Transcript ENE 429 Antenna and Transmission Lines

ENE 490
Applied Communication Systems
Lecture 2 circuit matching on Smith chart
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1
Review (1)
 High frequency operation and its applications
 Transmission line analysis (distributed elements)
– Use Kirchholff’s law to obtain general equations
for transmission lines
– Voltage and current equations are the combination
of incident and reflected waves.
V ( z )  V0 e jz  V0 e jz
V0  jz V0 jz
I ( z) 
e

e
Z0
Z0
where Z0 is a characteristic impedance of a transmission line.
16/11/50Assume the line is lossless.
2
Review (2)
 Terminated lossless line
– voltage reflection coefficient
V0  j 2d
( d )   e
V0
– impedance along a transmission line
Z L  jZ 0 tan d
Z (d )  Z 0
Z 0  jZ L tan d
or
1  ( d )
Z (d ) 
1  ( d )
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Review (3)
- voltage standing wave ratio
V (d ) max
V (d ) min

I (d ) max
I (d ) min
1 L

 VSWR
1  L
 source and loaded transmission line
Zin  Z 0
in  (d  l ) 
  0e 2 jl
Zin  Z 0
Z S  Z0
S 
Z S  Z0
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Review (4)
 power transmission of a transmission line
 
1
Pav  Re VI *
2
 for lossless and a matched condition
2
1 VS
Pin 
 Pavs
8 ZS
 power in decibels
P W 
P  dBm   10 log
1 mW
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Impedance matching network (1)
 The need for matching network arises because
amplifiers, in order to deliver maximum power to a
load or to perform in a certain desired way, must be
properly terminated at both the input and the output
ports.
Z1=50 W
+
VS
Input
matching
network
Output
matching
network
Transistor
Z2=50 W
-
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ZS
ZL
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Impedance matching network (2)
 Effect of adding a series reactance element to an
impedance or a parallel susceptance are
demonstrated in the following examples.
 Adding a series reactance produces a motion along a
constant-resistance circle in the ZY Smith chart.
 Adding a shunt susceptance produces a motion
along a constant-conductance circle in the ZY Smith
chart.
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Ex1 Adding a series inductor L (zL =
j0.8) to an impedance z = 0.3-j0.3.
z L= j0.8
z = 0.3-j0.3
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Ex2 Adding a series capacitor C (zC = j0.8) to an impedance z = 0.3-j0.3.
zC =-j0.8
z = 0.3-j0.3
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Ex3 Adding a shunt inductor L (yL = j2.4) to an admittance y = 1.6+j1.6.
y L=-j2.4
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y = 1.6+j1.6
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Ex4 Adding a shunt capacitor C (yC =
j3.4) to an admittance y = 1.6+j1.6.
yC=j3.4
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C
y = 1.6+j1.6
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Examples of matching network design
 Ex5 Design a matching network to
transform the load Zload = 100+j100 W to
an input impedance of Zin = 50+j20 W.
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Ex6 Design the matching network that provides YL
= (4-j4)x10-3 S to the transistor. Find the element
values at 700 MHz.
C
L
50 W
y L=(4-j4)x10-3S
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L
L matching networks
L
C
C
L
C
C
Z load
L
C
Z load
Z load
C
L2
Z load
C1
Z load
L2
Z load
C1
Z load
L1
Z load
L2
L
Z load
C2
L
C
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Z load
C2
C
Z load
Z load
L1
L
L1
C
Z load
Z load
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Forbidden regions
 Sometimes a specific matching network cannot be
used to accomplish a given match.
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Load quality factor
 The developed matching networks can also be viewed as
resonance circuits with f0 being a resonance frequency.
These networks may be described by a loaded quality factor,
Q L.
f0
QL 
BW
 The estimation of QL is simply accomplished through the use
of a so-called nodal quality factor Qn. At each node of the
L-matching networks, there is an equivalent series input
impedance, denoted by RS +jXS. Hence a circuit node Qn can
be defined at each node as
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Qn 
XS
RS
BP

GP
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Circuit node Qn and loaded QL

Qn
QL 
2
 L-matching network is not a good choice for a design
of high QL circuit since it is fixed by Qn.
 For more complicated configurations (T-network, Pi-
network), the loaded quality factor of the match
network is usually estimated as simply the maximum
circuit node quality factor Qn.
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The example of Q calculation
L=3.18 nH L=3.18 nH
50W
C=12.7 pF
VS
10 W
Z IN = 50 W
 At 500 MHz
Qn = 2
then QL = 1.
and
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f 0 = 500 MHz
BW 
QL
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Ex7 The low pass L network shown below was designed to
transform a 200 W load to an input resistance of 200 W.
Determine the loaded Q of the circuit at f = 500 MHz.
AC
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R = 200 W
20 W
C =4.775 pF
L = 19.09 nH
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Constant Qn contours
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The upper and lower part of Q contours
satisfy a circle equation.
 Since
1 
1U 2 V 2
2
z  r  jx 


j
1   (1  U )2  V 2
(1  U )2  V 2
 then
x
2U
Qn  
r 1U 2 V 2
 which can be written as
U 2  (V 
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1 2
1
)  1 2
Qn
Qn
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Contour equations
 The equations for these contours can be
derived from the general derivation of the
Smith chart. By following the derivation, Qn
contours follow this circle equation,
2
i2
 2 1 
1
  r 
  1 2
Qn 
Qn

where the plus sign is taken for positive reactance x and
the minus sign for negative x.
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Qn circle parameters
 For x > 0, the center in the  plane is at (0, -1/Qn).
 For x < 0, the center in the  plane is at (0, +1/Qn).
 the radius of the circle can be written as
1
r  1 2 .
Qn
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Ex8 Design two T networks to transform the load
impedance ZL = 50 W to the input impedance Zin
= 10-j15 W with a Qn of 5.
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Ex9 Design a Pi network to transform the load
impedance Zload = 50 W to the input impedance Zin
= 150 W with a Qn of 5.
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