1
\$\begingroup\$

I am having difficulty calculating the gain and input resistance of the circuit below.

Neglecting ro, capacitance, and considering Ibias as ideal (infinite resistance), would this mean that the current from M1 is 0 (gm1Vgs)?

And further, indicating that the output voltage is 0?

Regarding the input resistance, I am not sure how to treat small-signal vo. I found a similar case regarding a common source with source degeneration, where the author (Razavi) was calculating the output impedance, and he shorted the small-signal input. I am not sure if this applies to the circuit below. Also, similar to the case above, since gm1Vx = 0, would this mean the input resistance in 0?

M1 is causing me lots of problems. If the input resistance is 0 this would imply that the noise figure would be 0. If you consider a source with resistance Rs, the alpha value (Rin/(Rin+Rs)) is 0, thus the noise from Rs is zero. Meaning the total noise figure would be each noise contribution divided by 0.

enter image description here enter image description here

\$\endgroup\$
9
  • \$\begingroup\$ it seems that in both your circuits you are missing the incremental source resistance 1/gm \$\endgroup\$
    – S.s.
    Feb 20, 2018 at 19:39
  • \$\begingroup\$ Neglecting M1 ro, capacitances and Ibias output resistance you paint yourself in the corner of a nonsense circuit, a current source can't be left open. \$\endgroup\$
    – carloc
    Feb 20, 2018 at 21:57
  • \$\begingroup\$ I think your \$V_o\$ = 0 is incorrect. Sure no current flows but without the feedback path through M2, the AC gain would be infinite as there is no impedance at the output. Do you understand what this circuit actually does? If there's a small increase at /$V_i/$ what happens at \$V_o\$? \$\endgroup\$ Feb 20, 2018 at 21:58
  • \$\begingroup\$ @carloc a current source can't be left open Normally I would agree but this is small signal (so linearized) and there's feedback, which would "limit" what happens on the output. I think if we'd use the formula for circuits with negative feedback that might lead to a solution. \$\endgroup\$ Feb 20, 2018 at 22:01
  • \$\begingroup\$ @Bimpelrekkie mmm a kinda of ideal operational amplifier then ... vi=0 virtually, R1 can be neglected, yes it could make sense, all in all a transresistance amplifier with Rm=-1/gm2. Well a little tricky at least. \$\endgroup\$
    – carloc
    Feb 20, 2018 at 22:11

1 Answer 1

0
\$\begingroup\$

Not an answer yet, a few hints though

schematic

simulate this circuit – Schematic created using CircuitLab

If you consider just your M1 MOS and its bias source small signal model you may consider \$g_\mathrm{m}v_\mathrm{i}\$ current dependant generator.

It's open circuit, so the only consistent operation point can be "zero output current" i.e.

$$i_\mathrm{o}=g_\mathrm{m}v_\mathrm{i}=0\quad\Rightarrow\quad v_\mathrm{i}=0$$

So, looking at the input port we have \$(v_\mathrm{i}=0\,;\,i_\mathrm{i}=0)\$


In these conditions output voltage could be written as output current over output conductance (zero since it's open circuit)

$$v_\mathrm{o}=-\frac{i_\mathrm{o}}{g_\mathrm{o}}=\frac{0}{0}\;\rightarrow\;\mathrm{undefined}$$

Undefined just means that \$v_\mathrm{o}\$ could be any value.


Toghether they form one circuit which exhibits zero input voltage and current (sometimes called virtual ground in english speaking countries) and whatever output voltage.

This is similar to an ideal opamp, the only difference is that our circuit output current has to be zero. All the above is similar to a nullor

So now @user367640, any idea??

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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