case study: parallel coupled- line combline filter...edition, by michael steer. scitech publishing,...

MICROWAVE AND RF DESIGNMICROWAVE AND RF DESIGN

Based on material in Microwave and RF Design: A Systems Approach, 2nd

Edition, by Michael Steer. SciTech Publishing, 2014.

Presentation copyright Michael Steer

Case Study:Parallel Coupled-Line Combline Filter

Presented by Michael Steer

Reading: § 16.1–16.4

Index: CS_PCL_Filter

Case Study: Parallel Coupled-Line Combline Filter

Copyright 2013 M. Steer and IETADA

C b C bC 1 C 3

50

50 t2COutput

1 2

Input

40

11 S 21

60

20

10

30

S50

S21

(dB

)

0

4

8

12

16

20

11S

(dB

)

240.5 1.0 1.5

0

Frequency (GHz)

1

Specifications● Bandpass filter● Center frequency: 1 GHz● 10% Bandwidth● Steep filter skirts

– requires Chebyshev response, choose a ripple factor of 0.1

● Low loss in passband– Microstrip technology

(Also low fabrication cost and very good performance.)

2

Microwave filter design

● Combination of Art and Science● Art: knowing the structures that intrinsically

have the desired response.● Science: knowing how to use mathematics in a

synthesis process to obtain the required tailoring of the response.

3

Art: choice of topologyParallel microstrip lines in combline configuration.

0

dB

12111 30-80

-60

-40

-20

21S

Frequency (GHz)

4

50

50 Output

1 2

Input

Science: synthesis procedure

50

50

Output

1 2

Input

5

Vg

L 11L 21

C 21

C31

L 31

11C

11

How to go from

to

using mathematical synthesis,while maintaining desired electrical characteristics.

(perhaps more complicated)

Optimization: an alternative to synthesis

● Optimization given a final structure that almost has the right response, use optimization to get the exact final response.

● E.G. Adjust line widths and lengths; number of microstrip lines, capacitor values.

● Works if the starting solution is very close.

● Does not provide insight or lead to new solutions.

● Even then, optimization with more than 6 variables is a problem.

6

50

50 Output

1 2

Input

Summary● Filter design, as with most RF design, is a combination

of art and science.

● The art is identifying the structures that intrinsically have the desired response.

● The science is developing the mathematical procedure to go from the mathematical specification of the desired response to the final microstrip realization.

● Choose topology (art), use synthesis procedure (science), use optimization to almost perfect design, use fabrication and test to perfect design.

7

Case Study: Parallel Coupled-Line Combline Filter. Part B

8

Filter design is based on circuit transformations.

Vg

L11L 21

C21

C31

L31

11C

11

Outline● Begin with a lumped element filter.

● Consider circuit model of coupled lines● Work out the steps to go from the lumped

element circuit to a transmission line-based circuit.

9

Z0

Vg

L12 C12

L22 C21

L32 C32

Z0Vg

L11L 21

C21

C31

L31

11C

11 or

Third-order filter● The lumped element filter has three

LC resonators:

● So (perhaps) the transmission line equivalent has three resonators:

● Consider as two pairs of coupled lines:

Vg

L 11L 21

C 21

C31

L 31

11C

11

1

2

2

1

10

Network model of a pair of coupled lines

1

2

4

3

1: n

4

3

:1n

Z

1

2

02

01

Z

The equivalent circuit of a pair of coupled lines

Is obtained by equating symbolic ABCD equations

11

Combline section and network models

1

2

4

3

Combline section

1: n n

4

3

:1

1

201Z

02Z

1: n n

4

3

:1

1

201Z

02Z

12

Combline section and network models

1

2

4

3

Combline section

1: n n

4

3

:1

1

201Z

02Z

2

1: n

Z 02

Z 012

1

( fr = f0 ) Z1: n

02

Z011 2

Z

Z

012

1 2Z 011 022

13

Translation of a circuit with stubs to coupled lines

1

2

4

3

Z

Z

012

1 2Z 011 022

So if the following structure is seen in a circuit (a Pi arrangement of shorted stubs)

Then it can be replaced by a combline section

14

Comparison of lumped element filter and Pi arrangement of stubs

Z

Z

012

1 2Z 011 022

Pi arrangement of shorted stubs.

Three connected resonators.Vg

L 11L 21

C 21

C31

L 31

11C

11

Each stub is a resonator.

A shorted stub corresponding to a parallel LC resonator.

An open circuit stub corresponds to a series LC resonator.

15

So the conversion from a lumped element filter to Pi network of shorted stubs is not direct.

But The Idea is Starting to Come Through

Summary● The key idea is to begin with a lumped element filter

prototype and put the circuit in the form of a collection of shorted stubs in a PI configuration.

16

Z

Z

012

1 2Z 011 022

Want basic circuit structure to be

But cannot start from here (3rd order BPF)

(Model of two PCL in combline configuration.)

Case Study: Parallel Coupled-Line Combline Filter. Part C, Step 1:

Develop Lowpass Prototype Filter

17

L21

C11 C31Vg

11

Outline● Begin with a lumped element filter.

● Calculate element values.

18

L21

C11 C31Vg

11

19

TRANSMISSION2 REFLECTION2

This is the 6th order Chebyshev response

ε is called the ripple factor.

