Once the electrons reach the end of the channel, they experience the high electric field in the depletion region surrounding the drain junction and are rapidly swept to the drain terminal. Therefore, the device still conducts, but I can't understand why the current is unchanged.
8
-
\$\begingroup\$ The current is unchanged -- compared to what? Other parts of the channel away from the end? Compared to a different mode of operation for the MOSFET? The current can't change at different parts of the channel because charge must be conserved. \$\endgroup\$– nanofaradJul 28, 2021 at 16:13
-
\$\begingroup\$ Did you learn MOSFET's operation regions? @nanofarad. In reality, Id is still slightly increase, but in ideal it is unchanged. I just want to know why it is unchanged in physical. \$\endgroup\$– mi truongJul 28, 2021 at 16:22
-
\$\begingroup\$ I did learn them. I am asking clarification for your question because it was not worded clearly. Your comment does not help to clarify since it starts by stating that Id increases slightly (as a result of which effect that you care about?) but then asks why it's unchanged in the physical case. \$\endgroup\$– nanofaradJul 28, 2021 at 16:26
-
\$\begingroup\$ Okay, so why it is unchanged in physical case? @nanofarad \$\endgroup\$– mi truongJul 28, 2021 at 16:31
-
\$\begingroup\$ Again, unchanged compared to what base case? You've used the word "unchanged" multiple times, but you haven't stated what you're comparing it to. Compared to elsewhere in the channel? Compared to a different drain voltage? Compared to a different device geometry? \$\endgroup\$– nanofaradJul 28, 2021 at 16:33
|
Show 3 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
Example N-JFET: The width and the form of the depletion region is determined by VGS (general width) as well as VDS (unsymmetric form). The channel width becomes smaller in the drain region (equivalent to a broader depletion region).
Small voltages VDS (max 1 V): The channel behaves (nearly) like an ohmic region because the influence of VDS on the form of the depletion region is not yet remarkable. Hence, ID increases (nearly) linearly with VDS.
Rising VDS: The channel width becomes smaller close to the drain region and this effect counteracts (stops) the linear ID increase.
Saturation region: Now the influence of the rising VDS dominates the width of the depletion region: Any further VDS increase causes the remaining n-channel to become smaller and smaller (pinch-off effect). As the result, both effects (ohmic behaviour and pinch-off) nearly cancel each other and ID remains nearly constant.
As we know, in electronics nothing is ideal - and this applies, of course, also to the FET: Up to a certain extend, the ohmic behaviour of the channel cannot completely stopped by the pinch-off effect. Hence, there is still a small ID increase with rising VDS values.
