David New Jersey
- Jun 6, 2014
- 12
- Joined
- Jun 6, 2014
- Messages
- 12
Hey folks, wondering if anyone can point me either to the theory or a solution for an issue that comes up more and more for me with my older electronics - I have a DC component that goes bad in a circuit, lets take today's issue, a miniature DC filament bulb used for illuminating things from which the reflected light gets measured. I need to replace the bulb, but it's specs are no longer available in the same voltage, physical dimensions and current draw. It needs to be a filament bulb for spectra considerations so the vast range of drop in LED's aren't an option given their spectral incompleteness. The only other filament (more specifically incandescent) bulbs that will physically fit in place with the same wattage output are of different voltage and current ratings.
In my current case, according to my meter, my device is driving the bulb at 3.0VDC and it's drawing 1.0A. But modern replacement miniature bulbs only have that kind of wattage in higher voltage bulbs, like 6VDC or 12VDC.
If the bulb were always on or off, I'd simply power it with a supply independent of the circuit, but the circuit turns it on and off for testing and calibration purposes, checking it with a light sensor during these on off intervals, so it would be nice to have the replacement bulb be powered by the circuit so the light output is in sync with the photo diode.
What comes to mind for a tinkerer like myself are (say in the case of a 12VDC replacement bulb);
1) Use a sub 1A, 3V relay and power the other bulb with a separate 12V supply,
2) Somehow use a voltage regulator (suggestions?)
3) Go crazy and program an arduino, which seems like overkill.
Any help, in advance, is greatly appreciated - thanks! David
In my current case, according to my meter, my device is driving the bulb at 3.0VDC and it's drawing 1.0A. But modern replacement miniature bulbs only have that kind of wattage in higher voltage bulbs, like 6VDC or 12VDC.
If the bulb were always on or off, I'd simply power it with a supply independent of the circuit, but the circuit turns it on and off for testing and calibration purposes, checking it with a light sensor during these on off intervals, so it would be nice to have the replacement bulb be powered by the circuit so the light output is in sync with the photo diode.
What comes to mind for a tinkerer like myself are (say in the case of a 12VDC replacement bulb);
1) Use a sub 1A, 3V relay and power the other bulb with a separate 12V supply,
2) Somehow use a voltage regulator (suggestions?)
3) Go crazy and program an arduino, which seems like overkill.
Any help, in advance, is greatly appreciated - thanks! David
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