By

Professor Syed Idris Syed Hassan

Sch of Elect. & Electron Eng

Engineering Campus USM

Nibong Tebal 14300

SPS Penang

Microwave Circuit Design

Introduction

• 10 weeks lecture + 4 weeks ADS simulation

• Assessments :8 tests + 2 ADS assignments + 1 final examination

• Class : 9.00- 10.30 lecture

10.30-11.00 rest (tea break)

11.00-12.30 lecture

12.30- 1.00 test

Dates

• 06/04/02 Morning• 20/04/02 Morning• 27/04/02 Morning• 04/05/02 Morning• 11/05/02 Morning• 18/05/02 Morning• 25/05/02 Morning

• 08/06/02 Morning• 15/06/02 Morning• 22/06/02 Morning• 29/06/02 morning• 06/07/02 Morning• 20/07/02 Morning• 27/07/02 Morning

Syllabus • Transmission lines• Network parameters• Matching techniques• Power dividers and combiners• Diode circuits• Microwave amplifiers• Oscillators• Filters design• Applications• Miscellaneous

References

• David M Pozar ,Microwave Engineering- 2nd Ed., John Wiley , 1998

• E.H.Fooks & R.A.Zakarevicius, Microwave Engineering using microstrip circuits, Prentice Hall,1989.

• G. D. Vendelin, A.M.Pavio &U.L.Rohde, Microwave circuit design-using linear and Nonlinear Techniques, John Wiley, 1990.

• W.H.Hayward, Introduction to Radio Frequency Design, Prentice Hall, 1982.

Transmission Line

Equivalent CircuitR RL L

C

G

z

C

LZo

CjG

LjRZo

Lossy line

Lossless line

CjGLjR

LC

AnalysisLjR ),( tzI

dz

dIzI

),( tzV CjG dz

dVzV

zdt

dILzRIz

dz

dV

dt

dILRI

dz

dV

zdt

dVCzGVz

dz

dI

dt

dVCGV

dz

dI

From Kirchoff Voltage Law Kirchoff current law

zdt

dVCzGVz

dz

dIII

z

dt

dILzRIz

dz

dVVV

(a) (b)

Analysis

Let’s V=Voejt , I = Ioejt

Therefore

Vjdt

dV Ijdt

dI then

ILjRdz

dV VCjGdz

dI a b

Differentiate with respect to z

dz

dILjR

dz

Vd 2

2

VCjGLjRdz

Vd 2

2

Vdz

Vd 22

2

dz

dVCjG

dz

Id 2

2

ICjGLjRdz

Id 2

2

Idz

Id 22

2

AnalysisThe solution of V and I can be written in the form of

z zBe Ae V o

z z

Z

Be AeI

where

CjG

LjRZo

Let say at z=0 , V=VL , I=IL and Z=ZL

Therefore

B A VL o

LZ

B AI

and LL

L ZI

V

e f

c d

CjGLjRj and

AnalysisSolve simultaneous equations ( e ) and (f )

2o L LZ I V

A

2

o L LZ I VB

Inserting in equations ( c) and (d) we have

22)(

zz

oL

zz

Lee

ZIee

VzV

22)(

zz

o

Lzz

Lee

Z

VeeIzI

Analysis

2)cosh(

zz eez

But and

2)sinh(

zz eez

Then, we have

)sinh()cosh()( zZIzVzV oLL

)sinh()cosh()( zZ

VzIzI

o

LL

)sinh()cosh(

)sinh()cosh(

)(

)()(

zZV

zI

zZIzV

zI

zVzZ

o

LL

oLL

and

*

**

Analysis

)sinh()cosh(

)sinh()cosh()(

zZzZ

zZzZZzZ

Lo

oLo

)tanh(

)tanh()(

zZZ

zZZZzZ

Lo

oLo

Or further reduce

or

For lossless transmission line , = jsince

)tan(

)tan()(

zjZZ

zjZZZzZ

Lo

oLo

)cos()cosh( zzj

)sin()sinh( zjzj

AnalysisStanding Wave Ratio (SWR)

node

antinode

Ae-zBez

z

z

Ae

Be

1zzzL AeBeAeV

1o

z

o

zz

L Z

Ae

Z

BeAeI

Reflection coefficient

Voltage and current in term of reflection coefficient

oL

LL Z

I

VZ

1

1

1

1

o

L

Z

Zor

AnalysisFor loss-less transmission line = jBy substituting in * and ** ,voltage and current amplitude are

