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Design >> RF Design >> question about conversion gain in passive mixer
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Message started by KiamRF on Mar 13^th, 2010, 12:42pm

Title: question about conversion gain in passive mixer
Post by KiamRF on Mar 13^th, 2010, 12:42pm

Hi all,
I read some papers talking about the passive mixers. They mentioned that the single-balanced passive mixer has 6-db higher conversion gain than double-balanced passive mixer. Although, they provide some explanation but I still don't get it. Could anyone suggest me the books or any idea about this?

Thanks.

Title: Re: question about conversion gain in passive mixer
Post by aaron_do on Mar 14^th, 2010, 7:39pm

Well the math is correct, but intuitively it is due to the loading. Looking into the output node of the single-balanced mixer we see only one transistor, but looking into the output node of a double-balanced mixer we see two transistors.

Title: Re: question about conversion gain in passive mixer
Post by RFICDUDE on Mar 18^th, 2010, 3:24am

What are the assumptions underwhich the two are compared?

For instance, if the double balanced case has a fully differential driving source that is the equivalent of two of the single ended driving sources then I would think the conversion gain would be about the same. But of course the power consumption would be more because of the differential driving source.

In the single balanced case, the single ended driving source is only connected to one output pin (load) during each half cycle of the LO. The same is true for the double balanced case, but now there are two driving sources connected to each load during any given half cycle.

There is also a similar relationship for active mixers because for the same bias current and load resistance the input transconductance will be 1/2 the single balanced case for the double balanced mixer. But if you double the current for the double balanced mixer then the conversion gain is the same.

Let me know if I've missed something.

Title: Re: question about conversion gain in passive mixer
Post by aaron_do on Mar 18^th, 2010, 6:27am

Hi RFICDUDE,

I don't think you can treat the analysis in that way.

Firstly for active mixers, the output impedance of the switches is high due to the active biasing point. As a result, the load is primarily determined by the resistor load (which is the typical connection). Naturally the double-balanced architecture requires double the current for the same conversion gain.

As for down-conversion in passive mixers, the load is determined by the average conductance of the switches. For a single-balanced mixer, each output only sees one transistor and so the load is approximately the average conductance of that transistor. For the double-balanced passive mixer, each output sees two transistors and so the load is approximately two average conductances in parallel. Note that for single-balanced mixer, gain is calculated from single-ended input to differential output while for double-balanced it is from double-in to double-out.

Naturally practical designs are not so straightforward since we need to consider the loading to the LNA and the VCO, but there are works which have shown that ideally the single-balanced passive mixer has double the conversion-gain.

cheers,
Aaron