I am using the circuit below to indicate the status of a 230V AC line.
This works well when I apply power to it. The LED is on with a good brightness.
However if I switch the main on and off and back very quickly the 100 Ω resistor (R7) gets fried. I have made several circuit of the kind and checked all polarity etc. This is mounted on PCB. Replacing the resistor make the circuit good again but if I dare switching the power too quickly it does the same again.
Any idea what could cause this?
As an alternative to the answer by GodJihyo, here is another way of looking at the problem.
When the power is off, R8 discharges C3. So the voltage across it is zero. The voltage drop in the diodes is negligible when compared with 230V.
The peak voltage of 230V AC is about 325V.
So at the moment you turn on the power, pretty much the entire supply voltage is across R7. That voltage could be anything between 0V and 325V. Using P = V²/R, for a fraction of a second, the power dissipated in R7 could be anything between 0W and 1000W.
Once C3 reaches the supply voltage, its impedance limits the current, and R7 cools down.
But if you keep turning the power off and on very rapidly, R7 gets hit by a rapid succession of power surges, and fries.
At line frequency the capacitor has a high enough impedance to limit the current, but when switching on and off quickly you can create transients at higher frequencies where the capacitor impedance is low. This can allow enough current to flow to exceed the resistor's ratings, and eventually the LED may also be damaged.
At 60 Hz 0.22\$\mu\$ will have an impedance of around 12k\$\Omega\$, limiting the current to around 20 mA. If the resistor is 1/4 W the maximum current it can handle will be $$ \sqrt{\frac{0.25 W}{100\Omega}} = 50 mA $$ For a 1/2 W resistor it's around 70 mA, and for 1 W it's 100 mA. So it doesn't take a lot of extra current to burn up the resistor, just raising the frequency to 180 Hz would make the capacitor's impedance around 4k\$\Omega\$ and be enough to blow a 1/4 W resistor.
R7 needs to be a pulse-withstanding 3W part. Through-hole would keep the heat off the PCB. Nowadays several very good SMD pulse-withstanding pets are available so both should work. It will probably survive your tests then.
D1 could be in series with the LED, not across it. It should be UF4007. The rectifier you have there is too slow and not really rated for the application. The LED could be overstressed and fail because of it.
There still should be an anti-parallel rectifier across the LED, but one in series with the two is needed as well. Both can be UF4007 to keep the number of BOM lines down.
simulate this circuit – Schematic created using CircuitLab
