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double balance mixer conversion gain (Read 2062 times)
ZYC
Guest
double balance mixer conversion gain
Aug 01
st
, 2005, 8:14pm
Hi,
I am doing a double balance mixer. I found that there is 6dB difference of conversion gain while I use transcient or PAC to do simulation. For example, result from PAC is 12 dB and 6 dB from transcient. I am using ideal balun to convert single-ended port to mixer differential input and mixer differential output to single-ended output port. Can anybody tell me which result I should trust? Thank you very much.
BR,
ZYC
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uncle_ezra
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Posts: 90
Re: double balance mixer conversion gain
Reply #1 -
Aug 2
nd
, 2005, 12:33am
Well transient is always the most reliable simulation, but I am astonished by the difference. What accuracy are you running your PSS? Conservative? I heard there is some dB difference between conservative, moderate and liberal?
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pancho_hideboo
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Re: double balance mixer conversion gain
Reply #2 -
Aug 2
nd
, 2005, 2:45am
Hi.
What definition of gain do you use each for PSS/PAC and Transient Analysis ?
These are surely same ? ???
The definition of Voltage gain of PAC direct plot form is "Vout/(E/2)",
here E means open voltage of signal source.
It is not "Vout/Vin", here Vin means voltage at an input node of DUT.
Do you use gain definition of "Vout/Vin" for Transient Analysis ?
If S11 is far from matched, "Vout/(E/2)" and "Vout/Vin" are quite different.
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Kita━━━━━━(゚∀゚)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related
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ZYC
Guest
Re: double balance mixer conversion gain
Reply #3 -
Aug 2
nd
, 2005, 3:44am
I have tried three accuracy defaults in PSS. The conversion gain is very close to each other.
I am using "Vout/Vin" as transient gain.
My friend told me his PAC result was the same as transient result in single balance mixer simulation.
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ZYC
Guest
Re: double balance mixer conversion gain
Reply #4 -
Aug 2
nd
, 2005, 3:46am
I have tried three accuracy defaults in PSS. The conversion gain is very close to each other.
I am using "Vout/Vin" as transient gain.
My friend told me his PAC result was the same as transient result in single balance mixer simulation.
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pancho_hideboo
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Re: double balance mixer conversion gain
Reply #5 -
Aug 2
nd
, 2005, 4:23am
Hi.
For PAC, do you use gain definition of "Vout/Vin" which is same definition for Transient Analysis ?
For simulator accuarcy or difficulty , there is no difference between single or double balanced as far as it can converge appropriately.
In general, the ambiguity in gain definition or input voltage definition causes discrepancy between simulation results.
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Kita━━━━━━(゚∀゚)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related
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Andrew Beckett
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me about Life...
Posts: 1392
Bracknell, UK
Re: double balance mixer conversion gain
Reply #6 -
Aug 2
nd
, 2005, 5:13am
Problems tend to occur when people have ports which aren't matched, and then put (say) "1" as the value on the port, and expect the PAC output to give them gain. If you want voltage gain, look at the difference in dB between the two nodes (at the appropriate frequencies). It should tie up pretty well between transient and PAC, provided that the mixer is responding linearly (i.e. well below the 1dB compression point).
Regards,
Andrew.
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Eugene
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Posts: 261
Re: double balance mixer conversion gain
Reply #7 -
Aug 2
nd
, 2005, 11:00am
Some simple things to check:
1. Are you sure your transient analysis has reached steady state? If you only monitor peak voltage and the transient analysis has not reached stated state, your results will be misleading.
2. Are you sure you selected the right sideband options in your PAC analysis. I frequently require several attempts before I get it right.
3. A factor of EXACTLY 6 dB is very suspicious. It usually arises out of something simple like computing gain as the ratio of an output peak voltage to an input peak-to-peak voltage. I assume your transient and PAC circuits are one in the same so that you do not have a matching issue. For example, the port has a factor of two such that you only get the specified output when it drives a matched load. But if the circuits are the same, that should not be the issue.
4. Do you more than one PAC source? For example, if you cut and paste a source with PACmag=1, it is easy to forget that you now have two PAC sources when you meant to have only one. Both will contribute to the output.
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ZYC
Guest
Re: double balance mixer conversion gain
Reply #8 -
Aug 2
nd
, 2005, 6:45pm
Yes, the problem is caused by unmatched input.
My input port resistance is 50ohm. It is much smaller than mixer input impedance. So vin is almost equal to E(not E/2). That is why around 6dB difference. Thank you very much. :)
BR,
ZYC
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