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LNA Test Key (Read 5508 times)
new_guy
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LNA Test Key
Aug 04th, 2007, 1:20am
 
Hi,

I'm doing a standalone LNA for testing. I know the testing equipment is 50 ohms based, currently through literature, I found 3 kinds of termination.

1) Spara Matching, match S22 to 50 ohms (my view: safest but would putting a matching network degrade the power gain by 50% or -3 dB, maximum power matching theory)
2) Putting a common drain buffer  (my view: how to deemb the effect after measuring its Sparamaters???)
3) the LNA has a LC tank load, saw some literature, putting a dc blocking cap before the GSG bias tee  (my view:LC tank resonates at my opearting freq leaving inductor's series resistor, even if from calculation i make the series resistance 50 ohms, how would putting a dc blocking cap match series resistor to 50 ohms of bias tee. The series cap would cause a anti-clockwise shift on the smith chart. Right???)

Hope it makes sense. Thanks for anyone's views.

Regards
Newbie
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aaron_do
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Re: LNA Test Key
Reply #1 - Aug 4th, 2007, 2:10am
 
Hey,

this is what i did and it seemed to work very well.

1) Add a 950 ohm in series with 50 ohm load in parallel to your LC tank load (or whatever load you have). I'm assuming the 1k resistance won't really affect your LNA.

2) Tap the output in parallel with the 50 ohm.

3) measured S22 will tell you the actual resistance of the 950 and 50 ohm resistors.

4) use these values to de-embed the voltage gain. Voltage gain is roughly 975/25*S21, but again you have to use S22 to find the actual values of the 950 and 50 ohm resistor.

5) NF can be measured by directly tapping a separate GSG from the normal LNA output. As long as the 50 ohm output doesn't change the DC operating point of the LNA you're ok.

BTW from what you said all three are a little difficult to de-embed. With knowledge of S22 you should be able to de-embed the 50 ohm load as per you 3rd method....use S22 to find Zout. assume S21 is gm(Zout//50) and de-embed the 50.

cheers,
Aaron
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« Last Edit: Aug 4th, 2007, 6:49pm by aaron_do »  

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new_guy
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Re: LNA Test Key
Reply #2 - Aug 5th, 2007, 9:00pm
 
Hi Arron,

I'm not too sure wat u meant, have drawn out the diagram, was wondering if its correct?
Thanks

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Newbie
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aaron_do
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Re: LNA Test Key
Reply #3 - Aug 5th, 2007, 10:11pm
 
Yeah that's roughly what i did, but you need a blocking cap. In my case, i used a resistor load of 925 ohm, so i just split it up into 900+50//50. The advantage of this is the shape of the output in the frequency domain doesn't change. It would change with a buffer or LC match.

BTW if you want to measure noise, you need to add a separate output pad at the node between the transistor, the 950 and the LC tank. With a DC blocking cap too...

cheers,
Aaron
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manresa,spain
Re: LNA Test Key
Reply #4 - Aug 6th, 2007, 3:09am
 
Hi to all,
In testing I've used the methods 1 and 2. They work well if you've got the option of manufacturing the matching network or the buffer as separate circuits to characterize them, also if you use a buffer it's probably better to measure Voltage gain with a PSA instead of s-parameters with a VNA. The advantatge of using a pi-match or similar is that you can characterize it with the VNA, obtain a touchstone file and directly put it the simulator to see how it affects the real implementation, also using some math you can de-embed it more easily than a buffer.
Hope it helps,
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new_guy
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Re: LNA Test Key
Reply #5 - Aug 13th, 2007, 2:14am
 
Hi everyone,

Thanks for all your helpful insights. I'm not really sure on adding resistors in the RF path, keep having the feeling RF signals would be attenuated, right?

Thanks

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New_guy
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didac
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manresa,spain
Re: LNA Test Key
Reply #6 - Aug 14th, 2007, 6:29am
 
Hi,
I think that you are right, but this attenuation can be compensated easily by math when doing the measurements, what I'm not sure about is the effect of the resistors in the Q of the inductor and in the NF measurement.
Hope it helps,
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aaron_do
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Re: LNA Test Key
Reply #7 - Aug 14th, 2007, 5:42pm
 
Yes, better to make sure before you use it. For me i had no load inductor, so i conveniently split up the load resistor. If you add an extra resistor maybe you'll have more difficulty de-embedding.
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