MATCHING NETWORKSwebpages.iust.ac.ir/nayyeri/Courses/MCD/Lecture_4.pdf · 2018. 1. 13. · Matching networks provide a transformation of impedance to a desired value to maximize the

MATCHING NETWORKS

Matching networks provide a transformation of impedance to a desired value to maximize the power dissipated by a load. For example, the figure below illustrates the matching networks for a transistor amplifier. The matching networks ensure that the proper impedance is seen by the amplifier. One such matching method may be a conjugate match of the impedance.

We will discuss many useful types of matching networks in class, including: L-section matching Quarter wave transformers Single stub tuners

Microwave Circuits Design

http://webpages.iust.ac.ir/nayyeri/courses/mcd/

Dr. Vahid Nayyeri

01Highlight

L-Section Matching Networks

L-sections utilize purely reactive components such that no power is dissipated in the matching network

Smith Charts are an extremely useful manner by which to design L-section matching networks



Dr. Vahid Nayyeri

L-section design is best performed on an Admittance/Impedance Smith chart. Adding series reactive loads will modify the impedance by adding negative reactance (series C),

or positive reactance (series L) Adding shunt reactive loads will modify the admittance by adding negative susceptance (shunt

C), or positive susceptance (shunt L). Note that a solution for a given L-section is not guaranteed. The Smith chart provides visual

insight into the feasibility of design.



Dr. Vahid Nayyeri



Dr. Vahid Nayyeri

Example:A load ZL = 10 + j 10 is to be matched to a 50 line. Design L-section matching networks at 500 MHz using: a) a series L, shunt C b) a series C, shunt L

Solution: a) Given the normalized load: 0.2 + j 0.2, move along a constant-resistance circle until the unit

nH60.2 50

3.182 (500 10 )

Lp

⋅= =

´

pF62 1

12.7450 2 (500 10 )

Cp

= ⋅ =´

b) For the series capacitor, we move along the constant resistance circle in the opposite direction until

the unit conductance circle is intersected. Thus, the normalized impedance of the capacitor is –j 0.6. The normalized admittance at this point is 1 + j 2. Thus, the normalized admittance of the inductor must be –j 2. Thus, at 500 MHz:

pF61 1

10.60.6 50 2 (500 10 )

Cp

= ⋅ =⋅ ´

nH650 1

7.952 2 (500 10 )

Lp

= ⋅ =´

conductance circle is intersected. This adds j 0.2 normalized reactance. The normalized admittance at this point is 1 - j 2. Thus, the normalized capacitor admittance must be j 2. This brings us to the origin. Finally, the values for L and C are computed at 500 MHz:



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Analytical Solution for the L-Matching Network

A simple matching network can be achieved with only 2 reactive elements Transforms both the real and the imaginary part of the input impedance

A common configuration of the 2 reactive elements is referred to as the L-section matching network (or “el-section”)

Two types:

LZjB

jX

LZjB

jX

inZ inZ

Network #1 Network #2 where, L L LZ R jXConsider network #1:

1 1 1in L L

L L

L L

R jXZ jX jXjBR BXjB

R jX



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VahidTypewritten textRs < RL

VahidTypewritten textRs > RL

We desire that in sZ R . Therefore, we can write the equation:

1

L Ls

L L

R jXR jXjBR BX

We essentially have two degrees of freedom to solve for: B and X Multiply both sides by the denominator:

( ) ( )s L s L s L L L LR jBR R BX R jX XBR jXBX R jX Equate the real and imaginary terms:

Real:

Imag: 1L L s L s

L s L L

B XR X R R R

X BX BR R X

From the imaginary term, find X as a function of B:

1 L s s

L L

X R RXB R BR

Plug this into the real. Then, solving for B:

2 2

2 2

LL L L s L

s

L L

RX R X R RR

BR X

We can solve for B, and then calculate X from B. Note, that there are 2 solutions available Choose the design that makes sense physically, and is easiest to construct From Network #1:



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2 2

2 2

LL L L s L

s

L L

RX R X R RR

BR X

Note that in the radical term, the argument can be negative. Typically, Network #1 is only used in the case when L sR R .

