Why may a poor reverse recovery diode cause "shoot-through" in a inverter?

[onemilimeter]

Let's refer to a 3-phase inverter, which made up of six power MOSFETs. Each MOSFET has an intrinsic body diode with rather poor reverse recovery characteristics. Learned from several online articles that the poor reverse recovery diode may cause a "short-circuit" or "shoot-through" of one inverter leg. For example, if the upper MOSFET is turned on when the body diode of opposing MOSFET  in the same inverter leg is conducting, then an "effective short circuit" may occur. However, I do not understand how a poor reverse recovery diode may cause "short-circuit" in one inverter leg. Kindly comment. Thanks.

[Rerouter]

a diode is a short with a voltage drop while conducting forward, and if it is very slow to recover, it can even pass current of the opposite polarity through it before it recovers,

the voltage drop of such varies, and with a mosfet body diode i dont have the faintest on what it might be, just be aware while its recovering, its effectivly a bipolar voltage drop,

[jahonen]

If one of the body diodes is conducting (with neither mosfet turned on) due to external inductive load and the opposite side mosfet turns on in the same inverter leg, this produces a short (or huge spike of current) via body diode until diode recovers. These current spikes cause all kinds of problems, like EMI. Obviously, it is not a problem if load would be completely resistive (as there would be no current through the body diodes then).

Regards,
Janne

[onemilimeter]

If one of the body diodes is conducting (with neither mosfet turned on) due to external inductive load and the opposite side mosfet turns on in the same inverter leg, this produces a short (or huge spike of current) via body diode until diode recovers.

Thanks. Please refer to the 3-phase inverter shown on the attached image. Let's the Vdc is 35V, the Rd_on of the MOSFET is 0.3ohm, and the forward voltage of the MOSFET body diode is 0.7V. If the upper MOSFET (Q1) is turned on while the lower MOSFET body diode is conducting as described in jahonen's last post, the current though Q1 will be (35-0.7)/0.3 = 114A. Does it mean this large current of 114A will flow through the Q2 body diode too?

[onemilimeter]

a diode is a short with a voltage drop while conducting forward, and if it is very slow to recover, it can even pass current of the opposite polarity through it before it recovers...

Thanks. What will limit the opposite polarity current that passes through the diode? Will it exceed peak reverse recovery current stated in its datasheet?

[jahonen]

If one of the body diodes is conducting (with neither mosfet turned on) due to external inductive load and the opposite side mosfet turns on in the same inverter leg, this produces a short (or huge spike of current) via body diode until diode recovers.

Thanks. Please refer to the 3-phase inverter shown on the attached image. Let's the Vdc is 35V, the Rd_on of the MOSFET is 0.3ohm, and the forward voltage of the MOSFET body diode is 0.7V. If the upper MOSFET (Q1) is turned on while the lower MOSFET body diode is conducting as described in jahonen's last post, the current though Q1 will be (35-0.7)/0.3 = 114A. Does it mean this large current of 114A will flow through the Q2 body diode too?

Yes, basically that is correct. Also, since power supply rails have some inductance, large di/dt of the current spike tends to generate huge voltage spikes on the rails, even killing bridge transistors if inductance and/or di/dt is high enough.

Regards,
Janne