source pull simulation and input impedance match

Design one stage power ampifier. After doing the load pull and source pull simulation, I got the load and source impedance. Then I do the impedance matching for both the input and output of the transistor, but I get a very bad input matching? Why? Thanks.

Because Zo is not your Source or Load Impedance.Zo=50 Ohm..
You have to enter Zs and ZL into Impedance fields, not Zo.You make a mistake while designing matching circuit..

Thank you for your reply. [ATTACH=CONFIG]153282


The Zo in the red cycle is always there, and the blank space after it is not for entering the Zo. It is for the Zs or ZL input.

You can see the tuoturial:
https://www.youtube.com/watch?v=s_JIxy4obMs

In your matching template, the center of the Smith Chart has 5.5+j*2.8 Ohm, so you try to match an impedance to this value then you drive your amplifier with 50 Ohm source impedance.
So, there is no match at all.. You have to match this impedance 5.5+j*2.8 (whatsoever) to 50 Ohm.

In the smith chart utility tool it says ZS* and you entered 5.5+j*2.8. That means if you see the impedance at the input of the extracted matching network, you get 5.5-j*2.8 ohm. So your transistor sees 5.5-j*2.8 ohm.



I would like to know is it always necessary to keep 50 ohm in ZS*. We can also do like I said above right?

I may be wrong here, I am not sure. Somebody tell me why is it says ZS*, instead of Zs?. In the video ZS=25*j*30 and he uses this itself in the place of ZS* in the smith cahrt utility

Simply conjugate matching.

Then in that case, in the smith chart utility If I enter ZS* as 5.5+j*2.8 and build the circuit, the impedance looking at the input of the extracted matching network should be 5.5-j*2.8 ohm. But I get Zin as 5.5+j*2.8 ohm.

Can you understand conjugate matching.

Wrong.

It is natural result.
Zs=5.5-j*2.8, Zin=5.5+j*2.8=conj(Zs)

Surely learn.

I see.. makes sense. Thanks. Surely that guy made a mistake in the video then.