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I am trying to understand S parameters on RF transistors.

I chose the BFP840ESDH6327XTSA1 as my RF transistor, to operate it at 10 mA, 12GHz. The design should be optimal for simultaneously matching impedance and maximum gain. Lets say that the input/output loads are 50 Ω.

I am unsure as to how can I spot the S11 values from the Smith chart that the datasheet provides. For example, for the S11 I have the below figure from the datasheet. S11 is on the input side, where I have a 50 Ω source, so the input (source) is on point A.

My question is: How do I find the S11 parameter from this graph?

In the meanwhile I am posting my ideas since I am doing it as a homework project:

Now, point A is not being crossed by any line, so my ideas to find the S11 there are:

  1. Either the graph does not have information of S11 for a 50 Ω load (most probably?) so I need to match the impedance of 50 Ω that I have on the input to one of the drawn lines on the graph (green line) so that can I change my input impedance from 50 Ω to a known impedance and use this S parameter.
  2. I need to "walk" on the resistance path -blue line- (and select A2 point as S11) OR on the imaginary part -red line- (and select A1 point as S11), which both do not make sense to me.

How to find S-parameters

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    \$\begingroup\$ Just sharing a potentially useful resource: If you look up W2AEW on Youtube, he has published several videos addressing this topic. \$\endgroup\$
    – JYelton
    Jan 25, 2022 at 20:00

2 Answers 2

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Either the graph does not have information of S11 for 50 ohm load (most probably?)

It does. Graphically converting between impedance and reflection (\$S_{11}\$) is entirely the point of the Smith chart.

If you superpose cartesian coordinates over the top of the Smith chart, with the real part of \$S_{11}\$ the x-axis and imaginary part on the y-axis, then you can read \$S_{11}\$ off the chart directly:

enter image description here

If you are working on paper, then you can use a ruler (and possibly a protractor, if you prefer a magnitude-phase representation rather than real and imaginary) to measure the cartesian coordinates, and that gives you \$S_{11}\$.

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I am answering my own question

Okay I figured it out.

The S parameters are Independent from the Input/Output impedance. So my question does not make sense.

I was wondering on what does S parameters depend on? The answer (based on what I see from datasheets like this one where the S parameters are on a table) seems to be: the frequency, current and voltage.

On the graph shown on the question, ALL of the three dependencies are visible. The frequency (which I failed to see) is depicted on some spots on the smith graph on the lines drawn.

Since I want 12Ghz and 10mA, the point I am interested of, is at the top middle of the smith chart.

finding S parameter from Smith Chart

Therefore, $$ S_{11} = 0.29 + 1.3j $$

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  • \$\begingroup\$ The outer circle of the Smith chart is the unit circle. Any point with imaginary part 1.3j would be off the top side of the chart, 1.3 times the distance from the center to the outer edge. The point you indicated is (by eye) about 0.05 + 0.8j, definitely not 0.29+1.3j \$\endgroup\$
    – The Photon
    Jan 26, 2022 at 1:10
  • \$\begingroup\$ Wait, real part is between 0.2 and 0.3 on the smith chart. Imaginary part is between 1 and 1.5. You might be looking at another point? \$\endgroup\$ Jan 26, 2022 at 5:50
  • \$\begingroup\$ I'm talking about the point you labeled as "\$S_{11}=0,29+1,3j\$. But that number is not \$S_{11}\$. You multiply that value (0.29+1.3j) by the system \$Z_0\$ to get the input impedance. So your input impedance is \$50(0.29+1.3j)\$ or \$14.5+65j\$. You get the reflection coefficient by figuring out what the cartesian coordinates of the point would be (which I estimated as about 0.05+0.8j). \$\endgroup\$
    – The Photon
    Jan 26, 2022 at 5:57

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