Bern University of Applied Sciences Engineering and Information Technology Division of Electrical and Communication Technology
Smith V3.10
Prof. Fritz Dellsperger 5.2010
Content
Impedance Matching 1 Impedance, Admittance, Reflection Coefficient, VSWR and Return Loss 1 Matching arbitrary impedances to 50 Ohm 2
Example 1: Use 2 reactance elements, Highpass 2 Example 2: Use 2 reactance elements, Highpass 3 Example 3: Use 2 reactance elements, Lowpass 4 Example 4: Use 2 reactance elements, Lowpass 5 Example 5: Antenna Match with 3 or more reactance elements, Low Q, Highpass 6 Example 6: Antenna Match with 3 or more reactance elements, Low Q, Lowpass 7 Example 7: Match Ceramic Filter to 50 Ohm 8 Example 8: Antenna match using reactance and serie line element 10 Example 9: Antenna match using serie line and open stub 11 Example 10: Antenna match using serie line and shorted stub 12 Example 11: Antenna match using double stub tuner 13 Example 12: Nonsynchronous Transformer 15
Low Noise Amplifier Design 18 Example 13: Low Noise Amplifier, 2.0 GHz 18
Conjugate Matching 26 Example 14: Conjugate Match 26
Serial Transmission Line with Attenuation 29 Example 15: Match using transmission line with loss 29
Sweeps 30 Example 16: Input impedance of a Chebyshev lowpass filter 30 Example 17: Broadband load match 31
1
Impedance Matching
Impedance, Admittance, Reflection Coefficient, VSWR and Return Loss
Impedance: Z R jX R: Resistance X: Reactance
0
1Z Z
1
Admittance: Y G jB G: Conductance B: Susceptance
1
ZY
1
YZ
Reflection Coefficient: 0 0
0 0
Z Z Y Y
Z Z Y Y
0Z : Reference Impedance
RL
020
0
Z ZVSWR 110
VSWR 1 Z Z
Voltage Standing Wave Ratio:
0 0
RL20
0RL
020 Z Z Z Z
1 Z Z1 10VSWR
1 Z Z1 10
Return Loss: 0
0
Z ZVSWR 1RL 20 log 20 log 20 log
VSWR 1 Z Z
All values for cursor position in Smith-Chart are displayed in window “Cursor”.
2
Matching arbitrary impedances to 50 Ohm
Example 1: Use 2 reactance elements, Highpass Problem: Match an impedance of 10 j7 to 50 . Use 2 reactance (L,C) in a circuit topology
with highpass characteristic. Frequency: 150 MHz. Smith project file: Example1.xmlsc
3
Example 2: Use 2 reactance elements, Highpass Problem: Match an impedance of 100 j50 to 50 . Use 2 reactance (L,C) in a circuit
topology with highpass characteristic. Frequency: 500 MHz. Smith project file: Example2.xmlsc
4
Example 3: Use 2 reactance elements, Lowpass Problem: Match an impedance of 10 j7 to 50 . Use 2 reactance (L,C) in a circuit topology
with lowpass characteristic. Frequency: 150 MHz. Smith project file: Example3.xmlsc
5
Example 4: Use 2 reactance elements, Lowpass Problem: Match an impedance of 100 j50 to 50 . Use 2 reactance (L,C) in a circuit
topology with lowpass characteristic. Frequency: 500 MHz. Smith project file: Example4.xmlsc
6
Example 5: Antenna Match with 3 or more reactance elements, Low Q, Highpass Problem: Match an antenna impedance of 20 j12 to 50 . Use L and C in a circuit topology
with highpass characteristic and do not exceed a max
X 12Q 0.6
R 20 (for maximum bandwidth).
Frequency: 450 MHz. Smith project file: Example5.xmlsc
7
Example 6: Antenna Match with 3 or more reactance elements, Low Q, Lowpass Problem: Match an antenna impedance of 20 j12 to 50 . Use L and C in a circuit topology
with lowpass characteristic and do not exceed a max
X 12Q 0.6
R 20 (for maximum bandwidth).
Frequency: 450 MHz. Smith project file: Example6.xmlsc
8
Example 7: Match Ceramic Filter to 50 Ohm Problem: For measurement purposes match a 10.7 MHz 300 Ohm Ceramic filter to 50 Ohm using a parallel resonance circuit with capacitive voltage divider and L = 330 nH. Frequency: 10.7 MHz. Smith project file: Example7.xmlsc
f0 10.7 MHz R1 300 L 390 nH
B2 L f0
2
R1791.298 kHz
9
C2
C1
RL
Rres
For Q ≥ 10 following approximations can be used:
ofQ
B
res
1C
2 BR
2o
1L
C
res
L
RN
R
p
N 2
1
CC
N 1
2C NC
10
Example 8: Antenna match using reactance and serie line element Problem: Match an antenna impedance of 30 j40 to 50 . Use one reactance and one
serie line. Frequency: 430 MHz. Smith project file: Example8.xmlsc
11
Example 9: Antenna match using serie line and open stub Problem: Match an antenna impedance of 30 j40 to 50 . Use one serie line and an open
stub. Frequency: 430 MHz. Smith project file: Example9.xmlsc
12
Example 10: Antenna match using serie line and shorted stub Problem: Match an antenna impedance of 30 j40 to 50 . Use one serie line and a shorted
stub. Frequency: 430 MHz. Smith project file: Example10.xmlsc
13
Example 11: Antenna match using double stub tuner Problem: Match an antenna impedance of 30 j40 to 50 . Use a double stub tuner with
serie line length of 80 mm and r 1 . Frequency: 430 MHz.
