1. Use s parameters-determining_inductance_capacitance 2. Relationship Between Common Circuits and the ABCD Parameters 3. Converts Z-parameters to S-parameters 4. Relationships Between Two-Port S and ABCD Parameters 5. Via and equivalent circuit

1. 1
Use S-parameters
Determining Inductance and
Capacitance
Jay Chang

2. 2
Relationship Between Common Circuits and the ABCD Parameters
A B
C D
 
 
 

3. 3
Converts Z-parameters to S-parameters
• where U is the identity or unit matrix and Zn is the termination impedance of each port.
• The derivation assumes that each port is terminated in the same value.
• Equation above converts Z-parameters to S-parameters.
• conversion back from S-parameters to Z-parameters for an arbitrary-sized matrix:
• Derive the transformation of the ABCD to S-parameters.
• The final solution is summarized in
• where Zn is the termination impedance at the ports, which are all assumed to be equal.
• The final relationship between a two-port Z-matrix and the ABCD matrix is shown as:
[ ] [ ] [ ]A Z S→ →

4. 4
Relationships Between Two-Port S and ABCD Parameters
Zn is the termination impedance at the ports

5. 5
Via and equivalent circuit
Ex1: Extract an equivalent circuit for the via shown in Figure from the following S-parameter matrix
measured at 5 GHz assuming port impedance values (Zn) of 50 Ohm:
Step 1: Transform the S-matrix into ABCD parameters using the relations between 2-port S and ABCD
parameters.
Sol:
Step 2: Choose the form of the equivalent circuit. A signal propagating through the via will experience
the capacitance of the via pad, the inductance of the barrel, and then the capacitance of the second
pad. This configuration fits the pi model.
Pi model

6. 6
Step 3: Use the relations in Table to calculate the admittance of each segment in the pi model
Due to symmetry in the circuit, Y1 = Y2.
Step 4: Calculate the circuit values:

7. 7
The model can be extracted as either a Pi or a T network now we show T model.
The inductance values will include the L of the trace and the via barrel (it is assumed that the test
setup minimizes the trace length, and subsequently the trace capacitance is minimal).
The capacitance represents the via pads
Ex2: Assume the following s-matrix measured at 5 GHz
Convert to ABCD parameters
T model
11 12
21 22
0.110 0.153 0.798 0.572
0.798 0.572 0.110 0.153
S S j j
S S j j
− − −   
=   − − −  
0.827 20.08
0.0157 0.827
A B j
C D j
   
=   
   
Relating the ABCD parameters to the T circuit
topology, the capacitance and inductance is
extracted from C & A.
3
1
1 2
3
1 1
0.0157 0.5
1
2
2
1 0.827 1 0.35
1/ ( 2 )
VIA
VIA
VIA
C j C pF
Z
j fC
Z j fL
A L L nH
Z j fC
π
π
π
= = = ⇒ =
⋅
⋅
= + = = + ⇒ = =
⋅
1 1 2
1 2
3 3
2
3 3
1
0.827 20.08
0.0157 0.8271
1
Z Z Z
Z Z
Z Z j
Z j
Z Z
 
+ + +     =     + 
 

8. 8
Ex3: Using the two independently measured values of S-parameters for a via and a loss-free
transmission line at 5 GHz, calculate the, equivalent S-parameters that would be measured if the two
circuits were cascaded as shown Figure. Assume that the termination impedance is 50 Ohm.
(a) configuration of the independently measured via;
(b) transmission line; (c) via cascaded with the transmission line.

9. 9
Sol:
Step 1: Convert to ABCD parameters
Step 2: Multiply the ABCD matrices
Step 3: Convert back to S-parameters using Table where Zn = 50 Ohm (the termination impedance)

10. 10
Thank you for your attention

11. Chapter4:MicrowaveNetworkAnalysis
TABLE 4.2 Conversions Between Two-Port Network Parameters
S Z Y ABCD
S11 S11
(Z11 − Z0)(Z22 + Z0) − Z12 Z21
Z
(Y0 − Y11)(Y0 + Y22) + Y12Y21
Y
A + B/Z0 − C Z0 − D
A + B/Z0 + C Z0 + D
S12 S12
2Z12 Z0
Z
−2Y12Y0
Y
2(AD − BC)
A + B/Z0 + C Z0 + D
S21 S21
2Z21 Z0
Z
−2Y21Y0
Y
2
A + B/Z0 + C Z0 + D
S22 S22
(Z11 + Z0)(Z22 − Z0) − Z12 Z21
Z
(Y0 + Y11)(Y0 − Y22) + Y12Y21
Y
−A + B/Z0 − C Z0 + D
A + B/Z0 + C Z0 + D
Z11 Z0
(1 + S11)(1 − S22) + S12S21
(1 − S11)(1 − S22) − S12S21
Z11
Y22
|Y|
A
C
Z12 Z0
2S12
(1 − S11)(1 − S22) − S12S21
Z12
−Y12
|Y|
AD − BC
C
Z21 Z0
2S21
(1 − S11)(1 − S22) − S12S21
Z21
−Y21
|Y|
1
C
Z22 Z0
(1 − S11)(1 + S22) + S12S21
(1 − S11)(1 − S22) − S12S21
Z22
Y11
|Y|
D
C
Y11 Y0
(1 − S11)(1 + S22) + S12S21
(1 + S11)(1 + S22) − S12S21
Z22
|Z|
Y11
D
B
Y12 Y0
−2S12
(1 + S11)(1 + S22) − S12S21
−Z12
|Z|
Y12
BC − AD
B
Y21 Y0
−2S21
(1 + S11)(1 + S22) − S12S21
−Z21
|Z|
Y21
−1
B
Y22 Y0
(1 + S11)(1 − S22) + S12S21
(1 + S11)(1 + S22) − S12S21
Z11
|Z|
Y22
A
B
A
(1 + S11)(1 − S22) + S12S21
2S21
Z11
Z21
−Y22
Y21
A
B Z0
(1 + S11)(1 + S22) − S12S21
2S21
|Z|
Z21
−1
Y21
B
C
1
Z0
(1 − S11)(1 − S22) − S12S21
2S21
1
Z21
−|Y|
Y21
C
D
(1 − S11)(1 + S22) + S12S21
2S21
Z22
Z21
−Y11
Y21
D
|Z| = Z11 Z22 − Z12 Z21; |Y| = Y11Y22 − Y12Y21; Y = (Y11 + Y0)(Y22 + Y0) − Y12Y21; Z = (Z11 + Z0)(Z22 + Z0) − Z12 Z21; Y0 = 1/Z0.