2/12)2cos(21)( zAzV

2/12)2cos(21)( z

Z

AzI

o

Voltage at maximum and minimum points are

)1(max AV )1(min AVand

1

1sVSWRTherefore

For purely resistive load o

L

Z

Zs

g

h

Analysis

oL

oL

ZZ

ZZ

oo sZZI

VZ

1

1

min

maxmax s

ZZ

I

VZ o

o

1

1

max

minmin

Other related equations

From equations (g) and (h), we can find the max and min points

,4,2,02 z

,3,2 z

Maximum

Minimum

Important Transmission line equations

tanh

tanh

Lo

oLoin jZZ

jZZZZ

oL

oL

ZZ

ZZ

1

1SWR

ZoZL

Zin

Various forms of Transmission Lines

Two wirecable Coaxial

cable

Microstripeline

Rectangularwaveguide

Circularwaveguide

Stripline

Parallel wire cable

daforadoradL /ln2/cosh 1

daforad

orad

C /ln2/cosh 1

adZo 2/cosh1 1

Where a = radius of conductor d = separation between conductors

Coaxial cable

abC

/ln

2

abL /ln2

abZo /ln2

1

Where a = radius of inner conductor b = radius of outer conductor c = 3 x 108 m/s

r

cc

ckf

2

bakc

2

ab

Micro strip

whe r

t

t=thickness of conductor

Substrate

Conducted strip

Ground

Characteristic impedance of Microstrip line

hwwhZhwFor eeeff

o /25.0/8ln60

1/

444.1/ln667.0393.1/

3771/

hwhwZhwFor

eeeffo

25.0 /104.0/1212

1

2

1hwwh ee

rreff

5.0/1212

1

2

1

err

eff wh

1

2ln

t

htww e

e thhe 2Where

w=width of striph=height and t=thickness

Microstrip width

rr

rroZA

11.023.0

1

1

2

1

60

2/1

roZB

260

2)2exp(

)exp(8/

A

AhW

rr

r BlbBBhW

61.039.01

2

112ln1

2/

For A>1.52

For A<1.52

Simple Calculation

2

377

hw

Z

r

o

2377

/ orZ

hw

Approximation only

Microstrip components

• Capacitance

• Inductance

• Short/Open stub

• Open stub

• Transformer

• Resonator

Capacitance

Zo ZoZoc

1cZC

oc

12sin

2

cZC

oc

smc

r/

103 8

1

For 8

For8

Inductance

Zo ZoZoL

1c

ZL oL

1sin

c

ZL oL

smc

r/

103 8

1

For 8

For8

Short Stub

Zo

Z

Zo

Zo ZL

oL ZX /tan 1eff

o

360

tanoLsc jZXZ

Open stub

Zo

Z

Zo

Zo ZL

oc ZX /cot 1eff

o

360

cotococ jZXZ

Quarter-wave transformer

Zo ZoZT

Zmx/min

ZL

x

oL

oL

ZZ

ZZ

2

x in radian sZxZ omax)(

1

1s

sZsZZZZZ ooomxinT .