Smith project file: Example11.xmlsc
14
Use Edit Element (doubleclick on element in schematic) to adjust for desired line length.
Change value
Push Draw and see on Smith-chart and schematic how this affect transformation
OK when done
15
Example 12: Nonsynchronous Transformer Problem: Match an impedance of 10 j12 to 50 . Use an open stub and a nonsynchronous
transformer. Frequency: 2.4 GHz. Smith project file: Example12.xmlsc
16
Use Edit Element (doubleclick on element in schematic) to adjust for desired line length.
Change value
Push Draw and see on Smith-chart and schematic how this affect transformation
OK when done
17
Properties of Nonsynchronous Trafo: It uses two pieces of line with the same length. One line must have the line impedance of the source and the other line the impedance of the load. The total length depends on impedance ratio and is much shorter than / 4 .
L = Length of one line section
1 2k Z / Z 1
2 11r
L a tan
kk
Nonsynchronous-Transformer
Ltot
0.1 1 100
0.05
0.1
0.15
0.2
k
Lin
elen
gth
in L
ambd
a
L
0.1 1 1015
20
25
30
35
k
Lin
elen
gth
in D
egre
e
L
Line length as function of impedance ratio k
18
Low Noise Amplifier Design
Example 13: Low Noise Amplifier, 2.0 GHz
Problem: Design input and output matching network for a LNA using BFG33G. Frequency: 2 GHz Smith project file: Example13-input.xmlsc
Z0
Z0
Output matching network
Input matching network
11 12
21 22
S S
S S
Transistor
2
Z2
S
ZS
1
Z1
L = 2*
ZL = Z2*
If we choose a source impedance of ZS = (50+j30)Ohm we get a gain of approx. 12 dB and a NF of approx. 3.85 dB with only a serie inductor as input matching network.
19
With a few calculations we get:
S-parameters at 2000 MHz: Source impedance at 2000 MHz:
ZS 50 j 30( )
S0.121 e
j 149 deg
3.756 ej 78.9 deg
0.108 ej 59.4 deg
0.41 ej 54.5 deg
Z0 50
S
ZS Z0
ZS Z00.083 0.275j S 0.287
arg S 73.301deg
2 S2 2
S1 2 S
2 1 S
1 S1 1 S
0.136 0.392j
Z2 Z0
1 2
1 2 45.987 43.54j( )
For the output network we conjugately match Z2 (or 2 ) to 50 Ohm.
There are several possibilities to realize the output matching network.
20
Problem: Output matching network 1 Smith project file: Example13-output1.xmlsc
21
Simulation of LNA with Output matching network 1 versus Frequency in Agilent ADS:
CC1C=1.9 pF
LL2
R=L=3.8 nH
LL1
R=L=2.4 nH
S2PSNP1File="BFG33G.S2P"
21
Ref
TermTerm2
Z=50 OhmNum=2
TermTerm1
Z=50 OhmNum=1
m1freq=S(2,2)=0.015 / 119.981impedance = 49.220 + j1.312
2.000GHz
m2freq=S(1,1)=0.373 / 125.040impedance = 27.457 + j19.485
2.000GHzm5freq=dB(S(1,1))=-8.564
2.000GHz
m6freq=dB(S(2,2))=-36.259
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
-35
-30
-25
-20
-15
-10
-5
-40
0
freq, GHz
dB(S
(1,1
))
m5
dB(S
(2,2
))
m6
Returnloss vs Freq
m5freq=dB(S(1,1))=-8.564
2.000GHz
m6freq=dB(S(2,2))=-36.259
2.000GHz
m4freq=nf(2)=3.852
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
3.7
3.8
3.9
3.6
4.0
freq, GHz
nf(2
)
m4
NF vs Freq
m4freq=nf(2)=3.852
2.000GHzm3freq=dB(S(2,1))=12.005
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
10
11
12
13
14
9
15
freq, GHz
dB(S
(2,1
))
m3
Gain vs Freq
m3freq=dB(S(2,1))=12.005
2.000GHz
freq (1.500GHz to 2.500GHz)
S(2
,2) m1
S(1
,1)
m2
Output 1 Match
m1freq=S(2,2)=0.015 / 119.981impedance = 49.220 + j1.312
2.000GHz
m2freq=S(1,1)=0.373 / 125.040impedance = 27.457 + j19.485
2.000GHz
22
Problem: Output matching network 2 Smith project file: Example13-output2.xmlsc
23
Simulation of LNA with Output matching network 2 versus Frequency in Agilent ADS:
m1freq=S(2,2)=9.355E-4 / 17.504impedance = 50.089 + j0.028
2.000GHz
m2freq=S(1,1)=0.380 / 124.571impedance = 27.160 + j19.854
2.000GHz
freq (1.500GHz to 2.500GHz)
S(2
,2) m1
