Chapter 5 Impedance Matching and Tuning Chapter 5 Impedance Matching and Tuning • The matching network is ideally lossless and is usually The matching network is ideally lossless, and is usually designed so that the impedance seen looking into the matching network is Z0. • Only if Re[ZL] ≠ 0, a matching network can always be found. • The quarter-wave impedance transformer is for read load. • Important factors in selecting matching networks: (1)complexity , (2)bandwidth, (3)implementation, (4)adjustability . 1 Review of Smith Chart z L = Z L / Z 0 z L Z L / Z 0 = r L + jx L 2 5.1 Lumped Element Matching (L-networks) L-type or L-network matching is the simplest matching network. (1) z L = ZL/Z0 is outside the r=1 (1) z L = ZL/Z0 is inside the r=1 (1) z L ZL/Z0 is outside the r 1 circle (rL < 1). (2) Z i i ih B h (1) z L ZL/Z0 is inside the r 1 circle (rL > 1). (2) Z i h ih B h (2) ZL is series with jB, then shunt with jX. (2) ZL is shunt with jB, then series with jX. 3 Analytical Solutions (r L >1) • When r L > 1, let Z L = R L + jX L and z L = Z L / Z 0 = r L + jx L 0 1 1 in Z jX Z jB R jX 0 0 ( ) L L L L L R jX B XR XZ R Z 0 2 2 (1 ) L L L L X BX BZ R X R X R X ZR 0 0 2 2 L L L L L L X R X ZR Z B R X in Z 0 0 1 L L L XZ Z X B R BR 4 • Note that when r L > 1, the square root in B has real results.
Transcript
Chapter 5 Impedance Matching and TuningChapter 5 Impedance Matching and Tuning
• The matching network is ideally lossless and is usuallyThe matching network is ideally lossless, and is usually designed so that the impedance seen looking into the matching network is Z0.
• Only if Re[ZL] ≠ 0, a matching network can always be found.• The quarter-wave impedance transformer is for read load.q p• Important factors in selecting matching networks:
(1)complexity, (2)bandwidth, (3)implementation, (4)adjustability.( ) p y, ( ) , ( ) p , ( ) j y
1
Review of Smith Chart
zL = ZL / Z0zL ZL / Z0= rL + jxL
2
5.1 Lumped Element Matching (L-networks)p g ( )L-type or L-network matching is the simplest matching network.
(1) zL = ZL/Z0 is outside the r=1(1) zL = ZL/Z0 is inside the r=1 (1) zL ZL/Z0 is outside the r 1 circle (rL < 1).
(2) Z i i i h B h
(1) zL ZL/Z0 is inside the r 1 circle (rL > 1).
(2) Z i h i h B h (2) ZL is series with jB, then shunt with jX.
(2) ZL is shunt with jB, then series with jX.
3
Analytical Solutions (rL>1)
• When rL > 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
01
1inZ jX ZjB
R jX
0 0( )L L
L L L
R jXB XR X Z R Z
0
2 2
(1 )L L L
L
X BX BZ R X
RX R X Z R
00
2 2
L L L L
L L
X R X Z RZ
BR X
inZ
0 01 L
L L
X Z ZXB R BR
4• Note that when rL > 1, the square root in B has real results.
Analytical Solutions (rL<1)
• When rL < 1, let ZL = RL + jXL and zL = ZL / Z0 = rL + jxL
0
1 1 1( )in L L
jBZ R j X X Z
0 0
0
( )( )
L L
L L
BZ X X Z RX X BZ R
0Z
0
0
( ) /L LZ R RB
Z
inZ0
0( )L L L
Z
X X R Z R
• Two analytical solutions are physically realizable.
• One of the two solutions may result in smaller reactive elements and
5
One of the two solutions may result in smaller reactive elements, and may have better matching, bandwidth, or better SWR on the line.