In this case, the argument will always be positive. Network #2 is used when L sR R

Following the same procedure of equating in sZ R for Network #2, we derive:

L s L LX R R R X

/s L L

s

R R RB

R



Dr. Vahid Nayyeri

01Highlight



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The Quarter Wave Transformer

The quarter wave transformer is another useful narrow band matching technique that allows the use of a quarter-wavelength of transmission line with controllable impedance connected to a real load

Useful for waveguides for which one can control the dimensions and hence the characteristic line impedance (printed waveguides) when fabricating the network

Useful Theorem: 2/4 oZ x Z x Z

,q qZ

/4q

LZ,oZ

1

2inZ

1inZ



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Methodology Given a complex load

LZ , choose

1 such that

1inZ is purely real, i.e.,

1inZ R

Choose qZ such that 02inZ Z . Thus, 2

o q q oZ R Z Z Z R Note that in practice, one needs to design both qZ and q such that q oZ Z R and

/4q q



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Example Design a QWT to match a 100+j50 ohm load to a 50 ohm transmission line. To design the QWT, move along the SWR circle to the nearest real axis crossing (Max or Min?). Determine 0.4640.447 j

Le

Recall:

max 0.464 0.037 for 02 2 2 2d n

tan130.9

tanoL

oino L

Z jZZ Z

Z jZ

Finally,

80.9q oinZ Z Z Check:

1

1

tan50

tanqin

qinqq in

Z jZZ Z

Z jZ



Dr. Vahid Nayyeri

The Complete Smith ChartBlack Magic Design

,p:a, "'~ TOWARD LOAD->

General Design Rule Given a complex load

LZ , choose

1 such that

1inZ is purely real, i.e.,

1in

LZ , then

min

min

2

1for 0

2 2 2

1

1L

q o oinZ Z Z Z

If oR Z (typically, if Im 0LZ , then

min

max

2

for 02 2

1

1L

q o oinL

d n

Z Z Z Z

R Z (typically, if Im 0Z R

If o



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Bandwidth

The QWT is only truly matched at the resonant frequency. However, one can determine a frequency bandwidth over which the QWT yields a reflection coefficient below a desirable threshold.

Assume, LZ R . Then, then input impedance to the QWT is:

tan

tanq q q q

q qinq q q q

R jZ R jZ tZ Z Z

Z jR Z jRt

This results in the reflection coefficient:

2

2

q o q ooin

o q o q oin

Z R Z jt Z Z RZ Z

Z Z Z R Z jt Z Z R

Since, 2q oZ Z R , this reduces to:

2o

o o

R ZR Z j t Z R

It can be shown that:

2

2

1

41 seco

o

Z RR Z

, where q q



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Define a maximum allowable reflection coefficient m . Then, let m for which m . Namely,

1

2

2cos

1om

mom

Z RR Z

Finally, the bandwidth is defined as:

2 2 m Since 2 f

c , and 2 mm m

fc

22 2 2 4o m m m

o o o

f ff ff f f



Dr. Vahid Nayyeri

VahidTypewritten textFRACTIONAL BANDWIDTH

vs. / of f for various values of / oLZ Z



Dr. Vahid Nayyeri

Single Stub Tuning

An alternate method of matching is to reactively load the line with a shunt load rendering the net line impedance equal to the characteristic impedance.

A single stub tuner is best illustrated on a smith chart using the following procedure: Plot the normalized load impedance on the smith chart Draw the SWR circle and determine the line admittance Move toward the load until you cross the r = 1 circle At this point, the line admittance = 1 + jb Add in a shunt load with input admittance = -jb Note that a purely imaginary input admittance can be achieved by a short or open circuited

line with the proper line length At this point, the normalized admittance = 1. Thus, the line is matched to the load

As with the QWT, the geometry of the line is simply modfied to manipulate the line impedance and reach a matched condition. No lumped loads are needed.

