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Design >> RF Design >> kT/C noise in passive mixers
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Message started by carlgrace on Jan 9^th, 2007, 11:10am

Title: kT/C noise in passive mixers
Post by carlgrace on Jan 9^th, 2007, 11:10am

I am looking into switch-mode passive mixers and I've read most of the relevant papers. I am wondering, though, why isn't the output thermal noise density determined by kT/C noise of the sampling switch. It seems to me that there would be a lot of output noise because the reverse-biased diode formed my the drain diffusion of the MOS switch would sample the switch noise. This noise would be very high because the diffusion capacitance would be rather small. It seems from the papers, however, that this is not the case. Can anyone give me an intuitive explanation of why the output noise of the passive mixer isn't kT/C, where C is the parasitic capacitance of the drain diffusion?

thanks,
Carl

Title: Re: kT/C noise in passive mixers
Post by Ken Kundert on Jan 9^th, 2007, 4:05pm

kt/C noise is for sampling processes. passive mixers do not sample.

Title: Re: kT/C noise in passive mixers
Post by carlgrace on Jan 10^th, 2007, 7:27am

Ken,

Say that you are driving the switches with a less that 50% duty cycle clock. Then there are time slots where there is no resistive connection between the LNA and the VGA. In this case wouldn't you say the passive mixer is sampling? I imagine you would put a large filter cap at the IF port of the mixer, and that could lower the noise, but I am confused about this. I would think that during the time when all switches are OFF, the filter cap would be holding a sample of the rf input voltage. Am I totally off base here?

Carl

Title: Re: kT/C noise in passive mixers
Post by Ken Kundert on Jan 10^th, 2007, 9:27am

I was assuming a full wave bridge passive mixer. If you are using a track and hold as a mixer then kt/C noise would apply.

-Ken

Title: Re: kT/C noise in passive mixers
Post by mg777 on Jan 14^th, 2007, 12:56pm

Take thermal noise across a resistor R with a parasitic shunt capacitance C. The mean squared noise voltage is then 4kTR*(1/2piRC) so the R cancels and you get kT/C. You can integrate over the pole, it'll add some pi type factor from the tan^-1 function, that's all.

Point then is that even without sampling you get kT/C - sampling is just one form of integration. It may seem funny that the resistor is the source of the noise but its value disappears from the answer. Doesn't contradict the fluctuation-dissipation theorem in any way, though. The dissipation remains the mechanism for the system to remain in thermal equilibrium.

M.G.Rajan
www.eecalc.com

Title: Re: kT/C noise in passive mixers
Post by Ken Kundert on Jan 14^th, 2007, 11:37pm

Yes of course. My mistake.

-Ken

Title: Re: kT/C noise in passive mixers
Post by Croaker on Jan 15^th, 2007, 4:33am

Well, the low frequency portion of the PSD increases with R, but the 3db pole decreases with R. The increased low frequency noise is choked out and the RMS noise is independent of R.

mg777 wrote on Jan 14^th, 2007, 12:56pm:

It may seem funny that the resistor is the source of the noise but its value disappears from the answer.

M.G.Rajan
www.eecalc.com