I want to use an array of ceramic wirewound resistor in series and parallel as the load of an 80W boost converter with the output of 400volts dc. Is it suitable to use 8 numbers of 10W resistors as 80W load? I mean doesn't operating these kinds of resistors exactly in their rated wattage lead to overheating and failure?
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\$\begingroup\$ Yes, if resistors are good quality. Anyway it is better to consult its datasheet, at which conditions do they guarantee dissipated power. Good air convection will be required, and better to use fan, i think. \$\endgroup\$– VladimirJun 29 at 19:57
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\$\begingroup\$ Don't forget that, depending on the implementation, there may be some significant inductance which might affect the feedback (an extra zero). Best measure it beforehand to know what to expect. \$\endgroup\$– a concerned citizenJun 29 at 20:08
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\$\begingroup\$ @a concerned citizen , do you mean the inductance of resistors? What problem can this inductance cause? And what should I do? \$\endgroup\$– WeTechJun 30 at 3:37
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\$\begingroup\$ The answer may be yes (if you have the exact resistance), but the actual powwer disipated will depend on how the resistors are interconnected (the series and parallel) and the final resistance. \$\endgroup\$– StainlessSteelRatJun 30 at 14:05
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400 V , 80 W implies 0.2 A load of 2000 Ohms.
If you use all the rated load the temperature rise may be 100'C so at least 20% derating is recommended if not 50% from local ambient effects.
You can share power any way that adds up to 2kohm with equal values. It is more useful to use higher load R values in parallel so gradually increase the load before doing more stressful full load step tests. You can also use water cooled transistors in parallel with a 50 mV current sense for full range linear control. There are lots of options but beware of switched wiring inductance when turning off, so include clamp protection to the load.
