NMOS Common source DC-bias and gain

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SUMMARY

The discussion focuses on calculating the DC bias and AC gain for a common source NMOS amplifier. The participant correctly calculated the drain-source current (IDS) as 0.1 mA and the output current (IOUT) as 0.2 mA, leading to a total required current of 0.3 mA. The AC gain was derived using the transconductance (gm) formula, resulting in a gain of -10. The calculations were confirmed as accurate, emphasizing the importance of clearly defining the problem.

PREREQUISITES
  • Understanding of NMOS transistor operation
  • Familiarity with transconductance (gm) calculations
  • Knowledge of AC and DC analysis in amplifier circuits
  • Basic circuit analysis skills, including Ohm's Law and Kirchhoff's laws
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  • Study NMOS transistor characteristics and parameters
  • Learn about small-signal models for amplifiers
  • Explore advanced AC gain calculations in common source amplifiers
  • Investigate the effects of channel modulation on NMOS performance
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Electrical engineering students, circuit designers, and anyone involved in amplifier design and analysis will benefit from this discussion.

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Homework Statement



3PNOyHO.jpg

VOUT, they mean DC

Homework Equations


See picture, r0 is neglected also lambda (channel modulation)

The Attempt at a Solution



I calculate I DS with the formule given: 0.400*(0.5)²=0.1 mA

IOUT = 5/25000 = 0.2 mA

So the current source has got to be 0.1 + 0.2 = 0.3mA. Thats the DC current it has to be delivered right?

I also had to calculate the AC gain:

gm = K (VGS - VT) = 0.4mA/V

vin = vgs

vout / vgs = -gm * vgs * RD = -10This correct?
 
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If the thing in series with the +10V supply is indeed a current source labeled ID then you computed ID correctly.

A good first step is always to define the problem accurately.

You also have computed the ac gain correctly.
 
rude man said:
If the thing in series with the +10V supply is indeed a current source labeled ID then you computed ID correctly.

A good first step is always to define the problem accurately.

You also have computed the ac gain correctly.

Yes, I was also confused about the current source symbol. :) But it is one indeed.

Thanks for the confirmation that my calculations are correct.
 

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