Transcript Notes 18 - Multistage transformers

ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 18
Multistage Transformers
1
Single-stage Transformer
The transformer length is arbitrary in this analysis.
Step
 = 1
Z0
Z1 line
0
S21
Z1
ZL

0
S22
S110
, Z in

Z in - Z 0
Z in  Z 0
L
e- j1 Load
S120
Z L is real
e- j1
L 
Z L - Z1
Z L  Z1
From previous notes:
Step
Impedance
change
0
S110  - S 22

0
S 21
 S120 
Z1 - Z 0
Z1  Z 0
2 Z 0 Z1
Z0
1  S110  

Z1
Z1  Z 0
2
Single-stage Transformer (cont.)
From the self-loop formula, we have (as derived in previous notes)
- j 21
0
0
S
S

e
  S110  21 012 L - j 21
1 - S22  L e
For the numerator:
2
 2 Z 0 Z1 
4 Z 0 Z1
0 0
S21S12  


 Z1  Z 0   Z  Z 2
1
0


Next, consider this calculation:
1   S110 
2
2
2
Z1  Z 0 
 Z1 - Z 0 

Z12  Z 02  2 Z 0 Z1  Z1  Z 0    Z1  Z 0  2 Z 0 Z1 
 1 
 1

  1
2
2
2
Z

Z
 Z1  Z 0 
 Z1  Z 0 
 Z1  Z 0 
0 
 1
2
2
2
Z


2
1
 Z 02  2 Z 0 Z1    Z12  Z 02  2 Z 0 Z1 
2
 Z1  Z 0 
2
4 Z 0 Z1
 Z1  Z 0 
2
Hence

S S  1- S11
0
21
0
12
2
0

3
Single-stage Transformer (cont.)
We then have
  S110
1 S  


02
11
L
e- j 21
0
1 - S22
 L e- j 21
Putting both terms over a common denominator, we have
 
S110  S110 2  L e- j 21  1  S110 2   L e- j 21
1  S110  L e- j 21
or
S110   L e- j 21

1  S110  L e- j 21
4
Single-stage Transformer (cont.)
S110  L e- j 21

1  S110 L e- j 21
Assuming small reflections
S
0
11
 L  1
Note: It is also true that

0
  S110  S21
S120 L e- j 21
But S210 S120  1- S110 2   1
  S110  L e- j 21
e  j1
0
Denote 0  S11
, 1  L
 0  1 e- j 21
Z -Z
Z -Z
0  1 0 ; 1  L 1
Z1  Z 0
Z L  Z1
0  S110

L
e  j1

5
Multistage Transformer
i   i
1
Z0
Z1
3
2
Z2
i
 N -1
. . .
Z3
Z N -1
N
ZN
ZL
Assume 1   2   3    N  
Assuming small reflections:
e  j
0


e  j
e  j
1
e  j
e  j
2
e  j
e  j
3
 N -2
e  j
e  j
 N -1
N  L
e  j
6
Multistage Transformer (cont.)
Hence
    0  1 e- j 2  2 e- j 4  3e- j 6  .....   N e- j 2 N
Z n1 - Z n
n 
Z n1  Z n
Note that this is a polynomial in powers of z = exp(-j2).
i   i
1
Z0
Z1
2
Z2
3
Z3
. . .
i
 N -1
N
Z N -1
ZN
ZL
Assume 1   2  3    N  
7
Multistage Transformer (cont.)
    0  1 e- j 2  2 e- j 4  3e- j 6  .....  N e- j 2 N
If we assume symmetric reflections of the sections (not a symmetric layout
of line impedances), we have
0   N , 1   N -1 ,  2   N -2 , . . .
     e- jN 0  e jN  e- jN   1  e j ( N -2)  e- j ( N -2)   . . .
N odd  last term   N -1  e  e
j
- j

Last term
2
N even  last term   N
2
8
Multistage Transformer (cont.)
Hence, for symmetric reflections we then have
 - jN
 2e

    
2e- jN




1

cos
N



cos
N
2


...


cos
N
2
n


...


