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I was reading the data sheet for 1N4007 datasheet for 1N4007 and I noticed that the values for the current (it's not clear from the datasheet, but they probably mean the non-repetitive peak forward surge) is given for resistive and inductive loads only. For capacitive loads, it requires a 20% derating. What is the reason for this? Does it have something to do with the p-n junction?

From what I understand, for a given current I, the diode requires a voltage drop V(I) (as complex as it can be), thus the power dissipated is I*V(I) and that should determine the max allowed current. This has nothing to do with the external circuitry used to obtain the current I. The only possible explanation I can imagine is that when connected to a capacitor, there is charge leakage into the p-n junction that causes changes. Any thoughts?

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The problem is that you may have a hard time limiting the absolute maximum of the current flowing through the diode, if the load is capacitive, since only the resistance of the wires, switches and the internal resistances of the source and the capacitor limits this current (of course, unless there's some circuit for explicit current limiting). Therefore nobody sizes the diode according to the peak value (in time) of the current as that would result in a way oversized diode, instead, the average value of the current is used with some derating.

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  • \$\begingroup\$ Actually, I might have misunderstood you. Do you mean that a capacitive load will produce variable current over time, and thus only averaged values should be limited (in this case to 80% of the peak value that would otherwise be obtained with resistive loads only)? \$\endgroup\$ – Ivan Dec 15 '13 at 5:10
  • \$\begingroup\$ A capacitive load will result in a largely variable current, and the peak can be extremely large - for a very short time. The "better" quality wires/capacitors you use, the larger the peak can get, unless you employ some resistors or other circuit elements to artificially limit this peak current. What I say is that there's no point in trying to design your circuit according to the peak value (you cannot really calculate that, anyway), instead, you design your circuit according to the average value - but then the data sheet advises to derate, so you multiply your average value by 1/0.8=1.25. \$\endgroup\$ – Laszlo Valko Dec 15 '13 at 15:44

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