1
Notes 18
ECE 5317-6351 Microwave Engineering
Fall 2011
Multistage Transformers
Prof. David R. JacksonDept. of ECE
2
021S 1- je
011S 0
22S
012S 1- je
L
Single-stage Transformer
0
0
-in
in
Z ZZ Z
1
1
-LL
L
Z ZZ Z
0 0 1 011 22
1 0
0 10 0 0021 12 11
1 1 0
--
21
Z ZS SZ Z
Z ZZS S SZ Z Z
Step Impedance change
LZ1Z0Z
1=
, inZLZ is real
From previous notes:
Step Z1 line
Load
The transformer length is arbitrary in this analysis.
3 200 0
21 12 111-S S S
1
1
- 20 00 21 1211 - 20
221 -
jL
jL
S S eSS e
From the self-loop formula, we have (as derived in previous notes)
Single-stage Transformer (cont.)
2
0 0 0 1 0 121 12 2
1 0 1 0
2 4Z Z Z ZS SZ Z Z Z
22 2 2 22 22 1 0 1 0 0 10 1 01 0 1 0 0 111 2 2 2
1 0 1 0 1 0 1 0
22 2 2 21 0 0 1 1 0 0 1
21 0
0 12
1 0
2- 21 1 1 1
2 2
4
Z Z Z Z Z ZZ ZZ Z Z Z Z ZSZ Z Z Z Z Z Z Z
Z Z Z Z Z Z Z Z
Z ZZ Z
Z Z
Hence
For the numerator:
Next, consider this calculation:
4
1
1
- 2011
- 20111
jL
jL
S eS e
Putting both terms over a common denominator, we have
1 1
1
- 2 - 20 0 2 0 211 11 11
- 2011
11
j jL L
jL
S S e S eS e
or
Single-stage Transformer (cont.)
1
1
- 20 2110
11 - 2022
11 -
jL
jL
S eS
S e
We then have
5
011 1LS Assuming small reflections
1
1
- 2011
- 20111
jL
jL
S eS e
1- 2011
jLS e
00 11 1, LS Denote
1- 20 1
je
1 0 10 1
1 0 1
- -; L
L
Z Z Z ZZ Z Z Z
L
1je
1je
00 11S
Single-stage Transformer (cont.)
1- 20 0 011 21 12
jLS S S e
Note: It is also true that
But 0 0 0 221 12 111- 1S S S
6
Assuming small reflections:
LZ0Z 1Z 2Z 3Z . . . -1NZ NZ
1 2 3 -1N Ni i i
1 2 3 N Assume
je
0 1 2 3 -2N -1N N L
je je je je
je je je je je
Multistage Transformer
7
- 2 - 4 - 6 - 20 1 2 3
1
1
.....-
j j j j NN
n nn
n n
e e e eZ ZZ Z
Multistage Transformer (cont.)Hence
LZ0Z 1Z 2Z 3Z . . . -1NZ NZ
1 2 3 -1N Ni i i
1 2 3 N Assume
Note that this is a polynomial in powers of z = exp(-j2).
8
- 2 - 4 - 6 - 20 1 2 3 .....j j j j N
Ne e e e
0 1 -1 2 -2
- - ( -2) - ( -2)0 1
, , , . . .
. . .N N N
jN jN jN j N j Ne e e e e
--1
2
2
odd last term
even last term
j jN
N
N e e
N
Multistage Transformer (cont.)
If we assume symmetric reflections of the sections (not a symmetric layout of line impedances), we have
Last term
9
-0 1
2
-0 1 -1
2
12 cos cos - 2 ... cos - 2 ... ;2
2 cos cos - 2 ... cos - 2 ... cos ;
even
odd
jNn N
jNn N
e N N N n
e N N N
N
n N
Multistage Transformer (cont.)
Hence, for symmetric reflections we then have
Note that this is a finite Fourier cosine series.
10
Multistage Transformer (cont.)
Design philosophy:
If we choose a response for ( ) that is in the form of a polynomial (in powers of z = exp (-j2 )) or a Fourier cosine series, we can obtain the needed values of n and hence complete the design.
