appa impedance matching 1

8/13/2019 AppA Impedance Matching 1

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Appendix A Impedance Matching

In Chapter 6 we noted the deleterious effects of impedance mismatches on

transmission lines. Mismatches result in power being reflected back to the

source and in higher-than-normal voltages and currents that can stress the line

and connected equipment. In general, best results are obtained when the load

is matched to the characteristic impedance of the line. When this is not the

case, it is often useful to connect some sort of matching network to correct

the mismatch. Similarly, when either the load or the line is unbalanced,

and the other is balanced, it is highly desirable to install a device called a balun

network to convert one to the other. Matching networks can be constructed

using lumped constants (inductors, capacitors, and transformers), transmis-sion line sections, or waveguide components. Matching networks are some-

times broadband, so as to pass a wide range of frequencies, but often they are

made narrow in bandwidth to allow them to act as bandpass filters as well.

We noted earlier that the impedance looking into a transmission line

varies with its length. For a lossless line the impedance repeats every one-

half wavelength. If it is acceptable to have a mismatch on a short section of

line near the load, it is often possible to simplify a matching problem by

backing off a short distance (less than one-half wavelength) from the load

and installing the matching device at a point on the line where the imped-

ance is easier to work with. It should be obvious that this is a narrowband

technique, since the electrical length (in degrees) of a particular physicallength of line (in meters) varies with frequency.

The Smith Chart

For many years (since 1944 in fact), a special circular graph called a Smith

chart has been used to indicate complex impedances and admittances and

the way in which they vary along a line. It is more common nowadays to do

transmission line calculations with the aid of a computer than to use pencil

and compass, but many of the computer programs use the Smith chart to dis-

play their output. Test instruments, called vector network analyzers, thatmeasure complex functions, also tend to use this display format. Therefore

it is important to understand the layout of the Smith chart and to be able to

interpret itthe actual calculations can be safely left to a computer.

Figure A.1 shows a conventional paper Smith chart. A distance of one-

half wavelength on the line corresponds to one revolution around the chart.

Clockwise rotation represents movement toward the generator, and coun-

terclockwise rotation represents progress toward the load, as shown on the

543


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544 APPENDIX A

FIGURE A.1 Smith Chart (Courtesy of Analog Instruments Company)


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chart itself. For convenience there are two scales (in decimal fractions of a

wavelength) around the outside of the chart, one in each direction. Each

scale runs from zero to 0.5 wavelength.

The body of the chart is made up of families of orthogonal circles; thatis, they intersect at right angles. The impedance or admittance at any point

on the line can be plotted by finding the intersection of the real component

(resistance or conductance), which is indicated along the horizontal axis,

with the imaginary component (reactance or susceptance), shown above the

axis for positive values and below for negative.

Because of the wide variation of transmission line impedances, most

paper Smith charts use normalized impedance and admittance to reduce the

range of values that have to be shown. The value of 1 in the center of the

chart represents an impedance equal to Z0. Paper charts are also available

with an impedance of 50 in the center, and computerized displays canhave any value for Z0, with no normalization required.

To normalize an impedance, simply divide it by the characteristic

impedance of the line.

z Z

Z=

0

(A.1)

where

z = normalized impedance at a point on the lineZ = actual impedance at the same point

Z0 = characteristic impedance of the line

Since z is actually the ratio of two impedances, it is dimensionless.

EXAMPLEA.1 Y

Normalize and plot a load impedance of 100+ j25 on a 50-line.

SOLUTION

From Equation (A.1),

z = Z

Z0

= +100 25

50

j

= 2+ j0.5

Figure A.2 shows how this impedance would be plotted on a normalized

chart.

IMPEDANCE MATCHING 545


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546 APPENDIX A

FIGURE A.2 Use of Smith Chart (Courtesy of Analog Instruments Company)


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Once the normalized impedance at one point on the line has been plotted,

the impedance at any other point can be found very easily. Draw a circle with

its center in the center of the chart, which is at the point on the horizontal axis

where the resistive component is equal to one. Set the radius so that the circlepasses through the point just plotted. Then draw a radius through that point,

right out to the outside of the chart, as shown in Figure A.2. Move around the

outside in the appropriate direction, using the wavelength scale as a guide. Just

follow the arrows. If the first point plotted is the load impedance, then move in

the direction of the generator. Once the new location on the line has been

found, draw another radius. The normalized impedance at the new position

is the intersection of the radius with the circle. In other words, the circle is

the locus of the impedance of the line. Every point on the circle represents the

impedance at some point on the line.

The radius of the circle represents the SWR on the line. In fact, it is usu-

ally referred to as theSWR circle. The SWR can easily be found by reading thenormalized resistance value where the circle crosses the horizontal axis to

the right of the center of the chart. Another way is to mark the radius of the

circle on the standing wave voltage ratio scale at the bottom of the chart. If

required, the SWR can also be found in decibels by using the adjoining scale.

