Smith Chart - Impedance Matching

7/27/2019 Smith Chart - Impedance Matching

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Impedance Matching

& Smith Chart

Microwave EngineeringEE 172

Dr. Ray Kwok





Impedance Matching - Dr. Ray Kwok

Technician tuningeven for 50Ω designs

output 50 Ω

50 ΩΩΩΩ

50 ΩΩΩΩ

50 ΩΩΩΩ

input 50 Ω

50 ΩΩΩΩ

Think of a block of cracked glass…




Stub tuning ?

0385.0t7288.0t0

04.0t0123.0t0385.0t7692.0t

t

t2.004.0

t1923.01

7692.0tB

1923.0t)t1923.01(Bt9615.0

t

t2.004.0

t9615.0

t1923.00385.0B

9615.0Bt9615.0)t1923.01(

)1923.0t( j9615.0Bt9615.0)t1923.01(B jt9615.0 j)t1923.01(

t9615.0 j)t1923.01(

)1923.0t( j9615.0

t)1923.0 j9615.0( j1

jt1923.0 j9615.0B j1

tanY j1

tan jYY

1923.0 j9615.0Z1Y

2.0 j1Z

2

22

L

L1

L

L

L

−−=

+−−=−

−=

+

−=−

+=+−

−=

−=−

=+−

++=++−+−

+−

++=

++

++=−

β+

β+=

+==

−=

l

l

ZLShort piece of open stub → capacitor

e.g. ZL = 50 − j 10 Ω

Zin = 50Ω

50ΩC

Length = ?

Y1want Y1 = Yo - jB

ω=

===

−=−=−

λ=π

λ=

=β

β==

<

−=−==

=−=−

λ=π

λ=

=β

β==

0028.0C

0028.050 / 140.0YBB

140.02.0t / 04.0B

094.02

591.0

591.0

tan671.0t

0B

01.050 / 497.0YBB

497.02.0t / 04.0B

009.02

0573.0

0573.0

tan0574.0t

o

o

l

l

l

l

l

l

t = 0.0574 or 0.671

real

imaginary

need inductor !!!!




Matching methods

Stub tuning doesn’t always work

Many other ways to match Lumped elements

Transmission lines

Stubs (open, short, series, shunt) Single Stub

Double Stubs

Quarter-wave transformer

Many combinations Can’t just blindly optimize

Need Smith Chart




SmithChart

A chart of Γ

Γ = u + j v

u

v

relate to Z or Y

ρ = 1

max reflection circle




Constant Resistance Circles

2

2

2

2

22

22

22

2222

22

22

1R

1v

1R

Ru

0v1R

1R

1R

R

1R

Ru

0v1R

1Ru

1R

R2u

0)1R(v)1R(Ru2)1R(u

vu1)vuu21(R

v)u1(

v2 j)vu1( jXR

jv)u1( jv)u1(

jvu1 jvu1

11Z

1Z

1Z

+

=+

+

−

=++

−+

+−

+−

=++

−+

+−

=++−+−+

−−=++−

+−

+−−=+

+−+−⋅

−−++=

Γ −Γ +=

+

−=Γ

real

R & X are normalized

circle for each R

If R = 0, u2 + v2 =1

u

v

R=0

If R = 1, (u - ½)2 + v2 = (½)2

R=1

Family of Constant

Resistance Circles




Constant Reactance Circles

[ ]

222

22

22

22

22

X

1

X

1v)1u(

X

v2v)1u(

v2v)u1(X

v)u1(

v2 j)vu1( jXR

=

−+−

=+−

=+−

+−

+−−=+

imaginary

circle foreach X

If X = 1, (u-1)2

+ (v-1)2

=1

u

vX=1

If X = 2, (u - 1)2 + (v – ½)2 = (½)2

X=2

X=-1

X=-2ρ = 1 circle

Family of Constant Reactance Circles




Constant VSWRCircles

u

v

ρ = 1

max reflection circle

Not shown in Smith Chart

0 < ρ < 1

ρ

θ

θ ρ je=Γ

Any point on Smith Chart

has a definite Z and Γ




Locate Z

e.g. ZL = 50 + j 25 Ω

ALWAYSNORMALIZE

FIRST

50Ω ZL

5.0 j1ZL +=

R = 1circle

X = 0.5circle

ZL




Read off ρρρρand VSWRρ

VSWR =1.6

ρρρρ = |Γ ΓΓ Γ | =0.24




Phase of Γ ΓΓ Γ

76o

Γ = 0.24 (76o)




