OK, here's a topic that's been floating around here for a long time, but ...

I just did a trial install of some LED bulbs in tails and turn sigs. Since the LEDs don't draw much, I now have a bulb-out light on the dash. One option is to install big honkin' resistors on each LED bulb to simulate incandescent bulbs. Another is to disable the dash bulb-out light from the module behind the seats. Regardless of this, you still need a big honkin' resistor for the flashers to get the flash rate back to where it belongs.

1) So at this point, has anyone come up with a smaller size resistor solution (a bigger resistor value) for this? Since one of the advantages to LED bulbs is less current drain, I don't want to add in resistors that will merely draw current and get hot.
2) Is there an electronic flasher that is a direct substitute that can be used, one that will flash at the correct rate regardless of what bulbs are used?
3) One post from a few years back indicated that you could pull a connection from the lamp failure module and disable the dash indication. Does this work for any and all tail/brake bulb locations, or is there more than one connection that has to be disconnected from the lamp failure module?

Jerry.

 

1.) No idea.
2.) I think ur talkin about a load equalizer
3.) I think u need to just pull that single wire for the entire rear brake bulbs....I could be wrong.

 

On # 2), there are electronic flashers, as opposed to thermal flashers (current-based). I'm guessing that these may flash at the designed rate regardless of current load. I think a load equalizer is a different thing, prob just a big resistor.

 

I say your best solution is to ignore the light.
I've been ignoring my airbag light. Even though I know my airbag is working.

 

And exactly how do you tell that your airbag is working? Not really something I'd want to test.

 

if you increased the resistance you'd not only dim the load, but increase the need for a LARGER resistor due to higher heat output from higher resistance.

Lowering the resistance would lower the overall size of the resistor needed, however then you would not have enough resistance to trigger the sensor for the light.

 

These resistors are in parallel with the load, no? They must be, to cause MORE current to be drawn to fake out the bulb sensor.
What you said here only applies if the resistors are in series with the load. Raising the resistance causes less current to be drawn, so a smaller resistor (physical size & wattage) is needed. A lower value resistor draws more current and produces more heat.

Apparently, the standard fix for the flasher thing is a 6 ohm, 50 watt resistor that they tell you must be heatsinked against metal cause it gets hot. I'm trying to avoid this by coming up with a lower current way of faking out the flasher, or modifying the flasher, etc.

Jerry.

 

Right thought, wrong way of looking at it. Higher resistance will produce more heat. A resistor draws no current whatsoever, it's job is to RESTRICT current, therefore the more it restricts, the hotter it gets.

The lower the resistance, the more current it passes, the less it restricts, the cooler it will run.

 

I have a related question.

I have installed the LEDs for the turn signals, as well as the load equalizers/resistors. However, I have not attached them directly to any other metal to further dissipate heat. They are just hanging there.

Since turn signals are used fairly infrequently, is there really that big of a heat build-up? I would assume that current only flows when the turn signals are activated, and during that time, not much heat would be generated?

Am I missing something?

 

Yes, you're right, probably not too big of a heat build up. You might want to spot check how hot they are getting, just to be sure. A bigger thing for me is that, if you look at the resistor value of 6 ohms, at 12 volts, it's a 2 amp load. For tail and brake lights, adding those resistors in makes the overall current roughly equal to that of a bulb, so there's no savings of current to go to LEDs. The current you saved gets wasted in a resistor as heat, just to keep bulb-out indicator and flasher happy.

 

Thanks for the reply.

I don't know why others go with the LEDs, but for me it was aesthetics, not any "current savings" or anything like that. I went with large amber LEDs that are clear when not active, for that "stealth" look that others are also achieving by using those PIAA silver-coated/amber bulbs or their equivalents. The fact that the resistors keep the flashers and bulb-out indicator happy is exactly why I think they're great.

 

Sorry, O.L.T., I really don't mean to pee on your Cheerios, so I want to do this in a friendly way. What you've said is not correct, and I know this because I have an MSEE, just for perspective. We all have things that we've misinterpreted along the way, we're human. That's OK. Not looking to start any arguments, just trying to present proper, correct info, and correct any misperceptions.
A resistor IS a LOAD. If placed across a voltage source, it will DRAW current (cause current to flow) from that source. The LOWER the resistance, the MORE current that will be drawn from the source. Conversely, the HIGHER the resistance, the LESS current that will be drawn from the source. Heat is produced in a resistor proportional to the power dissipated in that resistor, which is equal to I2R (i-squared-times-r). The higher the resistance, the more the resistor is restricting, the less current flowing, and the less power dissipated in the resistor, so less heat. So a resistor passing MORE current (a lower resistance value) will run HOTTER due to the I2R drop and heat produced (in watts).
You can test this with a 9V battery and a 100 ohm resistor (or lower) and a 1K resistor (or higher). The 100 ohm resistor will get hotter over time. If you plug in some numbers into the P=I2R equation, using a constant voltage, it will also prove this. The reason is that because of the squared current term, the squared current in the circuit goes up faster than the resistance goes down, as you reduce the resistance.
Having said that, if you take two resistors that have the SAME current flowing thru them, the higher resistance will get hotter, since the power dissipated is greater. (this may have been what you were thinking of). But this is also because the voltage across it had to be jacked up to produce the same current (as is flowing in the smaller resistor).

 

Then we're both in the EE field and will agree to disagree. The resistor only functions as a load if it is the load in the circut, whereas on my HP meters, using an actual load with the resistor in line, it does not function as the primary load and therefore does not react the same as it would in a circut with a battery and a resistor only.