Passband ripple,PBR = (1+ ε2)

Ripple in dB,RdB = 10 log(PBR)

Steeper filter skirt for• Higher order• Larger ripple

19

Third-order Chebyshev filter

Coefficients of a Chebyshev lowpass prototype filter normalized to a radian corner frequency of ω0 = 1 rad/s and a 1 Ω system impedance (i.e., g0 = 1 = gn+1).

The ripple factor is ε. ε = 0.1 is a ripple of 0.0432 dB.

ω0 is the radian frequency at which the transmission response of a Chebyshev filter is down by the ripple. Here ω0 = 1 radian/s.

20

21

L21

C11 C31Vg

11

A third-order Chebyshev lowpass filter prototype

g0 = 1g1 = 0.85158g2 = 1.10316g3 = 0.85158g4 = 1

C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F

ω0 = 1 rad/s

21

22

Chebyshev filter coefficients from recursive formula

22

Summary● Step 1: developed 3rd order Chebychev lowpass

prototype filter.

23

L21

C11 C31Vg

11

C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F

ω0 = 1 rad/s

Case Study: Parallel Coupled-Line Combline Filter. Part D, Step 2:

Remove Series Inductor

24

L21

C11 C31Vg

11

C11 C21 C31

111 1

Outline

● Use an inverter(s) to replace series inductor.

● An inverter can be implemented using transmission lines.

● Where there are transmission lines it may be possible to equate them to an inverter (if they are /4 long).

25

Inverters

K

ZL

Zin

K

= K0Z

= K0ZinZ

LZImpedanceinverter

A /4 long line is an inverter

2

inL

KZZ

An inverter can be realized using a transmission line.

Wherever there are /4 long transmission lines an impedance inverter can be realized (probably).

Lossless telegrapher’s equation:

0in 0

0

tan( )tan( )

L

L

Z jZZ ZZ jZ

/ 2; tan( )

0in 0

tan( )tan( )L

jZZ ZjZ

20

inL

ZZZ

26

Consider combline section and network model

1

2

4

3

Combline section

1:n n

4

3

:1

1

201Z

02Z

A combline section of coupled lines /4 long inherently presents two impedance inverters.

27

Replacement of a series inductor by a shunt capacitor plus inverters

L

C

-1:1

KK

Equivalence is demonstrated using ABCD parameters.

1T

0 1L

sL

K

L

2

0T

/ 0jK

j K

1

1 0T

1sC

C

-1:1

3

1 0T

0 1

For the cascade2

CASCADE 2 1 2 3

0 1 0 0 1 0 1T T T T T

/ 0 1 / 0 0 1 0 1jK jK sCK

j K sC j K

2L CK

sL

28

Equivalence of a series inductor and a shunt capacitor plus inverters

L

C

-1:1

KK 2L CK

29

C KK

Drop negative unity transformer as it only affects phase and not filter response.

Inverter form of lowpass prototype filter

L21

C11 C31Vg

11

C11 C21 C31

111 1

C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F

ω0 = 1 rad/s C11 = 0.85158 FC21 = 1.10316 FC31 = 0.85158 F

ω0 = 1 rad/s

30

31

+1

( odd)

( even)

g2 g4

g1 g3

gn

g0

Vg

gn n

gn n

1

Vg

g1 g2 g3 g4

1 1 1 1 1

-1:1 -1:1 -1:1

1

Vg

g1 g2 g3 g4

1 1

g5

1 1 1

Ladder prototype filters using impedance inverters

31

Summary: Inverter form of lowpassprototype filter

C11 C21 C31

111 1

C11 = 0.85158 FC21 = 1.10316 FC31 = 0.85158 F

ω0 = 1 rad/s

32

Case Study: Parallel Coupled-Line Combline Filter. Part E, Step 3:

Bandpass Transformation

33

C11 C21 C31

111 1

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

34

First lumped element transformation to BPF, 1 GHz

L21

C11 C31Vg

11

Z 0

Vg

L12 C12

L 22 C 21

L32 C 32

Z 0

34

+11(rad/s)

(s)T 2

(rad/s)1

T s( ) 2

0 2

LPF HPF

35

BPF and center frequency transformation

1 11 22 2

1 10 1 22 2

0 1 2

1112 12

0 11 0

950MHz, 1050MHz

1000 MHzfractional bandwidth,

1, CC LC

R

dB

This assumes that the LPF corner frequency is 1 radian/s.

L21

C11 C31Vg

11

Z0

Vg

L12 C12

L22 C21

L32 C32

Z0

36

1

Vg

L12 C12

L22 C22

L32 C32

1

Transformation to BPF, 1 GHz

C12 = 1.35533 nF = C31L12 = 18.6894 pH = L32C22 = 14.4271 pFL22 = 1.75573 nH

10

20

30

40

60

50

S214

8

12

16

24

20

11S(d

B)

21S(d

B) S11

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

0

Frequency (GHz)

0

36

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

37

BPF and center frequency transformation

C11 C21 C31

111 1

1 11 22 2

1 10 1 22 2

0 1 2

/ /111 1

0 11 0

950MHz, 1050MHz

1000 MHzfractional bandwidth,

1, CC LC

R

dB

This assumes that the LPF corner frequency is 1 radian/s.

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

38

Prototype BPF and center frequency transformation

C1 = 1355.33 pF = C3

L1 = 0.0186894 nH = L3

C2 = 1755.73 pF

L2 = 0.0144271 nH

/

/

/

/

/

/

Z

Z

012

1 2Z 011 022

Recall: desired basic circuit structure

(Model of two PCL in combline configuration.)