Move alongTransmissionLinee.g. ZL = 50 + j 25 Ω

50Ω ZL

ZL

0.1λ

Zin

All 50 Ω, constant ρ

Starting phase

0.145λ

ending phase0.245λ

0.1λ along

the Constant

VSWR Circle

Zin(1.65, 0.1)

Zin = 50(1.65 + j 0.1) Ω

4o

Γ in = 0.24 (4o)

0.145 +0.1

= 0.245λ

clockwise




Quarter-Wave

e.g. ZL = 50 + j 25 Ω

50Ω ZL

ZL

λλλλ/4

Zin

Starting phase

0.145λ

ending phase0.395λ

0.25λ along

the Constant

VSWR Circle

Zin(0.8,-0.4)

Zin = 50(0.8 - j 0.4) Ω

0.145 +0.25

= 0.395λ

clockwise

L

L

in

Lin

Lino

YZ1Z

ZZ1

ZZZ

==

=

=

YL = 0.02(0.8 – j 0.4) Ω-1

Convert Z↔ Y easily !!




SeriesConnections

Z

Use Z

Add series L (clockwise)

Add series C (counter-clockwise)

ser-L

ser-C

Add series R (inward)

ser-R

xline

Add transmission line




ParallelConnections

Y

Use Y

Add shunt C (clockwise)

Add shunt L (counter-clockwise)

shunt C

shunt L

Add shunt R (inward)

shunt R

xline

Add transmission line




Move alongthe 3 circlese.g. ZL = 50 + j 25 Ω

ALWAYSNORMALIZE

FIRST

50Ω ZL

5.0 j1ZL +=

R = 1circle

X = 0.5circle

ZL




Exercisee.g. ZL = 20 - j 25 Ω

f = 159 MHz (ω = 109) Find Zin, Γ in, VSWR.

5.0 j4.0ZL −=

ZL

50Ω, λ/8 ZL20Ω

10nH

20pF

YL

B = ωC/Yo = 0.02Zo= 1

Y1

0.192λ

0.317λ

+0.125λ

Y2

Z2

X = ωL/Zo = 10/Zo= 0.2

Z3

Y3

G = G/Yo = Zo /R= 50/20 = 2.5

Yin

Zin

(0.28,0.13)

Zin = 50(0.28+j0.13) = 14 + j7 ΩΩΩΩ

ρ=0.57

164o

Γ ΓΓ Γ in = 0.57 (164o)SWR=7

VSWR = 7 at Y1 junction !!






5.0 j4.0ZL −=

Exercisee.g. ZL = 20 - j25 Ω

f = 159 MHz (ω = 109) Find Zin, Γ in, VSWR.

ZL

50Ω, λ/8 ZL20Ω

10nH

20pF

B = ωC/Yo = 0.02Zo= 1

0.443λ

0.068λ

+0.125λ

Z2

X = ωL/Zo = 10/Zo= 0.2

Z3

G = G/Yo = Zo /R= 50/20 = 2.5

Zin

(0.28,0.13)

Zin = 50(0.28+j0.13) = 14 + j7 ΩΩΩΩ

ρ=0.57

164o

Γ ΓΓ Γ in = 0.57 (164o)SWR=7

VSWR = 7 at Y1 junction !!

Z1




LOAD

Z=ID=

20-i*25 OhmZ1

CAP

C=ID=

20 pFC1

TLIN

F0=EL=Z0=ID=

0.159 GHz45 Deg50 OhmTL1

IND

L=ID=

10 nHL1

RES

R=ID=

20 OhmR1

PORT

Z=P=

50 Ohm1

Microwave Office (AWR)

0 1 .

