Topic 6 -- Smith Charts.pdf

12/15/2016

ECE 4380/5390 Spring 2013 Instructor: Dr. Raymond Rumpf Office: A-337 E-Mail: [email protected]

Topic #6

Smith Charts Smith Charts

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Outline • • • • • •

Construction of the Smith Chart Admittance and impedance Circuit theory Determining VSWR and  Impedance transformation Impedance matching

Smith Charts

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Construction of the  Smith Chart

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Polar Plot of Reflection Coefficient The Smith chart is based on a polar plot of the voltage reflection  coefficient .  The outer boundary corresponds to || = 1.  The  reflection coefficient in any passive system must be|| ≤ 1.

    e j



  radius on Smith chart

  angle measured CCW from right side of chart Smith Charts

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Normalized Impedance All impedances are normalized.  This is usually done with respect to  the characteristic impedance of the transmission line Z0.

Z z Z0

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Reflection Coefficient form  Normalized Impedance We can write the reflection coefficient in terms of normalized  impedances.

ZL





Z0

Z L  Z0 Z0 Z 0 zL  1   Z L  Z0 Z L  Z0 zL  1 Z0 Z0

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Derivation of Smith Chart: Solve for Load Impedance Solving the previous equation for load impedance, we get



zL  1 zL  1

  z L  1  z L  1

zL 

zL     zL  1 zL  zL   1  

z L 1     1   zL 

1  1 

1  1  Smith Charts

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Derivation of Smith Chart: Real and imaginary parts The load impedance and reflection coefficient can be written in  terms of real and imaginary parts.

z L  rL  jxL

   r  j i

Substituting these into the load impedance equation yields 1  1  1    r  j i  rL  jxL  1    r  ji  zL 

1   r   ji 1   r   ji

rL  jxL 

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Derivation of Smith Chart: Solve for rL and xL We solve or previous equation for rL and xL by setting the real and  imaginary parts equal. 1   r   ji 1   r   ji 1   r   ji  1   r   ji  1   r   ji 1   r   ji 1   r 1   r   j 1   r   i  j i 1   r   i2   2 1   r   i2

rL  jxL 



1   r2  j i  j r i  ji  j r i  i2



1      j 2 i



1   r 

2 r

2 i 2

1   r   

1   r2   i2

1   r 

2

  i2

2

xL 

 i2

2 i

j

rL 

1   2r  i2

1   r 

2

 i2

2 i

1   r 

2

  i2

2i

1   r 

2

 i2 Smith Charts

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Derivation of Smith Chart: Rearrange equation for rL We rearrange the equation for rL so that it has the form of a circle. rL 

1   2r  i2

1   r 

2

 i2 2

2

1   r 

2

  i2 

1   2r  i2 rL

 rL   rL  rL  1 2 0  r      i  rL  1   rL  1  rL  1 

1   r 

2

  i2 

1  2r  i2   0 rL rL rL

  rL  rL  1 rL  2  r    i     r 1  L    rL  1  rL  1

2 r   r2 

2  2r 1   i2  i  1   0 rL rL rL

2rL  r  rL  r2   r2  rL i2  i2  rL  1  0 2rL  r   rL  1  r2   rL  1  i2  rL  1  0 r 1 r 0  r2  2 L r   i2  L rL  1 rL  1 

2

2

  r  1 rL  1 rL  rL2 2  L  r    i  2 2 1 r  1 r  L   rL  1 L   2

2

 rL  rL2 r 2 1 2  L  r    i  2 2  1 r  rL  1  rL  1 L   2

 1 rL  2  r    i  2 r  rL  1 L 1 

can be factored

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Derivation of Smith Chart: Rearrange equation for xL We rearrange the equation for xL so that it has the form of a circle. xL 

2 i

1   r 

1   r 

2

2

 i2

 i2 

2 i xL

2 i  0 xL 

1   r   i2     2

swap terms

can be factored 2

 1  1   r  1   i    2  0 xL  xL  2

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Derivation of Smith Chart: Two families of circles Constant Resistance Circles 2

  1  rL  2  r    i    rL  1    1  rL 

Constant Reactance Circles

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2

 1   1    r  1   i      xL   xL  

These have centers at

r 

rL rL  1

2

2

These have centers at

r  1

i  0

Radii

i 

1 xL

Radii

1 1  rL

1 xL

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Derivation of Smith Chart: Putting it all together Lines of constant  resistance

Lines of constant  inductive reactance

Lines of constant  reflection coefficient

+

+

Superposition

=

Lines of constant  capacitive reactance

We ignore what is outside the || = 1 circle. We don’t draw the constant || circles. This is the Smith chart!