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

39

Summary prototype BPF

C1 = 1355.33 pF = C3

L1 = 0.0186894 nH = L3

C2 = 1755.73 pF

L2 = 0.0144271 nH

/

/

/

/

/

/

Case Study: Parallel Coupled-Line Combline Filter. Part F, Step 4:

Impedance Scaling

40

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

||

Vg

505050 50

LC C L C L1 1 2 32 3|| || || || ||

Principle of impedance scaling

● Every impedance in the circuit is scaled by the same amount

● So to go from 1 to 50 – The value of a resistor is increased by a factor of 50.– The value of an inductor is increased by a factor of 50.– The value of a capacitor is reduced by a factor of 50.– The value of an impedance inverter is increased by a factor

of 50.– The value of an admittance inverter is reduced by a factor of

50.

41

Summary, Step 4: BPF scaled to 50 .

' ' ' ' ' '

Vg

111 1

LC C L C L1 1 2 32 3

||

Vg

505050 50

LC C L C L1 1 2 32 3|| || || || ||

C1 = 1355.33 pF = C3

L1 = 0.0186894 nH = L3

C2 = 1755.73 pF

L2 = 0.0144271 nH

/

/

/

/

/

/

C1 = 27.1066 pF = C3

L1 = 0.934468 nH = L3

C2 = 35.1147 pF

L2 = 0.721359 nH

||

||

||

||

42

Case Study: Parallel Coupled-Line Combline Filter. Part G, Step 5:Conversion of Lumped-Element

Resonators

43

LC C0 Z 1Z 01

,

Z 1

Z 01

Outline

● Central idea: Obtain a broadband realization of the LC resonators in the BPF without using inductors.

● Realize the LC resonant circuit by a circuit with C and a stub.

● Equate admittances and the derivatives of admittances

44

LC C0 Z 1Z 01

45

LC Z 1Z 01

Narrowband resonator equivalence at ω0

Z01 is the characteristic impedance of the transmission line and Z1is the input impedance of the shorted transmission line.

2 degrees of freedom. 2 degrees of freedom.

Can only match admittance at one frequency.

45

46

LC C0 Z 1Z 01

,

Z 1

Z 01

Broadband resonator equivalence at ω0Z01 is the characteristic impedance of the line and Z1 is the input impedance of the shorted line.2 degrees of freedom. 3 degrees of freedom.

0

inin 0

at

and at .YY

Broadband match is obtained

by matching

Yin Yin/

r is, the radian resonant frequency of the stub (i.e. the frequency at which it is /4 long).

46

47

LC C0 Z 1Z 01

,

Z 1

Z 01

Broadband resonator equivalence at ω0

2 degrees of freedom. 3 degrees of freedom.

0

inin 0

at

and at .YY

Broadband match is obtained

by matching

Yin Yin/

Specific design choice r 0 (most common).

The admittance of the networks are equivalent(at 0) when:

Also (at 0)Z1 = jZ01

,

Z 1

Z 01The derivatives of the admittance of the networks are equivalent(at 0) when:

47

Z01 is the characteristic impedance of the line and Z1 is the input impedance of the shorted line.

48

The transmission line stubs present impedances Z1 = jZ01, Z2 = jZ02, and Z3 = jZ03 since the resonant frequencies of the stubs are twice that of the design center frequency.

Step 5. Bandpass combline filter with broadband realization of lumped-

element inverters/ / / /

1 3/ /2

01 03

02

21.0881 pF

27.3181 pF7.54713

5.82598

C C

CZ Z

Z

/ // // / Z 01 Z 02 Z 032C1C C 3

50 50

Convert LC resonators to hybrid C‐stub resonators.

• The commensurate frequency, fr, of the design is the resonant frequency of the stubs.

• By default all the stubs have the same fr.

• The design choice here is that fr =2f0. f0 is the center frequency of the design.

48

49

Summary, Step 5

/ // // / Z 01 Z 02 Z 032C1C C 3

50 50

• Broadband, but stubs have different characteristic impedances.

• Really want them to be the same as they will be realized by microstrip lines and we want them to have the same width.(Kind of, this is a little imprecise as the inverters are yet to be realized.)

49

/ / / /1 3/ /2

01 03

02

21.0881 pF

27.3181 pF7.54713

5.82598

C C

CZ Z

Z

Case Study: Parallel Coupled-Line Combline Filter. Part H

Step 6: Equalize Stub Impedances

50

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

Outline● Result of Step 5 (previous):

● Broadband, but want stubs to have the same characteristic impedances.

● Result of this step (Step 6):

/ / / /1 3/ /2

01 03

02

21.0881 pF

27.3181 pF7.54713

5.82598

C C

CZ Z

Z

/ // // / Z 01 Z 02 Z 032C1C C 3

50 50

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

/ / / / / / /1 3 2

/01 03 02

21.0881 pF

7.54713

C C C

Z Z Z

51

52

12

Target combline filter physical layout

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

Very approximately

52

Compare prototype with comblinenetwork model

53

• The impedances of the shunt stubs are mostly determined by the impedances of the individual microstrip lines.

• For manfacturability reasons we would like the microstrip lines to have the same width.

• Therefore we want the shunt stubs to have the same characteristic impedance.

• The impedances of the series stubs are mostly determined by the coupling of the individual microstrip lines.

