARNING EXTENSION
4
Determine the input impedance of the two-port
2221212
2121111
VhIhIVhIhV
1I
1V
2I
2V 22 4IV :constraint port output
1
1
IVRin
)4()4(
2221212
2121111
IhIhIIhIhV
122
212 41
Ih
hI
122
2112111 41
4 IhhhhV
][91
32
32,14
2221
1211
Shh
hh
23.15131614
)9/1(41)3/2)(3/2(414inR
Verification
1342127||612inR
TRANSMISSION PARAMETERS
The network contains NO independent sources
221
221
DICVIBIAVV
02
1
02
1
22
II V
ICVVA
02
1
02
1
22
VV I
IDIVB
ratio current circuit-short negative Dadmittance transfer circuit-open C
impedance transfer circuit-short negative Bratio voltage circuit openA
ABCD parameters
LEARNING EXAMPLEDetermine the transmission parameters
221
221
DICVIBIAVV
02
1
02
1
22
II V
ICVVA
02
1
02
1
22
VV I
IDIVB
jAV
j
jV
111
1
12
jVII
jV
2
112
1
02 I when02 V when
112 11
11
1
Ij
I
j
jI
jD 1
211 )1(12)1||1(1 Ij
jjI
jV
jB 2
LEARNING EXTENSIONDetermine the transmission parameters
221
221
DICVIBIAVV
02
1
02
1
22
II V
ICVVA
02
1
02
1
22
VV I
IDIVB
1I
1V
2I
2V
02 I when
3215.10
5.1012
AVV
3I
][71
5.10214242
2
1213 S
VICVII
02 V When
23
214242
12
DII
)23(1414)21||42( 2111 IIIV
21B
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
2
11
2221
1211
2
1
2
1
2221
1211
2
1
2221212
2121111
VV
zzzz
II
II
zzzz
VV
IzIzVIzIzV
2
1
2221
1211
VV
yyyy
1
2221
1211
2221
1211
zzzz
yyyy
1121
12221zzzz
Z
12212211 zzzzZ with
matrix ingcorrespondthe of tdeterminan the for stands symbol the table, conversion following the In
DCBA
hhhh
yyyy
zzzz
THYZ ,,,2221
1211
2221
1211
2221
1211
INTERCONNECTION OF TWO-PORTSInterconnections permit the description of complex systems in terms of simplercomponents or subsystems
The basic interconnections to be considered are: parallel, series and cascade
PARALLEL: Voltages are the same.Current of interconnectionis the sum of currents
CASCADE:Output of first subsystemacts as input for thesecond
The rules used to derive modelsfor interconnection assume thateach subsystem behaves in thesame manner before and afterthe interconnection
SERIES: Currents are the same.Voltage of interconnection is the sumof voltages
aaaba
aaa
a
aa
a
aa VYI
yyyy
YVV
VII
I
2221
1211
2
1
2
1 ,,bbb VYI manner similar a In
baba
baba
VVVVVVIIIIII
222111
222111
,,
:sconstraint ctionInterconne
ba
ba
VVVIII
ba YYY
VYYVYVYI babbaa )(
Parallel Interconnection: Description Using Y Parameters
YVIVV
yyyy
II
2
1
2221
1211
2
1
ndescriptioctionInterconne
Series interconnection using Z parameters
bbbaaa IZVIZV ,subsystem each of nDescriptio
ba
ba
VVVIII
sconstraint ctionInterconneIZZIZIZV baba )(
ba ZZZ
SERIES: Currents are the same.Voltage of interconnection is the sumof voltages
Cascade connection using transmission parameters
ba
ba
baba
IIIIVVVV
VVII
2211
2211
1212
:sconstraint ctionInterconne
a
a
aa
aa
a
a
IV
DCBA
IV
2
2
1
1
b
b
bb
bb
b
b
IV
DCBA
IV
2
2
1
1
221
221
DICVIBIAVV
2
2
1
1
IV
DCBA
IV
2
2
1
1
IV
DCBA
DCBA
IV
bb
bb
aa
aa
Matrix multiplication does not commute.Order of the interconnection is important
CASCADE:Output of first subsystemacts as input for thesecond
LEARNING EXAMPLEFind the Y parameters for the network
112
2j1I
1V
2I
2V
1I
1V
2I
2V
12
121 2II
IjVV
21,
21
21
1211 jySjj
y aa
SjySjy aa 21;
21
2221
212
211
32
IIVIIV
2113
51
3112 1
bY
][
21
52
21
51
21
51
21
53
Sjj
jjY
