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Tutorial-4 Impedance Matching

Electrical Engineering Department (ISY), Linkping University

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Problem-4.1 (Tutorial):
A 50 antenna is connected to the amplifier of low input impedance ZL = 20-j10. A possible interconnection using 50 transmission line is shown below, where an extra capacitor and quarter wavelength transformer (piece of transmission line) is used for matching. a) Calculate the reactance of C and the characteristic impedance ZOT needed for matching.

D=

b) Is this circuit applicable to any generic load? c) What will be the consequences of not using the match?

Solution: a).
For the quarter wave transmission line 2 Z2 Z OT Z inT = OT = Z L 20 j10

2 Z OT Z2 + j OT 25 50

which is inductive while the load is capacitive

2 The capacitor can cancel the inductive part of Z inT and Z OT should be

2 Z OT 50 = 50 Z 0T = 25 2 Z2 X C = OT = 25 50

b).
The circuit is suitable for capacitive loads.

c).
When matched, the impedance seen by antenna is Z 0 = 50 irrespective the signal frequency and line length. So the power of signal from antenna is: 2 EAnt ( Eant 2 ) P0 = 50 50 2 (E ) = ant 50 200
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When not matched, the impedance seen by the antenna is Z + Z 0 j tan l l Z in = Z 0 L , l = 2 Z 0 + Z L j tan l

Pdelivered

E Ant = Z o + Z in

Re( Z in )
2

But also Pdelivered = Po (1 ) irrespective of line length.

Z L Z 0 20 j10 50 3 j = = Z 0 + Z 0 20 j10 + 50 7 j

EAnt

Zo

l ZL

10 1 = = 50 5 2 Pdelivered = Po (1 ) =0.8P0
2

50
Ant

Amp

However, if f RF changes, then will change as well, accordingly.

Problem-4.2 (HW):
Use the Smith chart to calculate transmission line length and XL needed to achieve 50 input impedance.

Problem-4.3 (Tutorial):
L-match circuit is the simplest and most widely used for matching. Four possible configurations of the L-match are shown in figure below. The a) and b) are low-pass and c) and d) are high-pass configurations. a) Look at the circuits carefully and make out which circuits transform the impedance from high (load side) to low (source side) and vice versa. b) Assume that DC voltage has to be transferred from load to source and load impedance is higher than source impedance. Which circuit is suitable for matching in this case? c) Assume that the signal is to be AC coupled from source to load with low pass characteristic, which circuit is suitable for matching? d) Design a circuit to match a 100 source to 1000 load at 100MHz. Assume that the DC voltage must also be transferred from source to load.

Electrical Engineering Department (ISY), Linkping University

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Solution:
Q = Qs = Q p = RP 1 (1) RS

For L-match case general formula This basic equation applies to all four (a-d) cases

a).
Circuits a) & d) transform from High Z Load Low Z source Circuits b) & c) transform from Low Z Load High Z source
RS XS XP RP

b).
Circuit a

QS = X S R S

c).
None is suitable

Qp=Rp/Xp

d).
As we need DC Path and Z Load > Z source , the only choice is circuit a) (1) Q = QS = QP =
QS =

RP 1000 1 = 1 = 9 = 3 RS 100

R s = 100

LS CP R p = 1000

XS X S = QS RS X S = 300 RS 300 X S = L L = = 477 nH 2 100 M R 1000 QP = P = 333 XP 3

C=

1 1 C = = 4.8pF XP 2 (100M ) ( 333) & C = 4.8pF

L = 477nH

Problem-4.4 (Tutorial):
In L-match networks, there are two degrees of freedom and we are not free to select the Q of circuit once the resonant frequency is chosen. Q is proportional to the ratio of load and source resistance. and T networks give three degrees of freedom and we can choose the Q factor independently.
RS

RL

Electrical Engineering Department (ISY), Linkping University

rashad@isy.liu.se

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Design a -match network to match 1000 load resistance with 100 source resistance at 1GHz frequency and Q of 15.

Solution:
-Network Design Procedure:
RS XS XP2 -XS1 XP1 -XS2 R

XP1

RL

XP2

Virtual resistor
a) X S 1 , X S 2 are symbolic to indicate that these are opposite type of reactance than X P1 , X P 2 b) The -network can be split in to two L-Match Network. c) The design of each Network proceeds as separate L-Network in last exercise.

RH 1 (approximate Formula) Eq (1) R R = virtual resistance RH = RS or RL which is highest Q=

-XS1 XP1

-XS2 R XP2

T-Network:
Q= R 1 RSmall

Virtual resistor

R = virtual resistance RSmall = smallest terminating resistance used to match two low value impedances when high Q is desired. Design a -match network to match 1000 load resistance with 100 source resistance at 1GHz frequency and Q of 15.