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

Procedure/ / / /

1 3/ /2

01 03

02

21.0881 pF

27.3181 pF7.54713

5.82598

C C

CZ Z

Z

/ // // / Z 01 Z 02 Z 032C1C C 3

50 50

Want Z02 scaled so that new Z02 = Z01.

00

0y

2 31

J3J1 J J

0

Better to use admittance now as the analysis is based on building a nodal admittance matrix.

54

55

Element values are impedances except for y and y1, which are admittances.

x xy1

y = yx1

2

0 00

J3J1 J J

1 3

y J3J1

0

2 31

j/Jj/J

-j/J

j/J

-j/J

j/J

y = yx1

y1 J3J1

0

1 2 3

j/dj/d

-j/d -j/d

j/dj/d

Inverter impedance scaling

00

0y

2 31

J3J1 J J

0

Admittances are the same

if

Scaled original network

Original network

Procedure is:(A) Develop nodal admittance matrix of original network.(B) Develop nodal admittance matrix of scaled network.Then equate to find required parameters.

A

B

55

Jd = x

56

Realization of a series inductor as a shunt capacitor with 10 inverters.

1 nH

IMPEDANCE OR ADMITTANCEINVERTERS

y 11

1 nF

C

IMPEDANCE INVERTERS

10 10 1C

10 pF

xx

y = yx1

ADMITTANCE INVERTERS

1C= 0.1 S= 0.1 S

JJ

Example

Note the impact on the size of the capacitor!

56

57

Summary, step 6

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

/ / / / / / /1 3 2

/01 03 02

21.0881 pF

7.54713

C C C

Z Z Z

• The stubs now have the same impedance, and the capacitances are the same.

• After scaling so that Z01 = Z02:

57

Case Study: Parallel Coupled-Line Combline Filter. Part I

Step 7: Inverter Realization

58

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

Outline● The prototype filter from Step 6 is

● Realize inverters using stubs.

● Combine adjacent stubs.

● Result of this step (Step 7):

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9084 56.9084

59

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

60

Realization using short‐circuited stubs resonant at twice the passband center frequency.

56.9084

-j -j

j

56.910256.9084

56.9084

j

j

j

Inverter realization using stubs

Impedance inverter

Realization as alumped‐element circuit

Equivalence was established using ABCD parameters.

60

Inverter translation

/ / // / // / Z 01 Z 02 Z 032C1C C 3

56.9102 56.9102

/ / / / / / /1 3 2

/01 03 02

21.0881 pF

7.54713

C C C

Z Z Z

j

j

j

j

-j

Z 01= 56.9102Z x

7.54713 56.9102

= 7.54713 Stubs can be combined.

61

62

j

-j

Z 01= 56.9084Z x

7.54713 56.9084

= 7.54713

-jj 7.54713 56.9084

j 8.70106

j

Z 01= 8.70106

8.70106

Combining stubs

Represent impedance as one stub.

Represent parallel stubs as parallel impedances.

Convert to a single impedance.

62

63

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

Bandpass filter prototype without inverters

/ / / / / / /1 3 2

/01 03

012 023/ /02

21.0881 pF

8.70106 56.9084

10.2715

C C C

Z ZZ Z

Z

fr = 2f0f0 is the center frequency of the design.

Note that in many designs fr = 2f0.This is simply assumed sometimes. But fr could have another relationship.

63


1

2

4

3

Combline section

Z

Z

012

1 2Z 011 022

64

Model:

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

Z

Z

012

1 2Z 011 022

Z

Z

012

1 2Z 011 022

C 1 C32C

Output

1 2

InputWith capacitors:

An issue with resonant frequency

● So the f0’s are different!

● What do we do?

● We need to re-examine the development the lead to the assignment of fr .

1

2

4

3

Combline section

( fr = f0 )

Z

Z

012

1 2Z 011 022

Model:

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

( fr = 2f0 )

Here f0 is the center frequency of the match.

Here f0 is the center frequency of the bandpass filter.

65

Consider exact network model of combline section

1

2

4

3

Combline section

66

There is nothing here that depends on the relationship of fr and f0 .

Z 0 oo21: 2

1: 21: 2

1: 2

V1

V2

2 :1

2 :1 2 :1

2 :1

V4

V3

VX

WV

ZV

YVWI YI

ZIIX

I4

I3I1

I2

Z 0 e e2

Exact model:

Reconsider network models of comblinesection1

2

4

3

Combline section

67

There is nothing here that depends on the relationship of fr and f0 .

Z 0 oo21: 2

1: 21: 2

1: 2

V1

V2

2 :1

2 :1 2 :1

2 :1

V4

V3

VX

WV

ZV

YVWI YI

ZIIX

I4

I3I1

I2

Z 0 e e2

Exact model:

1: n n

4

3

:1

1

201Z

02Z

Approximate model:This is believed to be most accurate when fr = f0 .That is, when the lines are /4 long at the operating frequency.But it is a reasonably good model all frequencies, even when it is /8 long .

Here f0 is the operating frequency

Reconsider simplified network model of combline section

1

2

4

3

Combline section

1: n n

4

3

:1

1

201Z

02Z

2

1: n

Z 02

Z 012

1

( fr = f0 ) Z1: n

02

Z011 2

Z

Z

012

1 2Z 011 022

68

Pretty good model even when fr = f0 (e.g when it is /8 long).


