Microwave Circuit Design By Professor Syed Idris Syed

Microwave Circuit Design By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Engineering Campus USM Nibong Tebal 14300 SPS Penang

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- 2 nd 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 Circuit R L R C G Lossy line Lossless line L

Analysis Kirchoff current law From Kirchoff Voltage Law (a) (b)

Analysis Let’s V=Voejwt , I = Ioejwt then Therefore a b Differentiate with respect to z

Analysis The solution of V and I can be written in the form of c d where and Let say at z=0 , V=VL , I=IL and Z=ZL Therefore e and f

Analysis Solve simultaneous equations ( e ) and (f ) Inserting in equations ( c) and (d) we have

Analysis But and Then, we have * ** and

Analysis or Or further reduce For lossless transmission line , g= jb since a=0

Analysis Standing Wave Ratio (SWR) node antinode Reflection coefficient Ae-gz Begz Voltage and current in term of reflection coefficient or

Analysis For loss-less transmission line g = jb By substituting in * and ** , voltage and current amplitude are g h Voltage at maximum and minimum points are and Therefore For purely resistive load

Analysis Other related equations From equations (g) and (h), we can find the max and min points Maximum Minimum

Important Transmission line equations Zin Zo ZL

Various forms of Transmission Lines

Parallel wire cable Where a = radius of conductor d = separation between conductors

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

Micro strip Conducted strip he w t er Substrate Ground t=thickness of conductor

Characteristic impedance of Microstrip line Where w=width of strip h=height and t=thickness

Microstrip width For A>1. 52 For A<1. 52

Simple Calculation Approximation only

Microstrip components • • • Capacitance Inductance Short/Open stub Transformer Resonator

Capacitance Zo For Zoc Zo

Inductance Zo For Zo. L Zo

Short Stub Zo ZL Zo Zo Z

Open stub Zo ZL Zo Zo Z

Quarter-wave transformer l/4 Zo x ZT At maximum point q in radian Zo Zmx/min ZL

Quarter-wave transformer at minimum point q in radian

Resonator • • • Circular microstrip disk Circular ring Short-circuited l/2 lossy line Open-circuited l/2 lossy line Short-circuited l/4 lossy line

Circular disk/ring feeding a a * These components usually use for resonators

Short-circuited l/2 lossy line Zin Zo b a =nl/2 where = series RLC resonant cct

Open-circuited l/2 lossy line Zin Zo b a =nl/2 where = parallel RLC resonant cct

Short-circuited l/4 lossy line Zin Zo b a =l/4 where = parallel RLC resonant cct

Rectangular waveguide b a Cut-off frequency of TE or TM mode Conductor attenuation for TE 10

Example Given that a= 2. 286 cm , b=1. 016 cm and s=5. 8 x 107 S/m. What are the mode and attenuation for 10 GHz? Using this equation to calculate cutoff frequency of each mode

Calculation TE 10 a=2. 286 mm, b=1. 016 mm, m=1 and n=0 , thus we have Similarly we can calculate for other modes

Example TE 10 6. 562 GHz TE 20 TE 01 13. 123 GHz 14. 764 GHz TE 11 16. 156 GHz Frequency 10 Ghz is propagating in TE 10. mode since this frequency is below the 13. 123 GHz (TE 20) and above 6. 561 GHz (TE 10)

continue or

Evanescent mode Mode that propagates below cutoff frequency of a wave guide is called evanescent mode Wave propagation constant is Where kc is referred to cutoff frequency, g is referred to propagation in waveguide and b is in space When f 0< fc , But g = a +jb a=attenuation Since no propagation then The wave guide become attenuator b=phase constant

Cylindrical waveguide a TE mode Dominant mode is TE 11

continue a TM mode TM 01 is preferable for long haul transmission

Example Find the cutoff wavelength of the first four modes of a circular waveguide of radius 1 cm Refer to tables TE modes 1 st mode 2 nd mode TM modes 3 rd &4 th modes