I have some doubts concerning the calculation of the small-signal output resistance of a common-source stage with source-degeneration. This is related to another question I posted but I have refined my question in detail here as to exactly what is troubling me

The Problem

Calculate the small-signal output resistance of the following circuit including the effect of channel-length modulation and ignoring the body effect.

enter image description here

My Solution

I have two approaches to find the output small-signal resistance, they both involve,

  • Drawing the small-signal model of the circuit
  • Zero all independent sources (voltage sources = short, current sources = opens)
  • Applying a voltage Vx at the output and measure the resulting current Ix flowing.
  • Output resistance will then be Rout = Vx/Ix

This gives us the following circuit enter image description here

Now from here on, I have two approaches - the first approach gives me the correct answer, however the second approach gives me an incorrect answer.

Approach 1 (KVL)

enter image description here

Approach 2 (Resistors)

enter image description here

Why does my second approach give me an incorrect answer? I suspect it is something to do with the fact that the ro now only gets a current Ix flowing through it.? Something that didn't sit well with me in the second approach was that after decomposing the current source into a resistor, I had to get rid of the negative sign of the 1/gm. So current flow direction is strange to me too.

EDIT: Extra Info for Comments: enter image description here

enter image description here

  • \$\begingroup\$ I have already told you why: you can't make transform gm*Vs into a fixed resistor because Vs varies with I(Rs). Your approach implies a fixed Vs, but that's one possibility out of an infinity. If Vs is fixed then yes, you can equivalate gm*Vs with a fixed resistor, but that will only be true for that particular value of Vs. As long as there is current flowing through Rs, the voltage level at the source will change, with or without Vgs. \$\endgroup\$ Commented Mar 14, 2021 at 18:40
  • \$\begingroup\$ I agree with @aconcernedcitizen. When I got to the part in your derivation where you equate the a current source with a resistor, my mind went "huh?" \$\endgroup\$ Commented Mar 14, 2021 at 18:43
  • \$\begingroup\$ Not only, but even if the calculations with the first approach are OK, in the 3rd picture the current Ix circulating inside ro is wrong. \$\endgroup\$
    – barrow
    Commented Mar 14, 2021 at 21:15
  • \$\begingroup\$ Thank you for the responses. I think I understand it now. Apologies for so many questions. @barrow The current circulating in the loop is between the VCCS and ro is gmvgs before the sign of vgs has been taken into account. \$\endgroup\$ Commented Mar 14, 2021 at 21:31
  • \$\begingroup\$ @AlfroJang80, please, look attentively to your third picture. It shows a current Ix circulating inside ro. Well, this is completely wrong and confusing. \$\endgroup\$
    – barrow
    Commented Mar 14, 2021 at 21:50

1 Answer 1


You've written "Current source whose current depends on voltage across it = resistor".

This is true, as long as the current is linearly proportional to the voltage across the current source.

However, the voltage across that current source is \$V_{ds}\$, while its current is proportional to \$V_{gs}\$. Thus, the current is not proportional to the voltage across the current source, and replacing it with a resistor is invalid.

  • \$\begingroup\$ Got it! It seemed slightly weird to me as well because the new "resistor" I had gotten had a negative value. Thank you for the clarification. \$\endgroup\$ Commented Mar 14, 2021 at 22:43
  • \$\begingroup\$ @AlfroJang80 It'd be perfectly normal, in the general case, to get a negative value resistor from doing this. In the specific case of a transistor model, though, you certainly can't have it producing power out of nowhere. \$\endgroup\$
    – Hearth
    Commented Mar 15, 2021 at 2:39

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