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Alternate Way of Visualizing the  Smith Chart

Lines of constant resistance

Lines of constant reactance

Reactance Regions

L short circuit

Smith Charts

open circuit

C 15

3D Smith Chart The 3D Smith Chart unifies passive and active circuit design. 2D

3D

EE3321 ‐‐ Final Lecture

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Summary of Smith Chart

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Impedance and Admittance on the Smith Chart

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Admittance Coordinates We could have derived the Smith chart in terms of admittance. You can make an admittance Smith chart by rotating the standard Smith chart by 180.

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Impedance/Admittance Conversion The Smith chart is just a plot of complex numbers. These could be admittance as well as impedance. To determine admittance from impedance (or the other way around)… 1. Plot the impedance point on the Smith chart. 2. Draw a circle centered on the Smith chart that passes through the point (i.e. constant VSWR). 3. Draw a line from the impedance point, through the center, and to the other side of the circle. 4. The intersection at the other side is the admittance. admittance

impedance

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Visualizing Impedance/Admittance  Conversion

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Example #1 – Step 1 Plot the impedance on the chart z  0.2  j 0.4

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Example #1 – Step 2 Draw a constant VSWR circle z  0.2  j 0.4

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Example #1 – Step 3 Draw line through center of chart z  0.2  j 0.4

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Example #1 – Step 4 Read off admittance z  0.2  j 0.4 y  1.0  j 2.0

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Example #2 – Step 1 Plot the impedance on the chart z  0.5  j 0.3

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Example #1 – Step 2 Draw a constant VSWR circle z  0.5  j 0.3

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Example #2 – Step 3 Draw line through center of chart z  0.5  j 0.3

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Example #2 – Step 4 Read off admittance z  0.5  j 0.3 y  1.5  j 0.9

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Circuit Theory

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Adding a Series Capacitor Suppose we have an initial impedance of z = 0.5 + j0.7 And we add a series capacitor of zC = -j1.0. Since we do not change the resistance, we walk toward  capacitance (CCW) around the  constant resistance circle. The “angular” distance  covers X=-j1.0 around the constant R circle.

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Adding a Series Inductor Suppose we have an initial impedance of z = 0.8 - j1.0 And we add a series inductor of zL = j1.8. Since we do not change the resistance, we walk toward  Inductance (CW) around the  constant resistance circle. The “angular” distance covers X=j1.8 around the  constant R circle.

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Adding a Shunt Capacitor Here we use the Smith chart rotated  by 180° for admittance. In terms of admittance, capacitance  is a positive quantity, but the positive  direction is now downward on the  Smith chart.

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Adding a Shunt Inductor We again use the Smith chart rotated  by 180° for admittance. In terms of admittance, inductance is  a negative quantity, but the negative  direction is now upward on the  Smith chart.

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Summary

Series C

Shunt C

Series L

Shunt L

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Example #1 – Circuit Analysis

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Example #2 – Circuit Analysis 3.3157 nH



Z in  ?

||

1  R  j L  jC 1  R  j L  jC  1   R  j L  jC R  j L  1   2 LC  j RC

Z in 

1  1.5080 10

1.9894 pF

50 

f  2.4 GHz Z 0  50  50  j 1.5080 1010  3.3157 109 

  3.3157 10 1.9894 10   j 1.5080 10  50  1.9894 10 

10 2

9

12

12

10

 20  j 40   VSWR 

50   20  j 40   0.0769  j 0.6154  0.6282.9 50   20  j 40  1  0.0769  j 0.6154 1  0.0769  j 0.6154

 4.2654 Smith Charts

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Example #2 – Circuit Analysis Normalize Impedances j1.0 zin  ?

r L

1 j1.5

1

50   1.0 50  j 2  2.4 109  3.3157  109 

 j1.0 50  1 9 j 2  2.4  10 1.9894  1012  1 C  j1.5 50 

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Example #2 – Circuit Analysis Plot load impedance j1.0 zin  ?

1 j1.5

1

z 1

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Example #2 – Circuit Analysis Add series inductor j1.0 zin  ?

