planar wave guide

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Transcript of planar wave guide

Planar waveguides

Planar waveguides are an important subclass of

waveguides (transmission lines)

Planar waveguides can be used in integrated

circuits to connect the various microwave circuit

elements

Examples are shown in the next three slides

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Applications of planar

waveguidesAlso planar transmission lines can be used to

feed microwave energy from a microwave

generator to an antenna as shown

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Applications of planar

waveguidesPlanar waveguides can be used to

interconnect the various parts of a

microwave amplifiers

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Applications of planar

waveguidesPlanar wave guides also can be used to

design filters and phase shifters

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Filter Phase shifter

Types of planar wave guides

There are many types of planar waveguides

available. Examples are

Strip line

Microstrip line

Coplanar waveguide

Slotted lines

These waveguides support TEM, TE and TM

wave propagations

However only TEM will be considered

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Planar waveguide design

parametersWhen designning a given microwave circuit, it is

desired to know the characteristic impedance,

dispersion curves, phase velocity, phase delay,

capacitance, inductance and the attenuation per unit

length of the waveguide

Computing these parameters can be performed by

solving the Helmholtz equation

However, the computation can be greatly simplified

by the use of commercial microwave design tools

such as CST, AWR microwave office, Agilent ADS,

and HFSS

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Strip-line

The strip line is composed from a thin conducting

strip of width π‘Š centered between two conducting

ground planes of separation 𝑏

The region between the conducting planes is filled

with a dielectric material whose relative permitivity

is πœ–π‘Ÿ as shown

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Strip line

The strip line can be considered as a planar

coaxial cable

Exact solution for the electric field can be

obtained by a method known as conformal

mapping

However this avoided in practice and an

empirical formulas can be used to design the

strip line for specific characteristic impedance

𝑍0, phase velocity 𝑣𝑝, propagation constant 𝛽

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Formulas for Propagation Constant,

Characteristic Impedance

The phase velocity can be given by

𝑣𝑝 =1

πœ‡πœ–=

𝑐

πœ–π‘Ÿ

The propagation constant is given by

𝛽 =πœ”

𝑣𝑝= πœ” πœ‡0πœ–0πœ–π‘Ÿ = πœ–π‘Ÿπ‘˜0

The characteristic impedance can be approximated by

𝑍0 =30πœ‹

πœ–π‘Ÿ

𝑏

π‘Šπ‘’ + 0.441𝑏

Where π‘Šπ‘’ is the effective width of the center conductor

which is approximated by

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Strip line width for specific 𝑍0

The strip line width that can be used to design a

strip line with a specific characteristic impedance

can be determined from the following equation

π‘₯ =30πœ‹

πœ–π‘Ÿπ‘0βˆ’ 0.441

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Attenuation due to conductor

loss

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Example

Find the width for a 50 Ξ© copper stripline

conductor with 𝑏 = 0.32 π‘π‘š and πœ–π‘Ÿ = 2.2. If the

dielectric loss tangent is π‘‘π‘Žπ‘›π›Ώ = 0.001 and the

operating frequency is 10 GHz, calculate the

attenuation in 𝑑𝐡/πœ† . Assume a conductor

thickness of 𝑑 = 0.01 mm

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Solution

To find the strip line width we can rely on the

equations in slide 10

πœ–π‘Ÿπ‘0 = 2.2 50 = 74.2 < 120

Therefore

π‘₯ =30πœ‹

πœ–π‘Ÿβˆ’ 0.441 = 0.83

We can use the equationπ‘Š

𝑏= π‘₯ β‡’ π‘Š = 𝑏π‘₯ = 0.32 Γ— 0.83 = 0.266 π‘π‘š

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Solution

The total attenuation is the attenuation due to

the conductor loss plus the attenuation due to

dielectric material, therefore

𝛼 = 𝛼𝑐 + 𝛼𝑑

𝛼𝑑 =π‘˜π‘‘π‘Žπ‘›π›Ώ

2But

π‘˜ = πœ” πœ‡πœ– =2πœ‹π‘“ πœ–π‘Ÿ

𝑐= 310.6

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Solution

Now 𝛼𝑑 =310.6Γ—0.001

2= 0.155 𝑁𝑝/π‘š

The conductor loss can be determined from

The surface resistance for the copper is

π‘…π‘ πœ”πœ‡

2𝜎= 0.026 Ξ© π‘Žπ‘‘ 10 𝐺𝐻𝑧

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Solution

𝛼𝑐 = 2.6 Γ—10βˆ’3π‘…π‘ πœ–π‘Ÿπ‘0𝐴

30πœ‹(𝑏 βˆ’ 𝑑)= 0.122 𝑁𝑝/π‘š

The total attenuation is

𝛼 = 𝛼𝑑 + 𝛼𝑐 = 0.277𝑁𝑝

π‘šπ›Ό 𝑑𝐡 = 20 log10 𝑒𝛼 = 2.41 𝑑𝐡/π‘š

The wavelength at 10 𝐺𝐻𝑧 is

πœ† =𝑐

πœ–π‘Ÿπ‘“= 2.02 π‘π‘š

The attenuation in terms of wavelength is

𝛼 𝑑𝐡 = 2.41 Γ— 0.0202 = 0.049 𝑑𝐡/πœ†

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Microstrip line

The microstrip line is one of the most popular

type of transmission lines

The microstrip line is composed from a signal

conductor of width π‘Š and ground plane

separated by a dielectric material as shown

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Phase velocity, propagation

constant The phase velocity on microstrip line is given by

𝑣𝑝 =𝑐

πœ–π‘’

