EECS525 HW1∗

Wenhao Peng

February 22, 2019

Modelling MESFET TG13501

First, a biasing is tuned for the required condition where IDS = 15 mA and VDS = 3 V. Ideal lumpedelements are used as RF chokes.

Figure 1: Biasing

Then the S-Parameters from the datasheet are entered into a touchstone file and imported into ADS.The same small signal MESFET model is used. The cold device is obtained using the following biasing.

Figure 2: Biasing cold device

∗Due to a visit to ISSCC2019, a late submission (on Friday) was approved by Prof. Mortazawi.1Equations from the lecture hot/cold parameters extraction are used. All work done in ADS.

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Then the cold S-Parameters are converted to Z parameters and then the same formulas are used forextracing the cold parasitics.

Figure 3: sp to z

Figure 4: sp to y

These parameters are extracted.

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Figure 5: Relaxation time

Figure 6: R and C

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Figure 7: gm

Figure 8: L

The inductor extraction did not work out because they appear to be negative. This can be explainedby the datasheet schematic where the input has tapers and transmission lines and the output also hastransmission lines. Across this wide frequency range, it is not possible to model them as single inductors.Nevertheless, an initial result is obtained as follows, where negative components are omitted.

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Figure 9: Initial Guess

Figure 10: Initial Guess on port 9 and 10; S parameters from datasheet on port 1 and 2; ADS model on port3 and 4.

Two of the capacitances are close, and the drain-to-source resistance is close. Finally an optimizationis performed to make the small signal model closer to the ADS model, and the three inductors and gateresistance are included to allow additional tuning dimensions.

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Figure 11: Initial Values for Optimization

Figure 12: Device parameters from datasheet

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Figure 13: Optimization setup and final model parameters

Although numerically these values are not very close to each other, they are still within the same orderof magnitude. This is also reasonable because the actual datasheet is based on a schematic where input andoutput ports are more complicated than a vanilla passive lumped element. Furthermore, after tuning, the S

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Parameters of my model match those of the ADS model better. It is also reasonable that the ADS modeland the datasheet has some discrepencies because nonidealities are usually smoothed out when they createthat ADS model. On the Smith Chart, these values are compared.

Figure 14: S Parameters on Smith Charts. Datasheet SP on port 1 and 2; ADS model on port 3 and 4; andmy optimized model on port 11 and 12. They do resemble each other, although further improvements mightbe possible by tweaking the optimization knobs.

Problem 3.252

(a) Prescribed operating power gain of 10 dB

∆ = S11S22 − S12S21 = 0.3242 6 − 64.83◦

whose magnitude is less than 1, and

K =1− |S11|2 − |S22|2 + |∆|2

2|S12S21|= 1.033 > 1

2This problem was calculated step-by-step on CASIO fx-991ES PLUS by hand. Numerical errors apply.

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so this transistor is unconditionally stable.

gp =Gp

|S21|2=

10

1.922= 2.713

C2 = S22 −∆S∗11 = 0.76776 − 33.85◦

The constant power gain circle for this bilateral device is therefore given by center

Cp =gpC

∗2

1 + gp(|S22|2 − |∆|2)= 0.7813 6 33.85◦

and radius

rp =

√1− 2K|S12S21|gp + |S12S21|2g2p|1 + gp(|S22|2 − |∆|2)|

= 0.2141

Then select ΓL closest to the center (a.k.a. smallest distance from origin, which is magnitude of Cp minusradius rp)

ΓL = 0.5672 6 33.85◦

Γs = Γ∗IN =

[S11 +

S12S21ΓL

1− S22ΓL

]∗= 0.27606 93.35◦

Now

ΓOUT = S22 +S12S21Γs

1− S11Γs= 0.8597 6 − 33.85◦

so

ML =(1− |ΓL|2)(1− |ΓOUT |2)

|1− ΓLΓOUT |2= 0.6741

(V SWR)in = 1; (V SWR)out =1 +√

1−ML

1−√

1−ML

= 3.661

(b) Prescribed maximum power gain

gp,max =1

|S12S21|(K −

√K2 − 1) = 5.168

ΓML =gp,maxC

∗2

1 + gp,max(|S22|2 − |∆|2)= 0.95076 33.85◦

Γs = Γ∗IN =

[S11 +

S12S21ΓL

1− S22ΓL

]∗= 0.72836 135.4◦

which coincides (within tolerable numerical errors) the results obtained from simultaneous bilateral matching,where