1 j1.5

1

z  1 j

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Example #2 – Circuit Analysis Convert to admittance 1 j1.0

yin  ?

1

j1.5

y  0.5  j 0.5

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Example #2 – Circuit Analysis Add shunt capacitance 1 j1.0

yin  ?

j1.5

1

y  0.5  j1.0

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Example #2 – Circuit Analysis Convert to impedance j1.0 zin  ?

1 j1.5

1

z  0.4  j 0.8

We now know zin

zin  z  0.4  j 0.8

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Example #2 – Circuit Analysis Denormalize the impedance j1.0 zin  ?

1 j1.5

1

z  0.4  j 0.8

Z in  Z 0 zin  20  j 40 

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Example #2 – Circuit Analysis All together 3.3157 nH 1.9894 pF

50 

Z in  20  j 40 

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Determining VSWR and 

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The Horizontal Bar on the  Smith Chart

VSWR

Reflectance 

2

Reflection Coefficient 

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Determining VSWR 1. Plot the normalized load impedance on the Smith chart. 2. Draw a circuit centered on the Smith chart that intersections this point. 3. The VSWR is read where the circle crosses the real axis on right side. Example: 50  line connected to 75+j10  load impedance. z

Z L 75  j10   1.5  j 0.2 Z0 50 1 VSWR

impedance

VSWR = 1.55

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Example #1 – Circuit Analysis What is the VSWR? 3.3157 nH 1.9894 pF

50 

Z in  20  j 40 

VSWR  4.6

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Example #1 – Circuit Analysis What is the reflection coefficient?

  0.62 Smith Charts

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Impedance   Transformation

Normalized Impedance  Transformation Formula Our impedance transformation formula was

Z in  Z 0

Z L  jZ 0 tan   Z 0  jZ L tan  

zin 

z L  j tan   1  jz L tan  

We can write this in terms of the reflection coefficient. Z in  Z 0  Z0

0.5Z L  e j    e  j     0.5Z 0  e j    e  j    Z L cos    jZ 0 sin    Z0 Z 0 cos    jZ L sin   0.5Z 0  e j    e  j     0.5Z Z  e j    e  j   

 Z  Z 0  e j    Z L  Z 0  e j  Z L e j   Z L e j   Z 0e j   Z 0e j   Z0 L Z 0e j   Z 0e j   Z L e j   Z L e j   Z L  Z 0  e j    Z L  Z 0  e j 

 Z L  Z 0  e j   Z L  Z 0  e j   Z0 Z  Z0  e j   1 L  Z L  Z 0  e j  1

 Z0

1  e  j 2   1  e  j 2  

We normalize by dividing by Z0.

zin 

1  e  j 2   1  e  j 2   Smith Charts

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Normalized Admittance  Transformation Formula Our impedance transformation formula was

Z L  jZ 0 tan   Y  jY0 tan    Yin  Y0 L Z 0  jZ L tan   Y0  jYL tan  

Z in  Z 0

zin 

z L  j tan   y  j tan    yin  L 1  jz L tan   1  jyL tan  

We can write this in terms of the reflection coefficient.

zin 

1  e  j 2   1  e  j 2  

 yin 

1  e  j 2   1  e  j 2  

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Interpreting the Formula The normalized impedance transformation formula was

zin    

1  e  j 2   1  e  j 2  

Recognizing that  = ||ej, this equation can be written as

zin    

1   e j e  j 2   1   e j   2     1   e j e  j 2   1   e j  2   

Thus we see that traversing along the transmission line simply changes the phase of the  reflection coefficient. As we move away from the load and toward the source, we subtract phase from .  On the   Smith chart, we rotate clockwise (CW) around the constant VSWR circle by an amount 2l.   A complete rotation corresponds to /2.

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Impedance Transformation on the Smith chart 1. Plot the normalized load impedance on the Smith chart. 2. Move clockwise around the middle of the Smith chart as we move away from the load (toward generator). One rotation is /2 in the transmission line. 3. The final point is the input impedance of the line.