1

2

4

3

Combline section

Z

Z

012

1 2Z 011 022

69

Model:

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

Z

Z

012

1 2Z 011 022

Z

Z

012

1 2Z 011 022

C 1 C3

50

50 2COutput

1 2

Input

With capacitors:

These lines are /8 long at f0.

( fr = 2f0 )

70

Summary, Step 7

70

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

/ / / / / / /1 3 2

/01 03

012 023/ /02

21.0881 pF

8.70106 56.9084

10.2715

C C C

Z ZZ Z

Z

Case Study: Parallel Coupled-Line Combline Filter. Part J

Step 8:Scaling Characteristic Impedances of Stubs

71

t tt

tt

2C

Z0 23Z0 12

C 30Z t 1 0Z t 2 0Z t 3C 1

Outline● From Step 7

● Want the characteristic impedances of the shunt stubs to be between 30 and 80 .

/ / / / / / /1 3 2

/01 03

012 023/ /02

21.0881 pF

8.70106 56.9084

10.2715

C C C

Z ZZ Z

Z

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

72

Desired stub impedances

73

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

• The impedances of the shunt stubs are mostly determined by the impedances of the individual microstrip lines.

• For manfacturability reasons we would like the microstriplines to have reasonable width.

• On Alumina (r around 10) that means that we want the characteristic impedances of the stubs to be between 30 and 80 .

• The impedances of the series stubs are mostly determined by the coupling of the individual microstrip lines.

Scale Impedances

/ / / / / / /1 3 2

/01 03

012 023/ /02

21.0881 pF

8.70106 56.9084

10.2715

C C C

Z ZZ Z

Z

// / // // // / / Z 03C 3Z 022C1C Z 01

Z 012 Z 023

Scale to 80 .

Multiply impedances by a factor of 80/10.2715.

74

Want characteristic impedances of the stubs to be between 30 and 80 .

Scale Impedances/ / / / / / /

1 3 2/

01 03

012 023/ /02

21.0881 pF

8.70106 56.9084

10.2715

C C C

Z ZZ Z

Z

// / // // // / / Z03C3Z022C1C Z01

Z012 Z023

Multiply impedances by a factor of 80/10.2715.

t tt

tt

2C

Z0 23Z0 12

C 30Z t 1 0Z t 2 0Z t 3C 1

1 3 2/

0 1 03

0 12 0 23

0 2

2.70759 pF

67.7683 443.232

80

t t t

t

t t

t

C C C

Z ZZ Z

Z

75

76

Summary, Step 8

76

t tt

tt

2C

Z0 23Z0 12

C 30Z t 1 0Z t 2 0Z t 3C 1

1 3 2/

0 1 03

0 12 0 23

0 2

2.70759 pF

67.7683 443.232

80

t t t

t

t t

t

C C C

Z ZZ Z

Z

Case Study: Parallel Coupled-Line Combline Filter. Part K

Step 9: 50 Match

77

ttt

tt

139.404

139.404

Z0 23Z0 12

C 30Z t 1 0Z t 22C 0Z t 3C 1

78

Use Impedance Inverters

t tt

tt

2C

Z0 23Z0 12

C 30Z t 1 0Z t 2 0Z t 3C 1

Result of Step 8:

78

ttt

tt

139.404

139.404

Z0 23Z0 12

C 30Z t 1 0Z t 22C 0Z t 3C 1

79

Summary, Step 9

79

1 3 2/

0 1 03

0 12 0 23

0 2

2.70759 pF

67.7683 443.232

80

t t t

t

t t

t

C C C

Z ZZ Z

Z

ttt

tt

139.404

139.404

Z0 23Z0 12

C 30Z t 1 0Z t 22C 0Z t 3C 1

Case Study: Parallel Coupled-Line Combline Filter. Part L

Step 10: Implementing the Input/Output Inverters

80

t

tt Z0 23Z0 12

1C 3C

CbCb

0Z t 1 2C 0Z t 2 0Z t 3


● Implement input and output inverters.

81

ttt

tt

139.404

139.404

Z0 23Z0 12

C 30Z t 1 0Z t 22C 0Z t 3C 1

RL

Yin

C

C

a

b

K RL

Yin

An inverter as a capacitor network

K C

C

a

b

These are equivalent but only for resistive loads.

Equate admittances, note that Yin of inverter is real.

This is not the same as general matching which works with complex conjugate impedances.

It is the same as matching If input and output are resistances.

82

An inverter as a capacitor network

K C

C

a

b

These are equivalent but only for resistive loads.

This is can be shown by using a complex load and calculating the input impedance of the capacitive network with a complex load.

83

Derivation capacitor network (at 1 GHz)

84

in 22

50 1 1389.426 1139.540

La

b L

RY sCK sC R

1.06484 pF1.22170 pF

a

b

CC

C

C

a

b

389.426 50

inYLR

85

External inverters as capacitive networks

ttt

tt

139.404

139.404

Z0 23Z0 12

C 30Z t 1 0Z t 22C 0Z t 3C 1

Ca

Cb

Ca

Cb

85

Note that Ca and C1 are in parallel.