B1 = 1 + |S11|2 − |S22|2 − |∆|2 = 0.2525

B2 = 1 + |S22|2 − |S11|2 − |∆|2 = 1.537

C1 = S11 −∆S∗22 = 0.12036 − 135.4◦

C2 = S22 −∆S∗11 = 0.76776 − 33.85◦

Γms =B1 −

√B2

1 − 4|C1|22C1

= 0.7311 6 135.4◦

Γml =B2 −

√B2

2 − 4|C2|22C2

= 0.95546 33.85◦

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Problem 3.273

(a) 10dB constant gain circle

∆ = S11S22 − S12S21 = 0.5168 6 − 142.7◦

K =1− |S11|2 − |S22|2 + |∆|2

2|S12S21|= 0.5509 < 1

so this transistor is potentially unstable. Following the same procedure as the unconditionally stable.

gp =Gp

|S21|2=

10

2.432= 1.694

C2 = S22 −∆S∗11 = 0.33296 − 76.87◦

The constant power gain circle for this bilateral device is therefore given by center

Cp =gpC

∗2

1 + gp(|S22|2 − |∆|2)= 0.4094 6 76.87◦

and radius

rp =

√1− 2K|S12S21|gp + |S12S21|2g2p|1 + gp(|S22|2 − |∆|2)|

= 0.6417

Figure 15: Gp = 10 dB ciecle.

3Plotting was done in ADS, while numericals are obtained using formulas and CASIO.

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(b) ΓL and VSWR In order to choose the load reflection coefficient in the stable region, the load stabilitycircle is found as follows:

CL =(S22 −∆S∗11)∗

|S22|2 − |∆|2= 1.493 6 76.87◦

RL = | S12S21

|S22|2 − |∆|2| = 0.6868

CS =(S11 −∆S∗22)∗

|S11|2 − |∆|2= 1.302 6 115.2◦

RS = | S12S21

|S11|2 − |∆|2| = 0.4485

Because the magnitude of S11 and S22 are less than 1, load stable region is outside the stability circle.

Figure 16: ΓL and stability circle.

So is the source stable region. Then select ΓL in the stable region at 0◦, −60◦, −120◦, and −180◦, andusing

Γs = Γ∗IN =

[S11 +

S12S21ΓL

1− S22ΓL

]∗ΓOUT = S22 +

S12S21Γs

1− S11Γs

ML =(1− |ΓL|2)(1− |ΓOUT |2)

|1− ΓLΓOUT |2

(V SWR)out =1 +√

1−ML

1−√

1−ML

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The following table was obtained. Each selection ended up with a conjugate matching ΓS in the stableregion.

Figure 17: ΓS and stability circle.

Table 1: The several values from calculationΓL ΓS V SWRIN ΓOUT ML V SWROUT

0.8359 6 28.49◦ 0.6591 6 117.5◦ 1 0.85636 − 65.37◦ 0.2188 16.220.44266 − 20.78◦ 0.7130 6 103.3◦ 1 0.82376 − 73.08◦ 0.2187 16.220.27676 − 145.4◦ 0.7807 6 99.37◦ 1 0.8206 6 − 78.25◦ 0.2185 16.24

0.6782 6 144.0◦ 0.8537 6 97.70◦ 1 0.8347 6 − 84.27◦ 0.2184 16.25

(c) constant gain circles approaching stability circle

gp,15dB =15dB

|S21|2=

31.6

2.432= 5.351

gp,20dB =20dB

|S21|2=

100

2.432= 16.94

gp,40dB =40dB

|S21|2=

10000

2.432= 1693

C2 = S22 −∆S∗11 = 0.33296 − 76.87◦

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The constant power gain circle for this bilateral device is therefore given by center

Cp =gpC

∗2

1 + gp(|S22|2 − |∆|2)

and radius

rp =

√1− 2K|S12S21|gp + |S12S21|2g2p|1 + gp(|S22|2 − |∆|2)|

Table 2: The three circlesGp Cp rp

15 dB 0.81246 76.87◦ 0.399720 dB 1.181 6 76.87◦ 0.461940 dB 1.489 6 76.87◦ 0.6835

The 40 dB constant gain circle is almost the same as the load stability circle.

Figure 18: Gain circles approaching stability circle.

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