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Example #1 – Impedance Trans.  Normalized the parameters 0.67

Z 0  50 

Z L  50  j 25 

0.67

z L  1  j 0.5 

zin    

z L  j tan   1  j 0.5   j tan  2  0.67    1.299  j 0.485 1  jz L tan   1  j 1  j 0.5 tan  2  0.67 

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Example #1 – Impedance Trans.  Plot load impedance 0.67

zL  1  j 0.5 

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Example #1 – Impedance Trans.  Walk away from load 0.67 0.145 0.67

zL  1  j 0.5 

Since the Smith chart  repeats every 0.5,  traversing 0.67 is the  same as traversing 0.17. Here we start at 0.145 on  the Smith chart. We traverse around the  chart to  0.145 + 0.17 = 0.315. Smith Charts

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Example #1 – Impedance Trans.  Determine input impedance 0.67

zL  1  j 0.5 

Z in

Reflection at the load will be  the same regardless of the  length of line. Therefore the VSWR will the  same. The input impedance must  lie on the same VSWR plane. zin  1.3  j 0.5 Smith Charts

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Example #1 – Impedance Trans.  Denormalize 0.67

Z in

zL  1  j 0.5 

To determine the actual  input impedance, we  denormalize.

Z in  Z 0 zin   50  1.3  j 0.5   65  j 25 

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Example #2 – Calculate Line Length 3.5 cm

30  L  200 nH m

Z in

5 nH

C  163 pF m

f  1.5 GHz

Free Space Wavelength 0 

c0 3 108   20 cm f 1.5  109

Wavelength on the Line    LC 



2

  2 f







1 f LC

1

1.5 109 

 200 109 163 1012 

 11.68 cm

Note:  ≠ 0 so these are NOT the same.   Use  instead of 0 for Smith chart analysis. Smith Charts

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Example #2 – Calculate Z’s 3.5 cm

30  L  200 nH m

Z in

C  163 pF m

5 nH

f  1.5 GHz

Line Impedance Z0 

L 200 109   35.0  C 163  10 12

Load Impedance Z L  R  j L

 30  j 2 1.5  109  5  10 9   30  j 47.1 

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Example #2 – Normalize 3.5 cm

30  L  200 nH m

Z in

5 nH

C  163 pF m f  1.5 GHz

Normalize Load Impedance Z0 

L 200  109   35.0  C 163  1012

Load Impedance zL 

Z L 30  j 47.1   0.86  j1.34 Z0 35.0

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Example #2 – Transform 0.168

3.5 cm

30  L  200 nH m

Z in

C  163 pF m

5 nH

f  1.5 GHz

Line Length in Wavelengths 





3.5 cm  0.3 11.68 cm

Azimuthal Distance Start: 0.168 End: 0.168  0.30  0.468

0.468

Transformed Impedance zin  0.26  j 0.19

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Example #2 – Denormalize 0.168

3.5 cm

30  L  200 nH m

Z in

5 nH

C  163 pF m f  1.5 GHz

Denormalize Input Impedance Z in  Z 0 zin

  35.0   0.26  j 0.19   9.21  j 6.65  0.468

Exact Answer Z in  9.42  j 7.03

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Example #2 – Component Values 0.168

3.5 cm

30  L  200 nH m

Z in

C  163 pF m

5 nH

f  1.5 GHz

Component Values Z in  R 

1  9.21  j 6.65  jC

R  9.21  C

1  15.9 pF 2 1.5  109   6.65 

0.468

9.21  Z in

15.9 pF Smith Charts

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Impedance  Matching

Two‐Element Matching L Network The most common impedance matching circuit is the L network. Circuit Q is fixed and cannot be controlled.

Considerations: 1. Elimination of stray reactance. 2. Need for filtering. 3. Need to pass or block DC.

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Three‐Element Matching Pi and T Networks Three‐element matching networks are advantageous when it is desired to control the  circuit Q. Q must be greater than the Q possible with an L network. Pi Network

Q

T Network

max  Z s , Z L  fc  1 f R

Q

min  Z s , Z L  fc  1 f R

Virtual resistance R is used only during design.   It is not a physical component in the network.

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Two‐Element Impedance Matching with a  Smith Chart Z src  25  j15 

Z L  100  j 25 

Solution 1. Matching network must be low‐pass to  conduct DC. 2. This dictates series L and shunt C. 3. We walk along constant X circles until the  input  impedance is the source impedance. Z src  25  j15 

159 nH

38.8 pF Z L  100  j 25 

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What is Stub Tuning? (1 of 6) Z src Z0 , 

ZL

Power is reflected due to an impedance mismatch.



