Self installed dash indicator light burns out often

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#1
Make
Ford
Model
E350 Super Duty
Year
2006
Miles
175,000
Engine
5.4 liter
I installed a dual battery system, most of the ideas from this site. Van operates on a single battery when ignition is off and dual batteries when ignition is on. I installed a indicator light on the dash that lights when the dual battery system is activated. The problem is that the dash light burns out after a few months. I installed a 10 amp inline fuse to the light, in case I was getting a power surge. I am using a standard indicator light. A LED bulb would have to wired in, and I the NEON was only available in 110V. Is it possible to put a resistor in line to cut down the 12 volts a bit to increase bulb life? If so could you tell me which one? Thanks.
 

grcauto

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#2
I installed a dual battery system, most of the ideas from this site. Van operates on a single battery when ignition is off and dual batteries when ignition is on. I installed a indicator light on the dash that lights when the dual battery system is activated. The problem is that the dash light burns out after a few months. I installed a 10 amp inline fuse to the light, in case I was getting a power surge. I am using a standard indicator light. A LED bulb would have to wired in, and I the NEON was only available in 110V. Is it possible to put a resistor in line to cut down the 12 volts a bit to increase bulb life? If so could you tell me which one? Thanks.
That would be my approach to this. I would think a low ohm resistor would work fine. Maybe 25 ohms or so. Measure the resistance of the element on the bulb and use that as a reference. The thing about these is the element will be low resistance cold and nearly immediately get higher when on. If you know roughly what current it draws you can estimate it's hot value. I would think it should be less than 1/4 amp which would be about 45 ohms.
HTH
 
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#3
If a resistor is used under dash, will heat generated by the resistor be an issue? By an issue I mean the resistor touching other wires or anything under dash. Will it need to be isolated for safety reasons? I don't remember how warm a resistor gets, especially in a 12V application.
 

grcauto

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#4
If a resistor is used under dash, will heat generated by the resistor be an issue? By an issue I mean the resistor touching other wires or anything under dash. Will it need to be isolated for safety reasons? I don't remember how warm a resistor gets, especially in a 12V application.
Shouldn't be a problem. A 1 watt resistor is all you should need and they don't get hot enough to do anything.
 
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#5
The package on the Lamp Assembly that I am now using reads: 12VDC -100mA (I think that is 100/1000 amp or 0.1 amp)

I am also a bit confused with 25 ohm resistor 1 watt resistor 1/4 amp=45 ohms

I can turn a wrench, but when it comes to ohms, watts, resistors, not so good.
 

grcauto

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#6
The package on the Lamp Assembly that I am now using reads: 12VDC -100mA (I think that is 100/1000 amp or 0.1 amp)

I am also a bit confused with 25 ohm resistor 1 watt resistor 1/4 amp=45 ohms

I can turn a wrench, but when it comes to ohms, watts, resistors, not so good.
OK this is the formula. P=IxV and I=E/R ....P=watts....I=amps (current).....V=voltage......R=resistance in ohms.....We know we have 12 volts and the circuit is 1/10 amp. Use these to figure resistance. This would be TOTAL resistance without adding a resistor and we are at about 120 ohms (12/120) this is total circuit resistance. If we add a resistor of 120 ohms we will drop 1/2 (6 volts) across the resistor leaving 6 volts for the bulb. That's probably a little low. Especially since we have reduced the total current by half as well. So if we add a resistor of 60 ohms we will drop 3 volts across that resistor and have 9 for the bulb. This should be fine. Now for wattage. P=IxV.....First we need to calculate I. Total resistance of the circuit is now 180 ohms. The bulb plus 60.....that gives us a current of about .075 multiply that by 12 to get watts and we have under one. Use a 1 watt resistor and I would use a 50 ohm. You should be fine with a 50 ohm 1 watt resistor. That will light the bulb OK and save it's life. This is a rough design since I'm not sure how well the light will do with 9 volts. That's because the resistance of the element is not linear. Take a 9 volt battery and see how well it does for the bulb. If it's bright enough use the 50 ohm resistor if it's to dull use a 30 ohm resistor ect.
 
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#7
I'm no electrical genius, but it might work to wire in another bulb (in series) to act as a buffer/resistor. You wouldn't have to mount the bulb, it could be behind the dash or somewhere.
 
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#8
OK, I hooked up a 9V battery to a spare lamp, same as the one in the van, and found the lamp to still be too bright. I then remembered my old battery charger has a 6V and 12V setting, so I hooked up the lamp to 6V and find that would work just fine for brightness. So, do I need about a 1 watt
 

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OK, I hooked up a 9V battery to a spare lamp, same as the one in the van, and found the lamp to still be too bright. I then remembered my old battery charger has a 6V and 12V setting, so I hooked up the lamp to 6V and find that would work just fine for brightness. So, do I need about a 1 watt
You'll want a 1/2 watt 120 ohm resistor. The resistor will absorb about 3/10 of a watt so a half watt will work fine.
 
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#11
Well, I went to Radio Shack and in 1/2 watt resistor they had 100 or 150 ohm. No 120 ohm) I went with the 150 ohm. I may switch to the 100 ohm. I'll give the 150 ohm a try for a couple weeks. Thanks grcauto for all the tech info and Mobile Dan always a good idea!
Something you can do if you want it brighter is to add another 150 ohm resistor in parallel (both side by side) this would make the total resistance 75 ohms. If they are one followed by the other (this is in series) they add together for total resistance which would be 300 ohms. Do you have the 100 ohm? If so, just replace the 150 with it. Still want it a little brighter than that just parallel the two and you'll be at 62.5 ohms .
 
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#13
By adding two resistors in parallel the resistance goes down. I suppose each resistor allows X amount of current pass, so when you add a resistor you add to the current flow. Please see my high tech drawing attached to be sure I have it correct.
Correct...When in parallel, which is how you have them, resistance goes down and current goes up. If the resistors are of equal value it's one half total resistance.