Power Divider and Combiner
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Transcript of Power Divider and Combiner
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7/28/2019 Power Divider and Combiner
1/28
Power divider and combiner/coupler
divider combinerP1
P2= nP1
P3=(1-n)P1
P1
P2
P3=P1+P2
Divide into 4 output
Basic
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S-parameter for power divider/coupler
[ ]
=
333231
232221
131211
SSS
SSSSSS
SGenerally
For reciprocal and lossless network
jiforSSN
k
kjki ==
0
1
*1
1
*=
=
N
k
ki kiSS
1131211 =++ SSS
1232221 =++ SSS
1333231 =++ SSS
0*2313
*2212
*2111 =++ SSSSSS
0*3323
*3222
*3121 =++ SSSSSS
0*3313
*3212
*3111 =++ SSSSSS
Row 1x row 2
Row 2x row 3
Row 1x row 3
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ContinueIf all ports are matched properly , then Sii= 0
[ ]
=
0
0
0
2313
2312
1312
SS
SS
SS
S
For Reciprocal
networkFor lossless network, must satisfy unitary
condition
1
2
13
2
12 =+ SS
12
23
2
12 =+ SS
12
23
2
13 =+ SS
012*
23=SS
023*13 =SS
013*12 =SS
Two of (S12, S13, S23) must be zero but it is not consistent. If S12=S13= 0, then
S23 should equal to 1 and the first equation will not equal to 1. This is invalid.
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Another alternative for reciprocal network
[ ]
=
332313
2312
1312
0
0
SSS
SS
SS
S
Only two ports are matched , then for reciprocal network
For lossless network, must satisfy unitary
condition
12
13
2
12 =+ SS
12
23
2
12 =+ SS
12
33
2
23
2
13 =++ SSS 013*3312
*23 =+ SSSS
023
*
13
=SS
033*2313
*12 =+ SSSS
The two equations show
that |S13|=|S23|thus S13=S23=0
and |S12|=|S33|=1
These have satisfied all
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Reciprocal lossless network of two matched
S21=ej
S12=ej
S33=ej
1
3
2
[ ]
=
j
j
j
e
ee
S
00
00
00
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For lossless network, must satisfy unitary
condition
12
13
2
12 =+ SS
12
23
2
21 =+ SS
12
32
2
31=+ SS
032*
31=SS
023
*
21
=SS
013*
12=SS
Nonreciprocal network (apply for circulator)
[ ]
=
0
0
0
3231
2321
1312
SS
SS
SS
S
0312312 === SSS
0133221 === SSS
1133221 === SSS
1312312 === SSS
The above equations must satisfy the following either
or
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Circulator (nonreciprocal network)
[ ]
=
010
001
100
S
[ ]
=
001
100
010
S
1
2
3
1
2
3
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Four port network
[ ]
=
44434241
34
24
14
333231
232221
131211
SSSS
S
S
S
SSS
SSS
SSS
SGenerally
For reciprocal and lossless network
jiforSS
N
kkjki =
=
01
*
1
1
*=
=
N
k
ki kiSS
114131211 =+++ SSSS
124232221 =+++ SSSS
134333231 =+++ SSSS
0*2414
*2313
*2212
*2111 =+++ SSSSSSSS
0*4424
*4323
*4222
*4121 =+++ SSSSSSSS
0*3414
*3313
*3212
*3111 =+++ SSSSSSSS
R 1x R 2
R 2x R3
R1x R4
144434241 =+++ SSSS
0*4414*4313*4212*4111 =+++ SSSSSSSS
0*3424
*3323
*3222
*3121 =+++ SSSSSSSS
0
*
4434
*
4333
*
4232
*
4131 =+++ SSSSSSSS
R1x R3
R2x R4
R3x R4
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Matched Four port network
[ ]
=
0
0
0
0
342414
34
24
14
2313
2312
1312
SSS
S
S
S
SS
SS
SS
S
The unitarity condition become
1141312 =++ SSS
1242312 =++ SSS
1342313 =++ SSS
0*2414
*2313 =+ SSSS
0*3423
*1412 =+ SSSS
0*3414
*2312 =+ SSSS
1342414 =++ SSS
0
*
3413
*
2412 =+ SSSS
0*3424
*1312 =+ SSSS
0*
2423
*
1413
=+ SSSS
Say all ports are matched and symmetrical network, then
*
**
@@@
#
##
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To check validity
Multiply eq. * by S24
* and eq. ## by S13
* , and substract to obtain
02
142
13*14 =
SSS
Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain
02
34
2
1223 =
SSS
%
$
Both equations % and $ will be satisfy if S14 = S23 = 0 . This meansthat no coupling between port 1 and 4 , and between port 2 and 3 as
happening in most directional couplers.