Example 5 1 L-Section Impedance MatchingExample 5.1 L Section Impedance Matching
• No lumped element is required. Convenient for MIC fabrication.• Characteristic impedances of the lines and the stub can be different.• Keep the matching stub as close as possible to the load. • Solution procedure for shunt-stub and series-stub tunings.(1) Locate zL (yL) on the Smith Chart.(1) Locate zL (yL) on the Smith Chart.(2) Move along the const-Γ circle with 2d/λ wavelengths to
reach the g = 1 (r = 1) circle.(3) Sh t ith t i ith t t(3) Shunt with a susceptance or series with a reactance to
cancel the imaginary part. 10
Shunt-Stub Tuning Analytical Solution
0 00 tan
( )1 1L L Lin t dno stub
Z jZ t R j X Z tZ ZZ jZ t Y Z X t jR t G jB
0 0 0
220 0
022 2 2
( )( )(1 ) ,
L L L
L L LLin in
in inZ jZ t Y Z X t jR t
R t Z X t Z t XR tG B Y
G jB
022 20 0
2 2 20 0 0
2,( ) ( )
(1 ) ( ) real part
in inL L L L
in L L L
GR X Z t R X Z t
G Y Z R t R X Z t
0 0 0in L L L
00
If , and tan2
LL
XR Z t dZ
0
11 tan , 02 2
LL
X XZd
0
1
2 2
1 tan , 0LL
ZdX X
0
tan , 02 2 LX
Z
11
Shunt-Stub Tuning Analytical Solution0
2 2 20 0 0 0 0
If Z ,
( ) ( )( )L
L L L L L L
R
X Z X Z Z R Z R Z R Xt
0 0 0 0 0
0 0
2 2 2
( ) ( )( )( )
( ) [( ) ]
L L L L L L
L
L
tZ R Z
RX R Z X
00
0
( ) [( ) ]L L L
L
X R Z XZ
R Z
here tant lLet the stub susceptance Bstub (or Bs) = -Bin,
(a) open-stub: and0stubZ jZ t sstubstub jBjBtjYY 0
here tant l
1 1
0 0
1 1tan ( ) tan2 2
o s inl B BY Y
(b) Short-stub: and tjZZstub 0 sstubstub jBjBtjYY 0
1 10 01 1tan ( ) tansl Y Y tan ( ) tan
2 2s inB B
If lo < 0 or ls < 0, λ/2 must be added to have a realistic result. 12
Example 5.2 Shunt Single-Stub Tuning
Match ZL = 15 + j10 Ω to a 50-Ω line. Use a shunt open-stub.Sol: RL ≠ Z0,Sol: RL ≠ Z0,
2 2 20( ) ( )L
L L LRX R Z X 0
0
0
( ) ( )10 397.5
35
L L L
L
Zt
R Z
d
11 1
10 397.5 1 ( tan ) 0.38735 2
1
dt t
l B
1
0
1 ... tan2
10 397 5 1
o inin
l BBY
d
12 2
1
10 397.5 1 tan 0.04435 2
1o in
dt t
l BB
13
1
0
... tan2
o ininB
Y
Example 5.2 Shunt Single-Stub Tuning (Cont’d)Use a shunt open-stub to match ZL = 15 + j10 Ω line.
Solution 1: Solution 2:
14
5.3 Double-Stub Tuning – Analytical Solution
• In single-stub tuning, the length d is not tunable.
• If this causes difficulty in circuit implementation, use double-stub tuning.
Example 5.4 Double-Stub Tuning - PerformanceUse double-stub matching scheme to match ZL = 60 – j80Ω at 2.0 GHz to a 50-Ω line.
17
5.4 The Quarter-Wave Transformer• A single-section transformer may suffice for a narrow-band
impedance matching.• Single-section quarter-wave impedance matching λ= λ0 / 4 at the
desired frequency. (See Chap 2)
• Multisection quarter-wave transformer designs can be synthesized to yield optimum matching characteristics over a broad bandwith.