Solution:

L-Section:

Design a -network to match 100 source to 1000 ohm load with Q = 15. RH Q= 1 R RL = RH = 1000, RS = 100, Q = 15

RS Rp Rs -1

Xs XP RP

Q=

Q p =R p /X p Qs =X s /R s

Electrical Engineering Department (ISY), Linkping University

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R =

RH Q2 + 1 1000 R= = 4.42 226

Right Side:
QP = X P2 RP ( from L Section ) XP

Rs=100

-XS1 XP1

-XS2 RL=1000 R XP2

R 1000 = P = QP 15

X P 2 = 66.7 X QS = S RS X S 2 = QRS (XS2 & R are in series)

Virtual

X S 2 = 15 4.42 X S 2 = 66.3

Left Side:
QLEFT = RS 1 (From L-section formula) R
100 1 = 4.6 4.42

Please note that for left side RS is shunt leg

Rs=100

QP = RP X P
R 100 X P1 = P = = 21.7 QLEFT 4.6

-XS1 20.5 XP1 21.7

-XS2

66.3
XP2 66.7

RL=1000

QS =

XS X S1 = Q R RS

X S 1 = 4.6 4.42 = 20.33

Now X P1 , X P 2 , X S 1 & X S 2 can be either cap or inductor. Only constraint is X P1 & X S 1 are opposite type

X P 2 & X S 2 are opposite type

Electrical Engineering Department (ISY), Linkping University

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Possible circuits can be

20.5 21.7

66.9 66.7

87.4 21.7 66.7

1 = 7.33pF X c 2 f

C = 2.3pF

C = 1.82pF

20.5 21.7

66.9 66.7 21.7

87.4 66.7

L = 10.5nH L = 3.34nH

L = 7.32nH L = 3.34nH C = 2.3pF

20.5 21.7

66.9 66.7 21.7

46.4 66.7

C = 7.33pF C = 3.43pF L = 10.5nH

20.5 21.7

66.9 66.7

46.4 21.7 66.7

Note:
Now you can choose the circuit depending upon filtering and DC requirements.

Problem-4.5 (Tutorial):
Design procedure for T networks is exactly same as for networks. The T networks are usually used to match two low valued impedances when high Q is desired. Design a T match network to match 50 load resistance with 10 source resistance at 1GHz frequency and Q of 10.

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Solution:
Q = 10 (given) We adopt the same approach as in previous -network example. R Q= 1 10 RSmall -XS1=100 R = RSmall ( Q 2 + 1) = 10(100 + 1) = 1010 R(Virtual ) = 1010 X S 1 = QRS X S 1 = 10 (10 ) = 100 X P1 = R 1010 = 10 Q -XP1 101

-XS2=220 -XP2 231

50

X P1 = 101 Now calculate

QRIGHT = 4.4 X P2 = R QRIGHT = 1010 4.4

X P 2 = 230 X S 2 = QRIGHT .RL X S 2 = ( 4.4 )( 50 ) = 220

X S 2 = 220

Possible circuits can be

L=

XL 2f

100 101

220 231

100 79

220

1 C= 2fX C
15nH 35nH

Electrical Engineering Department (ISY), Linkping University

rashad@isy.liu.se

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1.59pF 0.723pF 11nH

100 101

220 231

100 79

220

100 0.723pF 15nH 28nH 101

220

100 179

220

231

1.59pF 35nH 28nH

100 101

220 231

100 179

220

Note:
The required network can be selected based upon filtering or DC biasing requirement.

Problem-4.6(Tutorial): a) Tapped capacitor or inductor is a simple kind of matching circuit widely used in b)
oscillator circuits. Derive the expression for its Rin (Zin to be more specific). Repeat the derivation for the tapped inductor circuit.

Solution:
1 j c R 1 2 2 + Z in = 1 + R j c1 2 j c2
j R2C1 + (1 + j R2C2 ) R2 1 = Zin = + jC1 2 R2C1C2 1 + jC2 R2 j c1 1 + jR2C2 Yin = jC1 1 + jR2 (C1 + C2 )

Electrical Engineering Department (ISY), Linkping University

rashad@isy.liu.se

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Re[Yin ] = Gin =

R (C1 + C2 ) 2 + 1
2 2 2

2 R2C12

Observe that Im[Yin] can be cancelled assuming that additional shunt inductor can be used (shown in the figure with dotted line). Further we concentrate on the real part. for >>
1 we have R2 (C1 + C2 )
2

Gin

C12 R2 (C1 + C2 ) 2

C + C2 or R in R2 1 C1 Hence, we see the circuit boosts the value of R2. At the same time
Im[Yin ]

C1C2
C1 + C2

which can be cancelled out by the inductive susceptance

BL =

1 = Im[Yin ] L

For the inductor tapped circuit a similar formula for Rin can be derived (HW)

Problem-4.7(Tutorial):
What is impedance transformation ratio of the coaxial balloon shown in figure below? Derive the expression for Zin.

Solution:
l = 2 l = 2 l

V ( x) = V (0) cos x Z 0 I (0) j sin x I ( x) = I (0) cos x V (0) j sin x Z0

(1) (2)

By inspection we find V (l ) = V1 I ( l ) = I 1
V ( 0) = V 2 I ( 0) = I 2

Electrical Engineering Department (ISY), Linkping University

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GND

Then from (1) and (2) V1 = V2 and I1 = I 2 . Moreover

x=0 V2 I0 I2 V1 I1 x = -l

2I1 = I 0
I1 = I0 V = 1 2 2Z 0

Zin =

V12 2V1 = = 4Z 0 I1 V1 2Z 0

So the impedance transfer ratio is 4 This can be used as e.g. 75 300 matching TV circuit. Observe that it is a so-called balun which transforms unbalanced impedance Z0 (connected to GND) into balanced impedance 4Z0.

Problem-4.8: (HW)
A 300 MHz power amplifier is connected to 100 antenna through L-match network, an inductance and 10 cm long 50 transmission line having velocity factor of 0.8. a) Calculate the value of inductance La. b) Draw the circuit of DC coupled, low-pass L-match network and calculate its component values.

Electrical Engineering Department (ISY), Linkping University

rashad@isy.liu.se