11
-- 2 - - 2 co1 sN Nj j j Nj N j N NAe e e A eA e
2 cos NNA
0
2
02 2
0 1,2, ..., -1
i
n
n
f f
d n Nd
Also, for
1N - 1st derivatives are zero maximally flat
Binomial (Butterworth*) Multistage Transformer
Consider:
*The name comes from the British physicist/engineer Stephen Butterworth, who described the design of filters using the binomial principle in 1930.
Choose all lines to be a quarter wavelength at the center frequency so that
(We have a perfect match at the center frequency.)
12
- 2 - 2
0
1NNj N j n
nn
A e A C e
We want to use a multistage transformer to realize this type of response.
- 2 - 4 - 6 - 20 1 2 3
- 2 - 2
0
......
1NNj N j
j j
n
j
n
n
j NN
A e A
e e e
C e
e
Use the binomial expansion so we can express the Butterworth response in terms of a polynomial series:
Binomial Multistage Transformer (cont.)
0
!1- ! !
NN N n N
n nn
Nz C z CN n n
where
A binomial type of response is obtained if we thus choose
Set equal
(Both are now in the form of polynomials.)
13
0
0
0 0
- 2L N
L
f
Z ZZ Z
A
zero length transmisison linesNote that as
, 1,2,.......,Nn nAC n N
0 0 1, N LZ Z Z Z
Equating responses for each term in the polynomial series gives us:
Binomial Multistage Transformer (cont.)
- 0
0
-2 N L
L
Z ZAZ Z
Hence
-1 0
1 0
- -2 N Nn n Ln
n n L
Z Z Z Z CZ Z Z Z
Hence
Note: A could be positive or negative.
This gives us a solution for the line impedances.
14
, 1,2,.......,Nn nAC n N
Note on reflection coefficients
Binomial Multistage Transformer (cont.)
Hence
!
- ! !Nn
NCN n n
! !
- ( ) ! ( )! ! ( )!N NN n n
N NC CN N n N n n N n
n N n
Although we did not assume that the reflection coefficients were symmetric in the design process, they actually come out that way.
Note that
15
Note: The table only shows data for ZL > Z0 since the design can be reversed (Ioad and source switched) for ZL < Z0 .
Binomial Multistage Transformer (cont.)
16
Example showing a microstrip line
Binomial Multistage Transformer (cont.)
A three-stage transformer is shown.
50 line100 line
1 / 4g2 / 4g
3 / 4g
1Z 2Z 3Z
0Z LZ
17
Binomial Multistage Transformer (cont.)
Figure 5.15 (p. 250)Reflection coefficient magnitude versus frequency for multisection binomial matching
transformers of Example 5.6. ZL = 50Ω and Z0 = 100Ω.
Note: Increasing the number of lines increases the bandwidth.
18
Use a series approximation for the ln function:
-1 1 ln ; 11 2
X X XX
-1 0
1 0
- -2 N Nn n Ln
n n L
Z Z Z Z CZ Z Z Z
-1
0
1 1ln 2 ln2 2
N Nn Ln
n
Z ZCZ Z
-1
0
ln 2 ln lnN N Ln n n
ZZ C ZZ
recursive
relationship
Binomial Multistage Transformer (cont.)
Hence
Recall
19
Bandwidth
1
-1 1cos2
2 cosN
mm
m
N
NmA
A
1
0 -1
0 0 0
- 4 4 12 2 - 2 2 - 2 2 - 2 - cos/ 2 2
Nm m m m mf f ff
f f f A
The bandwidth is then:mf 0f 02 - mf f
m / 2 - m
m
Binomial Multistage Transformer (cont.)
Maximum acceptable reflection
1
-1
0
4 12 - cos2
Nmf
f A
Hence
/ 2f / 2f
20
Summary of Design Formulas
Binomial Multistage Transformer (cont.)
1
-1
0
4 12 - cos2
Nmf
f A
-1
0
ln 2 ln lnN N Ln n n
ZZ C ZZ
- 0
0
-2 N L
L
Z ZAZ Z
- 2 - 2
0
1NNj N j n
nn
A e A C e
Reflection coefficient response
A coefficient
Design of line impedances
Bandwidth
0 2ff
!
