impedance matching and tuning presented by … matching maximum power is delivered when the load is...

Impedance Matching and Tuning

Presented by

EC67011

Presented by

Mrs.P.Rajeswari

VCET

Impedance Matching

Maximum power is delivered when the load is matched the line and the power loss in the feed line is minimizedImpedance matching sensitive receiver components improves the signal to noise

EC67012

Impedance matching sensitive receiver components improves the signal to noise ratio of the systemImpedance matching in a power distribution network will reduce amplitude and phase errors

ComplexityBandwidthImplementation

Adjustability

Two Approaches

Analytical solution

Graphical solution

EC67013

Matching with Lumped Elements (L Network)

Network for zL inside the 1+jx circle Network for zL outside the 1+jx circle

EC67014

Positive X implies an inductor and negative X implies a capacitorPositive B implies an capacitor and negative B implies a inductor

Possible configurations of two component

matching networks

EC67015

EC67016

EC67017

Matching with Lumped Elements (L Network)Smith Chart Solutions

Design an L-section matching network to match a series RF load with an impedance zL=200-j100Ω, to a 100 Ω line, at a frequency of 500 MHz.

EC67018

ZL=2-j1

yL=0.4+j0.5

B=0.29 X=1.22

B=-0.69 X=-1.22

EC67019

EC670110

Matching with Lumped Elements (L Network)Smith Chart Solutions

B=0.29 X=1.22

EC670111

B=-0.69 X=-1.22

Single stub matching

12 P.Rajeswari, AP/ECE VCETEC6701

Step-by-step Procedure


Drawbacks of single stub matching By using single stub impedance matching technique, thereflection losses are reduced considerably. Themain disadvantage is that this method is suitable forfixed frequency only. So as frequency changes thelocation of the stub will have to be changed.

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Another drawback is that , for adjusting the stub for finalposition along the line, the stub has to be moved orrepositioned. This is possible for open wire conductortransmission line. But in case of co-axial cable it isdifficult to locate the minimum point without aslotted section.

P.Rajeswari, AP/ECE VCETEC6701

Double stub impedance matching


Impedance matching using Smith chartSmith chart


SMITH CHART

33 EC6701

Applications of Smith chart Plotting an impedance

Measurement of VSWR

Measurement of reflection coefficient

Measurement of input impedance on the line

Impedance to Admittance conversion

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Impedance to Admittance conversion


Example : Plot the following impedance on the smith chart


Part-B


Design a single stub match for a load of 150+j255 ohmsfor a 75ohms line at 500MHz using Smith chart

SolutionsUsing admittances as short circuited stub is connected in parallel The normalized admittance can be obtained as,

SINGLE STUB


a). Locate point A at [0.1538-j0.2307] on chart as shown. It indicates

load point.

b). Draw SWR circle with centre at O(1,0) and radius equal to OA

c) The SWR circle cuts unity conductance circle in points B and C

which give possible stub locations.

d) Draw lines from O through A, B and C

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d) Draw lines from O through A, B and C

e) Consider point C first. Location of stub from load point is S1

S1=Distance between radial lines OA and OC going towards generator

in clockwise direction

=(0.5-0.465)λ+0.194λ=0.229λ


f) The line susceptance is inactive . Hence stub should

provide capacitive susceptance. The susceptance of line at

point C is +j2.4. the stub should provide –j2.4 curve

intersects it. Hence extend line OD. Then the length of the

stub is measuredfrom extremeright hand point on real

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stub is measuredfrom extremeright hand point on real

axis. Thus length of the stub is

L1=0.312λ-0.25λ=0.062λ


C

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A

B


Using double stub matching, match a complex load of ZL=18.75+j56.25 to a line with characteristic impedance Z0 75ohm

DOUBLE STUB

Determine the stub lengths, assuming a quarter wavelength spacing ismaintained between the two short circuited stubs.

A spacing of λ/4 is maintained between the stubs, stub 2 and stub 1. Forsmooth line operation of the transmission line the input admittance looking

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into the terminals 2,2 of the line should be,

Y2,2=1/Z0

i.e., the line beyond the point 2, 2 should appear to be a pure resistance of value‘Z0’ considering Z0 is purely resistive. Similar, to the single stub matching, theadmittance (normalized) at the point 2,2 must be,

Ys/G0 =1±j ba


The stub at 1. 1 must be capable to transformthe admittance at theterminating impedance end to the circle B which is displacedfrom the circleA; R = 1 by 'λ/4’

The quarter wavelength line will further transform the admittance into avalue at 2, 2 which will plot on the circle A. Thus the line to load distancebetween position 2, 2 is not required to be determined.


a) The normalized load impedance ZL=ZL/Z0=18.75+j56.25/75

Plotting the normalized impedance on the smith chart shown. T heimpedance circle is drawn with distance between the point (1 ,0) and thepoint of the norlized impedance as the radius (distance OA)

b) Moving by 180° (0.25 λ) on the impedance circle, i.e. at, a diametricallyopposite point to the point A, i.e. point B will give the norma lizedadmittance.From the Smith chart

yL = 0.4 -/1.2

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c) Circle A is the constant R circle for R =1. Circle B is the loc us of all thepoints on the circle A displaced by λ/4, quarter wavelength the stub 1 addsa susceptance (1/reactance) in parallel, this is done to cha nge the value ofy to such a value that it plots on the circle B.

Since stub 1 cannot alter the conductance (l/resistance). t o a point on thecircle B. point C,

y (at point C)=0.4-j0.5


d) Transfomiing the point C to the point D on the circle A. sinc e theline betwecn,1. 1 and 2. 2 is a quarter wave line that transfor ms theadmittance at 1, 1 to 2.2 such that the conductance equals thecharacteristic conductance. 1/Z 0 ,

y (at point D)=1.0 +j1.2

e) The stub length at2,2 should cancel the imaginary part of t he aboveadmittance. The susceptance of the stub at 2,2 must be –j1.2

f)To find the length of the stub with an admittance

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f)To find the length of the stub with an admittance(a) +j0.7 (b) –j1.2

The outside circle of teh smith chart (the circle, R=0), is mo ved aroundhaving a reference at point P, until

An admittance y=-1.2 is found at point E andAnd admittance y =0.7 is found at point F


A

P

D

F

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B

E

C

EC6701

g) From smith chart .Length of the stub 1 = Distance between P and Fls1=0.348λLength of the stub 2 = Distance between P and Els2=0.11λ


Microstripline matching networks

EC670156

Example

EC670157

EC670158

EC670159

EC670160

EC670161

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impedance matching and tuning presented by … matching maximum power is delivered when the load is...

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