The propagation constant is given by

𝛽 = πœ” πœ‡0πœ–0 πœ–π‘’ = π‘˜0 πœ–π‘’

Where πœ–π‘’ is the effective dielectric constant 1 <πœ–π‘’ < πœ–π‘Ÿ

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Effective dielectric constant πœ–π‘’

The effective dielectric constant can be

approximated by

πœ–π‘’ =πœ–π‘Ÿ + 1

2+πœ–π‘Ÿ βˆ’ 1

2

1

1 + 12𝑑/π‘Š

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Characteristic impedance

analysis formulaIf the dimensions of the microstrip are known, then

the characteristic impedance can be determined

from

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Characteristic impedance

Synthesis formulaIf the microstrip is designed to have a specific

characteristic impedance, then the ratio of its width

π‘Š to the dielectric thickness 𝑑 can be determined

form

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Microstrip attenuation

The attenuations for quasi TEM microstrip are

determined from the following two equations

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Example microstrip design

Design a microstrip line on a 0.5 mm alumina

substrate ( πœ–π‘Ÿ = 9.9, π‘‘π‘Žπ‘› 𝛿 = 0.001 ) for a 50 Ξ©characteristic impedance.

Find the length of this line required to produce a

phase delay of 270Β° at 10 GHz, and compute the

total loss on this line, assuming copper

conductors.

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Solution If we assume that

π‘Š

𝑑< 2, the width π‘Š can be

determined from

π‘Š

𝑑=

8𝑒𝐴

𝑒2𝐴 βˆ’ 2

𝐴 =𝑍060

πœ–π‘Ÿ + 1

2+πœ–π‘Ÿ βˆ’ 1

πœ–π‘Ÿ βˆ’ 10.23 +

0.11

πœ–π‘Ÿ= 2.142

π‘Š/𝑑 = 0.9654 which is less than 2

The required width of microstrip line is π‘Š =0.9654𝑑 = 0.483 π‘šπ‘š

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Solution

The length of the line that can produce a phase

shift of πœ™ = 270Β° can be determined from the

electrical length

πœ–π‘’ = 6.665πœ™ = 𝛽𝑙 = πœ–π‘’π‘˜0𝑙 = 270

π‘˜0 =2πœ‹π‘“

𝑐= 209.4 π‘šβˆ’1

𝑙 =270

πœ‹180

πœ–π‘’π‘˜0= 8.72 π‘šπ‘š

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Solution

By solving for 𝛼𝑑 and 𝛼𝑐 , we have a total

attenuation of 𝛼 = 𝛼𝑑 + 𝛼𝑐 = 0.022 + 0.094 =0.116 π‘‘π΅π‘š/π‘π‘š

The attenuation within the length of the phase

shifter is given by

π‘Žπ‘‘π‘‘π‘’π‘›π‘’π‘Žπ‘‘π‘–π‘œπ‘› = π›Όπ‘‘π΅π‘π‘š Γ— π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘™π‘–π‘›π‘’π‘Žπ‘‘π‘‘π‘’π‘›π‘’π‘Žπ‘‘π‘–π‘œπ‘› = 0.116 Γ— 0.872 = 0.101 𝑑𝐡

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Coplanar waveguide

The coplanar waveguide is composed

from a central conductor with two ground

planes as shown below

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Coplanar waveguide

The impedance of the coplanar waveguide can

be determined the width of the central conductor

π‘Š and the spacing from the ground planes 𝑆

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Commercial software

The dimensions of planar wave guide can

be adjusted for specific 𝑍0, 𝑣𝑝, 𝛽, π‘Žπ‘›π‘‘ πœ™ by

using CAD tool such on line tools or

Linecalc in ADS

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Linecalc

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You can find the line calculator program

under tools, then LineCalc

Starting the line calculatorIf you press the start LineCalc item, you may obtain a

window as shown below

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Entering the dielectric parameters

You can enter the

dielectric (substrate)

parameters in the

following section of

the LineCalc window

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Computing

𝑍0, π‘β„Žπ‘Žπ‘ π‘’ π‘ β„Žπ‘–π‘“π‘‘ πœ™ , πœ–π‘’ (Analysis)Enter the width and length of the transmission line and

click on the analyze window, the LineCalc program will

compute 𝑍0, πœ™, πœ–π‘’ as illustrated by

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Computing the π‘Š and 𝐿 for a

specific 𝑍0 and πœ™ (Synthesis)Enter the value of the desired 𝑍0 and the phase shift in

the section provided, press synthesize, the program will

compute the width and length of the transmission line

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Comparison of Common Transmission

Lines and Waveguides

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Microwave Frequency Bands (taken from wikipedia)

Letter Designation Frequency range

L band 1 to 2 GHz

S band 2 to 4 GHz

C band 4 to 8 GHz

X band 8 to 12 GHz

Ku band 12 to 18 GHz

K band 18 to 26.5 GHz

Ka band 26.5 to 40 GHz

Q band 33 to 50 GHz

U band 40 to 60 GHz

V band 50 to 75 GHz

E band 60 to 90 GHz

W band 75 to 110 GHz

F band 90 to 140 GHz

D band 110 to 170 GHz