In essence, the stub line cancels out the reactive power stored in the standing wave between the load and the line. Thus, all power is delivered to the load.



Dr. Vahid Nayyeri

Single stub matching problems can be solved on the Smith chart graphically, using a compass and a ruler. This is a step-by-step summary of the procedure:

(a) Find the normalized load impedance and determine the corresponding location on the chart.

(b) Draw the circle of constant magnitude of the reflection coefficient |Γ| for the given load.

(c) Determine the normalized load admittance on the chart. This is obtained by rotating 180° on the constant |Γ| circle, from the load impedance point. From now on, all values read on the chart are normalized admittances.



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0.5

-3

3

2

-2

zR

yR

(c) Find the normalized load admittance knowing that

yR = z(d=λ /4 ) From now on the chart represents admittances.

(a) Obtain the normalized load impedance zR=ZR /Z0 and find its location on the Smith chart

(b) Draw the constant |Γ(d)|circle 180° = λ /4



Dr. Vahid Nayyeri

(d) Move from load admittance toward generator by riding on the constant |Γ| circle, until the intersections with the unitary normalized conductance circle are found. These intersections correspond to possible locations for stub insertion. Commercial Smith charts provide graduations to determine the angles of rotation as well as the distances from the load in units of wavelength.

(e) Read the line normalized admittance in correspondence of the stub insertion locations determined in (d). These values will always be of the form

( )( )

stub

stub

d 1 top half of chartd 1 bottom half of chart

y jby jb

= +

= −



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0 5

-3

3

2

-2

zR

yR Load

location

First location suitable for stub insertion dstub1=(θ1/4π)λ

θ1

(d) Move from load toward generator and stop at a location where the real part of the normalized line admittance is 1.

Unitary conductance circle

(e) Read here the value of the normalized line admittance y(dstub1) = 1+jb

First Solution



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0 5

-3

3

2

-2

zR

yR Load

location

Second location suitable for stub insertion dstub2=(θ2/4π)λ

(e) Read here the value of the normalized line admittance y(dstub2) = 1 - jb

Unitary conductance circle

θ2

(d) Move from load toward generator and stop at a location where the real part of the normalized line admittance is 1.

Second Solution



Dr. Vahid Nayyeri

(f) Select the input normalized admittance of the stubs, by taking the opposite of the corresponding imaginary part of the line admittance

( )( )

stub stub

stub stub

line: d 1 stub: line: d 1 stub:

y jb y jby jb y jb

= + → = −

= − → = +

(g) Use the chart to determine the length of the stub. The imaginary normalized admittance values are found on the circle of zero conductance on the chart. On a commercial Smith chart one can use a printed scale to read the stub length in terms of wavelength. We assume here that the stub line has characteristic impedance Z0 as the main line. If the stub has characteristic impedance Z0S ≠ Z0 the values on the Smith chart must be renormalized as

0 00 0

' ss

Y Zjb jb jbY Z

± = ± = ±



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

2 1

-0.5

-3

3

2

-2

y = ∞ Short circuit

(f) Normalized input admittance of stub

ystub = 0 - jb

(g) Arc to determine the length of a short circuited stub with normalized input admittance - jb

0.5



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

-3

3

2

-2

y = 0 Open circuit

0.5


ystub = 0 - jb

(g) Arc to determine the length of an open circuited stub with normalized input admittance - jb



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

2 1

-0.5

-3

3

2

-2

y = ∞ Short circuit

0.5

(f) Normalized input admittance of stub ystub = 0 + jb

(g) Arc to determine the length of a short circuited stub with normalized input admittance + jb



Dr. Vahid Nayyeri

1

-1

0 0.2 0.5 5

0.2

-0.2

2 1

-0.5

-3

3

2

-2

0.5


ystub = 0 + jb

(g) Arc to determine the length of an open circuited stub with normalized input admittance + jb

y = 0 Open circuit



Dr. Vahid Nayyeri

1

1

1tan , ( 0)

