For standard LED's you only want about 3 volts and about 20 milliamps so you need to connect a resistor in series with the LED for a voltage drop of 11 volts since automotive voltage us about 14 volts(13.8) when the vehicle is running. Using Ohms Law, R = V/I so R = 11/20e-3 which is 550 ohms. A 470 ohm resistor would be a good enough standard value since the LED will take more than 20 milliamps. I know someone who usually uses 470 ohms for automotive purposes. I used 220 ohms for my six volt motorbike. You need 470 ohms for each LED you use. Connect the LED's in parallel with one another.
hello all. im interested in using some leds for automotive lighting and such. i have a BASIC understanding of electronics and electricity. i am familiar with diodes, but not leds. The very small information i have found for auto use talks about using resistors in parallel, but noone really goes into why. i believe its to drop the voltage? but i wouldnt bet my next paycheck on it. can anyone give me some basics on building led circuits? especially how to connect multipes and determine the proper resistors to use? Thanks for the help in advance.
Rodney said: "Connect the LED's in parallel with one another."
ehsjr said: "Do NOT wire LEDS in parallel."
Right now you might be a little confused, and who can blame you! Actually, what you want for automotive lighting is a combination, series-parallel circuit. Take ehsjr's example:
Realize that the +12VDC is variable between about +11.5 and +14.5, calculate your resistance based on optimal diode current at max auto supply voltage (don't forget to use the current *and* voltage specs of the LED that you choose in your calculations). You can now build the series-parallel circuit, adding legs to your heart's content. For this application, the auto's supply current is practically unlimited, unless you want to try to emulate aircraft landing lights or summat.
One and only one rule*: design the circuit to limit the current through the LED to some particular value. The simplest way is with a 1/2 watt (or larger) series resistor:
+Vcc-----[R]---[LED]---Gnd
Use the formula Resistance = (Vsupply - Vled)/Current If LEDS are used in series: Vled = Vled(1) + Vled(2) + ... Vled(n)
= until and unless you have a specific reason or reasons not to.
If you don't have specs on your LEDs, use these numbers for Vled: Red LED ~1.8 volts; white (or blue, blueish white) LED ~ 3.4 volts. and limit the current to about 20 mA (or less). (Other color LEDS will range between these Vled values)
If you have the specs for the LED, you can use them to chose the current limit, and you'll know Vled for each LED. Otherwise,
20 mA is generally a good value for the limit, and LEDS will light
*well* below 20 mA.
In a car, a nominal voltage of 14 for Vsupply can be used to compute the size resistor you need. With a red LED, that computes to 610 ohms, and I would recommend using 680 ohms (a standard value) for a little extra safety margin. With a white LED, it computes to 530 ohms. I'd recommend going up to at least 560 ohms, but you could use the 680 ohm resistor and still get over 15 mA through the LED. Probably won't see much change in brightness, either.
Do NOT wire LEDS in parallel. For multiple LEDS, use series like this, and re-compute R:
Vcc---[R]---[LED1]---[LED2]---[LED3]---Ground
In the case above, for red LEDs, it computes to 430 ohms. A 470 ohm standard resistor would be fine. With white LEDs, R computes to 190 ohms, and a standard 220 ohm resistor would be good.
If you want to use a parallel circuit, do this, and use the 680 ohm resistor mentioned earlier:
That's not the end of the story. The electrical "environment" in a car is hostile. There's all kinds of electrical transients, and it may be prudent to protect the LED's with a 15 volt TVS diode from Vcc to ground. In addition, the ambient temperature may be high, and you may want to limit the current to less than the ~20 mA (or the specs, if you have them) to compensate for high ambient temperature.
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The only thing I would add is to make sure that you calculate the power the resistor will have to dissipate (Current * Current * Resistance) and select a resistor with a power rating at least that large. Good practice is to use a resistor with 2X the required rating.
I can't argue with you, it just would depend on the application. I used a white LED to replace a high beam indicator on my motorbike and it works fine. It does happen to be in parallel with the other lights. Just had to solder a 220 ohm resistor to it. If I use LED's for the other lights I'll do the same. I realise you could hook up multiple LED's in series with a single resistor but if one LED fails you will have trouble knowing which one is dead when the whole string goes out. Just like Christmas lights eh? LED's don't go out too often but they can. They may be less reliable in automotive apps.
Right, I think that's the source of the confusion in the discussion. You are wiring resistor-LED-pair series circuits in parallel with other devices using the same source, and I think the ehj-someone is implying that non-current limited LED's should not be paralleled.
In pictures:
+V--[R]--[LED]--Rtn |---[circuit]---|
and not:
+V--[R]---[LED]--Rtn |-[LED]---| |-[LED]---|
which would probably be ok *if* R is chosen for the current used by the parallel group, and the LED's are closely matched for voltage drop.
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Since LEDs have a negative tempco, the common resistor is _never_
preferred because even small changes in diode characteristics can
cause one LED to hog current, overheat and, if it fails open, cause
the others to go, one by one.
Maybe I should explain that in the old days they used to make Christmas lights in series. When one light went out you had to test every bulb to see which one was out. Today they are making strings of LED Christmas lights which I believe are all connected in parallel.
Rodney wrote: Today they are making strings of LED Christmas lights
That makes sense only if each LED has it's own separate resistor to set the current:
| | |--/\\/\\/\\/--|>|--| | | |--/\\/\\/\\/--|>|--| | | |--/\\/\\/\\/--|>|--| | | |--/\\/\\/\\/--|>|--| | | where -/\\/\\/\\/- is a resistor and -|>|- is an LED.
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Frayed knot. ;)
1. Each resistor will have to dissipate about 2.4 watts
2. What\'ll happen when the mains polarity reverses and puts the
cathodes of the LEDs at 170V positive WRT the anodes?
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I tend to think of a string of Christmas tree lights as working off
the mains, so to get that low-voltage DC they\'d have to have a
power supply of some sort. Maybe a wall-wart? I don\'t know, I
haven\'t bought a string of lights in over ten years. Is that how
they do it?
I don't know either. I was originally responding to somebody who said they THINK that Christmas lights are available as LED's wired in parallel. My point was that IF that is the case, they would each need a separate series resistor. And of course run off a low voltage DC source.
When you think about it, the possible savings in efficiency are likely offset by the inconvenience of using a wall-wart, the power drain of the resistors, and the fact that they would only be used for 3 or 4 weeks out of the year. But I honestly don't know what is available.
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