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electrical question

Old 07-19-2010, 06:54 AM
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electrical question

I have relocated my oil pressure and neutral indicator lights to just 2 LEDs.

I have temporarily removed the gauge cluster, so if you look at the schematic you can see that for those 2 indicators, they each have their own wire to trip on the indicator and share the same (BK/BR) power supply. I tested the voltage of that BK/BR wire and with the ignition on, +12V like I thought.

On the other end of each indicator light there is the two separate trip wires, BL/RD for oil pressure and LG/RD for neutral which each go to their own sensor which is a switch and I am guessing they are normally closed when in neutral and when there is no oil pressure, thus why the lights are on while not running with ignition power.

However here lies the problem. When I wire up just as explained above the "Meter/tail/Illum." 10A fuse for the BR/BK blows which also powers the tail/brake lights and horn. Since there is basically nothing else on this circuit except for the two indicators I am using and horn/tail, why is it blowing this fuse? Are the LEDs drawing too much power or something?


Schematic (for reference):
https://www.superhawkforum.com/galle...00F-wiring.jpg

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Old 07-19-2010, 07:23 AM
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Start with the basics... The LED is oriented the right way and you have the correct resistors on there to make it work with 12V right?

A LED drawn a lot less power than a bulb and no LED will ever work right of the bat with 12V, it's basicly a short then, so that could be why the fuse blows...
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Old 07-19-2010, 07:50 AM
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Originally Posted by Tweety
Start with the basics... The LED is oriented the right way and you have the correct resistors on there to make it work with 12V right?

A LED drawn a lot less power than a bulb and no LED will ever work right of the bat with 12V, it's basicly a short then, so that could be why the fuse blows...
Ok that where I figured I needed to start. They're oriented the right way, yes. However unlike my LED tail, these didn't come with any resistors to limit the current draw. I know LEDs draw about 25mA and run off of ~2V, but I'm not 100% sure about those numbers for this exact application. Just going roughly, I need about a 10V drop across a resistor from the 12V battery that passes about 25mA to ONE LED, so 10/.025 = 400Ω. Think I should start there? And do you think it would be better for 400Ω for each LED or should I combine and go 10/(2 x.025) = 200Ω and use just 1 wire to supply them both? Thanks tweety
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Old 07-19-2010, 09:05 AM
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That's where to start yes...

Actually not too far off... Most small LED's draw between 10 and 30 mA, and 1.5-2V, so yeah your calculation is mostly right, a bit more margin might be a good thing though if you don't know for sure what spec they are...

Supplying the both from one resistor might prove fatal, since they will not always be turned on at the same time, so then the one turned on ends up getting to much and will probably stop working...

Also, since you have tried them with 12V it's more than possible that one or both of them are permanently fused now, making them into a wire/resistor, not a LED, so be prepared to replace them...
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Old 07-19-2010, 09:12 AM
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Yep, that could have happened hopefully not though. I have incandescent bulbs for the sockets that I removed, should have thought this out some more before I swapped them for LEDs. THanks for the help again Tweety.
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