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| RC Question |
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| DannyTheGhost:
--- Quote from: Yansi on June 16, 2019, 04:40:25 pm --- --- Quote from: DannyTheGhost on June 16, 2019, 04:12:27 pm ---If you integrate power over time you get energy, that's what you need to understand. As were stated above, to get your 'power dissipated' you simply need to get energy stored in capacitor (CU2/2) and divide it by 5*RC (how much time needed to 'fully' charge capacitor) energy/time=power --- End quote --- I don't understand your math. You either want POWER, which in this case is changing with time (then look at my exp equation), or you want the TOTAL ENERGY DISSIPATED, which does NOT have anything in common with your time constant, not even anything like 5*Tau. Total energy dissipated is equal the energy stored in the capacitor, which is exactly W = 0.5*C*U2 --- End quote --- Sorry, what I've calculated is average power dissipation in resistor during time when power can actually dissipate (ie there is current through resistor) |
| Yansi:
Current flows through the resistor infinitely from time t=0 to t=infinity. Average power dissipation may be relevant when supplying AC or pulsed voltage to the RC circuit. It has mostly zero meaning for designing one-shot charge or discharge circuits. Then the peak power and overall energy dissipated comes into play. For example when selecting an inrush current limiting thermistor (NTC), what you usually get from the datasheet is the pulsed energy as a relevant figure. Peak power dissipation is not that relevant, inrush NTCs are hefty device with a lot of thermal mass - heat is dissipated within almost all the volume of the device. The peak power they withstand is enormous. With switched safety discharge resistors you are usually after the allowed peak power dissipation and its duration so you can check the heat will spread fast enough within the resistor so no meltdown will occur. Contrary to the NTCs, in power resistors, heat is dissipated in the resistive material, which may be a wire (very small thermal mass on its own - temperature peaks very high locally), thin film carbon, etc. Which is why power resistors usually do not like too much power peaking. |
| bostonman:
The entire story to my original question is based on creating a reliability report. The resistor I stated in my original question needs to have a maximum power dissipation value based on some length of time. It's a 3W resistor, but we need to place a value for a short (instantaneous) value. So in other words, after a long time, the resistor dissipates nearly zero watts. In our report, we'd state it's a 3W, and, after a length of time, it dissipates approximately 0W. On another line, we need a specification for instantaneous wattage, and then how much the resistor dissipates. This is why I thought taking the integral from 0s to something like 1ms would give the total power. As an example, if the resistor can handle a peak of 1000W for 1ms (although I haven't seen such a spec), then I'd imagine taking the integral from 0s to 1ms would be the correct way to provide an "accurate" number. |
| The Electrician:
You might find some useful information here: https://www.vishay.com/resistors-fixed/high-pulse-load/ |
| RoGeorge:
--- Quote from: bostonman on June 17, 2019, 01:55:52 am ---The entire story to my original question is based on creating a reliability report... --- End quote --- Since you need to create a report for a resistor, I will assume that that resistor is part of a critical apparatus. The only way to approach this, is to contact the resistor's manufacturer, explain them exactly how do you plan to use the resistor, what are the implications if the resistor fails, and ask them for advice. It is more complicated than simply thinking only in terms of average and peak power. Again, forget about online advice from random people, and instead ask the engineers that are making that particular type of resistor you plan to use. |
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