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Discussion Starter · #1 ·
Hey everyone, hoping you can give me some information. I'm wanting to switch the color of my daytime running lights from yellow to white while staying a bulb and not going LED. I have looked around, and did some research. Found out that I need 194's, but not sure how to make them white. Is it as simple as buying a white bulb or is the yellow coming from the reflector/lens. Haven't found anything on different color bulbs, so I'm afraid it might be coming from the lens. And if so, not anything I want to mess with... :( Thanks for the help!
 

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yellow? really?? its a white bulb, maybe its just going dead and looks yellow?? LED's r the way to go, way brighter, more white, last longer... i put 6000k LED's in mine, have a slight blueish tint that contrasts nice with the headlight, making them more visible next to the headlight...
 

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there is no such thing as a "white bulb" the color is determined by the temperature of the filament/gas runs at in degrees Kelvin(10 grade chemistry anyone?....lol) Amber(yellow) is 2700K, Daylight 4100K and Super Bright White is 6000K.

Find a bulb that runs at the desired temp and you are done.

that said you won't find any incandescent that runs any hotter than 3000K, there fore you need to go LED
 

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Actually the colour temperatures are based on the light emitted by a black body at the requisite temperature, not the temperature of the bulb filament. There are lots of high colour temperature incandescents. It's usually done by using a blue filter in the envelope to take out the yellow component, but you can certainly get past 4K colour temperature without using a filter. LEDs run much cooler than incandescents. Neither do they have a filament. Don't confuse colour temperature with actual temperature.

That said, for daylight running lights yellow or amber is more easily seen by other vehicles and gives a clearer sense of approach speed and distance off. Going more white is something of a retrograde step.

Be careful of cheap leds for running lights. Some of them are very prone to early failure.

Rob
 

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Rob, our previous spat aside, maybe you could answer a question for me? How can you determine the proper resistance resistor to use to duplicate the correct flash rate of an incandescant turn signal in an LED turn signal? Is it simply a matter of measuring the resistance of the OEM turn signal, and then measuring the resistance of the LED turn signal and using resistors to close the gap? Is there something more to it? or am I just overthinking it?
 

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Rob, our previous spat aside, maybe you could answer a question for me? How can you determine the proper resistance resistor to use to duplicate the correct flash rate of an incandescant turn signal in an LED turn signal? Is it simply a matter of measuring the resistance of the OEM turn signal, and then measuring the resistance of the LED turn signal and using resistors to close the gap? Is there something more to it? or am I just overthinking it?
LEDs have a much higher resistance than incandescants, so you need to lower the effective resistance of the circuit back to somwhere near its original level. High resistance means less power dissipation (one of the advantages of using leds) but the relay depends on the heating effects of the power dissipated through it to operate the bi-metal strip which controls the flash rate.

The resistor sits in parallel between the led and ground. The easiest way is to measure the resistance of the original bulb and use a resistor of that value - the resistance of the led is so much higher that the small error won't matter. If you want to work it out, read on.

With resistors in parallel the effective resistance is derived from 1/Re = 1/R1 + 1/R2 + .........1/Rx

eg, for two 6 ohm resistors in parallel, 1/Re = 1/6+1/6 which equals 2/6 or 1/3.

if 1/Re = 1/3, then Re = 3 ohms

So in this case the effective resistance of the circuit is half of the value of each individual resistor.

In a DC circuit power (in Watts) can be taken as Volts * Amps.

If you have an 18W incandescant indicator bulb on a 12V circuit, then current throughh that bulb can be calculated from I = W/V.

I (amps) = 18 (Watts) /12 (Volts) = 1.5A.

Resistance = V/I = 12/1.5 = 8 ohm.

Given that the led power dissipation is probably about 1 or 2 watts, if you work out the resistance of the led then use that in the equation for the effective resistance of the circuit you'll see that the resistance of the led makes very little difference to the answer, so you can just ignore it.

As the resistor will be carrying about the same current as the original bulb it should ideally have a power rating equal to that bulb, so an 8 ohm 18W resistor would be the correct replacement for an 18W bulb. That's a heavy duty resistor, and as it's flashing and only used for short periods you can get away with an 8 ohm 3W unit probably.

Without working it out, for a 21W incandescant a 7 ohm resistor is probably about right.

Going from what you've seen about effective resistance in parallel circuits, if you replace both front and rear indies, then rather than using two resistors you can use a single resistor to serve both leds, of half the value in ohms but double the power rating. In most flsher circuits the outputr from the relay feeds the flasher switch, so you could actually fit a single resistor to ground between the relay output and the switch that would satisfy the loading for all four indies. Size it for one pair, as normally only one side is on at a time.

This won't work if you have a hazard warning option that flashes all 4 together unless you don't mind the flash rate being wrong in hazard warning mode.

Read up on Ohm's law, and Kirchoff's Theorem for the derivation of effective resistance in combination parallel and series circuits.

You've also seen that installing the resistors causes the overall circuit to draw just as much power as it did with incandescants, so you've lost the potential power saving of fitting the leds. An electronic relay designed for leds lets you run them without resistors and saves all the above work.

Rob
 

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Discussion Starter · #9 ·
Results of bulb upgrade

This is what I decided on. I bought Sylvania SilverStar 194's and Silverstar Ultra H7's. The 194's didn't turn the light from yellow to white, but it did take some of the yellow out IMHO. I really like the H7's though. They are noticeably brighter than stock. Personally, I didn't want to instal HID's or LED's, so this was a nice improvement change for me and I'm happy with it. :) Thanks for the help guys.

The first three pictures are before the swap, and the last two are after.
 

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