- ! !Nn
NCN n n
21
Example: three-stage binomial transformer
0
50 [ ]100 [ ]0.05
L
m
ZZ
-26.0 [dB]dBm -3
13-1
0
50 -1003 2 -0.041750 100
4 1 0.052 - cos2 0.0417
0.713
N A
fBWf
100 50 1Z 2Z 3Z
Example
Given:
71.3%BW
22
-3 31 0
1
0150ln ln 2 ln 4.519
10
91.
0
[
:
7 ]
Z Z C
Z
Z
-3 32 1
2
1250ln ln 2 ln 4.259
10
70.
0
[
:
7 ]
Z Z C
Z
Z
-3 33 2
3
2350ln ln 2 ln 3.999
10
54.
0
[
:
5 ]
Z Z C
Z
Z
30
31
32
3
C = 1
C = 3
C = 3
C = 13
Example (cont.)
!
- ! !Nn
NCN n n
-1
0
ln ln 2 lnN N Ln n n
ZZ Z CZ
23
1 91.685 [ ]Z
2 70.711 [ ]Z
3 54.585 [ ]Z
Example (cont.)
Using the table in Pozar we have:
0 1 2 3 0/ 2 : , , / 1.0907, 1.4142,1.8337LZ Z Z Z Z Z
(The above normalized load impedance is the reciprocal of what we actually have.)
1 2 3 0 0, , / 1.8337, 1.4142, 1.0907; 50[ ]Z Z Z Z Z
Therefore
Hence, switching the load and the source ends, we have
24Response from Ansoft Designer
Example (cont.)
-26
3.29 GHz 6.74 GHz
69.0%BW
50 line100 line
1 / 4g2 / 4g
3 / 4g
1Z 2Z 3Z
0Z LZ
11 1020logdB
S f
0 5.0GHzf
25
Chebyshev Multistage Matching Transformer
1
22
33
-1 -2
2 1
4 -3
2 -n n n
T x x
T x x
T x x x
T x xT x T x
-1 1: 1
1: 1n
n
x T x
x T x
For
For
1
1
cos cos , 1
cosh cosh , 1n
n x xT x
n x x
Chebyshev polynomials of the first kind:
We choose the response to be in the form of a Chebyshev polynomial.
26
Chebyshev Multistage Transformer (cont.)
Figure 5.16 (p. 251)The first four Chebyshev polynomials Tn(x).
27
m
0f0 -2ff
0 2ff
f
m - m
n increasing
/ 2
n=2
A B
n=1
-1 1
2
1
n=3
n=2
nT x
x
B A
Chebyshev Multistage Transformer (cont.)A Chebyshev response will have equal ripple within the bandwidth.
- sec cosjNN mAe T
m A
This can be put into a form involving the terms cos (n ) (i.e., a finite Fourier cosine series).
Note: As frequency decreases, x increases.
28
1
22
33
-1 -2
sec cos sec cos
sec cos sec 1 cos2 -1
sec cos sec cos3 3cos - 3sec cos
sec cos 2 sec cos sec cos - sec cos
m m
m m
m m m
n m m n m n m
T
T
T
T T T
Chebyshev Multistage Transformer (cont.)
We have that, after some algebra,
1
22
33
-1 -2
2 1
4 -3
2 -n n n
T x x
T x x
T x x x
T x xT x T x
Hence, the term TN (sec, cos) can be cast into a finite cosine Fourier series expansion.
29
-
-0 12 cos cos - 2 .... cos - 2 .....
sec cosjNm
jNn
N
e N
Ae
N N n
T
0
0
0
0
-
- 1
0 0
0
sec
secN mL
L
L
L N m
Z ZZ Z
Z ZAZ Z T
f
AT
As
Transformer design
Chebyshev Multistage Transformer (cont.)
From the above formula we can extract the coefficients n (no general formula is given here).
30
sec cos 1m
m m N m m
m
NA T A T
A
A
At
Chebyshev Multistage Transformer (cont.)
0sgn -L mA Z Z
Hence
0
0
0
0 :-0 sec
sec 0 sec 1
-
LN m
L
N m m
L
Z ZATZ Z
T
A Z Z
At
has the same sign as
31
Chebyshev Multistage Transformer (cont.)
Note: The table only shows data for ZL > Z0 since the design can be reversed (Ioad and source switched) for ZL < Z0 .
