Lost with Led's

130 to 150 ohm resistor.1/8  watt fine.

Thanks.  I have a 5.5k resistor .  It won't light up the LED.  Does that sound right?

For that LED, you want to drop 3 volts at 20ma, so that would be 150 ohms.  In truth, depending on the application, you may want to have much less current, there's nothing that says you have to run LED's at max power.  For markers or the like, I'd probably use something more along the lines of a 470 ohm resistor for that LED.

as well is your 5 volt power source regulated...typical 5 volt walwarts are 7volts plus unloaded .

Good point, one benefit of having a larger value resistor, plenty of margin for error.

My general LEDs rule of thumb is:

reds = 2.2V@ 20mA

greens = 2.5@20mA

Yellows = 2.4V@ 20mA

whites/blues = 4V@ 25mA

Others may disagree with values but these are general run of mill LEDs.  More exotic ones may require consulting the data sheets.  Don't get hung up on the slight 2.xx voltage differences, just when you use the white or blues that it matters.

So you subtract the LED voltage from your power supply voltage and divide by either 0.020 or 0.025 depending on LED color.  Use next larger value resistor. In fact, you can use larger values and you probably would not notice any appreciable brightness dropping.

So if I used the 5.5k ohm one the LED would not light up?

BTW, are you sure you have a 5.5k resistor? That's a somewhat unusual value to have just lying around...

Your LED will light up with 5.5k ohm. I don't have your exact DigiKey LED, but in above photo using a 5V supply and 150 ohm resistor, a similar LED is driven at 0.02 Amps or 20 mA (milli-Amps).  20 mA is a standard test current for this type of LED.

Next to it is the same LED with 5.5k ohm.  The current is about 0.6 mA and I suppose you could say it's lit up.  Photos are kind of tricky but it is quite dim and I'd say not usable in a train application like a marker light or whatever.  LEDs have an interesting property where the intensity is exactly proportional to the current...it's all about electrons flowing, so double the current, double the light.  The current in the left LED is about 35 times more than the right LED so it is 35 times brighter.  Kind of hard to gauge this difference with your eyeball.

Then, in the bottom photo I changed the resistor to drop the current in the left LED to only 0.06 mA.  I had to turn off the room light to take the photo but now your dim 5.5k ohm LED almsot seems like a searchlight as it is 10 times brighter (0.06 mA vs. 0.6 mA).  It's all relative and the human eye is remarkably sensitive.

Another point that needs to be considered. Simple application of marker lights, front or rear of a unit/locomotive is usually done as a Pair (2), most likely hooked in series. Simplest application with out getting into PCB reduced voltage and DC circuits is to hook these diodes to track voltage.   Any ideas what is a good normal application for this????   

Mike, if you power the LEDs from AC track voltage you not only need a dropping resistor but you will also need a diode to prevent excessive reverse voltage on the LED. LEDs typically have maximum reverse voltage ratings of around 5 volts or less. Wiring a pair in series saves you one resistor and one diode in this case.

Pete

Guess Mine is lighting up so dim I just don’t see it.  Yes I had the odd value laying around from some other project.  I just do not remember what.  Sourcing a few resistors locally is a real problem.

Pete, I hope you meant in parallel.  If you wire them in parallel back to back + to -, you don't need a diode as the lit LED provides the reverse voltage protection for the dark one.  Wiring them in series doesn't afford the same protection.  If you have LED's that have a 5V reverse voltage specification, 18V track power will subject the pair to around 13 volts reverse voltage.

If you wire them in parallel as I describe, you pick the resistor value for half the AC voltage and twice the power you'd use for a single LED at the same voltage.  That's because each LED is only seeing half the power, but the resistor is dissipating power on both halves of the AC cycle.

Atlas, Lionel and K-line use a simple 1K resistor for driving LED headlights and markers off track power.  So simple to mimic.

MTH use 150 ohm for series or parallel LEDs off 6V sources. Either CV or PS-2 driven.

Series wired was to ensure same intensity on each led, but if one fails both go out.  Parallel wiring ensures one stays on, but more current total flows.  Lionel, Atlas headlights wired this way.   G

John, it was Mike who seemed to ask about wiring them in series. If they are wired in series you would need one current limiting resistor and one diode to protect both LEDs from the reverse voltage. Personally, I add a diode and resistor to each LED regardless of how they are connected assuming the power source is AC.

Pete

My point is if you wire them in parallel across track power you can use one resistor and have reverse voltage protection for free.  You even balance the power draw on each side of the AC waveform, a minimal benefit, but better than unbalanced.

To clarify, for AC power, the LEDs are wired anode to cathode in parallel so that each LED lights on alternate half cycles of the AC waveform.