  1 
 
 

n
N  ; N even
 0
2 2




cos
N



cos
N
2


...


cos
N
2
n


...


cos











1
n
N -1
 0
 ; N odd
2


Note that this is a finite Fourier cosine series.
9
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.
10
Binomial (Butterworth*) Multistage Transformer
Consider:
    A 1  e- j 2 

N
 Ae- jN  e j  e- j   A e- jN 2 N cos N 
N
    A 2 N cos 
N
Choose all lines to be a quarter wavelength at the center frequency so that
f  f 0  i     

2
 
    0
2
(We have a perfect match at
the center frequency.)
dn
Also,
       0 for n  1, 2, ..., N -1
d n
2
1st  N - 1 derivatives are zero  maximally flat
*The name comes from the British physicist/engineer Stephen Butterworth, who
described the design of filters using the binomial principle in 1930.
11
Binomial Multistage Transformer (cont.)
Use the binomial expansion so we can express the Butterworth response in terms of
a polynomial series:
N
1  z 
N
  CnN z n
where CnN 
n 0
N!
 N - n ! n !
A binomial type of response is obtained if we thus choose
    A 1  e
- j 2

N
N
 A  CnN e- j 2 n
n 0
We want to use a multistage transformer to realize this type of response.
     0  1 e- j 2  2 e- j 4  3 e- j 6  ......   N e- j 2 N
 A 1  e
- j 2

N
N
 A CnN e- j 2 n
n 0
Set equal
(Both are now in the form of polynomials.)
12
Binomial Multistage Transformer (cont.)
Note that as f  0      0
    
 zero length transmisison lines 
Z L - Z0
 A 2N
Z L  Z0
Hence
 Z L - Z0 
A2 

Z

Z
0 
 L
-N
Note: A could be positive or negative.
Equating responses for each term in the polynomial series gives us:
n  ACnN , n  1,2,......., N
Hence
 Z -Z 
Z n 1 - Z n
 2- N  L 0  CnN
Z n 1  Z n
 Z L  Z0 
This gives us a
solution for the line
impedances.
Z0  Z0 , Z N 1  Z L
13
Binomial Multistage Transformer (cont.)
Note on reflection coefficients
n  ACnN , n  1,2,......., N
CnN 
N!
 N - n ! n !
Note that
CNNn 
N!
N!

 CnN
 N - ( N  n) ! ( N  n)! n! ( N  n)!
Hence
n   N  n
Although we did not assume that the reflection coefficients were
symmetric in the design process, they actually come out that way.
14
Binomial 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 .
15
Binomial Multistage Transformer (cont.)
Example showing a microstrip line
g1 / 4
50 line
Z1
g 2 / 4
Z2
g 3 / 4
100 line
Z3
ZL
Z0
A three-stage transformer is shown.
16
Binomial Multistage Transformer (cont.)
Note: Increasing the
number of lines
increases the
bandwidth.
Figure 5.15 (p. 250)
Reflection coefficient magnitude versus frequency for multisection binomial matching
transformers of Example 5.6. ZL = 50Ω and Z0 = 100Ω.
17
Binomial Multistage Transformer (cont.)
Use a series approximation for the ln function:
X -1 1
 ln  X  ;
X 1 2
X 1
Recall
Z n 1 - Z n
- N  Z L - Z0  N
2 
 Cn
Z n 1  Z n
 Z L  Z0 


 
 1 ln  Z n1   2- N CnN 1 ln  Z L 
2
 Zn 
2
 Z0 
Hence
ve
recursi
ship
n
o
i
t
a
l
e
r
 ZL 
ln  Z n 1   2 C ln    ln  Z n 
 Z0 
-N
N
n
18
Binomial Multistage Transformer (cont.)
Bandwidth

 m  2 N A cos N  m
1


N
-1  1  m
  m  cos
2 A 

 






Maximum acceptable reflection
m
m
The bandwidth is then:
f
f0
f -f 
2 0 m
f0
f / 2
fm
 /2
f / 2
f0

fm
m
4 m
4
1
 2-2
 2-2
 2 2 - cos -1   m
f0
 /2


2 A

Hence
1


N


f
4
1 
 2 - cos-1   m  
2 A

f0

 
 


 
 - m
2 f0 - fm

 
 
 

1
N
19
Binomial Multistage Transformer (cont.)
 f 

 f0  2
Summary of Design Formulas
    A 1  e
- j 2

N
  
N
 A  CnN e- j 2 n
Reflection coefficient response
n 0
CnN 
 Z L - Z0 
A2 

Z

Z
0 
 L
-N
A coefficient
 ZL 
ln  Z n 1   2 C ln    ln  Z n 
 Z0 
-N
N
n
1


N


f
4
1 
 2 - cos-1   m  
2 A

f0

 
 


