Power semiconductors

When one switch of an inverter turns on, which loss increases due to the reverse recovery of the diode in the opposite arm?

  1. AThe turn-on loss of the switch that turns on (the reverse recovery current is added to it)
  2. BOnly the loss in the gate drive circuit
  3. CThe loss in the heat sink
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Answer: A. The turn-on loss of the switch that turns on (the reverse recovery current is added to it)

When the freewheeling diode is switched to reverse bias, a reverse recovery current flows due to stored carriers. This current flows through the switch that is turning on, increasing its peak current and turn-on loss. Using SiC Schottky diodes, which have little reverse recovery, greatly reduces this loss.

  • A. Correct. The reverse recovery current adds to the switch's turn-on loss.
  • B. Gate drive loss is not the main part of the increase due to reverse recovery.
  • C. A heat sink does not generate losses.
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More Power semiconductors questions

  1. Among soft-switching techniques for power converters, which correctly describes ZVS (zero-voltage switching)?
  2. When a power MOSFET's gate is charged with a constant current, why does the gate voltage stay flat for a while (the Miller plateau)?
  3. Why is a negative gate voltage (for example, −3 to −5 V) sometimes applied to SiC MOSFETs or IGBTs in the off state?
  4. What maximum junction temperature (Tj max) is commonly specified for typical silicon power devices?
  5. When choosing power devices, why are devices with a voltage rating higher than the actual operating voltage used (derating)?