S(1
,1)
m2
Output 2 Match
m1freq=S(2,2)=9.355E-4 / 17.504impedance = 50.089 + j0.028
2.000GHz
m2freq=S(1,1)=0.380 / 124.571impedance = 27.160 + j19.854
2.000GHzm5freq=dB(S(1,1))=-8.408
2.000GHz
m6freq=dB(S(2,2))=-60.579
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
-35
-30
-25
-20
-15
-10
-5
-40
0
freq, GHz
dB
(S(1
,1))
m5
dB
(S(2
,2))
m6
Returnloss vs Freq
m5freq=dB(S(1,1))=-8.408
2.000GHz
m6freq=dB(S(2,2))=-60.579
2.000GHz
m4freq=nf(2)=3.852
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
3.7
3.8
3.9
3.6
4.0
freq, GHz
nf(
2)
m4
NF vs Freq
m4freq=nf(2)=3.852
2.000GHzm3freq=dB(S(2,1))=12.006
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
10
11
12
13
14
9
15
freq, GHz
dB
(S(2
,1))
m3
Gain vs Freq
m3freq=dB(S(2,1))=12.006
2.000GHz
24
Problem: Output matching network 3 Smith project file: Example13-output3.xmlsc
25
Simulation of LNA with Output matching network 3 versus Frequency in Agilent ADS:
m1freq=S(2,2)=0.013 / -67.850impedance = 50.471 - j1.197
2.000GHz
m2freq=S(1,1)=0.385 / 124.358impedance = 26.904 + j20.082
2.000GHz
freq (1.500GHz to 2.500GHz)
S(2
,2) m1
S(1
,1)
m2
Output 3 Match
m1freq=S(2,2)=0.013 / -67.850impedance = 50.471 - j1.197
2.000GHz
m2freq=S(1,1)=0.385 / 124.358impedance = 26.904 + j20.082
2.000GHzm5freq=dB(S(1,1))=-8.289
2.000GHz
m6freq=dB(S(2,2))=-37.854
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
-35
-30
-25
-20
-15
-10
-5
-40
0
freq, GHz
dB(S
(1,1
))
m5
dB(S
(2,2
))
m6
Returnloss vs Freq
m5freq=dB(S(1,1))=-8.289
2.000GHz
m6freq=dB(S(2,2))=-37.854
2.000GHz
m4freq=nf(2)=3.852
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
3.7
3.8
3.9
3.6
4.0
freq, GHz
nf(2
)
m4
NF vs Freq
m4freq=nf(2)=3.852
2.000GHzm3freq=dB(S(2,1))=12.005
2.000GHz
1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.41.5 2.5
10
11
12
13
14
9
15
freq, GHz
dB(S
(2,1
))
m3
Gain vs Freq
m3freq=dB(S(2,1))=12.005
2.000GHz
26
Conjugate Matching
Example 14: Conjugate Match
Problem: Conjugately match impedance Z1 (or Gamma 1 ) to 50 Ohm.
Method 1: Start at Z1 and transform with network to 50 Ohm. In this case Z1 is used as load impedance for the network and after transformation we would like to see 50 Ohm at the input of the network. Method 2: Start at 50 Ohm and transform with network to Z1* = conjugate Z1
In this case 50 Ohm is used as load impedance for the network and after transformation we would like to see Z1* into the input of the network. Both method result in the same network.
T1
L1
C1
50
Method 1 Method 2
1
1Z
*1
*1
Z
27
Example: 1Z 10 j10
Method 1: Smith project file: Example14-1.xmlsc
Z1
50
Start here
28
Method 2: Smith project file: Example14-2.xmlsc
Z1*
50
Start here
29
Serial Transmission Line with Attenuation
Example 15: Match using transmission line with loss
Problem: Match an impedance of (23.7 + j 101) Ohm to 50 Ohm using a lossy transmission line with an electrical length of about 2 wavelength, attenuation of 2 dB/m and a serial reactance. Frequency: 500 MHz Smith project file: Example15.xmlsc
30
Sweeps
Example 16: Input impedance of a Chebyshev lowpass filter
Problem: Plot input impedance of a 50 Ohm Chebyshev lowpass filter with n = 3, Ripple = 0.1 dB and cut-off frequency = 100 MHz Frequency: 10 MHz to 450 MHz Smith project file: Example16.xmlsc
31
Example 17: Broadband load match
Problem: Given: Load impedance = (100.8+j24.2) Ohm @ 140 MHz, (106.9+j41.7) Ohm @ 145 MHz, (121.2+j60) Ohm @ 150 MHz Find LC-lowpass network to match within VSWR of 1.2. Use standard component values as possible. Frequency: 140 MHz to 150 MHz Smith project file: Example17.xmlsc
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