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Thread: Load resistors for LEDs

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    Default Load resistors for LEDs

    So this weekend I installed taillights from a 2012 Jag into my 2009. Much nicer design with full LED, whereas the old design still had bulbs for the turn and reverse.

    ...and now, as expected, my information center tells me there is a bulb issue. So I will need to install resistors for both the turn and reverse circuits. (I find it a little odd that the car knows these are an issue the moment I turn it on. I would have thought it would only recognize it when I actually used the turn signal or reversed. Whatever. Clearly it knows.)

    Anyway, when I install these I've seen it done two different ways, but I'm guessing only one way is the proper way. The first way and the way I expect it SHOULD be done is to bridge the circuit between the incoming power wire and the ground wire with the resistor. In some cases, I've seen people add the resistor inline on the power lead only.

    My natural inclination is to think the first way actually wouldn't work. If I recall correctly in electronics class grade 11, electricity always flows to the path of least resistance. So if there is a resistor bridging the positive and negative, would it not just bypass that and only flow through the LED?

    So, for anyone who has added dummy loads for this reason, which is the correct way to wire it, or are either acceptable?

    Also, I had read I need 50 ohm resistors. Yet, almost everything I can find online is something like 6 ohm. Is it possible I really need a 50 ohm resistor? That seems really, really... resistant.

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    Last edited by Sugarphreak; 08-18-2019 at 01:05 PM.

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    Yeah, I am aware the objective is to match the same load level now that everything is LED.

    Ok, so the lights have several power leads in via the harness, but only one ground out. Could I attach the resistor in series to the ground only, even though it's just two of the five circuits I want to increase the load for?

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    Last edited by Sugarphreak; 08-18-2019 at 01:04 PM.

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    Quote Originally Posted by Sugarphreak View Post
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    That sounds right to me if it is a common ground for everything

    I am assuming you already looked into the proper sized resistor
    Well, I know how many watts it should be but am still unsure about the Ohms. Like I said, forums told me I should get a 50 Ohm resistor, but I know that typically a load resistor is something like 5-7 ohms so I'm not really sure which to use. I have both on order though as they were cheap.

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    Last edited by Sugarphreak; 08-18-2019 at 01:04 PM.

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    Quote Originally Posted by Sugarphreak View Post
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    The best way to figure out how to size your resistor would be to test your OEM resistance. Then test your LED lights, and just order a resistor to suit.

    Yeah, makes sense.

    So, I'll ignore your previous post but you still maintain these are to be attached in series right? So odd - most of the information online seemed to indicate parallel, and that's what I figured was the correct way. I see a lot of this sort of setup online:
    poison-spyder-led-resistor-kit-for-led-tail-lights-07-17-wrangler-jk-2.jpg

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    Ignore everything SP said hes wrong. The resistor has to be hooked up in parallel to the bulb. (the hot feeding it and the common ground) If you hook it up in series you are dropping the voltage that the bulb is going to get and its going to be dim or just not work. A resistor is only hooked up in series when you are using a lower voltage LED with a higher voltage (usually 5v leds on a 12vcar battery).

    Hooking it up in paralleled keeps the voltage at 12 which your bulb needs to work. You add the parallel resistor to add the load to make up the difference in power between the LED and the old bulb.

    If you can tell me the wattage of the old bulb and the new i can tell you what size resistor you need.

    That diagram you just posted is the only way to make this work and is 100% correct.

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    Quick diagram to help you understand.

    Top drawing would be stock bulb assuming 12W

    second drawing is assuming a 6W replacement LED bulb

    third drawing is assuming a 4W Replacement.



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    That's very helpful. I will assume it is a 12v circuit and not something weird like a 5v. (Wouldn't see why).

    That gives me a good idea, and I will read the bulbs; hopefully they will tell me printed on them as opposed to ripping the housing out again just to do the test.

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    Last edited by Sugarphreak; 08-18-2019 at 01:04 PM.

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    The sylvania bulb guys lists both the 09 and 12 as having a PSY19W for rear signal bulb. Its 19w 12v. Still need to know the power or current of the new led bulb.

    a 50Ω resistor does not make sense as it would only be 2.88W in a 12v circuit Meaning the Led bulb would have to be 16W. They are probably suggesting a 50w resistor. The watts when dealing with resistors is the max power they can dissipate without burning up. A higher wattage resistor is physically bigger but has no bearing on what its resistance(Ω) is going to be.

    More likely resistances are in the chart below. Again the idea is to have the LED and resistor pulling the same amount of current as the original incandescent bulb.

    10Ω - 14.4W (4.6W LED bulb)
    15Ω - 9.6W (9.4W LED Bulb)
    20Ω - 7.2W (11.8 W LED bulb)
    25Ω - 5.76W (13.24W LED bulb)

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    Quote Originally Posted by Sugarphreak View Post
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    That is fair


    I am a bit confused about what you are trying to do though. My understanding of a common issue with LED lights is they flash too quickly; which would be because they have less resistance and are allowing more amperage through. That would recharges the indicator capacitor more quickly resulting in hyper-flashing.

    In your above diagram, you are specifying the amperage of the LED light as 0.5a, which means it is 24 ohms (as opposed to the bulb, which you are showing to have only 12 ohms). Which if true, I totally agree with you, and hooking a resistor in parallel to increase the system amperage makes sense.

