I'm replacing the LED in a car tail-light. Naturally, it will run off 12V (battery) but the circuit I have uses just an LED and a 0.6w (440ohm) resistor. But the existing circuit includes two, larger resistors (the power ratings, that is, I assume are higher) and a diode. Can anyone tell me what functions these play in the circuit?

Is the diode for protecting from reverse current, as I'd assume? (not for voltage drop or anything like that), if so, why would there be a risk of this?

And why are the resistors so much larger, when the LED I put in should easily run from a 0.6 resistor? Would the existing LED be a higher wattage LED; It's just the same looking led, there's no heat-sink or anything.

  • \$\begingroup\$ Where is the stripe on the diode? \$\endgroup\$
    – endolith
    Jun 17 '10 at 21:24
  • 4
    \$\begingroup\$ Your question would be much easier to answer if you supplied a schematic. As is, we can only guess. If you don't have a normal schematic editor, just tab over to circuitlab.com, make a diagram of your three components, take a screenshot, and upload it to your favorite image host. \$\endgroup\$ Jun 17 '10 at 22:33
  • 1
    \$\begingroup\$ One thing the extra resistors are used for is to fool the electronics into thinking it is looking at normal globes, otherwise the globe check system can give you false reports. Try it without the resistors and see what appears on the dash. \$\endgroup\$
    – Greg Higginbottom
    Jun 18 '10 at 7:16
  • \$\begingroup\$ everything's just in series. \$\endgroup\$
    – Chris2048
    Jun 25 '10 at 17:55

It is likely a zener to keep the LED at a constant brightness. Car power is not constant, it's only loosely regulated, and since the LED has a very fast response time changes in vehicle voltage (say when you start your airconditioning) may be percieved as flashing your light.

Filament light bulbs take awhile (100-300mS) to visibly react to a change in voltage, so it's not an issue for them.

  • 2
    \$\begingroup\$ Aha! This is it. So the zener will be in the range of 5-10V, and we'll have one resistor (to limit the zener current) in series with a parallel combination of your zener and the series combination of a resistor and an LED. \$\endgroup\$ Jun 17 '10 at 22:50

The diode is likely a zener diode to short out voltage spikes generated inadvertently by the alternator.

  • \$\begingroup\$ Hmmm, possibly, but aren't zeners usually glass/transparent? This one is just a black cylinder. \$\endgroup\$
    – Chris2048
    Jun 25 '10 at 17:53
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    \$\begingroup\$ No, I've seen zeners that are black cylinders before. \$\endgroup\$
    – pingswept
    Jul 6 '10 at 16:09
  • \$\begingroup\$ I've seen zeners in all sorts of packages \$\endgroup\$
    – Tim
    Feb 6 '11 at 17:38

I don't have an answer to your question about the extra components without a schematic, although I'd agree with Pingswept's guess - Remember that your LED doesn't need a reverse current protector; it is a diode itself.

However, I will point out that you want to use 13.8V (The voltage output by the alternator) for your calculations, not 12V.

Small difference, but it's always good to remember. In this circuit, it means you can''t put in a 13V zener, or you'll find yourself sinking the alternator voltage for a short amount of time, and then your zener will release its magic smoke.

  • \$\begingroup\$ depends on how big the zener is <3 \$\endgroup\$
    – Kortuk
    Jun 18 '10 at 5:44
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    \$\begingroup\$ No, I don't think that has anything to do with it. A typical car alternator has 80-120 Amps of current with which it tries to reach 13.8V. If the zener can sink that kind of current, it would dissipate 1,300W and make a nice little space heater. If you have a source, I want one! \$\endgroup\$ Jun 18 '10 at 13:01
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    \$\begingroup\$ LEDs are sensitive to reverse voltage, typically with a reverse voltage rating of a few volts. They are very easily destroyed by reverse breakdown so it is common to fit an opposite polaritydiode across the LED if there is any risk of reversed polarity. \$\endgroup\$
    – Russell McMahon
    Aug 15 '11 at 13:11

A point about LED's - most led have a very low reverse breakdown voltage, so a standard led may fail if connected the wrong way round on about 12 volts (Yes I know - car batteries are 13.8 approx, but can be anything from 10 to 16 V, with a lot of 'noise' on the supply)

So to protect the LED from being put in the wrong way round use a standard diode...

The high watt resistors are there to 'disable' the blown bulb detection circuit in some cars... (They are just across the socket to 'waste' power)

Some cars will display a warning that a bulb is blown. To prevent this you must take about the current a bulb should take...

if your Led's just took 20ma (0.02A) at 14v (0.28W) and the original bulb was 5W - around 357ma (0.357A) - a circuit that expected .36A seeing .02 would trigger an error...

  • 1
    \$\begingroup\$ This question is a few months old... \$\endgroup\$
    – tyblu
    Dec 12 '10 at 23:29
  • \$\begingroup\$ @tyblu - Interestingly, it's still accruing answers almost two years later (some of which have been deleted). \$\endgroup\$ May 3 '12 at 20:18
  • \$\begingroup\$ Even if the bulb is installed correctly, it's possible for automotive electrical systems to see pretty large positive and negative swings when things like motors are switched in and out. I forget the exact specs for what automotive accessories are supposed to be able to tolerate without damage, but something like a tail light is apt to be considered "safety equipment" and have even more rigid specs. \$\endgroup\$
    – supercat
    Nov 26 '13 at 16:33

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