At maximum point

Quarter-wave transformer

oL

oL

ZZ

ZZ

2

x in radian

sZxZ o /)( min

1

1s

sZsZZZZZ oooinT //.min

at minimum point

Resonator

• Circular microstrip disk

• Circular ring

• Short-circuited/2 lossy line

• Open-circuited /2 lossy line

• Short-circuited /4 lossy line

Circular disk/ring

a

r

oa

2

841.1

feeding

a

4a

* These components usually use for resonators

Short-circuited/2 lossy line

n/2

Zin Zo

oZR

o

oZL

2

LC

o2

1

22

R

LQ o

2

where

= series RLC resonant cct

Open-circuited /2 lossy line

n/2

Zin Zo

22

RCQ o

= parallel RLC resonant cct

oZR

oo ZC

2

CL

o21

2where

Short-circuited /4 lossy line

/4

Zin Zo

oZR

= parallel RLC resonant cct

oo ZC

4

CL

o21

24

RCQ o

2

where

Rectangular waveguide

ab

22

2

1

b

n

a

mfcmn

Cut-off frequency of TE or TM mode

222111

cog

o

gTEZ

g

oTMZ

mNpabba

Ro

og

sc /2 232

3

Conductor attenuation for TE10

2o

sR

Example

Given that a= 2.286cm , b=1.016cm and xS/m. What are the mode and attenuation for 10GHz?

22

2

1

b

n

a

mfcmn

Using this equation to calculate cutoff frequency of each mode

Calculation

GHzfc9

28

10 10562.6002286.02

103

TE10

a=2.286mm, b=1.016mm, m=1 and n=0 ,thus we have

Similarly we can calculate for other modes

Example

Mode fcmn

TE10 6.562 GHz

TE20 13.123GHz

TE01 14.764GHz

TE11 16.156GHz

TE10TE20 TE01

TE11

6.562GHz 13.123GHz14.764GHz 16.156GHz

Frequency 10Ghz is propagating in TE10.mode since this frequency is below the 13.123GHz (TE20) and above 6.561GHz (TE10)

continue

026.02o

sR

12

05.158

mao

mNpabba

Ro

og

sc /0125.02 232

3

mdBedB cc /11.0log20)( or

Evanescent modeMode that propagates below cutoff frequency of a wave guide is called evanescent mode

22

112

oc

222ock Wave propagation constant is

Where kc is referred to cutoff frequency, is referred to propagation in waveguide and is in space

j =attenuation =phase constant

When f0< fc , 222ock

But

Since no propagation then

The wave guide become attenuator

Cylindrical waveguide

a

ap

f nmcnm 2

,

n p'n1 p'n2 p'n3

0 3.832 7.016 10.174

1 1.841 5.331 8.536

2 3.054 6.706 9.97

a

pk nmcnm

,

22cnmonm k

TE modeDominant mode is TE11

o

gTEZ

1'211

22

11p

ka

R oc

go

sc

continue

a

ap

f nmcnm 2

a

pk nmcnm

22cnmonm k

TM mode

g

oTMZ

n pn1 pn2 pn3

0 2.405 5.520 8.654

1 3.832 7.016 10.174

2 5.135 8.417 11.620

TM01 is preferable for long haul transmission

ExampleFind the cutoff wavelength of the first four modes of a circular waveguide of radius 1cm

Refer to tables

n p'n1 p'n2 p'n3

0 3.832 7.016 10.174

1 1.841 5.331 8.536

2 3.054 6.706 9.97

TE modes TM modes

n pn1 pn2 pn3

0 2.405 5.520 8.654

1 3.832 7.016 10.174

2 5.135 8.417 11.620

1st mode

2nd mode

3rd &4th modes3rd &4th modes

Calculation

nmcnm p

a 2

ap

f nmcnm 2

1st mode Pnm= 1.841, TE11mc 0341.0

841.1

01.0211

2nd mode Pnm= 2.405, TM01

1st mode Pnm= 3.832, TE01 and TM11

mc 0261.0405.2

01.0201

mcc 0164.0832.3

01.021101

Stripline

wb

bW

bZ

ero 441.0

30

35.0/35.0 2 b

WbW

b

W

b

We

35.0b

WWWe

Continue

120441.030

oror

ZforZb

W

120441.030

6.085.0

or

orZfor

Zb

W

On the other hand we can calculate the width of stripline for a given characteristic impedance

Continue

12016.0

12030

107.2 3

oro

s

orors

c

ZforBbZ

R

ZforAtb

ZR

t

tb

tb

tb

tb

WA

2ln

21

t

W

W

t

tW

bB

4ln

2

1414.05.0

7.05.01

Where

t =thickness of the strip