18
yield optimum matching characteristics over a broad bandwith.
2.5 The Quarter-Wave TransformerReview
A quarter-wave transformer is an impedance matching circuit
2tanR jZ l Z 1 11
12
tan tan
Lin
L Ll
R jZ l ZZ ZZ jR l R
Z Z
0
0
inin
in
Z ZZ Z
19
0 1 0If 0 is required, , then in in LZ Z Z Z R
Estimate the Bandwidth of a Single-Section Impedance Transformera Single Section Impedance Transformer
1 0 LZ Z Z
11
1
, tan tan Lin
L
Z jZ tZ Z tZ jZ t
Z jZ t
11 0 2
0 1 0 1 012
10 1 0 1 01 0
( ) ( )( ) ( )
L
in L LL
Lin L L
Z jZ tZ ZZ Z Z Z Z j Z Z Z tZ jZ t
Z jZ tZ Z Z Z Z j Z Z Z tZ Z
1 01
002 2
1 2 4( ) 4
L
LL
Z jZ tZ ZZ Z
Z Z j t Z Z Z ZZ Z t Z Z
2 20 0 00 02 4( ) 4 1L L LL L
L
Z Z j t Z Z Z ZZ Z t Z ZZ Z
22
0
sec
0LZ Z 0
0
0
When , cos2 2
L
L
L
Z ZZ Z
Z Z
0
0
Set a max , or cos2
Lm m m
L
Z ZZ Z
20
Bandwidth of the Matching Transformer
20
2( / 2 )41 1
m
LZ Z
202 2
0
: 1 sec( )
2
Lm m
m L
m m
Z Zff f f
00 0
0
, ,2 2
2
m mm m
L
ff f ff f
Z Z
0
20
cos1
Lmm
LmZ Z
F ti l b d idth10 or 0.1
0
Fractional bandwidth :2( ) 42m mf ff
f f
4 or 0.25
0 0
01 242 cos Lm
f f
Z Z
2 or 0.5
21
20
2 cos1 Lm
Z Z
Example 5.5Single Section Quarter Wave Transformer BandwidthSingle-Section Quarter-Wave Transformer Bandwidth
ZL = 10 Ω, Z0 = 100 Ω, SWR = 1.2S lSol:
1 0 31.6LZ Z Z
1 0.11m
SWRSWR
max 0 0
min 0 0
11
V V VSWRV V V
01
2
242 cos1
Lm Z Zff Z Z
20 0
1
1
4 0.1 2 31.6 62 cos 2 9%
Lmf Z Z
22
2 cos 2 9%900.99
5.5 Theory of Small Reflections• For applications requiring more bandwidth than that a single quarter-
wave section can provide, multi-section transformers can be used.
(1) Single-Section Transformer
212
121
ZZZZZZ
2
23
2ZZZZZ
L
L
1
12
2121
21
21
ZT
ZZZT
12
1212 1
ZZT
23
5.5 Theory of Small Reflections
(1) Multi-Reflections
2 2 41 12 21 3 12 21 3 2 ...j jT T e T T e
1 12 21 3 12 21 3 2
22 2 41 12 21 3 3 2 3 2(1 ...)j j jT T e e e
2 21 12 21 3 3 2
02 2
nj j n
nj j
T T e e
T T e e
12 21 3 1 31 2 2
3 2 1 31 1
j j
j j
T T e ee e
• If is small, • Γ the reflection from the initial discontinuity between Z1 and Z2 + the
312
1 3je
24
first reflection from the discontinuity between Z2 and ZL
Multisection Transformer
• N commensurate (equal-length) sections of transmission lines. Let ( q g )the total reflection be Γ.
• Assume that all Z increase or decrease monotonically ZL is real• Assume that all Zn increase or decrease monotonically, ZL is real, and “The theory of small reflections” holds.
