Fog light tech - resistor?

Huh?

Go to a friend's house or a parking lot and work on your car. Do the proper relay retrofit (the relay became OEM on 94+ fog light circuits).

And forget all about that resistor idea. If you'd go that far, simply don't use your fog lights. That would be a safer solution.

Good luck.

Here is the issue in a nutshell. The driving lights pull more current than the wiring and switch can sustain over time. The switch and wires get hot and burn. This is why most good aftermarket aux light makers add relays in their wiring. As mentioned earalirt in 94 Ford got wise. Just install a relay someplace after the switch and before the light. Get the power for the lights from the battery and use the factory switch to power on the relay.

And you can get a 30/40 amp SPST relay at Radio shack or your parts store (a relay for aftermarket driving lights will work just fine).

There are no exotic parts needed for the retrofit - you can get it all locally. The Corral has one of many write-ups on the web.
Good luck.

The resistor would cost more, generate more heat and be larger and harder to install than the relay. Not only that, but I can assure you that Radio Shack would not have a resistor capable of handling the current that an effective fog light system would require. I'd give you the math to prove it, but it probably would be meaningless to you.

SEE Mustang GT Fog Light Fix to fix the problem. The stock wiring isn't up to the job and is overheating. The headlight switch & turn/multifunction switch are affected by Ford's wiring problem. Sometimes it overheats so badly that the plastic shells of the wiring connectors start to melt. This will show you how to add a relay to the fog lights to relieve the overload on the headlight wiring.

I did mine differently, but I had to build my own wiring harness for the fog lights. This is more trouble than it is worth for most folks. I left all the wiring on the stock light switches in place and used the fog light wiring to power the relay coil. The other side of the relay coil is connected to ground. I have an inline fuse that picks up power from the battery side of the starter solenoid. It is connected to the relay contact. The other relay contact is connected to the new wiring harness I made for the fog lights.



The advantage of making your own fog light wiring harness is that you can run 100 watt fog light bulbs. The stock wiring harness will not use 100 watt bulbs without overloading and causing a fire.

Unless you are good at electrical wiring, have the skills and tools (crimp tool, soldering gun, heat gun for the heat shrink tubing, etc.) I recommend that you stick with the Corral method.

sgw_88FoxGT said:

Thanks for the responses guys.

jrichker, Re: Go ahead and write up the math if you don't mind. I've taken quite a few physics courses in college as well as an intro to electrical engineering course where we built everything from high and low pass filters to rectifiers and even a breadboard computer. I'm just rusty on all this stuff, haven't touched a circuit in a few years.

Click to expand...

You asked for it...
I= Currrent
E= Voltage
R= Resistance
W= Watts

Two 55 watt fog lamps =110 watts. Find the current in the circuit
I= W/E
110 watts/14 volts = 7.85 amps for fog lights alone.

Since the exterior lighting circuit supplies headlights, taillights, and parking lights, etc.
56 watts 2 each GE Part # L3156 corner light 28 watts each
90 watts 2 each GE Part # 9004 headlight 45/65 watts each (low beam)
63 watts 2 each GE Part # L194 parking light 31.5 watts each
56 watts 2 each GE Part # L3157 tail light 28 watts each
265 watts Total

Total other exterior lighting current
I=W/E
265 watts/14 volts = 18.92 amps
18.92 amps other exterior lighting current
+ 7.85 amps fog light current
26.77 amps with all exterior lights and fog lights on.

The 12 gauge power feed wire to the exterior lighting switch is rated at 20 amps
26.77
- 20.00
6.77 amps excess current

7.85 amps used by fog lights
-6.77 amps excess current
1.08 amps to run the fog lights left if you stay within the 20 amp limit of the wire.
With 1.08 amps of current, the fog lights probably won’t produce any useable light.

Added resistance required to reduce fog light current to permissible 20 amp limit
R=E/I
14 volts/6.77 amps = 2.06 ohms
Resistor wattage
W=EI
14 volts x 6.77 amps = 94.78 watts
You would need a 2 ohm, 100 watt resistor.

Light bulb ratings from http://www.roadparts.com/catalog/section30.pdf

Radio shack resistor catalog - no matches , and no combinations that could be used to make a 100 watt, 2 Ohm resistor.
See RadioShack.com

Ohm’s law – in case you have any questions about my formulas - Ohm's Law Calculators

sgw_88FoxGT said:

Thanks for the responses guys.

jrichker, Re: Go ahead and write up the math if you don't mind. I've taken quite a few physics courses in college as well as an intro to electrical engineering course where we built everything from high and low pass filters to rectifiers and even a breadboard computer. I'm just rusty on all this stuff, haven't touched a circuit in a few years.

Click to expand...

Hmmm...no reply....Must have swamped his boat in the sea of numbers that he asked for...

LOL, I dont post much, but I read alot and most of your responses lead me to belive I need heavier wiring myself. im always sure one of your extreme technical responses will cause me to overload. lol But it's great to have such a knowledgable stang guru on hand. hell, i dont post much because every question i think up is answered in depth by you already. the search feature reduces my posts. this was actually something I was planning on asking myself.
thanks u da man.

That was fun to read thats