Smith Chart & Matching Network
James Lu
James J. Q. Lu ECSE 2100 Fields & Waves I 1
Non matched Impedance (0)
• Reflections lead to Zin variations with line length and frequency • Power is wasted because of reactive power, which can also damage equipment during short circuit (for example) • Only partial power is delivered to the load. • SWR > 1: there will be voltage maxima on the line, voltage breakdown at high power levels • Noises (bounces or echoes)
James J. Q. Lu ECSE 2100 Fields & Waves I 2 Benefits of Matching (=0)
• Zin = ZO, independent of line length, and frequency (over the bandwidth of the matching network) • Maximum power transfer to the load is achieved • SWR = 1: no voltage peaks on the line • No bounces (echoes)
James J. Q. Lu ECSE 2100 Fields & Waves I 3 UQ
Load Matching
What if the load cannot be made equal to Zo for some other reasons? Then, we need to build a matching network so that the source effectively sees a match load.
Ps Z0 M ZL
0
Typically we only want to use lossless devices such as capacitors, inductors, transmission lines, in our matching network so that we do not dissipate any power in the network and deliver all the available power to the load.
James J. Q. Lu ECSE 2100 Fields & Waves I 4 Normalized Impedance It will be easier if we normalize the load impedance to the characteristic impedance of the transmission line attached to the load. Z z r jx Zo 1 z 1 Since the impedance is a complex number, the reflection coefficient will be a complex number u jv
2 2 2v 1 u v x r 2 2 1 u 2 v2 1 u v
James J. Q. Lu ECSE 2100 Fields & Waves I 5
Smith Charts The impedance as a function of reflection coefficient can be re-written in the form: 2 2 C r S2 1 1 u v Du T v2 r 2 1 u 2 v2 E 1 r U 1 r
2v C S2 x 2 1 1 2 2 u 1 Dv T 1 u v E x U x2
These are equations for circles on the (u,v) plane 2 2 2 x xo (y yo ) a
James J. Q. Lu ECSE 2100 Fields & Waves I 6 Smith Chart – Real Circles
1 Im 1 Circle
0.5
r=0 r=1/3 r=1 r=2.5 1 0.5 0 0.5 1 Re
0.5
1
James J. Q. Lu ECSE 2100 Fields & Waves I 7
Smith Chart – Imaginary Circles Im1
x=1/3 x=1 x=2.5 0.5
1 0.5 0 0.5 1 Re
x=-1/3 x=-1 x=-2.5 0.5
1
James J. Q. Lu ECSE 2100 Fields & Waves I 8 z L1 L Impedances, voltages, Smith Chart z L1 currents, etc. all repeat 1 every half wavelength zL r L jxL 1
1 l e j2 l zl l 1 l j 4 l Inductive e TG W R z=1+j 1 W S ro (atP max) S 1
z=1 ro Short P Open Pmin =0 Pmax (z=0) (z=)
= 1 =1
L y=1/(1+j)
T W =0.5-j0.5 Capacitive
Purely real impedances along Purely imaginary impedances the horizontal centre line along the periphery
James J. Q. Lu ECSE 2100 Fields & Waves I 9
Smith Chart Example 1
Given: - L 0.5 45 Z 50 o G
WT What is ZL?
ZL 501.39 j1.35
69.5 j67.5
L
T W
James J. Q. Lu ECSE 2100 Fields & Waves I 10 Smith Chart Example 2
Given: ZL 15 j25 Zo 50
G What is L? WT 15 j25 z L 50 0.3 j0.5
L 0.6- 123
z1 = 2 + j
z2 = 1.5 -j2 L
T z3 = j4 W z4 = 3 z5 = infinity z6 = 0 z7 = 1 z8 = 3.68 -j18
James J. Q. Lu ECSE 2100 Fields & Waves I 11
Smith Chart Example 3
Given: ZL 50 j50 Zo 50 6.78ns
G
WT What is Zin at 50 MHz? 50 j50 z L 50 1.0 j1.0 - L 0.445 64
e j2l e j4 l / e j2
in L L L L
T 2 244 W l f 504106 46.78410 9 0.339
in 180 - in 0.445 180
Zin 500.38 j0.0 19
James J. Q. Lu ECSE 2100 Fields & Waves I 12 Admittance
A matching network is going to be a combination of elements connected in series AND parallel. Z1 Impedance is well suited when working with series configurations. For example: Z2 V ZI ZL Z1 Z2
Impedance is NOT well suited when working with parallel configurations. Z1 Z2 Z Z Z 1 2 L Z1 Z2
For parallel loads it is better to work with admittance. 1 Y1 Y2 Y1 I YV Z1 YL Y1 Y2
James J. Q. Lu ECSE 2100 Fields & Waves I 13
Normalized Admittance Y y YZo g jb Yo 1 y 1 2 1 u2 v2 C g S 1 Du T v2 g D T 2 1 u 2 v2 E 1 g U 1 g
2v C S2 b 2 1 1 2 2 u 1 Dv T 1 u v E b U b2
These are equations for circles on the (u,v) plane
James J. Q. Lu ECSE 2100 Fields & Waves I 14 Admittance Smith Chart
Conductance Circles Susceptance Circles
1 1 Im Im b=-1 b=-1/3 b=-2.5 0.5 0.5
g=2.5 g=1 g=1/3 g=0 b= 1 0.5 0 0.5 1 1 g= 0.5 0 0.5Re 1 Re
b=2.5 b=1/3
0.5 0.5 b=1
1 1
James J. Q. Lu ECSE 2100 Fields & Waves I 15
Impedance and Admittance Smith Charts
• For a matching network that contains elements connected in series and parallel, we will need two types of Smith charts Z impedance Smith chart Z admittance Smith Chart • The admittance Smith chart is the impedance Smith chart rotated 180 degrees. – We could use one Smith chart and flip the reflection coefficient vector 180 degrees when switching between a series configuration to a parallel configuration.