0

1 .

0

- 1

.

0

1 0

. 0

10.0

- 1 0 . 0

5 .

0

5. 0

- 5 . 0

2 .

0

2. 0

- 2 .

0

3 .

0

3. 0

- 3 .

0

4 .

0

4. 0

- 4 . 0

0 .

2

0 . 2

- 0. 2

0 .

4

0 .

4

- 0

. 4

0 .

6

0 .

6

- 0

. 6

0 .

8

0

. 8

- 0

. 8

Graph 1Swp Max

0.159GHz

Swp Min0.159GHz

Z[1,1]

Smith Chart Example

S[1,1]Smith Chart Example

Freq ReZ11 ImZ110.159 0.27375 0.13908

One-port impedance

Freq MagS11 AngS11

0.159 0.5771 162.93

Reflection coefficient




Vmax & Vmin

ZL

100Ω, 1.5λ

80-j40ΩZg=100Ω

Vg=10V

θθθθ = 0

Vmax

θθθθ = ππππ

Vmin

0.394λλλλ

0.106λλλλ to Vmin

0.356λλλλ to Vmax




To what we match?

( ) ( ) ( )22

g

2

g

2

g

2

g

g

2

g

ininlossLoad

gg

2

XR4

RV

2

1

X2R2

RV

2

1RI

2

1PP

+=

+=== −

( ) 2

g

2

go

o

2

g

o

2

total

g

oLineLoadXRZ

ZV

2

1Z

Z

V

2

1ZI

2

1PP

2

++====

Case 1: match ZL = Zo = real → Γ L = 0, VSWR = 1 on the line

Zin lZo ZL

Zg=Rg + jXgVg

Zin

Zg

Vg

Case 2: match Zin = Zg → Γ in = 0, VSWR > 1

g

2

g

2

g

g

2

g

LoadR

V

8

1

R4

RV

2

1P ===

Case 3: match Zin = Z*g → Xin = -Xg

= max power available

conjugate matching

Ideally, match all Zo = Zg = ZL = real, then all 3 PLoad are the same = Pmax.




2-elementmatching

ZLmatchingnetwork

Zin = 50Ω

2 equations, 2 unknowns (Re, Im)

Single frequency matching

Bring ZL to center of chart→

Zo

Many choices

First element → to the “1” circlesR=1G=1

Choices:• lumped elements• transmission line (single, multiple)

• stubs (single, double)

• λ /4• multiple sections

ZL




LumpedElements

R=1G=1

If ZL is inside the G=1 circle,first element cannot be shunt

ZL (0.4, 1)YL (0.34, -0.87)

Z (0.4, 0.48)Y (1, -1.24)

=0.4+j1ZL

jB = +j1.24 = jZoωC

jX = -j0.52 = -j/ ωCZo

Z (0.4, -0.48)Y (1, 1.24)

=0.4+j1ZL

jB = -j1.24 = -jZo / ωL

jX = -j1.48 = -j/ ωCZo

Z (1, 1.4)Y (0.34, -0.48)


=0.4+j1ZL

jX = -j1.4 = -j/ ωCZo

Z (1, -1.4)Y (0.34, 0.48)


=0.4+j1ZL

jX = j1.4 = jωL/Zo

If ZL is inside the R=1 circle,first element cannot be in series




Using Stubs

ZL (0.4, 1)YL (0.34, -0.87)

Z (0.4, -0.48)Y (1, 1.24)

=0.4+j1ZL

short shunt stub jB = -j1.24 = -jcotβl

open series stub jX = -j1.48 = -jcotβl

Z (1, 1.4)Y (0.34, -0.48)

open shunt stub jB = +j0.39 = jtanβl

=0.4+j1ZL

open series stub

jX = -j1.4 = -jcotβl

l

l

l

l

β=β−=

β−=

β=

tan jYYcot jYY

cot jZZ

tan jZZ

oop

osh

oop

osh

One way, simply replace lumped elements with stubs

Zo can be

anything here

previous example




ZL (0.4, 1)YL (0.34, -0.87)

=0.4+j1ZL

TransmissionLine Matching

usually requires 1 more element

previous example

0.130λλλλ

0.185λλλλ

0.055λ

50 Ω

jX = -j1.85 = -j/ ωCZo

=0.4+j1ZL

0.435λλλλ

0.305λ

50 Ω

Y (1, 1.9)

short shunt stub jB = -j1.9 = -jcotβl

Z (1, 1.85)




Stub tuning ?

