Timeline for Why does reverse recovery behavior matter for half bridge bootstrap diode?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2022 at 5:29 | vote | accept | tobalt | ||
| Apr 1, 2022 at 5:04 | comment | added | Neil_UK | @tobalt No., As its forward bias voltage is limited, its forward current will be very very limited, so negligible charge will have been stored, so nothing to supply the reverse current when reverse biassed. | |
| Apr 1, 2022 at 4:03 | comment | added | tobalt | One thing I don't get though: If an IC integrates a pn diode and I add an extra Schottky externally in parallel: The pn will be still ~300mV forward biased during the low switch node phase. So it would still allow a large reverse current spike when the switch node goes high, no? | |
| Mar 31, 2022 at 21:27 | vote | accept | tobalt | ||
| Apr 1, 2022 at 4:19 | |||||
| Mar 31, 2022 at 15:33 | comment | added | Neil_UK | @DKNguyen With your experience, I can talk plainly. 'Vf' is a lie to children, silicon diodes will pass a current at any voltage. When you place a bias voltage on a diode, and read a current, it's generally reached equilibrium before you read the meter. This means that charge injection and recombination are balanced. When you've passed a large current through a diode for a while, a large amount of charge is stored. When you reverse bias it quickly, this stored charge flows out, leaving the diode appearing to conduct, see step recovery diode | |
| Mar 31, 2022 at 14:56 | comment | added | DKNguyen | So when a diode stops conducting because the voltage across it fails to exceed Vf, it is still forward biased? | |
| Mar 31, 2022 at 14:43 | comment | added | tobalt | Aha. I thought that the charge storage had more to do with the current through the diode. Apparently not. Thanks | |
| Mar 31, 2022 at 14:33 | history | answered | Neil_UK | CC BY-SA 4.0 |