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Voltage Gain Determination - RF Cafe Forums

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Measures
 Post subject: Voltage Gain Determination
Posted: Wed Jan 20, 2010 1:35 pm 
 
Captain

Joined: Tue Dec 15, 2009 10:22 am

Posts: 9

Hello

I'm trying to obtain the voltage transfer function employing an E5062A VNA.

From pages 24, 23 in:

http://cp.literature.agilent.com/litweb ... 2-1087.pdf

I see that it can be obtained in terms of the S parameters and the load reflection coefficient.

My question is: is it correct to assume that the source and load impedances are equal to the reference impedance (Zl=Zs=Zo=50 Ohm)?

In this case, the expression in page 24 of the document would reduce to Av=S21/(1+S11). Is it correct?

Thank you very much.


 
   
 
fred47
 Post subject: Re: Voltage Gain Determination
Posted: Fri Jan 22, 2010 11:49 pm 
 
General
User avatar

Joined: Wed Feb 22, 2006 3:51 pm

Posts: 113

When you measure a device with a VNA, the source impedance from the VNA is 50 Ohms (normally, unless you're in the TV industry, where they use 75 Ohms a lot). Likewise, the load impedance seen by the device you're testing is 50 Ohms, again due to the VNA.

Remember to use complex arithmetic when you use the equations!

Good luck!


 
   
 
Measures
 Post subject: Re: Voltage Gain Determination
Posted: Mon Jan 25, 2010 6:24 am 
 
Captain

Joined: Tue Dec 15, 2009 10:22 am

Posts: 9

So... can I assume that the reflection coefficients of the source and the load are zero and, then the referred expression is simplified as I said?

 
   
 
fred47
 Post subject: Re: Voltage Gain Determination
Posted: Mon Jan 25, 2010 1:27 pm 
 
General
User avatar

Joined: Wed Feb 22, 2006 3:51 pm

Posts: 113

Basically, yes.

But it's more precise to say that the source and load impedances (50 Ohms + j0 Ohms) determine the reference impedance.

If you connect a short piece of low-enough loss transmission line of characteristic impedance 50 Ohms between the two ports of the VNA, then calibrate out the length of the line, you'll get zero for S11 and S22, and one for S21 and S12 - which I think is the answer you're looking for, stated precisely.

Good luck!


Posted  11/12/2012

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