W.C.Chew
ECE 350 Lecture Notes
10. Impedance Matching on a Transmission Line.
We note that when the impedance of a load is the same as the characteristic impedance 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) Quarter-Wave Transformer
A quarter wave transformer, like low-frequency transformers, changes the
impedance of the load to another value so that matching is possible.
Z0
ZT
→
Z in
ZL
λ/4
A quarter-wave transformer uses a section of line of characterstic
impedance ZT of 4 long. To have a matching condition, we want Zin = Z0 .
From Equation (6.11) we have
2
Z
Z
L + jZT tan 2
T
(1)
Zin = ZT Z + jZ tan = Z T
L
L
2
since tan l = tan 2 4 = tan 2 = 1. In order for Zin = Z0 , we need that
p
(2)
ZT2 = Z0 ZL ) ZT = Z0 Zl :
If Z0 and ZL are both real, then ZT is real, and we can use a lossless line
to perform the matching. If ZL is complex, it can be made real by adding a
section of line to it.
Z0
ZT
Zin
Z0
Z1
λ/4
1
ZL
Example
Given that ZL = (30 + j 40), Z0 = 50, nd the shortest l and ZT so that
the above circuit is matched. Assume that ZT is real and lossless.
We want Z1 to be real and Zin to be Z0 = 50 in order for ZT to be real
and the matching condition satised. We nd that ZnL = 0:6+ j 0:8. In order
to make Zn1 real, the shortest l from the Smith Chart is 8 . Then Zn1 = 3:0,
and Z1 = 150. Since Zin = 50, we need
p
p
ZT = ZinZ1 = 50 150 = 86:6
in order for matching condition to be satised.
Note that the quarter wave transformer only matches the circuit at one
frequency. Often time, it has a small bandwidth of operation, i.e., it only
works in the frequencies in a small neighborhood of the matching frequency.
Sometimes, a cascade of two or more quarter-wave transformers are used in
order to broaden the bandwidth of operation of the transformer.
j1
j0.5
j2
Z nL
λ
l= 8
j0.2
0
0.2
0.5
1
2
Z n1
−j0.2
−j0.5
−j2
−j1
(b) Single Stub Tuning
Another device for performing matching is a single stub (either shorted or
opened at one end) which is shunted across the transmission line at z = ;d
from the load.
2
ZS
Y(–d)
VSWR > 1 ZL
Zin
VS
–d
Shorted
Stub
l, Z0
The location d is chosen so that the admittance Y (;d) looking toward
the load is Y0 + jB (Y0 = Z10 ). The length l of the shorted stub is chosen so
that its admittance is ;jB . Hence, when the stub is connected in parallel to
the transmission line at z = ;d, the impedance Zin = Z0, so that matching
condition is achieved.
A shorted stub has impedance and admittance given by
Zs = jZ0 tan l
(3)
Ys = ;jY0 cot l:
(4)
An open-circuited stub can also be used, and the impedance and admittance
are given by
Zop = ;jZ0 cot l
(5)
Yop = jY0 tan l:
(6)
j1
j0.5
j2
Y(-d)
j0.2
0
0.216λ
0.2
Z nL
0.5
1
Yshort
2
YnL
−j0.2
0.99λ
−j0.5
−j2
−j1
Example
3
Ystub
Let ZL = (100+ j 85), nd the minimum d and l that will reduce the VSWR
of the main line to 1. Assume that Z0 = 50.
We nd that the normalized load ZnL = 2 + 1:7j as shown on the Smith
Chart. Since this problem involves parallel connections, it is more convenient
to work with admittances. YnL = Z1nL is as shown. When we move toward
the generator, Yn(z) traces out a locus on the Smith Chart as shown. It
intersects the G = 1 circle as shown, after moving through 0:216. Therefore,
d = 0:216.
Now, Yn(;d) = 1 + j 1:4. Hence, Ynstub = ;j 1:4. From the Smith Chart,
we note that the admittance for a short is innity, and is at the right end of
the Smith Chart. To get a Ynstub = ;j 1:4, we move toward the generator for
0:099. Hence, l = 0:099.
Often time, it is not easy to change d, but quite easy to change l. We
note that both in the quarter wave transformer and the single stub tuner, we
have to change 2 parameters for tuning. We can provide these 2 degrees of
freedom by using two stubs, changing their length, but not their positions.
(c) Double Stub Tuning (optional reading)
Both single stub tuning and quarter wave transformer matching require
changing the location of the stub or the transformer. In practice, this is
dicult, and a double stub tuning removes the diculty.
3
Z0
Z0
A
Y1
B
Y2
1, Z 0
2, Z 0
Stub 1
Stub 2
4
ZL
All possible values of Yn2
by changing 2.
YnL
Rotation by
C2
All possible values of Yn1
by changing 1.
3
C1
Yn2
All possible values of Yn2 by
transforming from all possible
values of Yn1 by 3.
P
Yn1 = 1
Yn1 – Ynstub1
C3
Q
R
(1) In order to have a matched circuit, we should have Y1 = Y0 so that
Yn1 = 1. However, if we change l1 , the possible values of Yn1 trace out a
circle C1 as shown.
(2) If YnL is as shown, by changing l2 , the possible values of Yn2 trace out a
circle C2 as shown.
(3) When l3 is added, all the possible values of Yn1 at A is transformed to B
by a rotation according to the length of l3. This constitute a circle C3
which is all the possible values of Yn2 obtained from Yn1. There are only
two points, P and Q that the two circles C2 and C3 intersect. If we pick
P , then this point should correspond to the value of Yn2.
Yn2 = Ynl + Ynstub2
(25:1)
We can gure out Ynstub2 and hence the length l2 .
(4) The length l3 rotates the point P to the point R. Then R has the
impedance Yn1 ; Ynstub1 = 1 ; Ynstub1 . We can gure out Ynstub1 from
the Smith Chart and hence the length l1.
(d) Ferranti Eect
Z o = 50 Ω
R L = 25 Ω
VS
= 10 V
z=–
z=0
. Find VSWR on the line, and if l is allowed to vary arbitrarily, nd the
maximum voltage on the line.
5
We can nd VSWR from the Smith Chart or by calculator.
; 50 = ; 1 P (0) = Pv = 25
25 + 50
3
+ jPv j = 43 = 2:
VSWR = 11 ;
jPv j 23
|V(z)|
Vmax
λ/2
Vs
Vmin
z=–
–dmin
0
The voltage at Z = ;l is always xed to be Vs. Hence, we can see that
jV (z)j on parts of the transmission line can be longer than jVsj. If l is chosen
so that Vs is at Vmin, then
Vmax = VSWR Vmin = 10 volts 2 = 20 volts:
This amplication of voltage on a line is known as the Ferranti's eect. If
the VSWR on the line is very high, Vmax can be so large that it reaches the
breakdown voltage of the line. This is something one should be cautious of
in designing transmission line circuits.
6