Peak dissipation of a resistorPieter-Tjerk de Boer, PA3FWM email@example.com
(This is an adapted version of part of an article I wrote for the Dutch amateur radio magazine Electron, September 2021.)
While working on a digital clock that is directly powered from the mains, I saw a brief flash of light in the circuit when switching it on. Because the flash was so brief, it seemed like a spark, but the voltages in the circuit weren't high enough for that. No damage seemed to occur, because I could repeat the effect. That made it possible to make a movie, to see what really happened. See the figure, showing seven consecutive stills from the movie, at 1/25 second intervals.
We briefly see a flame out of the resistor, but so brief that the resistor survives. Well, there is some damage, because the smoke in the last few pictures must come from somewhere. According to its colour coding, the resistor originally was 470 ohms, but after going through this a couple of times it was about 2.2 kOhm.
The next figure shows the relevant part of the schematic: an electrolytic capacitor is charged directly from the mains (via an isolation transformer during these experiments, for safety) to about 300 volts, and the purpose of the resistor is to limit the charging current. During normal operation, the circuit uses about 10 mA DC; 10 mA in 470 ohms would dissipate 47 mW, well within the capabilities of this resistor. In reality the dissipation would be a bit larger, because the current only flows during the peaks of the mains sinewave, when the diodes conduct, and must thus be higher than 10 mA, but not yet too high for such a quarter-watt resistor.
So, how could this resistor go up in smoke? At initial switch-on, the capacitor is not yet charged. Potentially there's over 300 volts across the resistor then (peak voltage of the mains = 230 volts times square root of 2), which according to P = U2R implies a dissipation of about 200 watts (not milliwatts)!
Do we need a 200 W resistor then? That would be inconveniently large. Fortunately, that's not needed either, because the high power is dissipated only during a very short time: after half a period of the mains, the capacitor is already mostly charged. Ideally the manufacturer of the resistor would specify the energy (power times duration) that one is allowed to dissipate in it during a short time. But for most resistors this is not known, and definitely not for a resistor from the junkbox...
How much energy is dissipated in our resistor while charging the capacitor? One can mathematically show that if the source offers a constant voltage, equally much energy is dissipated in the resistor as ends up in the capacitor, here about 1 joule (½CU²). But in our case the voltage is not constant (it's the full-wave rectified mains), so the above doesn't quite apply; in fact, one can show that in such cases the dissipation will be less.
B.t.w., not just resistors can go up in smoke. When radio amateurs transmit at high power on the lower-frequency bands, they need a lot of voltage on a relatively short antenna, which can easily lead to sparks, melting, smoking or even fire. For education and enjoyment, images of that are collected in the "Hall of Flame" .