I was trying to derive the small-signal input impedance of a diode-connected MOSFET (figure included), and came up with this expression.

Is my method correct? I have been reading about methods of deriving the input and output impedances using a test voltage, and that is the method I have followed myself.

Some validation on the approach and the answer would help.

enter image description here

This is my solution.

enter image description here

  • \$\begingroup\$ Why is Rd there? What is it. Why is Rd in series in the first schematic but then to ground in the small signal model. That does not make sense. \$\endgroup\$ – Bimpelrekkie Oct 24 '18 at 6:54
  • \$\begingroup\$ Well, in the small signal model, we ground all DC voltages, right? VDD is a DC volatge, so grounding it basically makes Rd connected to the ground, if I'm not wrong. \$\endgroup\$ – Curiosity Oct 24 '18 at 6:55
  • \$\begingroup\$ I should have mentioned that VDD is a DC voltage. \$\endgroup\$ – Curiosity Oct 24 '18 at 6:55
  • \$\begingroup\$ So, Rd is just a resistor, used to control the DC current of the MOSFET. \$\endgroup\$ – Curiosity Oct 24 '18 at 6:56
  • 1
    \$\begingroup\$ I agree with the previous comment. Vt is constant so should be suppressed for small signal. The impedance should be 1/Gm in parallel with Rd. \$\endgroup\$ – Steve Hubbard Oct 24 '18 at 7:58

[edit] To avoid confusion, from the comments you state that you want to find the impedance looking into the source. In other words, this situation:


simulate this circuit – Schematic created using CircuitLab

I could find a few mistakes:

  • In the third step, you missed an \$R_D\$:

$$ v_D = \frac{g_mR_D}{1+g_mR_D}V_T $$

  • You also made an additional mistake in the second-last step. But it is irrelevant because of the first mistake.

The expression should resolve to:

$$ I_D = g_m\left(\frac{g_mR_D}{1+g_mR_D}-1\right)V_T $$

  • Then another mistake is that by your voltage/current convention, the impedance will be

$$ Z = -\frac{V_T}{I_D} $$

Don't forget the minus sign.

The result with these corrections will be:

$$Z = \frac{1}{g_m} + R_D$$

As expected. A diode-connected transistor has an approximate impedance of \$1/g_m\$ which is in series with \$R_D\$.

  • \$\begingroup\$ Thanks for the insight. But, why should Z be equal to the negative of the voltage to current ratio? \$\endgroup\$ – Curiosity Oct 24 '18 at 8:20
  • \$\begingroup\$ I think that 1/gm is in parallel with Rd so that would make your formula for Z incorrect! This might be due to the drawing, I was also confused thinking that Rd and /1gm are in series. They're not, the top of Rd is connected to Vdd. \$\endgroup\$ – Bimpelrekkie Oct 24 '18 at 10:29
  • \$\begingroup\$ @Curiosity That is because the impedance is the voltage across the nodes divided by the current flowing in the circuit. You defined the current as flowing out of the circuit. \$\endgroup\$ – Sven B Oct 24 '18 at 10:50
  • \$\begingroup\$ @Bimpelrekkie The author states in the comments that he tries to find the impedances looking into the source. This makes the transistor and resistor in series to AC ground. \$\endgroup\$ – Sven B Oct 24 '18 at 10:52
  • \$\begingroup\$ I see no mention of "looking into the source". OK, it's in the comments, then the schematic doesn't match. What a mess!! (the question, not your answer) All-in-all OP should be more clear on what he wants as now all of us are making assumptions. Literally: "derive the small-signal input impedance of a diode-connected MOSFET" and then the answer would be 1/gm. OP introduces the Rd resistor and complicates everything. \$\endgroup\$ – Bimpelrekkie Oct 24 '18 at 12:14

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