Hi ccurtis,
Thanks for your reply and advices.
I unfortunately don't know how long the 240v impulse lasts, probably less than half a second (but enough to trigger a 240v latching relay) and I have no tool to measure it even if it seems fairly constant. Can I use a trial & error technique with R1?
It's a rather unconventional circuit, but what the hey. I would dispense with R1 since it will only limit the minimum impulse time. Short it out.
The output relay contact closure length doesn't really matter, as long as it is above one second (5sec is fine as well). I don't have the specs of the relay yet but I suppose the capacity of the capacitor can be expressed in function of the relay coil resistance + R2. Could you help me with that?
Thanks !
Martin
First you determine from the relay spec sheet what the maximum coil voltage is. Call that voltage Vrmax. Then from the relay spec sheet you determine the relay drop-out voltage (the voltage at which the relay is de-energized). Call that voltage Vrmin. Choose R2 so that the voltage across the relay coil resistance is Vrmax with an initial capacitor voltage of 339 Volts (240×1.414). Add the coil resistance and R2. Call that sum R. Then you calculate the percentage the 339 voltage has to fall so that the voltage across the relay coil resistance is at Vrmin. Call that percentage, Pct. From the standard graph of percent voltage discharge for a capacitor vs. RC time constant (1 RC time constant is 63%, for example) you see how many RC time constants it takes to achieve Pct. From R, determine what C has to be so that for the number of time constants (RC) is greater than 1 sec.
BTW, I did this for one 12 volt coil relay I saw the specs for, and came up with 330 µF. A 330 µF 400 V capacitor is no small device, but it is widely available.
If the impulse voltage is going to be present for more than one second, or very frequently, the several watt power dissipation in R2 needs to be accounted for. To protect against burning up R2, use a fuse with a delay so that it will open in that case, or use a power resistor.
To be safe, and conservative, the diodes should have a minimum reverse breakdown voltage of 400 volts. The forward current rating should about 1 amp. The peak current will be higher, when charging a dead capacitor, but diodes can handle large peak currents. The 1N4005 is a good choice.
You could get by with a significantly smaller valued capacitor if you realize that a relay coil can withstand several times is rated coil voltage for a short period of time. This spec is not given in relay data that I have seen, however. Also, if that fact is to be exploited the impulse time must be limited, or the relay coil will be destoyed, unless some sort of protection is added.