# Why negative feedback (common source degeneration) increases linearity range?

From this online lecture (page 190-20), I see that the negative feedback (Rs in common source degeneration) helps increases linearity range.

I am wondering if anybody know an explanation for this intuitively or mathematically.

Another question is what is the difference between the left and right hand pictures in the image below? For small signal, they are exactly the same.

Thank you.

• The feedback provides a correction signal that you use to compare the actual output to the desired output. Then that error signal is used to improve the output in regard to how it tracks the input. Oct 16, 2016 at 5:54

Suppose you have a nonlinear transfer function $\frac{v_O}{v_{IN}} = a(v_{IN})$

If you add negative feedback 1/m such that:

$\frac{v_O}{v_{IN}} = a(v_{IN})(v_{IN} - \frac{ v_O}{m})$

so

$\frac{v_O}{v_{IN}} = \frac {a(v_{IN})}{1+\frac{a(v_{IN})}{m}}$

For $\frac{a(v_{IN})}{m} \gg 1, \frac{v_O}{v_{IN}} \approx m$ (linear gain of m)

For $\frac{a(v_{IN})}{m} \ll 1, \frac{v_O}{v_{IN}} \approx a(v_{IN})$ (nonlinear with negligible feedback)

As you can see, the closed loop gain must be reduced significantly from the open-loop gain: $m \ll a(v_{IN})$ for there to be a large reduction in nonlinearity.

• Thanks. I understand how negative feedback increases linearity but I would like to know why it increases "linearity range" not just linearity. Oct 16, 2016 at 6:36
• So the linearity range is the range over which the transfer function is linear to a certain degree. Increase the linearity and the range increases, at least until you run out of gain as I indicated above. Oct 16, 2016 at 6:43
• After you apply a local feedback the stage gain drops and this is why the "linearity range" increases. We know need a more voltage at the input to "move" out of his "linear region" because of a "less" gain.
– G36
Oct 16, 2016 at 11:11