James J. Q. Lu ECSE 2100 Fields & Waves I 16 Admittance Smith Chart Example 1 Given: y 1 j1
Find z G
WT • Procedure: Plot y •Plot 1+j1 on chart •vector =0.445-64 Flip 180 •Flip vector 180 degrees degrees 0.445- 116
Read L
T & z
z 0.5 j0.5 W
James J. Q. Lu ECSE 2100 Fields & Waves I 17
Admittance Smith Chart Example 2 Given: - 0.5 45 Zo 50 Find Y G • Procedure: WT •Plot Plot • Flip vector by 180 degrees • Read coordinate Read y Flip 180
degrees
L
y 0.38 j0.36 T W 1 Y 0.38 j0.36 50 7.6 j7.2 410 3 S
James J. Q. Lu ECSE 2100 Fields & Waves I 18 Matching Example
Ps Z0 50 M 100
0
Match 100 load to a 50 system at 100MHz
A 100 resistor in parallel would do the trick, but ½ of the power would be dissipated in the matching network. We want to use only lossless elements such as inductors and capacitors so we don’t dissipate any power in the matching network
James J. Q. Lu ECSE 2100 Fields & Waves I 19
Matching Example
We need to go from z=2+j0 to z=1+j0 on the Smith chart We won’t get any G closer by adding WT series impedance so we will need to add something in parallel. We need to flip over to the admittance
chart
L
T W
Impedance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 20 Matching Example
y=0.5+j0 Before we add the admittance, add a
mirror of the r=1 G
circle as a guide. WT
L
T W
Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 21
Matching Example
y=0.5+j0 Before we add the admittance, add a
mirror of the r=1 G circle as a guide WT Now add positive imaginary
admittance.
L
T W
Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 22 Matching Example
y=0.5+j0 Before we add the admittance, add a
mirror of the r=1 G circle as a guide WT Now add positive imaginary admittance jb = j0.5 jb j0.5 j0.5
j2100MHz C
L 50 T C 16pF W
16pF 100 Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 23
Matching Example
We will now add series impedance Flip to the
impedance Smith G Chart WT We land at on the r=1 circle at x=-1, i.e.
z = 1 – j1
L
T W
Impedance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 24 Matching Example
Add positive imaginary admittance to get to z=1+j0 G
WT jx j1.0 j1.0 50 j2 100MHz L
L 80nH
80nH L T W
16pF 100 Impedance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 25
Matching Example
This solution would have also worked
G
WT
32pF
L
T W
160nH 100 Impedance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 26 Matching Bandwidth
0 80 nH
5 16 pF 10 100 G T W 15 150 MHz
20
25 Reflection Coefficient (dB) 30
35
50 MHz L
40 T 50 60 70 80 90 100 110 120 130 140 150 W Frequency (MHz)
Because the inductor and capacitor impedances change with frequency, the match works over a narrow frequency range Impedance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 27
Single Stub Tuner
yl1 l 1
yl2 Z Z0 Z0 L
Z Goal: 0 0 z =1 (y =1) l in in 2 Open 1 1 l or yl Short zl 1 l 2 l = 1 l e yin = yl1 + yl2 = (gl1 + jbl1) + jbl2
gl1 = 1 (real-part condition) Stub length l = up=f bl1 = -bl2 (imaginary-part condition) Phase shift: =2l=2(2/)l=4(l/) (2 Degrees of freedom) Open: Z = -jZ cotl, or z = -jcotl in o l in l Short: Zin = jZotan , or zin = jtan James J. Q. Lu ECSE 2100 Fields & Waves I 28 Single Stub Tuner Match 100 load to a 50 system at 100MHz using two transmission lines connected in parallel
Flip to Admittance G chart WT y=0.5+j0 Adding length to Cable 1 rotates the reflection
coefficient
L
clockwise to g=1. T W
l1 = 0.152
y1l=1+j0.72 Admittance Chart James J. Q. Lu ECSE 2100 Fields & Waves I 29
Single Stub Tuner
The stub has to add a normalized admittance of -0.72
to bring the G trajectory to the WT center of the Smith
Chart
L
T W
Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 30 Single Stub Tuner
An short stub of zero length has an admittance=j By adding enough G WT cable to the short stub, the admittance of the stub will
reach to -0.72
L
T W
l2 = (0.401-0.25) = 0.151
Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 31
Single Stub Tuner
100 l1 = 0.152 0
G
l2 = 0.151 WT
L
T W
Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 32 Single Stub Tuner
l1 = 0.347 100 0
G
l2 = 0.097 WT
This solution would
have worked as well.
L
T W An open stub of zero length has an admittance=j0 Admittance Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 33
Single Stub Tuner Matching Bandwidth
l1 = 0.347 0 100
l = 0.097 G 2 T W 0
5
10
15
20 50 MHz
25
L
Reflection Coefficient (dB) T 30 W
35 150 MHz
40 50 60 70 80 90 100 110 120 130 140 15 Frequency (MHz) Because the cable phase changes linearly with frequency, the match works over a Impedance narrow frequency range Chart
James J. Q. Lu ECSE 2100 Fields & Waves I 34 Summary • Impedance matching is necessary to: –reduce VSWR – obtain maximum power transfer • Lump reactive elements and a single stub can be used. • A quarter-wave line can also be used to transform resistance values, and act as an impedance inverter. • These matching network types are narrow-band: they are designed to operate at a single frequency only.
James J. Q. Lu ECSE 2100 Fields & Waves I 36