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Assuming that the diode is rated for 20 kV and the pulse is only 250 V, will a 250 V 8 kA pulse that lasts 0.2 uS break a 20 kV 1 A diode? This also assumes that the diode recieves 10 of these pulses every second for 5 seconds.

For reference, terminal are axial leads and the diode is a HV rectifier diode.

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    \$\begingroup\$ Read the devices data sheet. Many will show limits for pulses of some duration. Outside of that, you've become a test engineer. 8000 x the rated current seems like a lo tho, even for a short duration. Will the bonded leads fail? Will the semiconductor fail? Hard to say. \$\endgroup\$
    – CrossRoads
    Feb 19, 2020 at 18:48
  • \$\begingroup\$ Semiconductors have imperfections in their substrate. That means that you can't just have n times the current if you reduce the pulse length by 1/n as n approaches infinite You eventually hit a limit where pulse is so short that the heat cannot dissipate fast enough to be evenly distributed and hotspots develop and those hotspots fail first. Where is this limit? I'm not sure but I'm quite certain 8000% of the rated continuous current is well beyond it. \$\endgroup\$
    – DKNguyen
    Feb 19, 2020 at 18:58
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    \$\begingroup\$ My guess is that you'd hit 250V Vf long before you hit 8000A. If 8000A is a violent explosion 20 series junctions (?) with 12V Vf across them sounds pretty bad too.You must have almost zero inductance to be able to get di/dt like that. \$\endgroup\$
    – Spehro Pefhany
    Feb 19, 2020 at 19:00
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    \$\begingroup\$ @dknguyen Not even 8000% but 800000% \$\endgroup\$
    – pericynthion
    Feb 19, 2020 at 19:29
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    \$\begingroup\$ The maximum voltage rating is for reverse bias and the maximum current rating is for forward bias. Relaxing the reverse bias voltage limit doesn't change the forward bias limits. \$\endgroup\$
    – Elliot Alderson
    Feb 19, 2020 at 20:02

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