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Design >> RF Design >> lo switch condition in normal active mixer
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Message started by baohulu on Feb 5^th, 2013, 11:45pm

Title: lo switch condition in normal active mixer
Post by baohulu on Feb 5^th, 2013, 11:45pm

hi, all
I have a quenstion about the lo switch condition in the mixer. for a gilbert cell, the on-switch of the lo is in the saturation. but why?
why not make the on-switch biased in linear region?
if the on-switch is biased in saturation, then the lo noise can be delivered to the mixer output, because the on-switch is a source follower, the switch source point parasitic cap will be charged or discharged according to lo noise. just like the book said (Analysis and Design of CMOS RF Integrated Circuits)
however, if the on-switch is biased in linear region, then, the source of the switch will be the same as its drain, will have no ralation with the lo noise, right?
of cource, I know for the IIP3, the linear region may be worse, but just say the noise figure, the linear region is better ?
thanks

Title: Re: lo switch condition in normal active mixer
Post by RFICDUDE on Feb 6^th, 2013, 4:42am

First, it does appear that the switch is biased in saturation, but it acts as a cascode (common gate) device when it is fully switched by the LO signal. Therefore it does not contribute extra noise from itself or from the LO when fully switched. Extra noise from switch transistors are mainly a problem when the mixer devices are transitioning from one LO phase to another.

Second, in the Gilbert Cell configuration it is not straight forward how you would switch the devices from the linear region. The drain voltage would have to be kept at a low enough voltage to insure the switch never enters saturation.

However, there is a solution. The passive current mode mixer is exactly what you suggest, but it does not pass DC current of the input transconductor. Instead, the AC current of the transconductor is AC coupled to a floating set of cross-coupled switches. The floating switches can only switch between off and Ron because VDS(dc)=0.

There are many papers on passive MOSFET mixers.

LO noise can transfer to the output signal in most mixers regardless of the mixer topology. The mixer is fundamentally a multiplier, so any noise on the LO will multiply with the input signal and showup on the output signal.

Title: Re: lo switch condition in normal active mixer
Post by baohulu on Feb 6^th, 2013, 7:30pm

hi , RFICDUDE
first, thanks for your reply.
I have seen the passive mixer that you said, but in this case, I don't want to talk about the passive mixer.
Let't return to the active mixer. the noise of the lo switch mainly comes from two places. one is during the transfer of LO phases, just as you said.
the other is the charge and discharge the parasitic cap (cp, not in the figure) at the source of M1 and M2 (M3 & M4). when M1 & M5 is on, and they are operating in saturation, then M1 & M5 is source follower (common gate), and the noise of Lo will charge cp, and at the other phase, cp will discharge the noise to the output. thus, LO noise is dilivered to the output.
but if I make the on-switch M1 & M5 operating at linear region, then they are not common gate, and the second noise will disappear. as to the implemetion, I think the following figure can do it. I can make M1 biased at linear region at dc. then when lo is added, the on-switch will operating at linear.

RFICDUDE wrote on Feb 6^th, 2013, 4:42am:

Title: Re: lo switch condition in normal active mixer
Post by rfidea on Feb 7^th, 2013, 9:20am

Yes, the LO noise voltage is be present at the sources of M1/M4 when those transistors are on. But that voltage noise will not change the current thru them and therefor it will not be present at the output.