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I'm trying to understand how MOSFET resistances work, but I'm seeing a lot of things that don't always fit together (probably due to my lack of understanding). Specifically the amplifier configurations (CS, CG, CD). Is this correct:

  1. Looking into the gate, resistance is infinity.
  2. Looking into the source, resistance is r0
  3. Looking into the drain, resistance is r_ds = 1/g_m

Is this view flawed somehow? Thanks!

Here is an example from Sedra Smith (which is a bit more involved but really, I have no idea what's going on here).

For the purpose of determining the close-loop gain of this amplifier with feedback, the A circuit is shown:

enter image description here

enter image description here

I understand why r02 is in parallel with Rf, but what is the 1/gm2 resistor doing there? Why is the first transistor a current source and a resistor, and the second a pair of resistors?

Edit: on second thought, I don't understand also why the first r0 is at the drain of the first transistor, and the second r0 is at the source of the second transistor.

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  • \$\begingroup\$ I don't think your values are quite right. I'm not sure it's an exact duplicate, but the answers to this question might steer you in the right direction: electronics.stackexchange.com/questions/79468/… \$\endgroup\$
    – Justin
    Aug 6, 2014 at 18:39
  • \$\begingroup\$ Mosho: Regarding the second diagram: According to the authors - is it the small signal eqivalent diagram of the first circuit? Or what is the relation between both figures? \$\endgroup\$
    – LvW
    Aug 6, 2014 at 20:28
  • \$\begingroup\$ @LvW it's a process to determine the open-loop gain of the amplifier. There are a few steps to follow from the initial circuit and the result is the second diagram. \$\endgroup\$
    – Mosho
    Aug 6, 2014 at 20:32

1 Answer 1

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Looking into the drain, the small-signal resistance is $$r_{id} = r_o = \frac{\lambda^{-1}+V_{DS}}{I_D}$$ if the source is at AC common (common-source configuration).

If the AC resistance from source to common is \$R_{ts} \ne 0\$, the small-signal resistance looking into the drain is

$$r_{id} = r_o \left(1 + \frac{R_{ts}}{r_s} \right) + R_{ts}$$

where

$$r_s = \frac{1}{g_m}$$

Looking into the source, the small-signal resistance is

$$r_{is} = r_s$$

The above assumes the body is connected to the source.


I understand why r02 is in parallel with Rf, but what is the 1/gm2 resistor doing there?

The lower right circuit is drawn oddly and further, seems to mix AC and DC sources which is an error.

If I were teaching this circuit, I would draw the AC circuit, with Q1 and Q2 replaced by their small-signal T-models, as follows

schematic

simulate this circuit – Schematic created using CircuitLab

Now, is it clear why \$r_{s2} = \frac{1}{g_{m2}}\$ is there?

Edit: on second thought, I don't understand also why the first r0 is at the drain of the first transistor, and the second r0 is at the source of the second transistor.

\$r_o\$ connects to the drain and source.

Since, for Q1, the source is grounded, \$r_{o1}\$ connects from D1 to ground.

Since, for Q2, the drain is AC grounded, \$r_{o2}\$ connects from S2 to AC ground.

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  • \$\begingroup\$ But I often see the transistor replaced or regarded as a 1/gm resistance, not sure what to make of it. \$\endgroup\$
    – Mosho
    Aug 6, 2014 at 19:45
  • \$\begingroup\$ I've edited my question with an example of something I'm having difficulty understanding. Would appreciate some guidance :) \$\endgroup\$
    – Mosho
    Aug 6, 2014 at 19:52
  • \$\begingroup\$ Mosho - you are right. The resistance into the source is 1/gm. Some people are using the inverse expression re=1/gm, even in formulas for the gain - for my opinion an unfortunate expression. It makes much more sense to think in terms of the quantity which connects the input and the output of the device, and that is the transconductance gm. \$\endgroup\$
    – LvW
    Aug 6, 2014 at 19:58
  • \$\begingroup\$ Mosho - I forgot to mention that point 2) and point 3) in your text is wrong. \$\endgroup\$
    – LvW
    Aug 6, 2014 at 20:02
  • \$\begingroup\$ @Mosho, I've updated my answer. \$\endgroup\$ Aug 6, 2014 at 21:05

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