Discussion
Why does the body diode's recovery matter in a bridge circuit?
Answer: D
When the opposite device turns on, the body diode's stored charge must be swept out, and that appears as a spike of current through both devices at once - a real dissipation, and a source of switching noise as well. The diode carries the freewheeling current perfectly well rather than preventing it, and it affects neither the channel resistance nor the threshold. Reverse recovery in a parasitic that cannot be removed is one reason synchronous converters need careful dead-time control, and it is why some designs add a faster external diode in parallel.
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