0
\$\begingroup\$

According to the Sedra/Smith Microelectronic Circuits textbook, 6/7 eds., increasing \$v_{DS}\$ beyond \$v_{OV}\$ has no effect on the channel shape and charge. Why is that? It seems as if (thinking naively) increasing \$v_{DS}\$ should move the point at which the channel is pinched towards the source, thereby decreasing the charge in the channel. Also, if \$v_{DS}\$ is large enough, wouldn't we have some sort of breakdown?

The textbook says further that any increase in v_{DS} above the saturation threshold appears as a voltage drop across the depletion region. Why is that?

A related question: why is the bottom of the channel a straight line?

\$\endgroup\$
1
  • \$\begingroup\$ I actually remember that the channel becomes trapezoid from the same book, but cannot remember the condition for it \$\endgroup\$
    – C K
    Feb 10, 2017 at 20:53

1 Answer 1

0
\$\begingroup\$

Velocity saturation. Above 1 volt/micron (SWAG), the carriers are not accelerated any more.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.