Post on 08-Apr-2018
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431-618 ( 2011)
Passive ComponentDesign & Simulation
Dr. Bo YangLecture 6: Impedance Matching
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At a glance
Last Class: Transmission Lines & SmithChartThis Class: Impedance Matching
B. Yang 2431-618, the University of Melbourne
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At Z Smith chart, curve from point Ato pint C indicates impedance
transformation from resistance 25 Ohm to inductive impedance (25
+j25) Ohm
At Y Smith chart, curve from point C to point D indicates admittance transformation from inductive admittance (20 - j20) mS to
conductance 20 mS (50 Ohm) 0 . 5
1 .
0
2 . 0
3 . 0
- 3
. 0
- 2. 0 - 0 .5
1 -10.51.02.05.0
B / Y = O
- 0. 2
B / Y O = 0 .2
x
C
D
G /
Y O = 0 . 2
-1 .0
0. 5 2 .0
3 . 0
-
3 . 0
- 2 . 0
- 1
. 0
- 0 . 5
0.5 1.0 2.0 5.0 r
x
X / Z O = 0 .2
X / Z O = - 0
. 2
-1
C
A1
R /
Z O
= 0
. 2
1 .0
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C
D A Y Smith chart
provides transformation from point C to point D
At combined Z-Y
Smith chart:
Z Smith chart provides
transformation from
point A to point C
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5
Load MatchingWhat if the load cannot be made equal to Z o for some otherreasons? Then, we need to build a matching network so that thesource effectively sees a match load.
0
LZ
sP
0Z
M
Typically we only want to use lossless devices such as capacitors,inductors, transmission lines, in our matching network so that wedo not dissipate any power in the network and deliver all theavailable power to the load.
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6
Matching Example
0
100sP 50Z0 M
Match 100 load to a 50 system at 100MHz
A 100 resistor in parallel would do the trick but of thepower would be dissipated in the matching network. We wantto use only lossless elements such as inductors andcapacitors so we dont dissipate any power in the matchingnetwork
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7
Matching Example
Add positiveimaginary admittanceto get to z=1+j0
ImpedanceChart
nH80L
LMHz1002 j500.1 j0.1 j jx
pF16 100
nH80
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8
Matching ExampleThis solution wouldhave also worked
ImpedanceChart
pF32
100nH160
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Matching Bandwidth
50 60 70 80 90 100 110 120 130 140 15040
35
30
25
20
15
10
5
0
Frequency (MHz)
R e
f l e c t
i o n
C o e
f f i c i e n
t ( d B )
pF16100
nH80
50 MHz
150 MHz
ImpedanceChart
Because the inductor and capacitorimpedances change with frequency, thematch works over a narrow frequency
range 431-618, the University of Melbourne 9B. Yang
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10
Single Stub Tuner
Match 100 load to a 50 system at 100MHzusing two transmission lines connected in parallel
0
100t 1
t 2
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11
Single Stub TunerAdding length toCable 1 rotates thereflection coefficientclockwise.
Enough cable isadded so that thereflection coefficientreaches the mirrorimage circle
nS49.3
MHz1003602251
1
1t
t
ImpedanceChart
2511
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12
Single Stub Tuner
The stub is going tobe added in parallelso flip to theadmittance chart.
The stub has to adda normalizedadmittance of 0.7 tobring the trajectory tothe center of the
Smith Chart
AdmittanceChart431-618, the University of Melbourne 12B. Yang
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13
Single Stub TunerAn open stub of zerolength has anadmittance=j0.0
By adding enough
cable to the openstub, the admittanceof the stub willincrease.
70 degrees will give
the open stub anadmittance of j0.7
AdmittanceChart
702
nS97.0
MHz100360270
2
2t
t
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14
Single Stub Tuner
AdmittanceChart
01003.5nS
0.97nS
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15
Single Stub Tuner
AdmittanceChart
01001.5nS
1.4nS
This solution wouldhave worked as well.
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50 60 70 80 90 100 110 120 130 140 1540
35
30
25
20
15
10
5
0
Frequency (MHz)
R e
f l e c t
i o n
C o e
f f i c i e n
t ( d B )
01003.5nS
0.97nS
Single Stub Tuner Matching Bandwidth
Because the cable phase changeslinearly with frequency, the matchworks over a narrow frequency range
50 MHz
150 MHz
Impedance
Chart431-618, the University of Melbourne 16B. Yang
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X 1
X 2
L-transformer
R1 Z 1 X 1 R2 Z 2
X 2
Equivalence when Z 1 = Z 2 : 21
21
12
12
12
1
211
22 X R X R
j X R
X R jX R
221 1 Q R R 221 1 Q X X
where Q = R 1 / X 1 = X 2 / R 2 - quality factor equal for series and parallel circuits
Impedance parallel and series circuits
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For conjugate matching with reactance compensation :
221 1 Q X X 2
21 1 Q R R
.1 /
/
21
22
11
R RQ
Q R X
Q R X
X 1
X 2
R2 Z in = R 1
Input impedance Z in will be resistive and equal to R 1 when :
where Q = R 1 / X 1 = X 2 / R 2 - quality factor equal for series and parallel circuits
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Matching with lumped elements
Bandwidth properties
Two L-type matching circuits
2211
2
1
22
11
/ 1
/ 1
/
RQC
Q R L
R
RQ
RQ L
RQC
L2
1 C 1 R2 R1
C 2
R2 L1
a ). b).