86

t

tt Z0 23Z0 12

1C 3C

CbCb

0Z t 1 2C 0Z t 2 0Z t 3

Summary, Step 10

1 1 3

2

0 1 0 3

0 12 0 23

0 2

1.64276 pF

2.70759 pF

1.22170 pF

67.7683 443.232

80

a t

t

b

t t

t t

t

C C C C

C

C

Z ZZ Z

Z

Case Study: Parallel Coupled-Line Combline Filter. Part M

Physical Design of Combline Filter

87

Cb CbC1 C3

50

50 t 2COutput

1 2

Input


● Capacitors stay as lumped-element capacitors● Implement the following in microstrip:

88

t

tt Z0 23Z0 12

1C 3C

CbCb

0Z t 1 2C 0Z t 2 0Z t 3

tt Z0 23Z0 12

0Z t 1 0Z t 2 0Z t 3

Key Concept

89

1

22

1

2

1

• Can treat three coupled lines as two pairsof coupled lines with the center line shared.

• Error is small.

• One transmission path that is missing is direct coupling of the first line to the third line. This coupling is very small.

90

w2w1

L

s1

w3w2

L

s2

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

tt Z0 23Z0 12

0Z t 1 0Z t 2 0Z t 3

tZ0 12

0Z t 1 0Z t 2

t

0Z t 3

Z0 23

0Z t 2

Physical design of the three coupled lines

90

91

w2w1

L

s1

tZ0 12

0Z t 1 0Z t 2

Implement one pair at a time.

91

Equivalent circuits for a combline section.

Z011 Z022

Z012

21

1: n

Z 02

Z01

1

2

1

2

4

3

92

Derivation of parameters

92

Equivalent circuits for a combline section:

Z011 Z022

Z012

21

1: n

Z02

Z01

1

2

012

011

011 02202 012 01

011 022 012

11 7.540 0.1326

3342 and 69.17

Zn KZ n

Z ZZ nZ Z nZ Z Z

From model theory:

93


93

Equivalent circuits for a combline section.1: n

Z02

Z01

1

2

1

2

4

3

Two estimates of coupled line system impedance:2

20 ,1 01 0 ,2 02 2

1 68.56 and 55.80 1

S SKZ Z K Z Z

K

This happened because the shunt stubs in the Pi arrangement of stubs is not symmetrical. So take mean:

0 0 ,1 0 ,2 63.8 S S SZ Z Z

94


94

Equivalent circuits for a combline section.

1: n

Z02

Z01

1

2

1

2

4

3

0

0

0 0

20 0

63.8

1 55.8 1

/ 72.9 S

S

o S

e o

Z

nZ Zn

Z Z Z

Dimensions of microstriplines determined using tables or iteratively solving coupled line equations.

95

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

Physical design of the three coupled lines

95

rh w

Conductor pattern

Stript

0

0

0

63.8 55.8 72.9

S

o

e

ZZZ

Choose alumina substrate with r = 10, h = 635 m.

Use lookup table for a 50 system impedance.

1 2 3

1 2

591 m (600 m rounded)635 m (650 m rounded)

7.24, 5.95

Take 6.56ee eo

e ee eo

w w ws s

96

Physical design

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = g/8 = 14.65 mm

w1

w2

w3

L

via

s

s1

2

Layout (to scale)

Capacitor values are unchanged (e.g. implement using surface‐mount capacitors).

96

0

0

30 cm @ 1 GHz

/g e

(recall fr = 2f0 )

97

Revisit Assumptions

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = 14.65 mm

These values were derived looking up a table for a50 system impedance.

However Z0S = 63.8

97

98

Revisit Assumptions, wCb C1 C3

t 2C Cb

w3w2w1

L

s s21

r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = 14.65 mmThese values were derived looking up a table for a50 system impedance.

Have three system impedances:

20 ,1 01

2

0 ,2 02 2

0 0 ,1 0 ,2

1 68.56

55.80 1

63.8

S

S

S S S

Z Z K

KZ ZK

Z Z Z

This choice mostly affects w1, w2, and w3.

Could optimize in EM simulation, but better to get closer now.

Choose Z0S = 55.8 .98

99

Update w

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

r = 10, h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 650 mL = 14.65 mm

Use Z0S = 55.8 .

99

100

Revisit Assumption, LCb C1 C3

t 2C Cb

w3w2w1

L

s s21

r = 10, h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 1150 mL = 14.65 mm

For e , used geometric mean of even and odd mode effective permittivity (affects L).

100

LC C0 Z 1Z 01

Instead of adjusting L in EM‐based optimization (to get Z1 right) we can tune capacitor (C0) .

101

Summary, physical design

1 3

2

1.64276 pF

2.70759 pF

1.22170 pFt

b

C C

C

C

101

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 650 mL = 14.65 mm

Case Study: Parallel Coupled-Line Combline Filter. Part N

Microwave Circuit Simulation

102

Cb CbC1 C3

50

50 t2COutput

1 2

Input

Outline● Physical Design

● First developed lumped-element BPF reference● Microwave circuit simulation

– Use microstrip coupled line element (MCLIN)– Compare and interpret response– Optimize

103

Cb C1 C3t 2C Cb

w3w2w1

L

s s21

w1

w2

w3

L

via

s

s1

2

104

Z0 = 50 ΩC13 = C33 = 27.107 pFL13 = L33 = 934.47 pHC23 = 288.54 fFL23 = 87.787 nH

Z0

Vg

L13 C13

L23 C23

L33 C33

Z0

Lumped-Element BPF for Reference

10

20

30

40

60

50

S214

8

12

16

24

20

11S(d

B)

21S(d

B) S11

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

0

Frequency (GHz)

0

104

105

0

Frequency (GHz)

60 24

20

10

30

40

50

4

8

12

16

20

11S(d

B)

S 21(d

B)

0.5 6.53.5

11S

S21

8.5

0

Wideband Response

Z0

Vg

L11 C11

L21 C21

L31 C31

Z0

105

106

0.80 GHz

1.20 GHz

0.90 GHz

0.94 GHz

0.92 GHz

1.10 GHz1.08 GHz

1.06 GHz1.02 GHz

0.98 GHz

S11 of the lumped-element BPF

Z0

Vg

L11 C11

L21 C21

L31 C31

Z0

10

20

30

40

60

50

S214

8

12

16

24

20

11S(d

B)

21S(d

B) S11

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

0

Frequency (GHz)

0

106

107

The zeros of the S11response, and hence the poles of the S21response, are at 0.96, 1.00, and 1.04 GHz.

