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Fixed the pilot light

A project log for AC Safety timer

Turns your soldering iron off after a half hour

nick-sayerNick Sayer 02/03/2018 at 22:050 Comments

As I reported last time, with no load, the pilot light stays on at about half brightness because of leakage from the triac. I was able to measure something like 60 volts with no load.

Well, the way to fix that is to add a parallel load to lower the voltage.

Today I grabbed some hefty resistors to try and figure out the highest resistance that would reliably turn the pilot off (after all, the lower the resistance, the more power is wasted when the circuit is turned on, and that power winds up being dissipated as heat). I had on hand 50kΩ, 91kΩ, 150kΩ and 300kΩ resistors from my work with the Hydra and OpenEVSE II designs - all of those rated for 1W and flameproof. The 91kΩ one worked and the 150kΩ didn't, so the 91kΩ one was selected. I just trimmed the leads a little shorter and jammed them into the output screw terminals on the board along with the output lead and pilot light wires.

It's worth mentioning that once you've picked a load resistance, you need to calculate the power to insure you use a resistor beefy enough. P=E^2/R, of course, so in this case this will dissipate just under 160 mW (for AC, you can use the RMS voltage for such calculations, and the result will be RMS power, but that's what's significant for measuring resistor dissipation).

It's probably not a great idea to put 3 wires into a single screw terminal, but there isn't going to be an unnatural amount of current flowing through this thing (not like an EVSE that switches 30 amps), so I have some confidence that it will be ok. Soldering it across the pilot lamp's terminals would have also been an option (perhaps a marginally better one).

But I'm happy now that the pilot lamp turns all the way off with nothing plugged into the socket.

P.s. The lesson EVERYONE should take away from this is that just because a non-mechanically switched circuit is ostensibly turned "off" that doesn't mean that there isn't a significant voltage available to give you a nasty surprise.

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