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I'm trying to obtain the expression of the noise figure defined for this circuit :

enter image description here

The transistor is in saturation, and strong inversion. The flicker noise can be ignored.

Here is my attempt:

enter image description here

For the output noise of \$R_s\$, I just multiplied the noise voltage of \$R_s\$ by the common source gain.

For the output noise of \$r_g\$, I just multiplied the noise voltage of \$r_g\$ by the common source gain. I'm not sure if I can do that.

For the output noise of the noise current \$i_{d,M1}\$ , I modeled that current source by a voltage source in series with the gate \$4kT\phi/g_{m1}\$ and just multiplied that noise voltage by the common source gain, can I do that?

And for the output noise of RD, it's the noise voltage source of RD itself.

The total noise at the output would be the sum of all of these. Is everything correct?

Sven B
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Scipio
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1 Answers1

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The question cannot really be answered as the boundaries are not stated clearly. For a full noise calculation you'd expect:

  • No gate, load or any parasitic capacitances seem to be considered. Perhaps this is not part of the assignment, but if it is you forgot it.
  • You did not consider flicker noise of the transistor. Perhaps this is not part of the assignment, but if it is you forgot it.
  • You did not consider the effect of the output impedance of the transistor (\$r_o\$) which is in most cases considered. Perhaps this is not part of the assignment, but if it is you forgot it.

If none of the above are an issue I don't see an immediate mistake with your reasoning (you'll have to be responsible for the actual math though).

Can I do that?

In all those cases, you have a noise voltage source that can be shifted to the front in place of \$v_s\$. Since you end up with the same circuit as the original, you can indeed simply multiply with the amplifier gain to get the output noise.

Is everything correct?

You can use the sympy library in Python or another CAS to validate your math if you're worried about it.

Sven B
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