nrk1 wrote on Oct 23rd, 2012, 8:14pm:BTW, a MOS in triode has a noise spectral density of 4kT*gds(This gds is measured in deep triode, and with square law, equals the gm in saturation, i.e. mu*Cox*(VGS-VT)) . There is no factor of 2/3.
Thanks nrk1, I correct myself, I suppose it is rather 4kT(gm+gds) in triode, with the gm being rather small.
nrk1 wrote on Oct 23rd, 2012, 8:14pm:sandman wrote on Oct 22nd, 2012, 2:18pm:".. neither does the other switch that is ON because its current is fixed by the RF input transconductance stage". The switch is driven by a current source (current-driven mixer). The switch noise current, when ON, flows to the low impedance Mixer output load.
This is not correct. The switch noise current will not flow into the low-Z load if the impedance on the driving side(transconductor) is very high.
Perhaps what you say is not completely true. There is a current divide at the output of the transconductor, much like what RFICDUDE also pointed out. If the intention is to use it in current mode, the load resistance in parallel with ro of the gm stage is typically small as no voltage gain is necessary here. I am referring to the load that connects it to the supply.
This prevents large voltage fluctuations at the input of the Mixer. (Furthermore, irrespective of upconversion or downconversion, the 'mixer' is AC coupled to the gm stage, in that there is no DC current from gm to output_load.)
This is not as small as the Ron (1/gds) of the Mixer-switch, but is definitely smaller than the ro of the gm. The current gain (if you will) of the switch noise current to the output load is
Iout,load Rswitch
---------- = -------------------------------------------
Iswitch Rswitch + ro//Rload_gm + Rout,load
My apologies for the poor illustration, but, you can see, this depends on the ratio between Rload_gm (the load of the gm stage) and the Rswitch (Ron of the switch in triode), although, this is actually Rswitch(t), assuming Rout,load is rather small!
As for the 'cascoding', I agree, that was only an abstraction, I guess it makes things a bit more complicated! I was only trying to figure out the assumption being the following statement in the paper for which it would be true ".. neither does the other switch that is ON because its current is fixed by the RF input transconductance stage"... which was actually my question.
Thanks for all your comments. Much appreciated!