    I thought they were less resistive than bulbs, and therefore the idea was to reduce reduce the system amperage back to OE levels?

    Current goes Down when Resistance goes up. Therefore LED bulbs have higher resistance than incandescent. LEDs are more efficient and draw less power (w) and because of that their current (A) is lower as well.

    basic formula E=IR therefore I=E/R

    EX
    12v / 12Ω =1A
    12v / 24Ω =.5A
    Last edited by Crazyjoker77; 05-14-2018 at 05:04 PM.

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    Last edited by Sugarphreak; 08-18-2019 at 01:04 PM.

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    Quote Originally Posted by Crazyjoker77 View Post
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    The sylvania bulb guys lists both the 09 and 12 as having a PSY19W for rear signal bulb. Its 19w 12v. Still need to know the power or current of the new led bulb.

    a 50Ω resistor does not make sense as it would only be 2.88W in a 12v circuit Meaning the Led bulb would have to be 16W. They are probably suggesting a 50w resistor. The watts when dealing with resistors is the max power they can dissipate without burning up. A higher wattage resistor is physically bigger but has no bearing on what its resistance(Ω) is going to be.

    More likely resistances are in the chart below. Again the idea is to have the LED and resistor pulling the same amount of current as the original incandescent bulb.

    10Ω - 14.4W (4.6W LED bulb)
    15Ω - 9.6W (9.4W LED Bulb)
    20Ω - 7.2W (11.8 W LED bulb)
    25Ω - 5.76W (13.24W LED bulb)
    Thanks for doing some homework. That is indeed the bulb.

    I know it sounds unlikely... But I promise that was mentioned as ohms. There was an adjacent post that mentioned the watts as not being super important as long as its rated higher... In the 25 watt range. The ohms seemed high to me as well.
    But these are not trained people... They are non-electricians who owned what was then a 80k car trying to save a few bucks for some reason.

    So forgive me for being an idiot here... What exactly do I need to measure with the new lamps? Seems to me that if I measure a circuit figure it would be on the car-side harness and the same as before. Or am I measuring the led resistance?

    Over my head.

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    Quote Originally Posted by Kloubek View Post
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    Thanks for doing some homework. That is indeed the bulb.

    I know it sounds unlikely... But I promise that was mentioned as ohms. There was an adjacent post that mentioned the watts as not being super important as long as its rated higher... In the 25 watt range. The ohms seemed high to me as well.
    But these are not trained people... They are non-electricians who owned what was then a 80k car trying to save a few bucks for some reason.

    So forgive me for being an idiot here... What exactly do I need to measure with the new lamps? Seems to me that if I measure a circuit figure it would be on the car-side harness and the same as before. Or am I measuring the led resistance?

    Over my head.
    You need the power(watts) or current(Amps). Resistance is not going to be easily measurable and only calculable only after you have those.

    Incandescent bulbs can't be measured for resistance because they have a positive temperature coefficient and if use your multi meter to a bulb not hooked up its resistance is going to be Extremely Low. As it heats up the resistance increases (very rapidly).

    LED bulbs also can't be directly measured for resistance for a much more complex reason that I'm not going to get into to (impedance). The only thing you can do is measure the current with an amp meter. or find the rated wattage posted somewhere.

    I have a hard time believing a 50Ω will work as I said it only adds 2.88W @ 12v. That 6Ω you ordered also sounds wrong as that adds more wattage than the original bulb. (24w)

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    Also you can buy new flasher modules that are designed to work with the LED bulbs. I put one in my sisters old jimmy when she bought LED taillights It was located behind the glove box but I have seen them in the underhood fuse panel as well. Although being a Jag I assume the signal function is probably built into the BCM and does not use a plugable flasher.

    Is this a aftermarket LED bulb that you bought or does it come with a factory LED. I'm just wondering why the sylvania guide lists them both the same.

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    Quote Originally Posted by Crazyjoker77 View Post
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    Is this a aftermarket LED bulb that you bought or does it come with a factory LED. I'm just wondering why the sylvania guide lists them both the same.
    No... Its the entire taillight assembly. No bulbs at all... Just wires going into the taillight itself. Its having problems because thr 2009 model had bulbs in addition to the factory leds in thr housing. The 2012 unit is all led.

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    CrazyJoker is the most correct so far.
    I'm not familiar with automotive LED applications, there's likely multiple LEDs and some circuitry (at least a current limiting resistor) inside an LED bulb.

    Speaking strictly for single LEDs, there's really no such thing as a 12v or a 5v LED. A LED has a voltage drop and a current rating. The resistor you use to make a LED work at a specific voltage is calculated based on the LED's voltage drop, the current required and the supply voltage. The resistor limits the current.
    It's calculated by subtracting the LED voltage drop from the supply voltage dividing by the desired current (Ohm's law)

    But again, automotive applications would be different, and the topic at hand is tricking the car's system into thinking there's enough current draw to stop it thinking there's a burnt out incandescent bulb, which means a resistor in parallel with the LED to increase the current draw.

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    As far as the ohms, I wonder if the guys on the forum experimented with different resistances and found the car didnt require a significant load to fool the car. I would imagine the lower the resistance means the least heat and the less chance the brightness would be adversely affected. (If it would be at all).

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