200.0B

03846.0B9615.0

00148.0B0074.0B9615.0B0074.003698.0

B9615.01923.0

0385.0

0385.0B1923.0

B1923.0t

1923.0t)t1923.01(Bt9615.0

B9615.01923.0

0385.0t

9615.0Bt9615.0)t1923.01(

)1923.0t( j9615.0Bt9615.0)t1923.01(B jt9615.0 j)t1923.01(

t9615.0 j)t1923.01(

)1923.0t( j9615.0

t)1923.0 j9615.0( j1

jt1923.0 j9615.0B j1

tanY j1

tan jYY

1923.0 j9615.0Z

1

Y

2.0 j1Z

2

2

L

L1

L

L

L

=

=

−=−+

−=

−

+=

+=−−

−=

=+−

++=+−−+−

+−

++=

++

++=−

β+

β+=

+==

−=

l

l

ZLShort piece of open stub→ shunt C

e.g. ZL = 50 − j 10 Ω

Zin = 50Ω

50ΩC

Length = ?

Y1want Y1 = Yo - jB

ω=

===

λ=

π≈β

=β=−

=

004.0C

004.050 / 200.0YBB4

2

6845tant

)200.0(9615.01923.0

0385.0t

o

l

l

l

real

imaginary

Quadratic equationspossible 2 solutions




Single transmission line matching

( )

( )??R50Z

ZR50ZX

X50)R50(Zt

)ZtX(ZRt50

ZRXt50Z50

) jZt jXR(Z)Xt jRtZ(50

t jXR jZ

jZt jXR

Z50

tan jZZ

tan jZZZZ

2

Lc

cLcin

≠

−=−=

+=

=−

++=−+

++

++

=

β+

β+=

l

l

( )( )

)R50(X50R50

)R50(

X50R50Z

R50

X50ZR50

2

22

22

−>

−−=

−=−

X50

)R50(Z

tan

)R50(

X50R50Z

c

22

c

−

=β

−−=

l

match with just 1 lossless transmission line?

In previous exampleZL = 50(0.4+j1) = 20 + j50

ZLZCZin = 50 Ω

2

X)R50(R

50R

>−

<

R(50 − R) = 600 < X2

Cannot tune match with

just one line !!!

conditions(real)

(imaginary)

If ZL = 25 + j20 Ω,

ZL = 0.5 + j0.4 normalized to 50 Ω

λ=π

λ=

=−

=−

=β

=

−−=−−=

>−

<

0776.02

)488.0(

53.04.0

)5.01(424.0

X

)R1(Ztan

424.0Z

)5.01(4.05.0

)R1(XRZ

X)R1(R

1R

c

c

22

2c

2

l

l

0.5 < 10.25 > (0.4)2

conditions ok

Z = Zc

t = tanβl

ok…. that means Zc is real, and

transmission line is a λ /4 transformer.

solve for Z

Yes, change Zo & length




In SmithChart ?

Last example

ZL = 25 + j20 Ω

Zo = 21.2 ΩLength = 0.078λ

Zin = 50 Ω

ZL = ZL /Zo = 1.19 + j0.94

Zin = 50/21.2 = 2.356

ZL (1.19, 0.94)

21.2 ΩΩΩΩ ZL

0.078λλλλ

Zin = 50Ω

0.172λ

0.250λZin(2.35, 0)

0.172 +0.078

= 0.250λ

clockwise

Matching doesn’t necessary

means center of the chart !!

Depends on Z of the line & system.