N!
 N - n ! n !
Design of line impedances
Bandwidth
20
Example
Example: three-stage binomial transformer
Given:
Z L  50 []
100   
N 3
Z1
Z2
Z3
50   
 50 -100 
A  2-3 
  -0.0417
 50  100 
1


3
f
4
-1  1  0.05  
 BW 
 2 - cos
 2  0.0417  
f0



 0.713

Z 0  100 []
m  0.05
dB
m  -26.0 [dB]
BW  71.3%
21
Example (cont.)
 ZL 
ln  Z n1   ln  Z n   2 C ln  
 Z0 
-N
Z1 :
N
n
 50 
ln  Z1   ln  Z 0   2-3 C03 ln 
  4.519
 100 
 Z1  91.7 []
Z2 :
 50 
ln  Z 2   ln  Z1   2-3 C13 ln 
  4.259
 100 
CnN 
N!
 N - n ! n !
C 30 = 1
C13 = 3
C 32 = 3
C 33 = 1
 Z 2  70.7 []
Z3 :
 50 
ln  Z 3   ln  Z 2   2-3 C23 ln 
  3.999
 100 
 Z 3  54.5[]
22
Example (cont.)
Using the table in Pozar we have:
ZL / Z0  2 :
 Z1, Z2 , Z3  / Z0  1.0907, 1.4142,1.8337
(The above normalized load impedance is the reciprocal of what we actually have.)
Hence, switching the load and the source ends, we have
 Z1, Z2 , Z3  / Z0  1.8337, 1.4142, 1.0907;
Z0  50[]
Therefore
Z1  91.685 []
Z 2  70.711 []
Z 3  54.585 []
23
Example (cont.)
g1 / 4
50 line
Z1
g 2 / 4
Z2
g 3 / 4
Z3
100 line
ZL
Z0

S11 dB  20 log10   f 
-26
3.29 GHz
f0  5.0GHz
6.74 GHz
BW  69.0%
Response from Ansoft Designer
24
Chebyshev Multistage Matching Transformer
Chebyshev polynomials of the first kind:
cos  n cos 1 x  , x  1

Tn  x   
1
cosh
n
cosh
x, x  1



T1  x   x
T2  x   2 x 2  1
For -1  x  1: Tn  x   1
T3  x   4 x 3 - 3 x
For x  1: Tn  x   1
Tn  x   2 xTn -1  x  - Tn -2  x 
We choose the response to be in the form of a Chebyshev polynomial.
25
Chebyshev Multistage Transformer (cont.)
Figure 5.16 (p. 251)
The first four Chebyshev polynomials Tn(x).
26
Chebyshev Multistage Transformer (cont.)
A Chebyshev response will have equal ripple within the bandwidth.
    Ae- jN TN secm cos 
This can be put into a form
involving the terms cos (n ) (i.e.,
a finite Fourier cosine series).

Tn  x 
m
1
B
A
A
B
n=
2
m  A
n=
1
2
n
ni
n=
2
n=
3
g
sin
a
e
cr
x
-1
1
f0 -
f
2
m
f0
 /2
f0 
f
2
 f
 - m  
Note: As frequency decreases, x increases.
27
Chebyshev Multistage Transformer (cont.)
We have that, after some algebra,
T1  sec  m cos    sec  m cos 
T2  sec  m cos    sec 2  m 1  cos 2  -1
T3  sec  m cos    sec3  m  cos 3  3cos   - 3sec  m cos 
Tn  sec  m cos    2  sec  m cos   Tn-1  sec  m cos   - Tn-2 sec  m cos  
T1  x   x
T2  x   2 x 2  1
Hence, the term TN (sec, cos)
can be cast into a finite cosine
Fourier series expansion.
T3  x   4 x 3 - 3 x
Tn  x   2 xTn -1  x  - Tn -2  x 
28
Chebyshev Multistage Transformer (cont.)
Transformer design
    Ae- jN TN  sec m cos 
 2 e- jN  0 cos N  1 cos  N - 2    ....  n cos  N - 2n    .....
From the above formula we can extract the coefficients n (no general formula is given here).
As f  0      0 
  0   ATN  sec  m  
Z L - Z0
Z L  Z0
 Z -Z 
1
 A L 0 
 Z L  Z 0  TN  sec  m 
29
Chebyshev Multistage Transformer (cont.)
At    m
  m    m  A TN  sec  m cos  m   A TN 1  A
 A   m
At   0 :
   0   ATN  sec  m  
 TN  sec  m   0
Z L - Z0
Z L  Z0
 sec  m  1
 A has the same sign as  Z L - Z 0 
Hence
A  sgn  ZL - Z0  m
30
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 .
31
Chebyshev Multistage Transformer (cont.)
Bandwidth
BW 
4
f
 2- m
f0

