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In a back-to-back MOSFET configuration, if the single MOSFET is maximum-rated at BVDSS=100 V (like the IRF530 in the example), does that mean that the series of both has a breakdown voltage of 200 V or rather 100 V?

My argument in favor of 100 V would be that the lower MOSFET's forward-biased body diode has a maximum voltage of 0.4-0.8V if current starts to flow, so only the upper MOSFET is truly blocking. But I am not sure if this is the correct way of looking at it.

schematic

simulate this circuit – Schematic created using CircuitLab

oliver
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    Yes: the total breakdown voltage is the sum of (Reverse-Bias) + (Forward Bias). So 100V + 0.6V. – Dwayne Reid Apr 17 '23 at 17:46
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    Think about it this way: if M1 were to conduct, then that puts 150V on the gate of M2, which will definitely cause M2 to fail (source-gate voltage limit is typically +/-20V or less.) – rdtsc Apr 17 '23 at 19:34
  • Your schematic has the gates shorted to the sources. This short should be replaced with a gate driving circuit. – PStechPaul Apr 18 '23 at 02:41
  • The shorted gates are only for artificially forcing the MOSFET into off-state (i.e. only for demonstration). – oliver Apr 24 '23 at 09:55

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