prof. ji chen notes 13 transmission lines (impedance matching) ece 3317 1 spring 2014
TRANSCRIPT
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Prof. Ji Chen
Notes 13 Transmission Lines
(Impedance Matching)
ECE 3317
Spring 2014
ZLZ0
ls
Z0s
d
2
Smith Chart
Impedance matching is very important to avoid reflected power, which causes a loss of efficiency and interference.
Zg
z
Sinusoidal source ZL
z = 0
Z0
S
L
0
0
LL
L
Z Z
Z Z
We will discuss two methods: Quarter-wave transformer
Single-stub matching
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Quarter-Wave Transformer
Quarter-Wave Transformer: First consider a real load.
ZL = RLZ0 Z0T
Zin
/ 4l
0T0T
0T
tan
tanL
inL
Z jZ lZ Z
Z jZ l
2tan tan tan
2l l
20T
inL
ZZ
Z
20T
inL
ZZ
R realHence
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20T
0L
ZZ
R
ZL = RLZ0 Z0T
Zin
/ 4l
Set 0inZ Z
This gives us 0T 0 LZ Z R
Hence
Quarter-Wave Transformer (cont.)
Example:
0
0
50
100
50 100 70.71
L
T
Z
Z
Z
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Quarter-Wave Transformer (cont.)
Next, consider a general (complex) load impedance ZL.
Shunt (parallel) susceptance
s s LY jB jB Choose
YL = 1 / ZLZ0 Z0T
/ 4l
L L LY G jB
YLTOT = GLZ0 Z0T
/ 4l New model:
ZLTOT = 1 / GL (real)
Bs = -BL
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Quarter-Wave Transformer (cont.)
Summary of quarter-wave transformer matching method
0T 0 / L
s L
Z Z G
B B
YL = GL + j BLZ0 Z0T
/ 4l
Ys = jBs
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Quarter-Wave Transformer (cont.)
Realization using a shorted stub
0 cots s s sB Y l
YL = GL + j BLZ0 Z0T
/ 4l
Bs = -BL
ls
Z0s
(An open-circuited stub could also be used.)
0 tans s s sX Z l
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Quarter-Wave Transformer with Extension
ZLZ0 Z0T
Zin(-d)
/ 4l
Z0
We choose the length d to make the input impedance Zin (-d) real.
We then use a quarter-wave transform to change the impedance to Z0.
d
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Quarter-Wave Transformer with Extension (cont.)
ZLZ0 Z0T
/ 4l
Z0
d
Example
0 50[ ]Z
50 75 [ ]LZ j
1 (1.5)NLZ j
10
NLZ
NinZ d
0.176
0.250
0.250 0.176d
0.074d
0
4.3NinZ d SWR
Quarter-Wave Transformer with Extension (cont.)
Wavelengths towards generator
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4.3 50 4.3 215[ ]Nin inZ d Z d
0.074d
0 50 215TZ
0 103.7[ ]TZ
Z0T
/ 4l
Z0
d
0 50[ ]Z 50 75 [ ]LZ j
Quarter-Wave Transformer with Extension (cont.)
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Single-Stub Matching
A susceptance is added at a distance d from the load.
ZL Y0 = 1 / Z0
d
s sY jB
1) We choose the distance d so that at this distance from the load
0in inY Y jB
2) We then choose the shunt susceptance so that
s inB B
(i.e., Gin = Y0)
inY
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Single-Stub Matching (cont.)
ZL Y0
d
s sY jB
0inY Y
The feeding transmission line on the left sees a perfect match.
s inB B
0in inY Y jB
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Realization using a shorted stub
ZL Z0
ls
Z0s
d
(An open-circuited stub could also be used.)
Single-Stub Matching (cont.)
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ZL Z0
ls
Z0s
d
We use the Smith chart as an admittance calculator to determine the distance d.
1) Convert the load impedance to a load admittance YL.
2) Determine the distance d to make the normalized input conductance equal to 1.0.
3) Determine the required value of Bs to cancel Bin.
4) If desired, we can also use the Smith chart to find the stub length ls.
Single-Stub Matching (cont.)
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ZL Z0
ls
Z0s
d
Example0 50[ ]Z
100 100 [ ]LZ j
2 2NLZ j
10.25 0.25
2 2N
LY jj
/6 o0.62 0.62 30jL e 0
0
1
1
NL L
L NL L
Z Z Z
Z Z Z
Single-Stub Matching (cont.)
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2 2 NLZ j
X
X
0.178
Y 0.25 0. 5 2NL j
1 1.57j
1 1.57j
0.322
0.363
0.219
0.041
0.219 - 1.57
1.57 0.3
63
Ns
Ns
jY
Y
d
dj
at
at
Solution :
Add
or
0.170. 0.2198041
0.320. 0.3632041
Wavelengths toward loadWavelengths toward generator
Smith chart scale:
Use this one
Single-Stub Matching (cont.)
Re Rez z or
Im Imz z or
1inG 1NinG
X
X
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Next, we find the length of the short-circuited stub: 1.57NsB
Rotate clockwise from S/C to desired Bs value.
S / C
0-j0.5
0-j1
0+j0.5
0+j1
0+j0
nY
Im z
Re z
0+j2
0-j2
0 1.57NsY j
Assume Z0s = Z0
Otherwise, we have to be careful with the normalization
(see the note below).
Admittance chart
Single-Stub Matching (cont.)
0 0
0 0
/
1.57 /
N Ns in s
s
B B Y Y
Y Y
Note: In general,
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0
0
1
tan
cot
cot
1.57 cot
1cot 1.57; tan 0.637
1.572
tan 0.637 0.567 [radians]
s s s
s s s
Ns s
s
s s
s s
Z jZ l
Y jY l
B l
l
l l
l l
Hence :
0.340 0.250sl
From the Smith chart:
0.0903sl
Analytically:
Single-Stub Matching (cont.)
0.09sl
S / C
X0 1.57j
0.09
O / C
Admittance chart
0.250
0.340
0
20
ZL
z0
UNMATCHED
Single-Stub Matching (cont.) 1+ 1.62L
1- 0.38L
z
1.621.55
0.0420.219
0.38
0.292
V z1.0
0.78
V 1
ZNL
X
1.62
0.78
0.219
0.042 (0.25 0.178 )
1.55Crank diagram
0.178
0.397
0
0.62L
z
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ZL
z0
ZL
z
jBs
0.219 0
MATCHED
UNMATCHED
z
1.621.55
0.78
0.0420.219
SWR = 1.0
1+ 1.62L
1- 0.38L
z
1.621.55
0.0420.219
0.38
0.292
V z1.0
V z
0.78
Single-Stub Matching (cont.)
SWR = 4.26
V 1