At f  0 : ATN  sec  m  
Z L - Z0
1  Z - Z  1 Z L - Z0
 TN  sec  m    L 0  
Z L  Z0
A  Z L  Z0  m Z L  Z0
 cosh  N cosh -1  sec  m  
1
 1 Z L - Z0
-1
 sec  m  cosh  cosh 
 m Z L  Z0
 N

 
 
Hence
X -1 1
 ln  X  ;
X 1 2
X 1
1
 1
 Z L  
-1
sec m  cosh  cosh 
ln   


N
2


 m  Z 0  
32
Chebyshev Multistage Transformer (cont.)
 f 

 f0  2
Summary of Design Formulas
    Ae- jN TN secm cos 
  
Reflection coefficient response
1
 1
 Z L  
-1
secm  cosh  cosh 
ln    


N
2


 m  Z0   
A  sgn  ZL - Z0  m
A coefficient
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 ).
f
4
 2 - m
f0

m term
Design of line impedances
Bandwidth
33
Example
Example: three-stage Chebyshev transformer
Given
Z L  100[]
50[]
Z1
Z2
Z3
100[]
Z0  50[]
m  0.05
Assumed symmetry : Γ3 = Γ0 , Γ 2 = Γ1
A  sgn  Z L - Z 0   m  A  0.05
N  3      A e- j 3 T3  sec  m cos  
 Ae- j 3 sec3  m  cos 3  3cos   - 3sec  m cos  
 2 e- j 3  0 cos 3  1 cos  
(finite Fourier cosine series form)
Equate
34
Example: 3-Section Chebyshev Transformer
Equating coefficients from the previous equation on the last slide, we have
2  0  A sec3  m
2 1  3 A sec3  m - 3 A sec  m
1
1
 100   
ln 
  
2
0.05
50



 


Also, sec  m  cosh  cosh -1 


3
1.408
  m  44.7 o  0.780[rad]  BW  1.007 100.7% 
BW 
 0 
1 
1
3
 0.051.408 
2

f
4
 2 - m
f0

  0  3  0.0698

1
3
3  0.05 1.408  - 3  0.05 1.408 
2
 1   2  0.1037
35
Example: 3-Section Chebyshev Transformer
Next, use
Z n 1 - Z n
 n
Z n 1  Z n
 Z n 1  Z n
1   n 
1-  n 
1  0.0698   57.5 
 
1- 0.0698 
1  0.1037   70.8 
Z 2  57.5
 
1- 0.1037 
1  0.1037   87.2 
Z 3  70.8
 
10.1037


 Z1  50
Z1  57.5   
Z 2  70.8   
Z 3  87.2   
Checking consistency :
Z 4  Z L  87.2
1  0.0698  100.3 
 
1- 0.0698 
36
Example: 3-Section Chebyshev Transformer
Alternative method:
Z n 1 - Z n 1  Z n 1 
n 
 ln 
  ln  Z n 1   ln  Z n   2 n
Z n 1  Z n 2  Z n 
 ln  Z1   ln  Z 0   2 0
 ln  50   2  0.0698
 4.051
 Z1  57.49 
ln  Z 2   ln  Z1   21
 4.259
 Z 2  70.74 
ln  Z 3   ln  Z 2   2 2
 4.466
 Z 3  87.05 
37
Example: 3-Section Chebyshev Transformer
From Table   m  0.05, N  3, Z L / Z 0  2, Z 0  50   
 Z1  1.1475  50  57.4   
Z 2  1.4142  50  70.7   
Z 3  1.7429  50  87.1   
38
Example: 3-Section Chebyshev Transformer
g1 / 4
50 line
Z1
g 2 / 4
Z2
g 3 / 4
Z3
100 line
ZL
Z0

S11 dB  20 log10   f 
-26
2.51 GHz
7.5 GHz
f0  5.0GHz
BW  99.8%
Response from Ansoft Designer
39
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.
40
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.
41