32
0 0 0
0 0 0
-
0
0
1
-1
0
- - -1 10 : sec sec
cosh co
-1 1sec cos
42 -
sh sec
h cosh L
L L LN m N m
L L m L
mL
m
m
m
Z Z Z Z Z Zf AT T
Z Z A Z Z Z
Z ZN Z Z
fWf
N
B
Z
At
Chebyshev Multistage Transformer (cont.)Bandwidth
-1
0
1 1sec cosh cosh ln2
Lm
m
ZN Z
Hence -1 1 ln ; 1
1 2X X XX
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Summary of Design Formulas
0
42 - mf
f
Reflection coefficient response
A coefficient
Design of line impedances
Bandwidth
Chebyshev Multistage Transformer (cont.)
0sgn -L mA Z Z
- sec cosjNN mAe T
No formula given for the line impedances. Use the Table from Pozar or generate (“by hand”) the solution by expanding ( ) into a polynomial with terms cos (n ).
-1
0
1 1sec cosh cosh ln2
Lm
m
ZN Z
m term
0 2ff
34
0
100[ ]50[ ]0.05
L
m
ZZ
3 0 2 1Γ = Γ , Γ = ΓAssumed symmetry :
- 3 3
- 3
- 3
0 1
0
3
sec cos
sgn - 0.05
3 sec co
3 3cos - 3sec cos
2 cos3 s
s
co
j
L
m
j
j
m
m
m
A Z Z
N e
e
A
Ae
A
T
50[ ] 1Z 2Z 3Z 100[ ]
Given
Equate(finite Fourier cosine series form)
Example: three-stage Chebyshev transformer
Example
35
30
31
30 0 3
31
1
o
1
2
-
44.7 0.780[rad
2 sec
2 3 sec -3 se
] 1.00
1 1 100sec cosh
7 100.7
c
1 0.05 1.408 0.06982
1 3 0.05 1.408 -3 0.05 1.4082
0.1
cosh ln3 2 0.05 50
1
037
.40 %8
m
m
m m
m
A
A
B
A
W
Also,
Example: 3-Section Chebyshev TransformerEquating coefficients from the previous equation on the last slide, we have
0
42 - mfBW
f
36
1
3
4
11
1
2
1 0.069850 57.5
1- 0.0698
1 0.103757.5 70.8
1- 0.1037
1 0.103770.8 87.2
1- 0.1037
1 0.069887.2 100.3
1- 0.
11-
9
-
06 8
n nn
n
nn n
n
L
n
Z ZZ Z
Z
Z
Z
Z Z
Z Z
Checkin
N
g con
ext, use
sistency :
1
2
3
57.5
70.8
87.2
Z
Z
Z
Example: 3-Section Chebyshev Transformer
37
1 1
1
1
1
0 0
2 1 1
3
1
2
3
2
2
ln ln 2
ln 50 2 0.069
57.49
8
4.051
ln ln 2
4.259
ln l
70.74
87.0
n
5
2
- 1 ln2
ln n
4
l 2
.466
nn n n
nn
nn
nn
Z Z
Z Z
Z Z
Z
Z
Z
Z
ZZ Z
Z ZZ
Z
Alternative method:
Example: 3-Section Chebyshev Transformer
38
1
2
3
0 0
1.147 50 57.4
0.05
5
1.4142
1.
50 70.7
50 8742
, 3, / 2,
9 7.1
50m L
Z
Z
N Z Z Z
Z
TablFrom e
Example: 3-Section Chebyshev Transformer
39
Example: 3-Section Chebyshev Transformer
-26
99.8%BW
2.51 GHz 7.5 GHz
11 1020logdB
S f
Response from Ansoft Designer
50 line100 line
1 / 4g2 / 4g
3 / 4g
1Z 2Z 3Z
0Z LZ
0 5.0GHzf
40
Example: 3-Section Chebyshev Transformer
Comparison of Binomial (Butterworth) and Chebyshev
The Chebyshev design has a higher bandwidth (100% vs. 69%).
The increased bandwidth comes with a price: ripple in the passband.
Note: It can be shown that the Chebyshev design gives the highest possible bandwidth for a given N and m.
41
Tapered Transformer
The Pozar book also talks about using continuously tapered lines to match between an input line Z0 and an output load ZL. (pp. 255-261). Please read this.
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