-1
\$\begingroup\$

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

\$\endgroup\$
15
  • \$\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\$ Oct 24, 2018 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, 2018 at 6:55
  • \$\begingroup\$ I should have mentioned that VDD is a DC voltage. \$\endgroup\$
    – Curiosity
    Oct 24, 2018 at 6:55
  • \$\begingroup\$ So, Rd is just a resistor, used to control the DC current of the MOSFET. \$\endgroup\$
    – Curiosity
    Oct 24, 2018 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\$ Oct 24, 2018 at 7:58

1 Answer 1

3
\$\begingroup\$

[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:

schematic

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\$.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for the insight. But, why should Z be equal to the negative of the voltage to current ratio? \$\endgroup\$
    – Curiosity
    Oct 24, 2018 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\$ Oct 24, 2018 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, 2018 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, 2018 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\$ Oct 24, 2018 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.