Timeline for MOSFET fast turn-off not enough
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2015 at 18:27 | comment | added | a concerned citizen | If I do that, then I need to do it for M4, too, which has fixed gound reference, and then for M3+M2 when the negative voltage comes around. It's clear that this is the cause so the driving must be changed, and it will be costly, no way around it. Maybe with the slower turn on the dead-time can be reduced, too. Thank you for the "light", I marked this answer as the answer. | |
| Dec 29, 2015 at 18:22 | vote | accept | a concerned citizen | ||
| Dec 29, 2015 at 18:08 | comment | added | jp314 | No, 160 mV shouldn't really turn on the FET. Fundamentally, you need stronger turn-off then turn-on strength. Try putting the GND end of your VCCS's for gate drive to ~ 3 V. This will turn on with less current, and drive VGS negative for off. Then 'just' come up with a circuit that does something similar. | |
| Dec 29, 2015 at 17:21 | comment | added | a concerned citizen | V(gh,mid)@6.05us=~160mV, is that enough to turn on M1? Still, I changed the gate drive to be a 100Ohm anti-parallel with a diode and the spikes lowered a bit. Increasing R to 1kOhm drastically reduced the spikes and dissipated power, but the gate drive is weak and I fear for low pulse widths. I'll have to concoct some sort of constant current with two slopes, slow before Miller and fast afterwards, while keeping the turn-off diode, but I suspect complications (both electronic and money). Thank you for the explanation. I'll still wait a while to see what other answers come up. | |
| Dec 29, 2015 at 16:59 | history | answered | jp314 | CC BY-SA 3.0 |