For this example, the SWR is read as 2.16.

Figure A.3 on page 548 is a computer printout from a program called

WinSmith, which is one of many that perform Smith chart calculations.The output is provided on a Smith chart, as well as in tabular form.

We noted before that the SWR depends on the magnitude of the reflec-

tion coefficient. There is another scale on the chart that allows this magni-tude to be found directly. See the reflection coefficient voltage scale at the

lower right of the chart. Marking off the radius of the SWR circle on this scale

will give the magnitude of, which can be read as 0.37, on both the paperand computerized charts. (On the computer chart,is represented by G andVSWRby V.)

It is possible to use the Smith chart to find the impedance at any point

along a line. The next example illustrates this.

EXAMPLEA.2 Y

A 50-line operating at 100 MHz has a velocity factor of 0.7. It is 6 m longand is terminated with a load impedance of 50+j50. Find the input imped-ance for the line.

SOLUTION

Now that we have seen how much the computer display resembles the paper

chart, let us use the computer. See Figure A.4.



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We need to know the length of the line in degrees. First find the wave-

length using Equation (6.1).

= vp

= v c

=

0 7 300 10

100 10

6

6

.

= 2.1 m

Now we can find the length in degrees with Equation (6.7).

= 360L

= 360 6

2 1.

= 1029

548 APPENDIX A

FIGURE A.3Printout fromwinSmith software

(Courtesy of NoblePublishing)


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From the plot in Figure A.4, we can see that the input impedance (repre-

sented by the small circle) is 19.36+ j5.44.

X

Now that we understand how impedances and lines are plotted on the

Smith chart, it should be possible to use the chart as an aid for impedance

matching. Since the center of the chart always represents the characteristic

impedance of the system, matching a line involves moving its input imped-

ance into the center of the chart. The progress of the solution to a matching

problem can be monitored by observing the input impedance: the closer it is

to the center of the chart, the better the match.Probably the best way to see how this works is to try a few examples

using various matching techniques. Each technique will be described,

then followed with an example. Smith charts will be used to illustrate the

technique.



(Courtesy of Noble

Publishing)


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Matching Using a Transformer

A transformer can be used to match impedances, provided the load imped-

ance is real at the point where the transformer is inserted. RF transformersare usually toroidal, using ferrite or powdered iron cores as illustrated in Fig-

ure A.5, although air-core transformers are also common. Recall from basic

electrical theory that the impedance ratio is the square of the turns ratio,

that is:

Z

Z

N

N1

2

1

2

2

=

(A.2)

Transformers are also useful for connecting balanced lines to unbal-anced lines, since there is no electrical connection between windings. A

transformer used in this way is called a balun transformeror just a balun.

A good example of a balun transformer can be found on the back of many

television sets, where it adapts 75- coaxial (unbalanced) cable-televisioncable to an antenna input designed for 300- twin-lead (balanced). Fig-ure A.6 is a photo of a typical TV balun. Since the required impedance ratio is

4:1, the turns ratio must be 2:1.

550 APPENDIX A

FIGURE A.5

Toroidaltransformer

FIGURE A.6TV balun


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A transformer can be used to match loads with complex impedances by

installing it a little distance from the load in the direction of the source. The

next example illustrates this.

EXAMPLEA.3 Y

Find the correct turns ratio and location for a transformer that is required to

match a 50-line to a load impedance of 75+ j25.

SOLUTION

We solve the problem by moving along the line in the direction of the gener-

ator until the impedance looking into the line is resistive. The arc represent-

ing the line crosses the horizontal axis of the chart at a point 16.8 from the

load. See Figure A.7(a).

Now the requirement is simply to match the 50-line to the impedance atthis point on the line, which is 88.38 , resistive. The required turns ratiocan be found from Equation (A.2):


FIGURE A.7(a)Printout fromwinSmith software



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Z

Z

N

N1

2

1

2

2

=

NN

ZZ

1

2

1

2

50

8838

0752

=

=

=

.

.

Figure A.7(b) shows what happens when we add a transformer with

this ratio at the point found above. The line is matched. Note, however,

that a change in frequency will change the electrical length of the line

and destroy the match. This can be seen by changing the frequency from30 MHz to 40 MHz while leaving everything else alone. See the result in

Figure A.7(c) on page 553. There is no longer a perfect match; however, a

closer look shows that the SWR is still a very reasonable 1.12, so the setup is

still usable.

552 APPENDIX A

FIGURE A.7(b)Printout fromwinSmith software



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A one-quarter wavelength of transmission line can also be used as a

transformer. A quarter wavelength along a line represents one-half a rota-

tion around the chart, so a one-quarter wavelength of line of the right im-

pedance can transform one real impedance into another. The characteristic

impedance Z0 of the line for the transformer can be found from the follow-ing equation:

=Z Z ZL0 0 (A.3)

EXAMPLEA.4 YSolve the problem of the previous example using a quarter-wave trans-

former.

