Why does channel length modulation occurring in mosfets decrease the gain of amplifiers?
Electrical – Channel length modulation
mosfetnmos
Related Solutions
Have a look at the picture below. The green lines show the drain current of a transistor without channel length modulation (resistance is inifinite) and the black lines are for a transistor with channel length modulation.
The current is obviously not zero, but the change of current (and therefore the slope of the curve) in the saturation region is zero, if no channel length modulation is present.
Long ago I went through a similar search for BJTs with the highest Early voltage. My finding was that higher voltage devices had a higher Early voltage. Turns out there is a fundamental reason for this that I won't go into.
Instead of using a high voltage device, I switched to a cascode design, which greatly increased the output impedance.
It is possible to stack more transistors into the cascode and increase the output impedance even more.
With FETs, a dual-gate device does the same thing as a cascode, but in one device.
Best Answer
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).