21

tan , ( 0)2

t td

t t

Analytical Solution

A Smith chart will solve the SST approximately. Simple analytic solutions can also be derived. Assume a load impedance (admittance):

; 1/L L L L LZ R jX Y Z

A distance d from the load, the line admittance is given as:

o L L

o oL L

Z j R jX tY G jB

Z R jX jZ t

where tant d Evaluating the real and imaginary parts:

2

22

1L

oL L

R tG

R X Z t

,

2

22

o oL L L

o oL L

R t Z X t X Z tB

Z R X Z t

Solving for d such that / 1oG Y

2 2 /, if

, if 2

o oL L L L

oLoL

LoL

o

X R Z R X ZR Z

R ZtX

R Z

Finally, there are two solutions for d:



Dr. Vahid Nayyeri

At this point, the normalized line admittance = 1 / ojB Y Thus, we need to add in a shunt stub tuner to cancel out the reactive part. Assume that the stub is terminated by a short circuit. Then:

1tan cotsc sco sc scin in

o

Z jZ Y jZ

It is desired that: scin

Y jB

Therefore:

11 1 1cot tan2

scsc

o o

jB jZ Z B

Similarly, for an open circuit stub:

1cot tanoc oco oc ocin in

o

Z jZ Y jZ

Hence

11 tan2

ocoBZ



Dr. Vahid Nayyeri

Double stub impedance matching

Impedance matching can be achieved by inserting two stubs at specified locations along transmission line as shown below

YA = Y01 dstub1

YR = 1/ZR Y01 = 1/Z01

Lstub1

Y0S1

Lstub2

Y0S2

dstub2

There are two design parameters for double stub matching:

The length of the first stub line Lstub1

The length of the second stub line Lstub2

In the double stub configuration, the stubs are inserted at pre-determined locations. In this way, if the load impedance is changed, one simply has to replace the stubs with another set of different length.

The drawback of double stub tuning is that a certain range of load admittances cannot be matched once the stub locations are fixed.

Three stubs are necessary to guarantee that match is always possible.

VahidHighlight

The length of the first stub is selected so that the admittance at the location of the second stub (before the second stub is inserted) has real part equal to the characteristic admittance of the line

Y’A = Y01 + jB dstub1

YR = 1/ZR Y01 = 1/Z01

Lstub1

Y0S1

dstub2

The length of the second stub is selected to eliminate the imaginary part of the admittance at the location of insertion.

YA = Y01 + jB – jB = Y01 dstub1

YR = 1/ZR

Lstub1

Y0S1

dstub2

Lstub2

Y0S2

Y01 = 1/Z01

Ystub = -jB

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0 5

0.5

-3

3

2

-2The normalized admittance that we want at locationdstub2 is on this circle

At the location where the second stub is inserted, the possible normalized admittances that can give matching are found on the circle of unitary conductance on the Smith chart.

YA = Y01

Lstub2

Y0S2

Y01 = 1/Z01

YR

dstubThink of stub matching in a unified way.

Single stub

YR

Lstub1

Y0S1

Double stub

The two approaches solve the same problem

dstub2

If one moves from the location of the second stub back to the load, the circle of the allowed normalized admittances is mapped into another circle, obtained by pivoting the original circle about the center of the chart.

At the location of the first stub, the allowed normalized admittances are found on an auxiliary circle which is obtained by rotating the unitary conductance circle counterclockwise, by an angle

( )aux stub2 stub1 214 4d d dπ πθ = − =λ λ

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0 5

The normalized admittance that we want at location dstub1is on this auxiliary circle.

Pivot here

This angle of rotation corresponds to a distanced12 = dstub2 -dstub1

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5

This is the auxiliary circle for distance between the stubs d21 = λ/8 + n λ/2.

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5


1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5

This is the auxiliary circle for distance between the stubs d21 = 3 λ/8 + n λ/2.

