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.

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