Timeline for Can I Exceed a N Channel MOSFET's -Vgs Rating When Using It as a High Side Switch
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 25 at 6:29 | history | became hot network question | |||
| May 25 at 3:28 | answer | added | Fabio Barone | timeline score: 1 | |
| May 24 at 21:27 | answer | added | Jos Bergervoet | timeline score: 0 | |
| May 24 at 17:57 | answer | added | winny | timeline score: 1 | |
| May 24 at 16:59 | comment | added | user319836 | What do "load is on" and "load is off" mean? | |
| May 24 at 16:51 | comment | added | Vera Fodor | @RussellH thank you for the advise! I added it. | |
| May 24 at 16:50 | history | edited | Vera Fodor | CC BY-SA 4.0 |
added 266 characters in body
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| May 24 at 16:49 | comment | added | Dave Tweed | If the load can store charge and hold its own input at +24V, then yes, this is something you need to deal with. Normally, as the gate voltage falls, the source voltage falls with it. | |
| May 24 at 16:44 | comment | added | user319836 | Instead of describing the circuit, use the Circuit Lab schematic tool in the toolbar of the question editor. Many of us can read a diagram easier than the text. | |
| May 24 at 16:30 | comment | added | Michal Dudka | Described situation cant happen. You assume that potential on gate does not change. And therefore MOSFET stays open :) In other words, you have to change Vgs first and then transistor reacts by closing or opening. You cannot "magicaly" turn transistor OFF without any change on gate.... | |
| May 24 at 16:28 | history | edited | Vera Fodor | CC BY-SA 4.0 |
added 183 characters in body
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| May 24 at 16:25 | comment | added | The Photon | You haven't told us enough about your circuit for us to know what voltage will be on the source, so there's no way for us to answer this. | |
| May 24 at 16:17 | history | asked | Vera Fodor | CC BY-SA 4.0 |