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(Topic ID: 277173)

#89 Bulbs in 39V FLASH flashing circuit?


By JethroP

44 days ago



Topic Stats

  • 5 posts
  • 3 Pinsiders participating
  • Latest reply 43 days ago by JethroP
  • Topic is favorited by 1 Pinsider

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35946919-1032-409C-BC31-1C34B9546666 (resized).jpeg
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39V Flashers (resized).jpg

#1 44 days ago

I found two burned out #89 bulbs in the flash circuit shown in the attachment. Also found the 330Ω resistor broken, presumably why the bulbs burned out. I have replaced the resistor, measures 330Ω and the 1Ω in series resistor also measures 1Ω. So before I replace the bulbs I checked the voltage across the two bulbs shown and measures 39V. This is in game started mode. Seems way too high, and won't it go higher when the "flash" signal is given?

#89 bulbs are 12V bulbs. I suppose they are good for 14V, but in series isn't that only total 28 voltage drop? Afraid to install new bulbs.

39V Flashers (resized).jpg
#2 44 days ago

they're 12V bulbs, but they're 'flashers' for a reason. They're bright because they're overvolted, and they flash because they can't be overvolted for two long. I think WPC runs its 89s at 20V iirc?

With that in mind, you've got 39V. and two bulbs in series, each intending to get ~20V, so I think that makes sense?

#4 44 days ago
Quoted from SarverSystems:

How would one replace these with LEDs?

That's exactly what I have in mind. You can buy #89 LED's and disconnect the 330Ω warming resistor, but I am planning to make my own LED lights using 5050 SMD's, cost about 2 cents each. See photo, using 4 SMD's I've assumed 3V drop across each one, so what I wired up so far should be good for 12V. I'll add 4 more SMD's and give it a try.

IMG_6208 (resized).JPG
#5 43 days ago

Made a loop of 9 SMD’s in series. Removed the 330 ohm resistor in the flash circuit. Works like a charm!! Very bright!!

35946919-1032-409C-BC31-1C34B9546666 (resized).jpeg
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