# Channel length modulation

Why does channel length modulation occurring in mosfets decrease the gain of amplifiers?

Channel length modulation causes a finite ratio d(Vds)/d(Id) which resembles a finite otput resistance ro. With other words: The drain current Id is not completely independent on Vds (Id rises slightly with Vds). This output resistance ro acts in parallel to the gain determining external ohmic resistor Rd. As a consequence, the ouput signal voltage is determined by the value ro||Rd (which is smaller than Rd).

When you increase voltage between drain and source (Vds), because of strong vertical electric field (from drain to bulk) channel on drain side become thinner. As a result total lenght of channel become shorter.

Normally drain current:

$I_D=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{TH})^2$

After shortening of channel

$I_D=\frac{1}{2}\mu_nC_{ox}\frac{W}{L-\Delta&space;L}(V_{GS}-V_{TH})^2$

At the end:

$I_D=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{TH})^2(1+\lambda&space;V_{DS})$

• Yes but the question was why does this reduce amplifier gain, not why does this change effective channel length. – Joel Wigton Apr 7 '17 at 22:48