Single Stub Tuningrefers to sliding a stub (any kind)

along a transmission line.

previous example

0.130λλλλ

0.064λλλλ

=0.4+j1ZL

0.435λλλλ

0.305λ

50 Ω

Y (1, 1.9)

short shunt stub

jB = -j1.9 = -jcotβl

In practice, usually shunt stubs,short stub for waveguides,

open stub for microstrip.

=0.4+j1ZL

0.434λ

50 Ω

open shunt stub

jB = j1.9 = jtanβl

ZL (0.4, 1)YL (0.34, -0.87)

Y (1, -1.9)




Any Stubtanβl can be “+” or “-”

previous example

0.130λλλλ

0.064λλλλ

=0.4+j1ZL

0.435λλλλ

0.305λ

50 Ω

Y (1, 1.9)

open shunt stub jB = - j1.9 = jtanβlβl = -1.086 + π = 2.055length = 0.327λ > λ/4

can use any type depends on realization.e.g. use shunt open stub only…

=0.4+j1ZL

0.434λ

50 Ω

open shunt stub

jB = j1.9 = jtanβlβl = 1.086

length = 0.173λ

ZL (0.4, 1)YL (0.34, -0.87)

Y (1, -1.9)

I d M t hi D R K k




Double-Stub Tuning• Preferable because it’s not sensitive to the initial line length.• Useful in tuning (especially in waveguide) with variable load.• 2 adjustable stubs connected to a fixed-length transmission line.

• In principle, can use any series or shunt stubs.• In practice, mostly shunt short stubs for waveguide.

• Impedance of the stubs are arbitrary, and they don’t have to be the same.• Impedance of the connecting line doesn’t have to be Zo of the system.• 2 parameters to tune: d1 & d2

ZL

fixed L

Zo

Z2 Z1a d j u s t a b l e d 1

a d j u s t a b l e d 2

Impedance Matching Dr Ray Kwok




Double Stub Tuninge.g. 2 shunt short stubs (50Ω)

separated by a 50Ω line of 0.2λ

0.3λλλλ

0.5λλλλ

– 0.2λλλλ

• rotate the 1-circle by line length

• adjust d1 along constant-G circle• stop at the rotated blue-circle• xline will bring it to the green circle• adjust d2 along the green circle to Zo

• not for all ZL !! Forbidden zone.

d1d2

ZL

0.2λ

50 Ω =0.4+j1

ZL(0.4,1)YL(0.34,-0.87)

Y (0.34,-0.2)

Y (1,1.22)

-cotβd1 = +0.67

βd1 = -0.98 +π = 2.16d1 = 0.344λ

-cotβd2 = −1.22βd

2= 0.687

d2 = 0.109λ

5 0 Ω

5 0 Ω

Impedance Matching Dr Ray Kwok




Quarter-Wave Transformer• Zc

2 = ZoZ1

• Transformer real-to-real, complex-to-complex impedance.

• Usually needs one more element to match (before λ /4)(lumped elements, stubs or transmission line).

• normalized to Zo,

ZL

λ /4

Zo

matching

elementZ1Zc

1c ZZ =

Impedance Matching - Dr Ray Kwok



Impedance Matching Dr. Ray Kwok

e.g. Quarter-Wavee.g. shunt stub (100Ω) then λ /4

ZL = 20 + j 50 Ω

• move Z to the real axis

• normalized Zc = √Z = √3.1; Zc = 1.76

• Zc = 50(1.76) = 88 Ω

ZL(0.4,1)YL(0.34, -0.87)

Z (3.1,0)Y(0.34,0)

(1/100) tanβd= (0.87)(1/ 50)

βd = 1.05d = 0.167λ

d

ZL

λ /4

Zc =0.4+j150 Ω1 0 0 Ω




Impedance Matching Dr. Ray Kwok

Advanced Impedance Matching

So far….• single frequency• 1-port network

Need:• wideband matching – multiple sections• multi-ports (simultaneously tuned)

require knowledge of multi-port network analysis




p g y

Homework

Smith Chart Exercise

Matching ExerciseDouble Stub Exercise

Date post:	02-Apr-2018
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Smith Chart - Impedance Matching

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