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Directional coupler
[ ]
=
00
0
00
00
0
3424
34
24
13
12
1312
SS
S
S
S
S
SS
S
If all ports matched , symmetry and S14=S23=0 to be satisfied
The equations reduce to 6 equations
11312 =+ SS
12412 =+ SS
13413 =+ SS
13424 =+ SS
0*3413
*2412 =+ SSSS
0*3424
*1312 =+ SSSS
2413 SS =By comparing these equations yield
*
*
**
**
By comparing equations * and ** yield 3412 SS =
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Continue
[ ]
=
00
0
00
00
0
j
j
j
j
S
Simplified by choosing S12= S34= ; S13=ej and S24= e
j
Where + = + 2n
[ ]
=
00
0
00
00
0
S
1. Symmetry Coupler = = /2
2. Antisymmetry Coupler =0 , =
2 cases
Both satisfy 2 +2 =1
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Physical interpretation
|S13 |2 = coupling factor = 2
|S12 |2 = power deliver to port 2= 2 =1- 2
Characterization of coupler
Directivity= D= 10 log
dB
P
Plog20
3
1=Coupling= C= 10 log
dBSP
P
144
3 log20
=
Isolation = I= 10 log dBSP
P
14
4
1 log20=
I = D + C dB
1
4 3
2
Input Through
CoupledIsolated
For ideal case |S
14
|=0
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Practical coupler
Hybrid 3 dB couplers
Magic -T and Rat-race couplers
= = /2[ ]
=
010
1
0
00
001
10
2
1
j
j
j
j
S
[ ]
=
0110
1
1
0
001
001
110
2
1S
=0 , =
= = 1 / 2
= = 1 / 2
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T-junction power divider
E-plane TH-plane T
Microstrip T
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T-model
jB
Z1
Z2
Vo
Yin
21
11
ZZjBYin ++=
21
11
ZZYin +=
Lossy line
Lossless line
If Zo = 50,then for equallydivided power, Z1 = Z2=100
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Example
If source impedance equal to 50 ohm and thepower to be divided into 2:1 ratio. Determine Z1
and Z2
ino
PZ
VP
3
1
2
1
1
2
1 ==
ino
P
Z
VP
3
2
2
1
2
2
2 == == 752
3
2
oZZ
== 15031 oZZ
o
oin
Z
VP
2
2
1=
== 50// 21 ZZZo
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Resistive divider
V2
V3
V1
Zo
Zo
P1
P2
P3
Zo V
oo
ZZ
Z +=3
Zo/3Zo/3
Zo/3
ooo
in ZZZ
Z =+=3
2
3
VVZZ
ZV
oo
o
3
2
3/23/
3/21 =
+
=
VVV
ZZ
ZVV
oo
o
2
1
4
3
3/32==
+
==
o
inZ
VP
2
1
2
1=
( )in
o
PZ
VPP
4
12/1
2
1 21
32 ===
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Wilkinson Power Divider
50
50
50
100
70.7
70.7
/4
Zo
/2 Zo
/2 Zo
2Zo
Zo
Zo
/4
2
2
Te ZZ
in=
oT ZZ 2=
For even mode
Therefore
For Zin =Zo=50== 7.70502TZ
And shunt resistor R =2 Zo = 100
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Example
Design an equal-split Wilkinson power divider for a 50 W systemimpedance at frequency fo
The quarterwave-transformer characteristic is
== 7.702 oZZ
== 1002 oZR
r
o
4
=The quarterwave-transformer length is
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Wilkinson splitter/combiner
application
/4
100
70.7
50
matching
networks
/4
100 50
70.7
70.7
70.7
Splitter combiner
Power Amplifier
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Unequal power Wilkinson
Divider
3
2
03
1
K
KZZ o
+
=
)1( 2032
02 KKZZKZ o +==
+=
KKZR o
1
R2=Z
o/K
R
R3=Z
o/K
Z02
Z03
Zo
2
3
2
32
==
portatPower
portatPower
P
PK
1
2
3
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Parad and Moynihan power divider
4/1
2011
+
=
K
KZZ o
2
3
2
32
==
portatPower
portatPower
P
PK
+=
KKZR o
1( ) 4/124/302 1 KKZZ o +=
( )4/5
4/12
03
1
K
KZZ o
+
=
KZZ o=04 K
ZZ
o=05
Zo
Zo
Zo
Z05
Zo4Zo2
Zo3
Zo1
R1
2
3
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Cohn power divider
/4
98 241 50
82 61
82 61
50
50 /4
VSWR at port 1 = 1.106VSWR at port 2 and port 3 = 1.021
Isolation between port 2 and 3 = 27.3 dB
Center frequency fo = (f1 + f2)/2
Frequency range (f2/f
1) = 2
1
2
3
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Couplers
/4
/4
Yo
Yo
Yo
Yo
Yse
Ysh Y
sh
Branch line coupler 2sh
2se Y1Y +=
2
se
2
sh
sh
2
3
YY1
2Y
E
E
+
=
( )20
1
310
E
E x=
x dB coupling
23
22
21 EEE +=
2
1
3
2
1
2
E
E
E
E1
+
=or
E1E2
E3
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Couplers
input
isolate
Output3dB
Output3dB 90o out of phase
3 dB Branch line coupler
/4
/4
Zo
Zo
Zo
Zo
2/Zo
2/Zo
Zo Zo
32 EE =
1Ysh=
2Y1Y22
se =+= sh
1.414Yse =
= 50oZ
= 50sh
Z
= 5.35seZ
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Couplers9 dB Branch line coupler
( )355.010
209
1
3==
E
E
( )22
1
2355.01 +
=
E
E
( ) 935.0355.01 2
1
2==
E
E
38.0935.0
355.0
2
3==
E
E
8.0=shYLet say we choose
38.0
8.01
8.02
1
2
2222=
+
=
+sesesh
sh
YYY
Y
962.136.038.0
6.1==seY
= 500Z
== 5.628.0/50shZ
== 5.25962.1/50seZ
Note: Practically upto 9dB coupling
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Couplers
/4
/4
/4
3/4
Input
Output in-phase
Output in-phase
isolated
1
2
3
4
Can be used as splitter , 1 as input and 2 and 3as two output. Port is match with 50 ohm.
Can be used as combiner , 2 and 3 as inputand 1 as output.Port 4 is matched with 50 ohm.
Hybrid-ring coupler
OC
1
21
2
OC
1/2
1/2
2
2
2
2
2
2
/8
/8
/4
/4
3/8
3/8
Te
To
e
o