0.80 GHz

1.20 GHz

0.90 GHz

0.94 GHz

0.92 GHz

1.10 GHz

1.08 GHz

1.06 GHz1.02 GHz

0.98 GHz

0.96 GHz1.00 GHz1.04 GHz


Z0

Vg

L11 C11

L21 C21

L31 C31

Z0

107

zeros

108

0.80 GHz

1.20 GHz

0.90 GHz

0.94 GHz

0.92 GHz

1.10 GHz

1.08 GHz

1.06 GHz1.02 GHz

0.98 GHz

0.96 GHz1.00 GHz1.04 GHz


108

S214

8

12

16

24

20

11S(d

B)

S11

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5Frequency (GHz)

0

zeros

zeros

109

Details: 6 μm gold metallization

Cb CbC1 C3

50

50 t2COutput

1 2

InputCb CbC1 C3

50

50 t 2COutput

1 2

MCLIN

Input

1 2 3

Circuit model using MCLIN element

1 3

2

1.64276 pF

2.70759 pF

1.22170 pFt

b

C C

C

C


109

110

Response with MCLIN element

10

20

30

40

60

50

S214

8

12

16

24

20

11S(d

B)

21S(d

B)

S11

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

0

Frequency (GHz)

0

Response of lumped‐element BPF

110

(a) s1 = s2 = 650 m (b) s1 = s2 = 1150 m

20

10

30

40

50

4

8

12

16

20

11S(d

B)

S 21(d

B)

1.060 24

0

0.5 1.5

0

Frequency (GHz)

S21(a)

S21(b)

S11(b)

1.06 GHz

0.88 GHz

1.20 GHz

1.12 GHz

1.08 GHz

1.10 GHz

1.04 GHz

1.00 GHz0.98 GHz

0.96 GHz

0.94 GHz 0.92 GHz

0.90 GHz

1.02 GHz

0.80 GHz

111

S11 response with MCLIN element

Locus gets close to origin twice.

11S(d

B)

4

8

12

16

20

S11

0

0.5 1.0 1.5Frequency (GHz)

24

Frequency (GHz)

40

11 , MCLIN

S 21, Lumped

S21 , MCLIN60

20

10

30

S50

S21

(dB

)

0

4

8

12

16

20

11S

(dB

)

240.5 1.0 1.5

0

112

1 3

2

1.22170 pF

1.64276 pF

2.70759 pF

b

t

C

C C

C


1 3

2

1.9076 pF

2.8748 pFt

C C

C

Optimized S11 response with MCLIN element

Values for optimized response. Account for error in L.

Cb CbC1 C3

50

50 t 2COutput

1 2

Input

L

Frequency (GHz)

40

11 , MCLIN

S 21, Lumped

S21 , MCLIN60

20

10

30

S50

S21

(dB

)

0

4

8

12

16

20

11S

(dB

)

240.5 1.0 1.5

0S11 response with optimized MCLIN element

Cb CbC1 C3

50

50 t 2COutput

1 2

Input

Path 1Path 2

Path 2 not considered in synthesis.

At 1.2 GHz Path 1 and Path 2 cancel.

At 0.8 GHz Path 1 and Path 2 reinforce.

There is partial reinforcement below 0.9 GHz.

There is partial cancellation above 1.1 GHz.

113

1.06 GHz

0.94 GHz

0.92 GHz

0.90 GHz0.80 GHz

1.20 GHz1.04 GHz

1.02 GHz

1.00 GHz0.98 GHz

0.96 GHz

1.10 GHz1.08 GHz

114

S11 response with optimized MCLIN element

1.06 GHz

0.94 GHz

0.92 GHz0.90 GHz

0.80 GHz

1.20 GHz1.04 GHz1.02 GHz

1.00 GHz0.98 GHz

0.96 GHz

1.10 GHz1.08 GHz

115

Comparison of S11 response

0.80 GHz

1.20 GHz

0.90 GHz

0.94 GHz

0.92 GHz

1.10 GHz

1.08 GHz

1.06 GHz1.02 GHz

0.98 GHz

Lumped BPF BPF with optimized MCLIN

1.06 GHz

0.94 GHz

0.92 GHz0.90 GHz

0.80 GHz

1.20 GHz1.04 GHz1.02 GHz

1.00 GHz0.98 GHz

0.96 GHz

1.10 GHz1.08 GHz

116

S11 response with with optimized MCLIN

40

11 , MCLIN S21 , MCLIN60

20

10

30

S50

S21

(dB

)

0

4

8

12

16

20

11S

(dB

)

240.5 1.0 1.5

0

During manual tuning look at both rectangular S11 plot and Smith chart plot.

