Timeline for answer to Dual mosfet 8205A - lithium battery protection circuit by rdtsc
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| S Sep 24, 2024 at 18:20 | history | suggested | Abdull | CC BY-SA 4.0 |
update to working link
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| Sep 24, 2024 at 17:26 | review | Suggested edits | |||
| S Sep 24, 2024 at 18:20 | |||||
| Feb 18, 2017 at 16:27 | comment | added | vini_i | @Atmega328 i think you confusion is that the active and saturation regions for a mosfet and bjt are opposite. Another thing to understand is that when Vgs rises Vds falls. With an increasing Vgs the mosfet turns on and as it turns on Vds decreases. As long as Vgs is not too high then Vgs-Vth is < Vds and the mosfet is in the saturation region. When Vgs gets high enough Vds will drop below Vgs-Vth and the mosfet will go into the linear region. | |
| Nov 30, 2015 at 9:00 | comment | added | user83582 | sorry for my dullness. I still don't understand it. I have read but not comprehend that wiki page many times. according to it, When VGS > Vth and VDS ≥ ( VGS – Vth ), the fet is in Saturation or active mode; and since Vds is supposed to be 0 zero, how can it be bigger or equal to Vgs-vth? more doubtful than before, that you mentioned this internal parasitic body diode. since both side are negative, how do I know which side the current goes?? and a diode should have a forward voltage? how can there be enough forward voltage for that diode, both ends are ground, zero voltage?? | |
| Nov 30, 2015 at 6:32 | history | answered | rdtsc | CC BY-SA 3.0 |