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I am looking to find short circuit current over input voltage to find transconductance Gm for a common source amplifier with source degeneration.

enter image description here

To do this, I first shorted the output terminal and ignored resistor D because the voltage across it is zero. Then the small circuit model seems to be shown below. Let Vx = 0 and Isc = ix. enter image description here

Looking at sources online, it seems like the next step is to do use KVL such following the path through ro and Rs. Something like:

(Isc - gm(vgs))ro + (Rs)Isc = 0

and

Vgs = Vin - Vs where Vs = Isc(Rs)

Solving for this gives an expression for Isc/Vin:

gmro / ro + Rs + gm(ro)(Rs)

However, I'm confused as to why I can't seem to get the correct answer when making the assumption that ro and Rs is in parallel. From my understanding, both resistors are connected at the same node and to ground, and thus they should paralle. When solving it this way, I set

gmVgs = Isc

and

Vgs = Vin - Vs where Vs = Isc(ro//Rs)

Using these steps gives an expression for Gm where

Gm = (1/gm + Rs//ro)^-1

I'm not quite sure why these two methods give different answers. Am I making an assumption that cannot be made? I feel like I'm misunderstanding the conditions for parallel resistance, but I'm not quite sure what it is.

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  • \$\begingroup\$ Gm is the input voltage to output current transfer (thus, the output is grounded). Your small-signal schematic is not the right answer for this (it's more suited to compute the output impedance of the MOS). Therefore, you do need to ground the top side of ro and, also, you must include the input voltage in your calculation. \$\endgroup\$
    – Designalog
    Commented Jun 17, 2023 at 9:28
  • \$\begingroup\$ My small signal model isn't quit clear but I'm considering that Vx is grounded. If the top side of ro is grounded, is it a proper assumption to say that ro and Rs are in parallel? \$\endgroup\$
    – YS KIM
    Commented Jun 17, 2023 at 9:32
  • \$\begingroup\$ if ro is grounded on the top, then ro||Rs as you say. \$\endgroup\$
    – Designalog
    Commented Jun 17, 2023 at 9:52

1 Answer 1

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Vs = Isc(ro//Rs) is not correct. You cannot club both the resistors here because you are interested in the current through Rs which is Isc. In other words, the current through the parallel combination of ro & Rs is not Isc.

Vs = Isc*Rs is correct.

Edit:

As you can see below, the current through ro||Rs is gmVgs

enter image description here

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  • \$\begingroup\$ Could you explain a bit more as to why the current through Rs is not Isc? Sorry, I'm not quite understanding what's happening \$\endgroup\$
    – YS KIM
    Commented Jun 17, 2023 at 9:49
  • \$\begingroup\$ Current through Rs is Isc \$\endgroup\$
    – sai
    Commented Jun 17, 2023 at 9:51
  • \$\begingroup\$ Current through ro//Rs is not Isc \$\endgroup\$
    – sai
    Commented Jun 17, 2023 at 9:53
  • \$\begingroup\$ Sorry, I meant why is the current through the parallel combination of ro and Rs not isc? \$\endgroup\$
    – YS KIM
    Commented Jun 17, 2023 at 9:56
  • \$\begingroup\$ This is because the current through the parallel combination of ro & Rs is gmVgs as per your diagram. \$\endgroup\$
    – sai
    Commented Jun 17, 2023 at 10:00

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