impedance matching and tuning presented by … matching maximum power is delivered when the load is...
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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
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Matching with Lumped Elements (L Network)
Network for zL inside the 1+jx circle Network for zL outside the 1+jx circle
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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
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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.
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ZL=2-j1
yL=0.4+j0.5
B=0.29 X=1.22
B=-0.69 X=-1.22
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Matching with Lumped Elements (L Network)Smith Chart Solutions
B=0.29 X=1.22
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B=-0.69 X=-1.22
Single stub matching
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Step-by-step Procedure
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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
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Impedance matching using Smith chartSmith chart
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SMITH CHART
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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
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Example : Plot the following impedance on the smith chart
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Part-B
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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
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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
P.Rajeswari, AP/ECE VCETEC6701
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
P.Rajeswari, AP/ECE VCETEC6701
C
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A
B
P.Rajeswari, AP/ECE VCETEC6701
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
P.Rajeswari, AP/ECE VCETEC6701
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.
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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
P.Rajeswari, AP/ECE VCETEC6701
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
P.Rajeswari, AP/ECE VCETEC6701
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
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Microstripline matching networks
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Example
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When you enjoy what you do, work becomes play...
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What we learn pleasure, we never forget.....
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