Timeline for Reason of saturation region in MOSFET
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2021 at 16:29 | comment | added | J. Murray | @Ben You may be interested in this online textbook which goes into some of the models used for MOSFETs. To answer your question, there is still an inversion region throughout most of the body, but it closes before it reaches the drain. The charge carriers flow from the source, though the carrier-filled inversion layer, across the small pinched off region, and into the drain. | |
| Jan 21, 2021 at 17:30 | comment | added | Ben | That was the answer I was looking for all the time! I couldn't understand what $ V_{DS} > V_{GS} - V_{th} $ means, now it makes sense! When there is no channel anymore, where do the electrons come from? | |
| Jan 21, 2021 at 12:21 | comment | added | J. Murray | @Ben It is removed. When $V_{DS}\neq 0$, the potential varies across the length of the channel. When $V_{DS}>V_{GS}-V_{th}$, then there will be a region near the drain where the potential difference between the gate and the body is less than $V_{th}$, which means that the conducting channel will not form. Note that the pinched-off region has no mobile charge carriers of its own, but it does not prevent charge carriers from passing through it. | |
| Jan 21, 2021 at 7:12 | vote | accept | Ben | ||
| Jan 21, 2021 at 7:12 | comment | added | Ben | interesting, that the current doesn't decrease there, nevertheless. Is it correct to see a pinched-off channel as "removed" in this small certain area? Or just very thin? Probably the latter? | |
| Jan 20, 2021 at 19:46 | comment | added | J. Murray | @Ben In the small region where the channel is pinched off, the electric field is very high, so electrons can still be pulled through it. | |
| Jan 20, 2021 at 19:00 | comment | added | Ben | Ok, I understood your explanation now, I guess. To boil it down: V_DS changes the positive charge at gate which affects the n-channel. So far, so good. But why is there still a current flowing, even when reaching this pinch-off limit? | |
| Jan 20, 2021 at 18:43 | comment | added | Ben | Sorry for my pedantry but then it should be "when source potential is higher..". Is there a difference if the potential from Drain is higher or from Source? I need to ask anything which is unclear as I try to understand all this for some hours meanwhile... | |
| Jan 20, 2021 at 18:26 | comment | added | J. Murray | @Ben (1) $V_{DS}$ is the voltage between the source and drain. If the source and drain are at the same potential, then $V_{DS}=0$. (2) Yes | |
| Jan 20, 2021 at 18:22 | comment | added | Ben | Thank you very much! How exactly do I have to understand "when source voltage is higher than drain voltage"? How can they be different when it is about V_DS? Another aspect towards V_GS - V_th: Isn't V_th fixed? | |
| Jan 20, 2021 at 17:01 | history | answered | J. Murray | CC BY-SA 4.0 |