117

0

Frequency (GHz)

60 24

20

10

30

40

50

4

8

12

16

20

11S(d

B)

S 21(d

B)

0.5 6.53.5

11S

S21

8.5

0

Wideband response of lumped-element BPF

Z0

Vg

L11 C11

L21 C21

L31 C31

Z0

117

10

30

40

50

4

8

12

16

20

11S

(dB

)

S21

(dB

)

0.5 6.53.5

S 21

0

8.5

0

Frequency (GHz)

11S

60 24

20

118

Optimized S11 response with MCLIN element

Cb CbC1 C3

50

50 t 2COutput

1 2

Input

Recall that fr = 2f0.So transmission lines look the same atfr , 3fr , 5fr …, i.e. f0 , 6f0 , 10f0 …

BUT impedance of capacitance is not the same, hence the spurious passbands are shifted.

Spurious basebands at f0 , 4f0 , 7.5f0 , …

fr = 2f0

10

30

40

50

4

8

12

16

20

11S

(dB

)

S21

(dB

)

0.5 6.53.5

S 21

0

8.5

0

Frequency (GHz)

11S

60 24

20

119

Effect of higher fr ?

Cb CbC1 C3

50

50 t 2COutput

1 2

InputWhat if fr = 3f0?

Spurious passbandswould be shifted higher in frequency. BUT diminishing returns, design becomes more sensitive.

fr = 2f0

Frequency (GHz)

40

11 , MCLIN

S 21, Lumped

S21 , MCLIN60

20

10

30

S50

S21

(dB

)

0

4

8

12

16

20

11S

(dB

)

240.5 1.0 1.5

0

120


1 3

2

1.22170 pF

1.9076 pF

2.8748 pF

b

t

C

C C

C

Summary, optimized physical design

Cb CbC1 C3

50

50 t 2COutput

1 2

Input

LOnly adjusted C1, Ct2, and C3.

Case Study: Parallel Coupled-Line Combline Filter. Part O

EM Simulation

121

Cb CbC1 C3

50

50 t2COutput

1 2

Input

400 μm× 400 μm tantalum vias6 μm gold metallization. EM enclosure has perfect conducting walls withXDIM = 22 mm, YDIM = 20 mm and height = 5.635 mm.

Cb CbC1 C3

50

50 t2COutput

1 2

Input

Cb CbC1 C3

50

50 t 2COutputInput

EM Subcircuit

231

1

3

2

w1

w2

w3

XDIM

YDIM

L

via

Enclosure

s

s1

2

122


1 3 2

1.22170 pF

1.9076 pF , 2.8748 pFb

t

C

C C C

Use optimized MCLIN‐based BPF values

S 21

(dB

)

10

20

30

40

60

50

0

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

4

8

12

16

24

20S11, EM

S21, MCLIN

S21, EM 11S(d

B)

Frequency (GHz)

0

123

Comparison of responses

Could further optimize . . .

S 21

(dB

)

10

20

30

40

60

50

0

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

4

8

12

16

24

20S11, EM

S21, MCLIN

S21, EM 11S(d

B)

Frequency (GHz)

0

124

Comparison of responses

• Bandwidth is smaller• Indicates lower overall coupling

• Notch above passband has shifted lower.

• Overall response is almost the same as with MCLIN‐based analysis

• Perhaps slight mismatch at center of passband.

• Use MCLIN‐based analysis to optimize design.

• Gridding in EM analysis (50 m used here could have resulted in EM analysis differences).

• Some subtle effects are captured in EM Simulation not in MCLIN analysis

• E.G. via coupling.

S11 response on a Smith chart

EM Analysis

0.80 GHz

1.00 GHz

0.98 GHz

0.96 GHz

0.94 GHz

0.92 GHz

1.08 GHz

0.90 GHz

1.06 GHz

1.20 GHz

1.04 GHz

1.02 GHz

1.10 GHz

1.06 GHz

0.94 GHz

0.92 GHz

0.90 GHz0.80 GHz

1.20 GHz1.04 GHz

1.02 GHz

1.00 GHz0.98 GHz

0.96 GHz

1.10 GHz1.08 GHz

126

S11 response (optimized)

MCLIN Analysis

S11 response (optimized)

MCLIN Analysis

EM Analysis

127

128

S11 response on a Smith chart

S 21

(dB

)

10

20

30

40

60

50

0

1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5

4

8

12

16

24

20S11, EM

S21, MCLIN

S21, EM 11S(d

B)

Frequency (GHz)

0

129

Wideband response

Frequency (GHz)

60 24

20

10

30

40

50

4

8

12

16

20

11S(d

B)

S 21(d

B)

0.5 6.53.5

S21

11S

0

8.5

0

Same as with MCLIN‐based analysis

130

Manufactured filter considerations

Cb CbC1 C3

50

50 t 2COutput

1 2

Input

L

w1

w2

w3

XDIM

YDIM

L

via

Enclosure

s

s1

2

It will be necessary to tune every filter manufactured.

Fabrication tolerances are about 1%.

Greater accuracy than that is required.

Tuning done by adjusting capacitor values.

131

Summary, Parallel Coupled-Line Combline Filter

Cb CbC1 C3

50

50 t2COutput

1 2

Input

L

Filter synthesized using a methodical process.

Microwave simulation required to optimize design.

EM simulation as a check as there are coupling mechanisms that cannot be captured otherwise.

Every filter manufactured will require tuning.

case study: parallel coupled- line combline filter...edition, by michael steer. scitech publishing,...

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