• It can be validated that
NL ZZZZZZZZ
23
212
101
0
25
NLN ZZZZZZZZ
...,,,23
212
101
0
Multisection Transformer
2 2
2 4 20 1 2
[ ( ) ( ) (if s mmetr]
.
)
..j j j
j N j NjN jN jN
NNe e e
0 1
0 1 /2
[ ( ) ( ) ... (if symmetry]
cos cos 2 ... , 2
)j jjN jN jN
NjN
e e e e e
N N N evene
0 1 1 /2
2cos cos 2 ... cos ,
j
N
eN N N odd
• We can synthesize any desired response as a function of frequency• We can synthesize any desired response as a function of frequency or θ, by properly choosing the Γn’s and enough number of sections N.
• The most commonly used passband responses are:
(1) Binomial or maximally flat response, and(2) Chebyshev or equal-ripple response. 26
H Fi d h C di R fl i ?How to Find the Corresponding Reflection?
2 4 20 1 2
Theory of Samll Reflection: ( ) ...j j j N
Ne e e 0 1 2( )
Binomial Multi-Section Transformer:
N
2
Binomial Multi-Section Transformer:(1 )j NA e
Chebyshev Multi-Section Transformer:
(sec cos )n mAT
27
5.6 Binomial Multisection Matching Transformer• Maximally flat response: the response is as flat as possible near the design
frequency. Also called Binomial matching transformer.• Let 2(1 ) A d tifi i llj NA • Let
2(1 ) Assumed artificially
2 cos
j N
NN
A e
A
0
0
00 2
2N L
NL
Z ZA AZ Z
0
2 2
0
!(1 ) ,( )! !
LN
j N N j n Nn n
n
NA e A C e CN n n
2 4 20 1 2
( ) ...
(0)
j j j NN
N N
e e e
AC C
N! ("enn factorial“)
Properly
( )2
configure the i
N Nn n nNAC C
mpedances to synthesize the needed
N! ( enn factorial )“Four factorial" is written as "4!”
reflections .n28
Binomial Multisection Matching Transformer
nnn xxxZZZ 11 1112lnln1
LNN
LNN
Nn
nnnn
ZCZZCACZ
xx
xZZZ
01
1
1
ln1122ln
11,1
2ln,ln2
NNnC
Ln
nNL
nNnnn
ZZ
ZC
ZZCAC
Z2/
1
001 ln
2222ln
n ZZ 0
• Exact results for Z 1 / Z0 for N = 2 thru 6 are given in Table 5 1• Exact results for Zn-1 / Z0 for N = 2 thru 6 are given in Table 5.1.
A 5th d bi i l f f Z 6Z• A 5th order binomial transformer for ZL = 6Z0
31• Also note that 4055.1
2689.46,0596.1
6625.56
E ample Binomial Transformer Design Z /Z <1Example Binomial Transformer Design ZL/Z0 <1
ZL/Z0
N=5
Z1 /Z0 Z2 /Z0 Z3 /Z0 Z4 /Z0 Z5 /Z0
6.0000 1.0596 1.4055 2.4495 4.2689 5.6625
0.1667 0.9438 0.7115 0.4082 0.2343 0.1766
• A 5th order binomial transformer for ZL = Z0 / 6
32
Bandwidth of the Binomial Transformer,cos2 m
NNm A i.e, tolerable max over the passband.m
N/1
m
m A1
21cos
N
mmm
Afff
ff
/1
1
0
0
0 21cos4242)(2
Example 5.6 N = 3, ZL = 2Z0 = 100 Ω, find the BW for Γm = 0.05.Sol: From Table 5.1, the required impedance Zn can be found to be
1 2 354.5 , 70.7 , 91.71 100 502 (0) 0 4167N
Z Z Z
A
11 3
2 (0) 0.41678 100 50
4 1 0.052 cos ( ) 70%
A
f
33
3
02 cos ( ) 70%
2 0.4167f
Binomial Transformer’s Frequency Response
Reflection coefficient magnitude versus frequency for multisectionbinomial matching transformer of Ex. 5.6 with ZL = 2Z0 = 100 Ω and
34
binomial matching transformer of Ex. 5.6 with ZL 2Z0 100 Ω and Γm = 0.05.
