Timeline for Determine MOSFET switching speed
Current License: CC BY-SA 4.0
16 events
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| Sep 25 at 22:54 | comment | added | Jason S | "Maybe it is not written because it depends on Vg, Ig, Qg" -- yes, it depends on the driving circuit. There's some inherent speed limitations, but you usually don't want to try to drive things that fast, just because it depends on hard-to-model parasitics that may vary from part to part. | |
| Sep 25 at 22:53 | history | edited | Jason S | CC BY-SA 4.0 |
added 332 characters in body
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| Sep 25 at 22:48 | comment | added | Jason S | I was sloppy about the term "max switching time". I'll fix.I have since written a blog entry about this, so I'll post a link. | |
| Sep 25 at 15:12 | comment | added | KaleM | @JasonS hi! by "max switching time" you mean the maximum time taken by the mosfet to turn on, right? If the answer is yes, it is not clear to me where in the datasheet is written the maximum frequency at which the mosfet can switch. Maybe it is not written because it depends on Vg, Ig, Qg . If the "max switching time" you calculated is ton, I should calculate toff and finally find f=1/(ton+toff) ? | |
| Feb 24, 2020 at 23:33 | history | edited | Jason S | CC BY-SA 4.0 |
added 43 characters in body
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| Feb 24, 2020 at 23:33 | comment | added | Jason S | Archive.org has it, whew. | |
| Feb 24, 2020 at 23:32 | comment | added | Jason S | Grr. Bit rot. No idea where to find this. | |
| Feb 22, 2020 at 14:22 | comment | added | Kyle Hunter | @JasonS The link of the note from TI, slup097 is dead. Care to share an updated source? I am unable to find it. | |
| Feb 18, 2019 at 13:22 | history | rollback | Jason S |
Rollback to Revision 1
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| S Apr 19, 2013 at 3:06 | history | suggested | Dirceu Rodrigues Jr | CC BY-SA 3.0 |
Complementing
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| Apr 19, 2013 at 1:28 | review | Suggested edits | |||
| S Apr 19, 2013 at 3:06 | |||||
| Oct 9, 2011 at 1:31 | comment | added | Jason S | you don't want to drive the MOSFET source, you want to control Vgs using a voltage source (not to be confused with the MOSFET source pin) via a series resistor. | |
| Oct 8, 2011 at 23:45 | comment | added | Dor | (1) OK but why would you like to drive the source? Why not driving the drain, which is more common? (2) As can be seen from the graphs, Vgs = Vgsth for a moment, but continues to rise to the applied Vgs voltage (which is higher). It stays there until Vds equals zero. What am I missing here? :s | |
| Oct 7, 2011 at 11:56 | comment | added | Jason S | because I = (Vs=5V - Vgate)/R, and most of the time the gate sits at Vgsth = 1-2V. | |
| Oct 7, 2011 at 7:26 | comment | added | Dor | Great & detailed answer, thank you. "If I have time, I'll write a blog entry about that and post a link to it" - sure :). One thing - why the gate charging current is 30-40mA ? Why not 5v/100ohm = 50mA ? | |
| Oct 6, 2011 at 13:40 | history | answered | Jason S | CC BY-SA 3.0 |