Notes 18 ECE 5317-6351 Microwave Engineering Fall 2011 Multistage Transformers Prof. David R. Jackson Dept. of ECE 1

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 Z

Z Z

1

1

-LL

L

Z Z

Z Z

0 0 1 011 22

1 0

0 10 0 0021 12 11

1 1 0

--

21

Z ZS S

Z Z

Z ZZS S S

Z 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 eS

S 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 S

Z Z Z Z

22 2 2 22 22 1 0 1 0 0 10 1 01 0 1 0 0 1

11 2 2 21 0 1 0 1 0 1 0

22 2 2 21 0 0 1 1 0 0 1

2

1 0

0 12

1 0

2- 21 1 1 1

2 2

4

Z Z Z Z Z ZZ ZZ Z Z Z Z ZS

Z Z Z Z Z Z Z Z

Z Z Z Z Z Z Z Z

Z Z

Z Z

Z Z

Hence

For the numerator:

Next, consider this calculation:

4

1

1

- 2011

- 20111

jL

jL

S e

S e

Putting both terms over a common denominator, we have

1 1

1

- 2 - 20 0 2 0 211 11 11

- 2011

1

1

j jL L

jL

S S e S e

S e

or


1

1

- 20 2110

11 - 2022

1

1 -

jL

jL

S eS

S e

We then have

5

011 1LS Assuming small reflections

1

1

- 2011

- 20111

jL

jL

S e

S 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 Z

Z Z Z Z

L

1je

1je

00 11S


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 e

Z Z

Z 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


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


Hence, for symmetric reflections we then have

Note that this is a finite Fourier cosine series.

10


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 cosNNA

0

2

02 2

0 1,2, ..., -1

i

n

n

f f

dn N

d

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 C

N 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 Z

Z ZA

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:


- 0

0

-2 N L

L

Z ZA

Z Z

Hence

-1 0

1 0

- -2 N Nn n L

nn n L

Z Z Z ZC

Z 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


Hence

!

- ! !Nn

NC

N n n

! !

- ( ) ! ( )! ! ( )!N NN n n

N NC C

N 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 .


16

Example showing a microstrip line


A three-stage transformer is shown.

50 line100 line

1 / 4g2 / 4g

3 / 4g

1Z 2Z 3Z

0Z LZ

17


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 1ln ; 1

1 2

XX X

X

-1 0

1 0

- -2 N Nn n L

nn n L

Z Z Z ZC

Z Z Z Z

-1

0

1 1ln 2 ln

2 2N Nn L

nn

Z ZC

Z Z

-1

0

ln 2 ln lnN N Ln n n

ZZ C Z

Z

recursive

relationship


Hence

Recall

19

Bandwidth

1

-1 1cos

2

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 m

f f ff

f f f A

The bandwidth is then:mf 0f 02 - mf f

m / 2 - m

m


Maximum acceptable reflection

1

-1

0

4 12 - cos

2

Nmf

f A

Hence

/ 2f / 2f

20

Summary of Design Formulas


1

-1

0

4 12 - cos

2

Nmf

f A

-1

0

ln 2 ln lnN N Ln n n

ZZ C Z

Z

- 0

0

-2 N L

L

Z ZA

Z 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 2

f

f

!

- ! !Nn

NC

N n n

21

Example: three-stage binomial transformer

0

50 [ ]

100 [ ]

0.05

L

m

Z

Z

-26.0 [dB]dBm -3

1

3-1

0

50 -1003 2 -0.0417

50 100

4 1 0.052 - cos

2 0.0417

0.713

N A

fBW

f

100 50 1Z 2Z 3Z

Example

Given:

71.3%BW

22

-3 31 0

1

01

50ln ln 2 ln 4.519

10

91.

0

[

:

7 ]

Z Z C

Z

Z

-3 32 1

2

12

50ln ln 2 ln 4.259

10

70.

0

[

:

7 ]

Z Z C

Z

Z

-3 33 2

3

23

50ln 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

NC

N n n

-1

0

ln ln 2 lnN N Ln n n

ZZ Z C

Z

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: 1

n

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 -2

ff

0 2

ff

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


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 Z

Z Z

Z ZA

Z Z T

f

AT

As

Transformer design


From the above formula we can extract the coefficients n (no general formula is given here).

30

sec cos 1

m

m m N m m

m

NA T A T

A

A

At


0sgn -L mA Z Z

Hence

0

0

0

0 :

-0 sec

sec 0 sec 1

-

LN m

L

N m m

L

Z ZAT

Z Z

T

A Z Z

At

has the same sign as

31


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 Z

N Z Z

fW

f

N

B

Z

At

Chebyshev Multistage Transformer (cont.)Bandwidth

-1

0

1 1sec cosh cosh ln

2L

mm

Z

N Z

Hence -1 1

ln ; 11 2

XX X

X

33

Summary of Design Formulas

0

42 - m

f

f

Reflection coefficient response

A coefficient

Design of line impedances

Bandwidth


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 ln

2L

mm

Z

N Z

m term

0 2

f

f

34

0

100[ ]

50[ ]

0.05

L

m

Z

Z

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

3

0 0 3

3

1

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

10.05 1.408 0.0698

2

13 0.05 1.408 -3 0.05 1.408

20.1

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 - m

fBW

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.

1

1-

9

-

06 8

n nn

n

nn n

n

L

n

Z Z

Z 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

- 1ln

2ln n

4

l 2

.466

nn n n

nn

nn

nn

Z Z

Z Z

Z Z

Z

Z

Z

Z

Z

Z ZZ Z

Z

Z

Alternative method:


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


39


-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


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.

Documents

Notes 18 ECE 5317-6351 Microwave Engineering Fall 2011 Multistage Transformers Prof. David R. Jackson Dept. of ECE 1