Example (2nd Midterm)Design a Butterworth transformer of N = 2 for ZL = 4Z0. Let Γ0, Γ1 and ΓL be
DIYg L 0 0, 1 L
respectively the reflection coefficients at the Z0 – Z1, Z1 – Z2 and Z2 – ZLjunctions, and Γ0 = ΓL. (a) Based on the “theory of small reflection,” find the Γ terms of Γ Γ and θ (b) If Γ = 0 and are required when θ0/ fΓin terms of Γ0, Γ1 and θ. (b) If Γin = 0 and are required when θ= λ / 4, find a and b.
0/ fin
2cos2 104
22
10 ee jjin
202
2cos2 0110
10210
in
4444444
11
02sin4 0
abbababababb
aa
in
359/141141
0434
31)(2
112
412322
aaaabaababaabaa
abab
ba
2751.0,5707.1
4321.08135
99/14,9273.1
271
27112
94
61
3 233 23
rqrrqr
qr
The “exact” solution in Table 5.1 is a = 1.4142 and b =2.8285. The error is from
358435.2/4
4067.12751.05707.191
ab
a the approximation used in the “theory of small reflection.”
5.7 Chebyshev Multisection Matching Transformers
Chebyshev polynomials: Recurrence formula:)()(
1)(
1
0
xxTxT
34)(24
33 xxxT
2)()(2)( 21 nxTxxTxT nnn
12)( 22 xxT 188)( 24
4 xxxT
Tn(x)345
n=1
2
n=1
xpassband
36
p
5.7 Chebyshev Multisection Matching Transformers(1) mapped to passband
Let x = cosθ, it can be shown that (2) t id th b d
1)(,1 xTx n
nTn cos)(cos 1)(1 T(2) outside the passband
(3) In general, 1)coscos()( 1 xxnxTn
1)coshcosh( 1 xxn
1)(,1 xTx n
1)coshcosh( xxn
Tn(x)345
n( )
1
2
n=1
xpassband
37
passband
Magnitude of Chebyshev Polynomials
|Tn(x)|345
22
n=1
xx
passband
38
passband
Chebyshev Responses
39
Mapping of Passband and Stopband
Define is mapped to passbandthe upper passband edgethe lower passband edge
1cos/cos xx m,1 xm1 x the lower passband edge ,1 xm
1
1
( ) cos [cos (cos / cos )] (sec cos )(sec cos ) sec cos
n m n mT x n TT
1
22
3
(sec cos ) sec cos
(sec cos ) sec (1 cos 2 ) 1
(sec cos ) sec (cos3 3cos ) 3sec cos
m m
m m
TTT
40
3
4 24
(sec cos ) sec (cos3 3cos ) 3sec cos
(sec cos ) sec (cos 4 4cos 2 3) 4sec (1 cos 2 ) 1m m m
m m m
TT
Design of Chebyshev Transformer
jN NNN ]22[2)(
LmN
L
njN
ZZZZ
TAAT
ZZZZ
nNNNe
)(sec1)(sec)0(
...]2cos...2coscos[2)(
00
10
mmNm
LmNL
ATAZZTZZ
)sec(cos)(sec 00
L
L
mm
L
L
mL
LmN
f
ZZZZ
NZZZZ
ZZZZ
AT 1cosh1coshsec,11)(sec
0
01
0
0
0
0
41m
ff 420
Table 5.2 Chebyshev Transformer Design
42
Example 5.6Design a Chebyshev transformer of N=3 for ZL = 2Z0 = 100 Ωwith Γm = 0.05.
