# Why does reverse recovery behavior matter for half bridge bootstrap diode?

I have seen several recommendations in the past to use a schottky bootstrap diode in half bridge drivers, even if the driver integrates a Si junction diode for the purpose, because supposedly the reverse recovery of that diode would be really bad. Random reference.

simulate this circuit – Schematic created using CircuitLab

But why is the reverse recovery behavior of diode D1 even important?

My thinking is as follows:

• When the switch node turns low, current flows through D1, Cboot and M4 to replenish the charge in Cboot.
• This is only a very short pulse, as Cboot will soon be "full" limited by the logic voltage. Then, D1 will be essentially off, having a very small forward bias. This small bias will slowly raise the bootstrap voltage a little more.
• Much later, the half bridge toggles. Current flows from Cboot and M2 into M1's gate. D1 doesn't take part because the bootstrap voltage is still too high for it to develop appreciable forward bias.
• The switch node jumps up. So when the diode becomes reverse biased, it is off already. Then from what would it even need to recover?

• When the Switch node turns low, current flows through D1, Cboot and M4 to replenish the charge in Cboot

Which stores charge in the diode junction, it's how silicon junction diodes work.

• This is only a very short pulse, as Cboot will soon be "full" limited by the logic voltage. Then, D1 will be essentially off, having a very small forward bias. This small bias will slowly raise the bootstrap voltage a little more.

The diode stays 'on' while it has charge stored in it. The bias across the diode is still forward, and so is doing nothing to sweep charge out of the junction, in fact charge is still being injected by the forward current. Charge is also recombining slowly within the junction, so the diode will head to a new equilibrium stored charge level where the recombination equals the injection rate.But it's only started to go there, it doesn't happen instantly. In a 'fast' diode, recombination happens more quickly than in a slow one like a 1N4004.

• Much later, the half bridge toggles. Current flows from Cboot and M2 into M1's gate. D1 doesn't take part because the bootstrap voltage is still too high for it to develop appreciable forward bias.

As far as D1 is concerned, not a lot has changed round it. This state is fairly brief.

• The switch node jumps up. So when the diode becomes reverse biased, it is off already. Then from what would it even need to recover ?

Now finally the bias across D1 reverses. It's still on, as there's plenty of charge remaining in the junction. The remaining charge that hasn't recombined now starts to get swept out of the junction. This sends a large reverse current through the diode heating it, and dragging the voltage on Cboot down a bit.

A schottky diode doesn't have this charge storage mechanism, so behaves more like the sequence in your question.

• Aha. I thought that the charge storage had more to do with the current through the diode. Apparently not. Thanks Mar 31, 2022 at 14:43
• So when a diode stops conducting because the voltage across it fails to exceed Vf, it is still forward biased? Mar 31, 2022 at 14:56
• @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 15:33
• 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? Apr 1, 2022 at 4:03
• @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 5:04

A Schottky is not usually necessary or worthwhile.

If D1 is implemented in an integrated circuit, it is likely quite fast and reverse recovery current & charge -- while they can be large -- won't exceed the original charge, so the energy loss would be minimal.

After the initial charging of CBOOT, D1 only has to replenish the charge lost by supplying gate charge to M1 (via M2) (and possibly some small additional currents consumed by the driver or maybe current-sensing circuits also powered from VBOOT). D1's own capacitance\charge will be negligibly less than that.

If in an IC system, there could be a (very) slight risk using a Schottky -- during the M1,M4 dead time, the switch node is at -0.7 V (or lower because of inductance). This can mean that D1 could charge to more than VLOGIC. If not careful, this could exceed the safe VBOOT..VSWITCH rating.

• Ah interesting. I didnt think about the dead time! This also means that the boot diode actually conducts strongly right before the switch node goes high, namely in the dead time. So that certainly is a case why reverse recovery would matter! Apr 3, 2022 at 4:33