Now, consider this charging was very fast and that no heat was lost. In a RC circuit, charging the capacitor from 0 to the source voltage always dissipates half of the energy given by the source in the resistance and the other half is stored in the capacitor, which also means these energies are equal. The power dissipation being large in the beginning won't necessarily make it perceptibly hot in a single charging, you have to take in account its thermal capacity in account. Yet 5mm LEDs do not as the micron size gold whisker wire bond can fail(fuse open) at some level after a peak current that is usually 30mA for a 20mA thermal rating.
Semiconductors have a SOA Power vs pulse width curve that spans a wide range of Pk/avg powers It might be expected to rise 11 deg C butĬonvection and lead conduction cooling would reduce that to barelyįollow absolute max specs with a 50% safety margin In 1ms when the Vcap=64% Vin leaving 36% Voltage across the resistor and the square of that yields 13% Pmax at t=RCīut assume that the thermal mass requires X seconds to reach 1/2 it’s rated 100’C max temp rise above 25’C the power in 1ms will result in a case temperature rise of 0.1% of X temperature rise.īut what if the part is only rated for 1/8W or a peak power of 115 This means a start current of 1.2A, 14.4W and after 350us it is 7.2W Well, according to Ohm's law we know that:
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