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what is a Gm stage?

Hi,

I'm trying to figure out how to implement a Gm stage. I understand the basic idea - voltage in, current out - and I've seen some figures like (a) and (b) shown in the following diagram, ie. a differential pair with no load representing a Gm stage:

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But I'm not sure what exactly to do with this differential pair with no load. I Can I just use the iout+/- from the differential pair directly as the inputs to the next stage? Or do I need a load for the differential pair? And what could I use as the load? I thought about a current mirror, as in (c), but that means it's just an voltage amplifier - the current never goes "out" of this stage, it just gets redistributed between the input transistors. This seems wrong (???).

Perhaps some DC current sources (with common mode feedback for the biasing), as in (d). Then the iout current due to a change in Vin is forced out to the load? But isn't this just a differential amplifier?

Thanks for your help,

Ted

Reply to
tedthornton11
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It is wrong. The point about a current mirror is that both of the transistors at the top of the diagram are sourcing the same current, so the difference between the currents being sunk by the two input transistors has to flow into, or out of, the output connection.

If you used resistors as loads on each of the drains, then - and only then - you can run the stage as a source of voltage gain with very little current flowing in the output connection. Do pay attention.

The Gm stage in a differential amplifier ....

-- Bill Sloman, Nijmegen

Reply to
bill.sloman

A Gm stage is what you have stated but not realised; a forward transconductance stage. As you correctly surmise, it is an amplifier, but specified by it's Io vs. Vin characteristics rather than some other parameter.

I didn't open your link, but maybe that helps.

Cheers

PeteS

Reply to
PeteS

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