Z L  Z0 Z L  Z0

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What is Stub Tuning? (2 of 6) dA

Z src Z0 , 

ZL

A We want to add a short circuit stub to match the impedance.

0

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Stub Tuning Concept (3 of 6) dA

Z src Z0 , 

ZL YA  Z A1  Z 01  jBA

We back off from the load until the real part of the  input admittance is 1/Z0. At this point, the real part of admittance is matched to  the transmission line.

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Stub Tuning Concept (4 of 6) dA

zsrc Z0 , 

YSA

zL

Yin  Z 01 We can match perfectly to this admittance by  introducing a shunt element with the conjugate  susceptance.

YSA   jBA

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Stub Tuning Concept (5 of 6) A YSA   jBA

Z0 , 

To realize this shunt susceptance with a short‐circuit  stub, we back off some distance        from a short  A circuit load until the input admittance is –jBA. This is our stub.

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Stub Tuning Concept (6 of 6) dA

zsrc Z0 , 

zL

A Last, we add the stub at position dA from the load to  cancel the susceptance of the load. 1

Yin  Z 0

The load is matched and we have zero reflection!

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Voltage on Transmission Line  Before and After Stub Tuning zsrc Z0 , 

zL

V(z) without stub tuner

V(z) with stub tuner

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Single‐Stub Tuning Procedure Step 1 – You will be given Z0, ZL, n and f.  Normalize the impedances and calculate . zL 

ZL Z0



c0 nf

Step 2 – Plot zL and then invert to find the corresponding admittance yL. Step 3 – Walk CW around the constant VSWR circle until the R=G=1 circle intersects it.   There will be two intersections (A=closest, B=farthest). Step 4 – Pick A or B.  They will be complex conjugates.  Assume A for the following steps. Step 5 – How far in the CW direction did you travel to get to A and B?  These are dA and dB. Step 6 – yA is the admittance where the stub is about to be placed.  This is probably point A.   We need to cancel the reactive component A of this. yA  g A  j  A

 ySA  1  j  A

Step 7 – Find the –jA circle on the chart and follow it to the outside of the chart. Step 8 – Start at the far right side of the chart and move CW to the point above. Step 9 – Determine the distance (in wavelengths) this represents.  This is lA.

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Single‐Stub Tuning Example – Step 1 Problem: A 50  transmission line with an air‐core operates at 100 MHz and is connected to a  load impedance of ZL = 27.5 + j35 .  Design a single‐stub tuner. Step 1 – Normalize impedance and calculate . zL 



Z L 27.5  j 35    0.55  j 0.7 50  Z0 c0 3 108 m s  3m nf 1.0  100  106 Hz 

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Single‐Stub Tuning Example – Step 2 Plot impedance and invert to  find admittance. We read

yL  0.72  j 0.87

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Single‐Stub Tuning Example – Step 3 Walk CW around the constant  VSWR circle until the R=G=1 circle intersects it. We read

A

A  1  j1.1 B  1  j1.1

B

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Single‐Stub Tuning Example – Step 4 Pick A or B. We will pick A because it leads to  the shortest stub. A

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Single‐Stub Tuning Example – Step 5 0.165

How far CW did we traverse to  get to A? First part:  0.5 – 0.364 = 0.136 A

Second part: 0.165 Total: 0.136 + 0.165 = 0.301

d A  0.301

0 or 0.5

d A  90.3 cm

Smith Charts

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Single‐Stub Tuning Example – Step 6 yA is the admittance where the  stub is about to be placed.  We  chose point A.

 A  1  j1.1

A

We need to cancel the reactive  component A of this.

ySA   j1.1

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Single‐Stub Tuning Example – Step 7 Find the –jA circle on the chart  and follow it to the outside of the  chart.

ySA   j1.1

We are setting up to do an  admittance transformation in the  stub to realize a –j1.1 input  admittance.

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Single‐Stub Tuning Example – Step 8 Start at the far right side of the  chart and move CW to the point  above (move away from short). Here we are doing an admittance  transformation to realize –j1.1. The far right side of the Smith  chart is a short circuit for  admittances.  Admittance  transformation

A

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Single‐Stub Tuning Example – Step 9 Determine the distance (in  wavelengths) this represents.   This is lA. Line starts at 0.25. Line ends at 0.367. Total length: 0.367 – 0.25 = 0.117

 A  0.117  A  0.117  3 m   0.351 m  35.1 cm Smith Charts

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