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0 5This is the auxiliary circle for distance between the stubs d21 = n λ/2. NOTE: this is not a good choice for double stub design!

Given the load impedance, we need to follow these steps to complete the double stub design:

(a) Find the normalized load impedance and determine thecorresponding location on the chart.

(b) Draw the circle of constant magnitude of the reflectioncoefficient |Γ| for the given load.

(c) Determine the normalized load admittance on the chart. This isobtained by rotating -180° on the constant |Γ| circle, from theload impedance point. From now on, all values read on the chartare normalized admittances.

(d) Find the normalized admittance at location dstub1 by movingclockwise on the constant |Γ| circle.

(e) Draw the auxiliary circle

(f) Add the first stub admittance so that the normalized admittancepoint on the Smith chart reaches the auxiliary circle (twopossible solutions). The admittance point will move on thecorresponding conductance circle, since the stub does not alterthe real part of the admittance

(g) Map the normalized admittance obtained on the auxiliary circleto the location of the second stub dstub2. The point must be onthe unitary conductance circle

(h) Add the second stub admittance so that the total paralleladmittance equals the characteristic admittance of the line toachieve exact matching condition

VahidHighlight

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0 5

0.5

-3

3

2

-2

zR

yR

(c) Find the normalized load admittance knowing that

yR = z(d=λ /4 ) From now on the chart represents admittances.

(a) Obtain the normalized loadimpedance zR=ZR /Z0 and findits location on the Smith chart

(b) Draw theconstant |Γ(d)|circle180° = λ /4

(d) Move to thefirst stub location

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0.5

-3

3

2

-2

yR

(e) Draw the auxiliary circle

(f) Second solution: Addadmittance of first stub toreach auxiliary circle

(f) First solution: Addadmittance of first stub toreach auxiliary circle

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0.5

-3

3

2

-2

(g) First solution: Map the normalized admittancefrom the auxiliary circle to the location of thesecond stub dstub2.

First solution: Admittance at location dstub2 before insertionof second stub

(h) Add secondstub admittance

1

-1

0 0.2 0.5 5

0.2

-0.2

21

-0.5

0.5

-3

3

2

-2

(g) Second solution: Map the normalizedadmittance from the auxiliary circle to the locationof the second stub dstub2.

Second solution: Admittance at location dstub2 before insertionof second stub

(h) Add secondstub admittance

As mentioned earlier, a double stub configuration with fixed stub location may not be able to match a certain range of load impedances.

This is easily seen on the Smith chart. If the normalized admittance of the line, at the first stub location, falls inside a certain forbidden conductance circle tangent to the auxiliary circle (and always contained inside the unitary conductance circle), it is not possible to find a value for the first stub that can bring the normalized admittance to the auxiliary circle. Therefore, it is impossible to position the normalized admittance of the second stub location on the unitary conductance circle.

When this condition occurs, the location of one of the stubs must be changed appropriately. Alternatively, a third stub could be added.

Examples of forbidden regions follow.

01Highlight

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5


The normalized conductance circle for the normalized admittance does not intersect the auxiliary circle.

Forbidden conductance circle. If the admittance at the first stub location falls inside this circle, match is not possible with the given two stub configuration.

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5


Forbidden conductance

circle

1

-1

0 0.2 0.5 5

0.2

-0.2

21

0.5

-3

3

2

-2

θaux

-0.5

This is the auxiliary circle for distance between the stubs d21 = 3 λ/8 + n λ/2.

Forbidden conductance

circle

1.pdf (p.1-8)2.pdf (p.9-12)3.pdf (p.13-15)4.pdf (p.16)5.pdf (p.17-21)6.pdf (p.22-31)7.pdf (p.32-33)

Documents

MATCHING NETWORKSwebpages.iust.ac.ir/nayyeri/Courses/MCD/Lecture_4.pdf · 2018. 1. 13. · Matching networks provide a transformation of impedance to a desired value to maximize the