notes 13 3317
Download
Report
Transcript notes 13 3317
ECE 3317
Prof. Ji Chen
Spring 2014
Notes 13
Transmission Lines
(Impedance Matching)
d
Z0
ZL
Z0s
ls
1
Smith Chart
Impedance matching is very important to avoid reflected power, which causes a
loss of efficiency and interference.
Z L Z0
L
Z L Z0
Zg
S
Sinusoidal source
L
Z0
ZL
z=0
We will discuss two methods:
z
Quarter-wave transformer
Single-stub matching
2
Quarter-Wave Transformer
Quarter-Wave Transformer: First consider a real load.
l /4
Z0
Z0T
ZL = RL
Zin
Z L jZ 0T tan l
Z in Z 0T
Z
jZ
tan
l
L
0T
2
tan l tan
2
Z 0T
Zin
ZL
l tan
2
2
Z 0T
Hence Zin
real
RL
3
Quarter-Wave Transformer (cont.)
l /4
Z0
Z0T
ZL = RL
Zin
Set
Example:
Zin Z 0
Hence
This gives us
2
Z 0T
Z0
RL
Z 0T Z 0 RL
Z 0 50
Z L 100
Z 0T
50 100 70.71
4
Quarter-Wave Transformer (cont.)
Next, consider a general (complex) load impedance ZL.
l /4
Z0
Z0T
YL = 1 / ZL
YL GL jBL
Shunt (parallel) susceptance
Choose Ys jBs jBL
Bs = -BL
l /4
New model:
Z0
Z0T
YLTOT = GL
ZLTOT = 1 / GL (real)
5
Quarter-Wave Transformer (cont.)
Summary of quarter-wave transformer matching method
l /4
Z0
Z0T
YL = GL + j BL
Ys = jBs
Z 0T Z 0 / GL
Bs BL
6
Quarter-Wave Transformer (cont.)
Realization using a shorted stub
(An open-circuited stub could also be used.)
l /4
Z0
Z0T
YL = G L + j B L
Bs = -BL
Z0s
ls
X s Z0s tan sls
Bs Y0 s cot sls
7
Quarter-Wave Transformer with Extension
Z0
l /4
d
Z0T
Z0
ZL
Zin(-d)
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.
8
Quarter-Wave Transformer with Extension (cont.)
Example
Z0 50[]
Z L 50 j 75 []
Z0
l /4
d
Z0T
Z0
ZL
Z LN 1 j (1.5)
9
Quarter-Wave Transformer with Extension (cont.)
0.176
Z LN
Wavelengths
towards generator
0
d 0.250 0.176
ZinN d
0.250
ZinN d SWR 4.3
d 0.074
10
Quarter-Wave Transformer with Extension (cont.)
Z0 50[]
l /4
d
Z0T
Z0
Z L 50 j 75 []
d 0.074
ZinN d 4.3
Z0T
Zin d 50 4.3 215[]
50 215
Z0T 103.7[]
11
Single-Stub Matching
A susceptance is added at a distance d from the load.
d
ZL
Y0 = 1 / Z0
Ys jBs
Yin
1) We choose the distance d so that at this distance from the load
Yin Y0 jBin
(i.e., Gin = Y0)
2) We then choose the shunt susceptance so that
Bs Bin
12
Single-Stub Matching (cont.)
d
Ys jBs
Y0
Yin Y0
ZL
Yin Y0 jBin
Bs Bin
The feeding transmission line on the left sees a perfect match.
13
Single-Stub Matching (cont.)
Realization using a shorted stub
(An open-circuited stub could also be used.)
d
Z0
ZL
Z0s
ls
14
Single-Stub Matching (cont.)
We use the Smith chart as an admittance calculator to determine the
distance d.
d
Z0
ZL
Z0s
ls
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.
15
Single-Stub Matching (cont.)
Example
Z0 50[]
Z L 100 j100 []
d
Z0
ZL
Z LN 2 j 2
Z0s
ls
YLN
Z L Z 0 Z LN 1
L
N
Z L Z0 Z L 1
1
0.25 j 0.25
2 j2
L 0.62 e j /6 0.62 30o
16
Single-Stub Matching (cont.)
0.041 0.322 0.363
Solution :
Im z or Im z
Add YsN - j1.57 at d 0.219
or
0.041 0.178 0.219
0.178
YsN j1.57 at d 0.363
Use this one
1 j1.57
Z LN 2 j 2
0.219
Re z or Re z
0.363
0.041
GinN 1
Gin 1
1 j1.57
YLN 0.25 j 0.25
Smith chart scale:
0.322
Wavelengths toward load
Wavelengths toward generator
17
Single-Stub Matching (cont.)
Next, we find the length of the short-circuited stub: BsN 1.57
Rotate clockwise from S/C to desired Bs value.
Im z
0+j1
Assume Z0s = Z0
Otherwise, we have to be
careful with the normalization
(see the note below).
0+j2
0+j0.5
S/C
Yn
0+j0
Re z
0-j0.5
Note: In general,
0-j2
0-j1
BsN BinN Y0 / Y0 s
1.57 Y0 / Y0 s
YsN 0 j1.57
Admittance chart
18
Single-Stub Matching (cont.)
From the Smith chart:
Admittance chart
ls 0.340 0.250
ls 0.09
Analytically:
0
S / C 0.250
O/C
Z s jZ 0 s tan ls
Ys jY0 s cot ls
0.09
BsN cot ls
0 j1.57
Hence :
0.340
1.57 cot ls
cot ls 1.57; tan ls
ls
2
1
0.637
1.57
ls tan 1 0.637 0.567 [radians]
ls 0.0903
19
Single-Stub Matching (cont.)
UNMATCHED
1+ L 1.62
1.62
1.55
1.0
ZL
V z
0.78
V 1
1- L 0.38
0.38
z
0
z
0
0.178
Z LN
1.55
0.292 0.219
0.042
z
0.042 (0.25 0.178 )
Crank diagram
1.62
0.78
L 0.62
0.219
0.397
20
Single-Stub Matching (cont.)
1+ L 1.62
1.62
V 1
UNMATCHED
1.55
1.0
V z
ZL
0.78
SWR = 4.26
0
1- L 0.38
0.38
z
0.292 0.219
z
0.042
MATCHED
1.62
ZL
jBs
1.55
SWR = 1.0
V z
0.78
0.219
0
z
0.219
0.042
z
21