WCChew

ECE Lecture Notes

Imp edance Matching on a Transmission Line

We note that when the imp edance of a load is the same as the character

istic imp edance of the transmission line there is no reected wave and all

the forward going power is dissipated in the load There are various ways to

achieve this impedance matching and we will discuss some of them below

a QuarterWave Transformer

A quarter wave transformer likelowfrequency transformers changes the

imp edance of the load to another value so that matching is p ossible

Z0 ZT

→ ZL Zin

λ/4

A quarterwave transformer uses a section of line of characterstic

imp edance Z of long To have a matching condition we want Z Z

T in

From Equation we have

Z jZ tan

Z

L T

T

Z Z

in T

Z jZ tan Z

T L L

tan In order for Z Z we need that since tan l tan

in

p

Z Z Z Z Z Z

l L T

T

If Z and Z are b oth real then Z is real and we can use a lossless line

L T

to p erform the matching If Z is complex it can be made real by adding a

L

section of line to it

Z0 ZT Z0

Zin Z1 ZL

λ/4

Example

Given that Z j Z nd the shortest l and Z so that

L T

the ab ove circuit is matched Assume that Z is real and lossless

T

Wewant Z to b e real and Z to b e Z in order for Z to b e real

in T

and the matching condition satised We nd that Z j In order

nL

to make Z real the shortest l from the Smith Chart is Then Z

n n

and Z Since Z we need

in

p

p

Z Z Z

T in

in order for matching condition to b e satised

Note that the quarter wave transformer only matches the circuit at one

frequency Often time it has a small bandwidth of op eration ie it only

works in the frequencies in a small neighb orho o d of the matching frequency

Sometimes a cascade of two or more quarterwave transformers are used in

order to broaden the bandwidth of op eration of the transformer

j1

j0.5 j2

Z nL

j0.2 λ l = 8

0 0.2 0.5 1 2 Zn1

−j0.2

−j0.5 −j2

−j1

b Single Stub Tuning

Another device for p erforming matching is a single stub either shorted or

op ened at one end which is shunted across the transmission line at z d

from the load  ZS Y(Ðd) Z VS in VSWR > 1 ZL

Ðd Shorted Stub

l, Z0

The lo cation d is chosen so that the admittance Y d lo oking toward

the load is Y jB Y The length l of the shorted stub is chosen so

Z

that its admittance is jB Hence when the stub is connected in parallel to

the transmission line at z d the imp edance Z Z so that matching

in

condition is achieved

A shorted stub has imp edance and admittance given by

Z jZ tan l

s

Y jY cot l

s

An op encircuited stub can also b e used and the imp edance and admittance

are given by

Z jZ cot l

op

Y jY tan l

op

j1

j0.5 j2

Y(-d) λ j0.2 0.216 Z nL

0 0.2 0.5 1 2 Yshort

Y −j0.2 nL 0.99λ

−j0.5 −j2

−j1 Ystub

Example 

Let Z j nd the minimum d and l that will reduce the VSWR

L

of the main line to Assume that Z

We nd that the normalized load Z j as shown on the Smith

nL

Chart Since this problem involves parallel connections it is more convenient

to work with admittances Y is as shown When we move toward

nL

Z

nL

the generator Y z traces out a lo cus on the Smith Chart as shown It

n

intersects the G circle as shown after moving through Therefore

d

Now Y d j Hence Y j From the Smith Chart

n nstub

we note that the admittance for a short is innity and is at the right end of

Y j wemovetoward the generator for the Smith Chart To get a

nstub

Hence l

Often time it is not easy to change d but quite easy to change l We

note that b oth in the quarter wave transformer and the single stub tuner we

have to change parameters for tuning We can provide these degrees of

freedom by using two stubs changing their length but not their p ositions

c Double Stub Tuning optional reading

Both single stub tuning and quarter w ave transformer matching require

changing the lo cation of the stub or the transformer In practice this is

dicult and a double stub tuning removes the diculty

3 Z0 A Z0 B ZL

Y1 Y2

1, Z0 2, Z0

Stub 1 Stub 2  All possible values of Yn2 by changing 2 .

YnL Rotation by C2 All possible values of Yn1 3 by changing 1 . C1 Yn2 P

All possible values of Yn2 by Yn1 = 1 transforming from all possible values of Yn1 by 3 . Yn1 Ð Ynstub1 C3

Q R

In order to have a matched circuit we should have Y Y so that

Y However if we change l the p ossible values of Y trace out a

n n

circle C as shown

If Y is as shown by changing l the p ossible values of Y trace out a

nL n

circle C as shown

When l is added all the p ossible values of Y at A is transformed to B

n

by a rotation according to the length of l This constitute a circle C

which is all the p ossible values of Y obtained from Y There are only

n n

two points P and Q that the two circles C and C intersect If we pick

P then this p oint should corresp ond to the value of Y

n

Y Y Y

n nl nstub

We can gure out Y and hence the length l

nstub

The length l rotates the p oint P to the p oint R Then R has the

imp edance Y Y Y We can gure out Y from

n nstub nstub nstub

the Smith Chart and hence the length l

d Ferranti Eect

Zo = 50 Ω

Ω VS RL = 25 = 10 V

z = Ð z = 0

Find VSWR on the line and if l is allowed to vary arbitrarily nd the

maximum voltage on the line 

We can nd VSWR from the Smith Chart or by calculator

P P

v

jP j

v

VSWR

jP j

v

|V(z)| λ/2 Vmax Vs

Vmin

z = Ð Ðdmin 0

The voltage at Z l is always xed to be V Hence we can see that

s

jV z j on parts of the transmission line can b e longer than jV j If l is chosen

s

so that V is at V then

s min

V VSWR V volts volts

max min

This amplication of voltage on a line is known as the Ferrantis eect If

the VSWR on the line is very high V can be so large that it reac hes the

max

breakdown voltage of the line This is something one should be cautious of

in designing transmission line circuits