Resistance R 1connected to
parallel reactive element must be
greater than resistance R 2 connected to
series reactive element
Z in
R1
0.707 R1
1 f / f 0 0
2 f 0 0 / 2Q f f
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Matching with lumped elements
- transformer
for each L-transformer, resistances R 1 and R 2 are transformed to some intermediate resistance R 0 with value of R 0 < (R 1, R 2 )
X 1
X 3
X 2
X 3
X 2 X 1
X 1
X 3
R0 R1 X 2 R2
X 3
X 1
X 3 R0 R1
X 2
R2 X 3
Connection of two L-transformers
T- transformer
for same resistances R 1 and R 2 , T- and -transformers have better filtering properties, but narrower bandwidth compared with single L-transformer
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Matching with lumped elements
-type matching circuits
111 / RQC 222 / RQC
212113 1 / QQQ R L
1 1 211
22 Q
R
RQ 1
2
121
R
RQ
L3
R1 C 1 R2 C 2
widely used as matching circuit
useful for interstage matching when active device input and
output capacitances can be easily
incorporated inside matching circuit
provides significant level of harmonic suppression
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111 / Q R L 222 / RQC
221223 1 / QQQ R L ,
1 1 222
11 Q
R
RQ 1
1
222
R
RQ
Matching with lumped elements
-type matching circuits
111 / Q R L
222 / RQC
212223 / 1 QQ RQC
1 1 211
22 Q
R
RQ 1
2
121
R
RQ
L3
R1 L1 R2 C 2
C 3
R1 L1 R2 C 2
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Matching with lumped elements
111 / 1 Q RC 222 RQ L
222123 1 / Q RQQC
1 1 221
21 Q
R RQ 1
2
122
R R
Q
L2
R1 R2 C 3
C 1
T-type matching circuits
widely used as input,interstage and output matching
circuits in high power amplifiers
can incorporate active device lead and bondwire
inductances within matching circuit
provides significant level of harmonic suppression
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111 / 1 Q RC 222 RQ L
212223 / 1 QQQ R L ,
1 1 212
12 Q R
RQ 1
1
221
R
RQ
L2
R1 L3 R2
C 1
Matching with lumped elements
T-type matching circuits
L2
R1 R2 C 3
L1 111 RQ L 222 RQ L
222213 1 / Q RQQC
1 1 221
21 Q R
RQ 1
2
122 R
RQ
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Matching with lumped elements Matching design example
L3 + L in
Rsource = R1
Rin C 3
L2 C in
Z in
C 1
L1 C 2 R2 R3
132-174 MHz 150 W MOSFET power amplifier:
three-section input matching
MHz152174132c f
2.1 9.0in j Z
Q = 152/(174 - 132) = 3.6
in
3
3
2
2
1 R R
R R
R R
For R in = 0.9 Ohm and R 1 = 50 Ohm: R 3 = 3.5 Ohm, R 2 = 13 Ohm
Q = 1.7
Two low-pass and one high-pass L-sections
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Matching with transmission lines
For conjugate matching with reactance compensation when Z S = Z in * :
Transmission-line transformer
For quarter-wave transmission line with = 90 :
tan tan
L0
0L0in jZ Z
jZ Z Z Z
Z 0,
Z L Z in Z S
L20in / Z Z Z
L
j j
Z Z
2exp 12exp 1
L
L
0
in
Impedance at input of loaded transmission line:
L
L
0
L
1 1
Z Z
Input impedance for loaded transmission line with electrical length of , normalized to its characteristic impedance Z
0 , can be found by rotating this impedance point clockwise by
2 around Smith chart center point with radius L
SL
2S
2SL
2L
2LS
0 R R
X R R X R R Z
LSLS
LS0
1
tan
R X X R R R
Z
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Matching with transmission lines For pure resistive source
impedance Z S = R S :
Z L Z S
Z 01, /8 Z 02, /4 Z 03, /8
0 tan tan 1 2L2 L2 020L R X Z Z X
2L
2LL0 X R Z Z
L0
0LS X Z
Z R R
For electrical length = 45
Any load impedance can be transformed into real source impedance using /8-transformer whose impedance is equal to magnitude of load impedance
To match any source impedance Z S and load impedance Z L, matching circuit can be designed with two /8-transformers
and one /4-transformer
L Z 0,
< 90 C
Z 0, < 90
tan 0 Z
L
Lumped and transmission line single-frequency equivalence
0
tan
Z C
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Conjugate matching: L-type transformer
Second implicit equation : numerical or graphical solution
Matching with transmission lines
R1 Z 1C R2 Z in
Z 0,
21
21
1212
12
1
211inin X R
X R j X R
X R jX R
2in1 1 Q R R 2in1 1 Q X X Real and imaginary parts of
tan tan
20
020in jR Z
jZ R Z Z
2220
2
220in tan
tan 1
R Z R Z R
2220
22
20
0intan
tan
R Z
R Z Z X
1 / RQC
2202
2
22
2
0
2
2
0
2
1
sin / cos
cos sin 1
Z R
Z R
R Z
R R
where X 1 = -1/ C
Matching for any ratio of
R 1 /R 2
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Matching design example 470-860 MHz 150 W LDMOSFET power amplifier:
three-section input matching
Q = 635/(860 - 470) = 1.63
in
3
3
2
2
1 R R
R R
R R
For R in = 1.7 Ohm and R 1 = 50 Ohm: R 3 = 5.25 Ohm, R 2 = 16.2 Ohm
Q = 1.2
For Z 01= Z 02 = Z 03 = 50 Ohm 1 = 30
, 2 = 7.5 , 3 = 2.4
Matching with transmission lines
MHz635860470c f
3.1 7.1in j Z
Lin Rsource = R1
R in C 3 Z inC 2 R2 R3
Z 03, 3
in
C 1
Z 02, 2 Z 01, 1
For 1 =
2 =
3 = 30
Z01 = 50 Ohm Z 02 = 15 7 Ohm Z 03 = 5 1 Ohm
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