Put LED’s in, simple Canadian tire BS. Put a 6ohm resistor in, but the car still flashes fast! Any idea? It’s 6ohm too little? I have access to any type of resistor I want thankfully, so any suggestions would be good!

 

Update: 18ohm was STILL not enough! Going to my computer store next time I can to try and sort this out

 

Figure out what the stock bulb resistance is, voltage drop across led's, the compute what resistance you need using ohm's law. Don't forget proper wattage for resistor else you'll cook 'em,

 

Doing some guesstimating, theoretically I should need 24ohms. Not even 30 ohms was enough, so I’m running to my shop later and picking up a 100ohm resistor. Not f****** around anymore!

 

Another update, still nothing good. Resistor has been placed in front of and behind the bulb, still continues to flash fast regardless of resistor strength. At this point, I've set up my bulb sockets with bullet connectors, making them totally modular. I have been placing the regular bulb in, and it still works 100% fine. At a total loss, I should be able to figure this out but I just can't seem to wrap my head around it

 

Getting real tired of this...

hooked up a potentiometer and spun it, nothing. Checked resistance, got 0-27.4ohms. Hooked up incandescent and LED in series, the LED was flashing while the incandescent was not..

hooked up original incandescent bulb and measured current, 2A. LED checks out at 175mA. Here’s my next question, should I be trying to replicate the current being used? I can hardly draw anywhere near close to that using resistors alone.

Or, can we even DO led lights?? Has anyone done it? Why am I failing at this?

 

With the led bulb in series with the incandescent, what is the voltage across each? Test with car running. The 2A above is with the car off or running? With this info it's then possible to figure out approx resistor value.

In a series circuit, current remains the same across each load, voltage varies. In a parallel circuit, voltage remains the same, current varies. The trick to fooling the bcm is to provide an equivalent current load. 2A means ~ 24-28 watt rating (based on 12-14 V).

 

JSolo, you are right. I’m a moron, I learned this in school even!

put the resistor in parallel, guess what’s working? 6ohm resistor in PARALLEL with the lightbulb, and everything is up and running!

Next is the fronts!

 

I have led turn signals no resistors no issues. Weird

 

Lets see if I can remember this. In a series circuit, the apparent load (as seen at the +, - connected to that load) is the quotient of voltage divided by the sum of all resistances. In case of the 6Ω resistor, that's ~2Amps. The led/diode bulb is expecting full voltage (12V) but is getting less because of the voltage drop across the resistor. The bulb itself does have some resistance, but is likely seen more as a short. I doubt the bulb's diodes are rated as 12V so it has some circuitry to reduce voltage. You'd have to measure to see what the drop across the bulb was in the this scenario. Keeping in mind the bcm is expecting a certain current draw to allow for the proper flash rate.

Now in a parallel circuit, the voltage across all loads is the same but the current through each leg differs. Say you have two identical loads (6Ω) in parallel. With a 12 V source, current through each would be 2A, or 4A total. So at the source, the two 6Ω in parallel are actually seen as a 3Ω load (or [6x6]/[6+6] ).

 

Holy smokes it’s math.

Reading this you are right, (bringing back the electrical math in me!) you’ve said it all!

And to the other post, I bet those bulbs have a resistor/load equalizer built in. I know for a fact mine don’t...

 

Blah.. That's basic math. I better know this stuff. Went to school for it.

Calc 1, and 2. (can't remember if I took 3), really kicked my @ss. Now that's math. There's a whole course dedicated to Fourier transform. Want to get your brain blown, read the first first paragraphs - https://en.wikipedia.org/wiki/Fourier_transform .