EE 205 Dr. A. Zidouri
E
e
l ctric Circuits II
Two-Port Circuits
Interconnected Two-Port Circuits
Lecture #32
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o
o
o
o
Terminated Two-Port circuit
Terminal Behavior
The Six characteristics of the terminated two-port circuit in terms of z parameters
Interconnected Two-Port Circuits
After finishing this lecture you should be able to:
¾
¾
¾
¾
Analyze The Terminated Two-Port circuit
Determine The characteristics of the terminated circuit in terms of z parameters
Recognize the different Interconnection of the Two-Port Circuits
Analyze The Cascade Connection
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EE 205 Dr. A. Zidouri
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The Circuit is driven at port 1 and loaded at port 2
A typically terminated two-port model is shown in Fig. 44-1
i1
i2
Zg
v1
Vg
v2
ZL
Fig. 44-1 A Terminated Two-Port Model
Zg represents the internal impedance of the source
ZL represents the Load impedance
Vg represents the internal voltage of the source
Analysis of this circuit involves expressing the terminal currents and voltages as function of Vg, ZL,
and Zg.
Terminal Behavior:
Six characteristics of the terminated two-port circuit define its terminal behavior.
1. The input impedance
2. The output current
Z in =
V1
I
or admittance Yin = 1
I1
V1
I2
3. Thevenin voltage and impedance
(V
Th
, Z Th ) with respect to port 2
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EE 205 Dr. A. Zidouri
I2
I1
V
5. The voltage gain 2
V1
V
6. The current gain 2
Vg
4. The current gain
The six characteristics in Terms of the z Parameters:
We develop the expressions using the z-parameters to model the two-port portion of the circuit.
Expressions involving other parameters y, a, b, h and g can be found in tables in text books.
The derivation of any one of the desired expressions involves the algebraic manipulation of the
two-port equations along with the two constraint equations.
These four equations using z parameters are:
(44-1)
i. V1 = z11 I1 + z12 I 2
V2 = z21 I1 + z22 I 2
iii. V1 = Vg − I1 Z g
iv. V2 = − I 2 Z L
ii.
To find the input impedance
Z in =
(44-2)
(44-3)
(44-4)
V1
we proceed as follows:
I1
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EE 205 Dr. A. Zidouri
In (44-2) we replace V2 from (44-4) we solve for I2 we get:
We then substitute in (44-1) and solve for Zin we get:
− z21 I1
Z L + z22
z z
Z in = z11 − 12 21
z22 + Z L
I2 =
(44-5)
(44-6)
To find I2 we first solve (44-5) for I1 after replacing V1 with the RHS of (44-3) the result is:
I1 =
Vg − z12 I 2
(44-7)
Z g + z11
We now substitute (44-7) into (44-5) and solve for I2:
I2 =
(Z
− z21Vg
g
+ z11 ) ( Z L + z22 ) − z12 z21
(44-8)
The Thevenin voltage with respect to port 2 equals V2 when I2 = 0. Therefore:
V2
But V1
= Vg − I1 Z g , and I1 =
I2 =0
Vg
Z g + z11
= z21 I1 = z21
V1
z11
(44-9)
when I2=0; therefore by substitution into (44-9) the open circuit
value of V2 is:
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EE 205 Dr. A. Zidouri
z21
Vg
(44-10)
I =0
Z g + z11
V
The Thevenin or output impedance is the ratio Z Th = 2 when Vg is zero (short circuit). Thus (44-3)
I2
becomes V1 = − I1 Z g
(44-11)
− z12 I 2
(44-12)
Substituting gives I1 =
Z g + z11
V2
Therefore
V2
I2
= Z Th = z22 −
Vg = 0
2
= VTh =
z12 z21
Z g + z11
The current gain comes directly from (44-5):
(44-13)
I2
− z21
=
I1 Z L + z22
(44-14)
V2
we replace I2 in (44-2) with its value from (44-4):
V1
⎛ −V ⎞
V2 = z21 I1 + z22 ⎜ 2 ⎟
(44-15)
⎝ ZL ⎠
To derive the voltage gain
Next we solve (44-2) for I1 in terms of V1 and V2:
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EE 205 Dr. A. Zidouri
⎛ −V ⎞
z11 I1 = V1 − z12 ⎜ 2 ⎟
⎝ ZL ⎠
Or
I1 =
V1 z12V2
+
z11 z11 Z L
(44-16)
Replace I1 in (44-15) and solve:
V2
z21 Z L
z21 Z L
=
=
(44-17)
V1 z11 Z L + z11 z22 − z12 z21 z11 Z L + ∆z
V
To derive the voltage gain 2 we first find I1 in terms of V1 and V2 by combining
Vg
(44-1), (44-3) and (44-4):
I1 =
Vg
z11 + Z g
+
z12V2
Z L ( z11 + Z g )
(44-18)
Then using (44-3) and (44-18) with (44-2) we derive an expression involving only V2 and Vg:
V2 =
zV
z21 z12V2
z
