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25th September 2019, 12:50 #1
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Loadpull for Differential PA
There are few posts on this but I just clarify it properly for my problem.
https://www.edaboard.com/showthread....ht=ideal+balun
https://www.edaboard.com/showthread....ht=ideal+balun
https://www.edaboard.com/showthread....ht=ideal+balun
First of all I would like to know whether the set up is correct in the figure.
Here is my concern. If I have single ended configuration, I will have Zopt from loadpull as, say, Ropt+j*Xopt. And the impedance looking into the output of this single ended is ,let, Rlj*Xl.
Now, I have a Differential configuration. I connected the outputs of the two singleended configurations to ideal_balun. So the impedance looking into the ideal_balun is 2*( Rlj*Xl) (in this case, output side)
should the value I get from the loadpull, ideally, be 2*(Ropt+j*Xopt)?

25th September 2019, 15:24 #2
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Re: Loadpull for Differential PA
Simply your thoughts are all wrong.
Differential mode can not exist in your circuit.
Surely learn basic linear circuits theory before EDA Tool Play.

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25th September 2019, 20:36 #3
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Re: Loadpull for Differential PA
Ideal Balun does not do Impedance Transformation.It acts simply as DifftoSingle Ended so the Impedances ( Input/Output) are same..
This structure has high Output Impedance ( I don't know the details but looks like it) so it's not convenient for PA.A PA should have more or less an Output Impedance closer to Load to match easily..( I presume the PA will work @+20dBm or more )

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25th September 2019, 22:15 #4
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Re: Loadpull for Differential PA
1. Your circuit is flawed and must be corrected, you have shorted the output.
2. Not sure if the corrected circuit is exactly equivalent to "two singleended configurations". If it is, your calculation would be correct.
3. I don't see an advantage of using a balun in simulation. Why not simply use a differential load?

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26th September 2019, 00:56 #5

26th September 2019, 10:28 #6
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Re: Loadpull for Differential PA
I inferred that you are doing mmwave from your previous posts. If that is the case, that current source will eat away unnecessary voltage headroom. I read somewhere that people don't use the current source beyond 5GHz (I have forgotten the paper). Just ground your sources. You will get a pseudodifferential pair but larger voltage headroom.
And as someone already pointed out, your outputs are shorted. Also use 'dcblock' and 'dcfeed' else you will can run into convergence issues while doing HBPSS (atleast I did in 45nm).
But you don't even need all of that. Connect your DC sources directly to the center tap of the transformers. True for both input and output.
What you get from the loadpull is the impedance that the output port wants needs to have. Its is as simple as that.
1 members found this post helpful.

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12th October 2019, 14:13 #7
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Re: Loadpull for Differential PA

12th October 2019, 14:28 #8
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Re: Loadpull for Differential PA
Can you understand differential circuit ?
https://www.edaboard.com/showthread.php?386764#2
No. There is no coupling between L.
Can you understand actual transformer ?
Ideal transformer is not mutual L.
Surely learn very basic things before EDA Tool Play.

13th October 2019, 06:45 #9
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Re: Loadpull for Differential PA
How are you saying that output is shorted

13th October 2019, 19:58 #10
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