question on equivalent transconductance

[analogic]

look at the following two circuits:
1) NMOS source follower with R as load. drain connected to vdd
2) NMOS common-source amplifier, with R as source degeneration and Rd as load.

the equivalent transconductance of the first one is gm, while the second one is around 1/R

as far as i know, the reason is that to calculate equivalent transconductance, we should ground output. but why should we ground output node? to reduce loading effect? why the two similar circuits give different transconductance? what is the physical meaning in it?

Thanks.

Added after 3 hours 50 minutes:

i know why Gm are different. the reason is that one Vin is applied to Vgs, while another one is not Vgs. but i still dont know why we have to ground Vout. can anyone explain it? thanks.

 

[Vamsi Mocherla]

Well, when you are measuring the transconductance, you have to short the output to ground because it is a simple two port networks concept.

Gm = iout/vin with vout=0. Here vout is between the Drain and source. When we have to get vout to 0, we short the source and the drain terminals. And the source is connected to ground anyways in the common source configuration.