Hi everyones thanks to read. Since a few weeks, i'm using the LightKeeper Pro from Ulta-Lit Tree Company (
formatting link
) to test and repair miniature lights sets (series-wired). I've found the Quick Fix Trigger utility very useful. as you can see on their website, it's based on a piezo igniter that sends a high voltage pulse through the lights set to activate (short) the shunt inside the bulb, that didn't do so when the bulb's filament burned out. I wanted to see how this system is built so i opened the unit. Here's two photos plus a schematic i've drawed for the igniter section. (Sorry, the symbol for the piezo igniter is probably wrong...i don't know the correct one).
formatting link
formatting link
formatting link
As i can see, the four diodes allow a peak reverse voltage of approximately4Kv, wich is sufficient to break down the shunt's insulation inside the burned bulb(s) and complete the series circuit. Once completed, half of an AC cycle can flow through the four diodes, allowing the set to glow (dimly) and show which bulbs are burned and needs replacement. What leads me to my question (please look at the schematic) is: why the reverse voltage doesn't seem to damages those diodes? According to the piezo igniter's website (
formatting link
), the output voltage is at least 15Kv. With the four 1n4007 wired in series, i should get a peak reverse voltage of about 4Kv isn't? For what i know, a diode is destroyed once it's PIV is exceeded. I think i missed something when studying the basic operation theory of a diode. ;-) TIA for any useful reply.-- Alain(alias:Kilowatt) Montréal Québec PS: 1000 excuses for errors or omissions, i'm a "pure" french canadian! :-) Come to visit me at:
formatting link
(If replying also by e-mail, remove "no spam" from the adress.)