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Scottychaos

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Discussion starter · #1 ·
I have just the headlight from a Bachmann davenport:

Image


Just the light housing, with its LED light, and two bare wires sticking out.
that's it... No Bachmann circuitry.

I want to add the light to a different locomotive, power the light with a battery, with just an on/off switch.
I know you cant just hook up a LED to any power, it needs "protection" from too much voltage.. somehow.
What would I need to convert this to a working light?

thanks!
Scot
 
Actually it needs the CURRENT limited, not the voltage.

access a LED calculator:

Greg
 
Discussion starter · #3 ·
Thats greg! that looks useful.
I went here:
So let's say "supply voltage" = 9 (a 9 volt battery)
I believe its a yellow LED, so based on the "typical Vf range" chart I enter "2" in the calculator.
That's 2 out of 3 entries in the calculator. How do I know what the "forward current" is for the third entry in the calculator?

thanks,
Scot
 
Scot,
20ma is max for most leds. So 9v battery,3v forward voltage, 20ma would be full brightness. If that is too bright, then just increase the resister value a couple hundred ohms at a time until the brightness is where you want. Or just substitute a lower ma value like say 15ma in the formula. That will give you a higher resister value to use. The resister value doesn't need to be an exact match. Just don't go any less resistance than the formula says for 20ma.
 
I agree, in the old days 20 ma was the standard, nowadays I go for 10ma, and see how it looks... I got an accucraft K4 wit a surface mount led at the bottom of a nice reflector, it was perfect, brightness, looks, etc... it was running 7 ma...

Greg
 
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