SOLUTION

Since a quarter-wave transformer, like a conventional transformer, can only

match real impedances, it is necessary to place the transformer at the same


FIGURE A.7(c)Printout fromwinSmith software



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Series Capacitance and Inductance

Where the resistive part of the load impedance is correct, the reactive part

can be canceled by adding a series reactance of the opposite type. If the resis-tive part is not correct, the reactance can be installed at the correct distance

from the load to bring the resistive part of the impedance to the cor-

rect value.

EXAMPLEA.5 Y

Use a series capacitor or inductor to match a 50-line to each of the follow-ing loads at a frequency of 100 MHz:

(a) 50+ j75

(b) 150+ j75

SOLUTION

(a) Since the real part of the impedance is correct, we need only add a capac-

itive impedance ofj75, that is, a capacitor with 75-reactance. Recallfrom electrical fundamentals that

Xc = 1

2C

whereXc = capacitive reactance in ohms

= frequency in hertz

C = capacitance in farads

In this case, we know fand Xc, so we rearrange the equation:

C = 1

2Xc

=

1

2 100 10 756

= 21.2 pF

It was not necessary to use the Smith chart, whether on paper or on

a computer, to solve this problem, but the computerized version of the chart

verifies the match. See Figure A.9 on page 556, and notice that the arc on the

chart follows one of the circles of constant resistance to the center of

the chart.



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(b) Since the real part of the load impedance is not equal to 50, it is neces-

sary to add enough line to reach a point on the circle representing a resis-tance of 50 , that is, the circle that passes through the center ofthe chart. Figure A.10(a) on page 557 shows that this takes a length of

35 electrical degrees. At this point the resistive component of the imped-

ance is correct but there is a capacitive reactance of 72.6 . Fig-ure A.10(b) shows that the system can be matched by using a series

inductance of 115 nH.

556 APPENDIX A




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Stub Matching

As noted earlier, shorted transmission line stubs are often used instead of

capacitors or inductors at VHF and above. Usually these are placed in parallelwith the main line, rather than in series. In this case it is easier to handle the

problem from an admittance rather than an impedance point of view, since

parallel admittances add. Shorted transmission lines can compensate only

for the imaginary component of the load admittance, so if the load admit-

tance is complex it is once again necessary to install the matching compo-

nent at some distance from the load. In this case we back off until the real

component of the admittance is equal to the characteristic admittance of

the line.

Smith charts are available with admittance as well as impedance coor-

dinates. It is also possible to use a conventional chart and convert from

impedance to admittance. Computer programs invariably offer a choice ofcoordinates. The WinSmith program introduced earlier allows for either

or both to be displayed. An example will demonstrate transmission line

matching with a single shorted stub.

EXAMPLEA.6 Y

Match a line with a characteristic impedance of 72to a load impedance of120 j100 using a single shorted stub.

SOLUTION

The stub must be inserted at a point on the line where the real part of the

load admittance is correct. This value is

1

7200139

= . S

Actually, we can simply find the conductance circle that passes through

the center of the chart. Figure A.11(a) on page 559 shows a chart with

both impedance and admittance circles. This is a little confusing, so inFigure A.11(b) the impedance circles have been removed. It can be seen

from either figure that a distance of 42 is required between the load and the

stub.

Next we add a shorted stub and adjust its length to bring the input im-

pedance to the center of the chart. It can be seen from Figure A.11(c) on page

560 that this requires a stub length of 40.

558 APPENDIX A


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(Courtesy of Noble

Publishing)




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X

The foregoing is only a sample of the ways in which lines can bematched to loads.

The problems that follow can be solved using a paper Smith chart or a

computer program, if available.

(Problems

1. A 75- transmission line is terminated with a load having an imped-ance of 45 j30. Find:

(a) the distance (in wavelengths) from the load to the closest place atwhich a quarter-wave transformer could be used to match the line

(b) the characteristic impedance that should be used for the quarter-

wave transformer

560 APPENDIX A

FIGURE A.11(c)Printout fromwinSmith software



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2. A 50- coaxial transmission line with a velocity factor of 0.8 is con-nected to a load of 89j47. The system operates at 300 MHz. Find theproper position and component value to match this line using:

(a) a series capacitor

(b) a series inductor

(c) a transformer

3. Find the SWR for each setup of problem 2, at a frequency of 350 MHz.

4. A transmitter supplies 100 W to a 50- lossless line that is 5.65 wave-lengths long. The other end of the line is connected to an antenna with

a characteristic impedance of 150+ j25.

(a) Find the SWR and the magnitude of the reflection coefficient.

(b) How much of the transmitter power reaches the antenna?

(c) Find the best place at which to insert a shorted matching stub on

the line. Give the answer in wavelengths from the load.

(d) Find the proper length for the stub (in wavelengths).



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appa impedance matching 1

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