1I
1V
2I
2V
xV
520
2112121 VVVVVVVV
xxxx
22
21
1222
2111
jVVVVI
jVVVVI
x
x
Find the Y parameters for the network using a direct approach
rearrange and V Replace x
LEARNING EXAMPLEFind the Z parameters of the networkNetwork A
Network BUse direct method,or given the Y parameters transform to Z… or decompose the network in a seriesconnection of simpler networks
jj
j
jjj
Za
2342
232
232
2322
1111
bZ
jj
jj
jj
jj
ZZZ ba
2365
2325
2325
2345
LEARNING EXAMPLEFind the transmission parameters
jjjj
DCBA
121
By splitting the 2-Ohm resistor,the network can be viewed as thecascade connection of two identicalnetworks
jjjj
jjjj
DCBA
121
121
2
2
)1()2())(1()1()1)(2()2)(1()2()1(
jjjjjjjjjjjjjj
DCBA
22
22
24122264241
jjjj
DCBA
sentreceivedPP
pfIVP
VVkVVV
LL
LL
||||3
0||300|| 22 voltage) (line
LEARNING by APPLICATIONGiven the demand at the receiving end, determinethe conditions on the sending end
221
221
DICVIBIAVV
Transmission parametersare best suited for thisapplication
In the next slide we show how to determine the transmission parameters for theline. Here we assume them known and use them for analysis
pfVPIIL
L 3|||| 2 flinelinetotal
flinelinetotal
IVQ
IVP
sin||||3
cos||||3
lineV
lineI
Power factor anglef
pfpfV
PIL
12 cos
3
221
221
DICVIBIAVV
end sending
the at Conditions
sendingsending PpfIV
1
1
Determining the transmission parameters for the line221
221
DICVIBIAVV
2
12 0
VVAI
27.09590.0C
CL
ZZZR
2
12 0
VICI
TI CCL
CT Z
VIZZR
ZI 212
SZ
ZZRCC
CL 13.9010.97522
2
12 0
IVBV
84.8400.100)( 21 LL ZRBIZRV
2
12 0
IIDV
27.09590.0
C
CL
ZZZR
LEARNING EXAMPLEDetermine the effect of the load on the voltage gain kRkRRMRA oi 49,1,500,1,000,20 21
5011
2 RRgain Ideal
2221212
2121111
VhIhIVhIhV
Using the hybrid parameters
21 VI for solving and gEliminatin
21122211
2111212 hhhh
IhVhV
22 IRV L :port output at Constraint
111
21122211
212
2
V
Rhhhhh
hV
V
L
for Solving
LL RRhhhhh
hG 247.11
88.4911
21122211
21
Effect of load resistance
Hybrid parameters are computed in next slide
Computing the hybrid parameters for non-inverting amplifier (repeat earlier example)
Non-inverting amplifier Equivalent linear circuit2221212
2121111
VhIhIVhIhV
21
21111211 )||(
RRRRRhIRRRV ii
2RI
DSI
o
iDSR R
IARIRR
RIII 11
21
122
21
121 RR
RRARho
i
21
1122
21
11 RR
RhVRR
RV
)(||0
21
22 RRR
VIVo
i
)( 21
2122 RRR
RRRho
o
LEARNING BY DESIGNGain required = 10,000 on a load of 1kOhm
5001
000,20
o
i
RMR
Aparameters Amplifier
000,1011
2 RRgain Ideal
kRMR
1999.9
1
2
For the final solution we will need to cascadeamplifiers. Hence the transmission parameterswill prove very useful
2121
22
21
11
21
22112112
21
1h
DhhC
hhB
hhhhh
hA H
table conversion the From
221
221
DICVIBIAVV
22 IRV L
LL RRBAV
VI
7.1661
66671
1
2
1
geliminatin
mShh
hMh
0.1,100.4
100.1,001.1
227
21
41211
Analysis of solution:-Even with infinite load the maximum gainis only 6,667Likely causes:-R2 is higher than input resistance Ri-Desired gain is comparable to the maximumgain, A, of the Op-Amp
Proposed solution:-Cascade two stages, each with ideal gain of100. 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
baba
baba
DBBABCBAAA
2
2
2
2
1
1
IV
DCBA
DCBA
IV
DCBA
DCBA
IV
bb
bb
aa
aa
bb
bb
aa
aa
LRBAV
V
1
1
2 Effect of load resistance. G=10,000
Two-Ports
Identical stages