+ 21 g − 22 V2
Z L ( z11 + Z g ) z11 + Z g Z L
(44-19)
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EE 205 Dr. A. Zidouri
After manipulation we get:
V2
z21 Z L
=
Vg ( z11 + Z g ) ( z22 + Z L ) − z12 z21
(44-20)
Example 44-1
The two-port circuit of Fig. 44-2 is described in terms of its b parameters, the values of which are:
b11 = −20 , b12 = −3k Ω , b21 = −2mS , and b22 = −0.2
a) Find the phasor voltage V2
b) Find the average power delivered to the 5kΩ load
c) Find the average power delivered to the input port
d) Find the load impedance for maximum average power transfer
e) Find the maximum average power delivered to the load in (d)
500Ω
500∠0o
5kΩ
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EE 205 Dr. A. Zidouri
Solution:
a) To find V2, we can either use (44-4) or from the voltage gain in (44-20), using the latter approach
and conversion tables:
∆b = ( −20 )( −0.2 ) − ( −3000 )( −0.002 ) = −2
V2
z21 Z L
=
Vg ( z11 + Z g ) ( z22 + Z L ) − z12 z21
( −2 )( 5000 )
∆bZ L
=
b12 + b11 Z g + b22 Z L + b21 Z g Z L −3000 + ( −20 )( 500 ) + ( −0.2 )( 5000 ) + ( −0.002 )( 500 )( 5000 )
10
10
=
Then, V2 =
500 = 263.16∠0o V
19
19
=
5000Ω load is
263.162
P2 =
= 6.93W
2 × 5000
b) The average power delivered to the
c) To find average power delivered to the input port, we first find the input impedance Zin,
From tables,
Z in =
b22 Z L + b12 ( −0.2 )( 5000 ) − 3000
=
= 133.33Ω
b11 + b21 Z L ( −0.002 )( 5000 ) − 20
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EE 205 Dr. A. Zidouri
500
= 789.47 mA
500 + 133.33
0.78947 2
The average power delivered to the input port is P1 =
133.33 = 41.55W
2
I1 =
Therefore I1 is
d) The total impedance for maximum power transfer ZL is the conjugate of Thevenin impedance.
Z Th =
b11 Z g + b12
b22 + b21 Z g
∗
Therefore Z L = Z Th = 10833.33Ω
=
( −20 )( 500 ) − 3000 = 10833.33Ω
( −0.002 )( 500 ) − 0.2
e) To find the maximum average power delivered to the load, we first find V2 from the gain
expression when Z L = 10833.33Ω
V2
1 416.67 2
= 8.01W
= 0.8333 thus V2 = 0.8333 × 500 = 416.67V and P2max =
V1
2 10833.33
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EE 205 Dr. A. Zidouri
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Two-port circuits may be interconnected in five
ways:
1) In Cascade Fig. 44-3a
2) In Series Fig. 44-3b
3) In Parallel Fig. 44-3c
4) In Series-Parallel Fig. 44-3d
5) In Parallel-Series Fig. 44-3e
a
b
a11' a12'
a21' a22'
c
a11" a12"
a21" a22"
d
e
Fig. 44-3 Five Basic Interconnections (a) Cascade (b) Series
(c) Parallel (d) Series-Parallel (e) Parallel-Series
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EE 205 Dr. A. Zidouri
We analyze only the Cascade connection because it occurs frequently in the modeling of
large systems.
The a-parameters are best suited for describing cascade connection
We seek the pair of equations:
V1 = a11V2 − a12 I 2 (44-1)
I1 = a21V2 − a22 I 2
(44-2)
From Fig. 44-4 we have
V1 = a11' V2' − a12' I 2' (44-3)
I1 = a21' V2' − a22' I 2' (44-4)
'
'
'
'
From interconnection V2 = V1 and I 2 = − I1 , then substituting yields:
V1 = a11' V1 ' − a12' I1' (44-5)
I1 = a21' V1 ' − a22' I1' (44-6)
From the second circuit
V1 ' = a11" V2 − a12" I 2
I 2' = a21" V2 − a22" I 2
(44-7)
(44-8)
By substitution we generate
V1 = ( a11' a11" + a12' a12" )V2 − ( a11' a12" + a12' a22" ) I 2
I1 = ( a21' a11" + a22' a21" )V2 − ( a21' a12" + a22' a22" ) I 2
(44-9)
(44-10)
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EE 205 Dr. A. Zidouri
By comparison we get the desired expressions:
a11 = a11' a11" + a12' a12"
a12 = a11' a12" + a12' a22"
a21 = a21' a11" + a22' a21"
a22 = a21' a12" + a22' a22"
(44-11)
(44-12)
(44-13)
(44-14)
Self Test 44:
Find the transmission parameters for the circuit in Fig. 44-5
Fig. 44-5 The Circuit for Self Test 44
Answer:
⎡5 44 ⎤ ⎡ 1 6 ⎤ ⎡ 27
T
T
T
=
=
=⎢
[ ] [ 1 ][ 2 ] ⎢
⎥
⎢
⎥
⎣1 9 ⎦ ⎣0.5 4 ⎦ ⎣5.5S
206Ω ⎤
42 ⎥⎦
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EE 205 Dr. A. Zidouri
i.e.
a11 = 27 a12 = 206Ω a21 = 5.5S a22 = 42
PC = VCN . I cC .Cos (θVC − θ iC )
a12 = 206Ω
a12 = 206Ω
a12 = 206Ω
a12 = 206Ω
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