2leds=1resistor?

Yes.  Wire 2 LED in series and add a 150ohm or so resistor.  G

In general, a problem with putting two components in parallel on one resistor to drop voltage to an appropriate level is that if one is disconnected, voltage drop across the resistor will lessen and you may burn out the other component. 

 I'm using 300 ohm resistor on the headlight LEDs. I thought if I put the resistor inline with the LEDs in parallel, it would work?

If one burnt out or became disconnected, I would think the other would hold?

I wouldn't bet on it.  You can get real tiny resistors, 1/8 watt, so space shouldn't be an issue.

If the LED's are identical units, 99.99% of the time parallel connections work fine.  Obviously, mixing types and/or colors is not fine, each color has a slightly different operating voltage.  I've also seen different styles and manufacturers of the same color LED have problems in parallel.

I also tend to run LED's below their maximum ratings.  If you check the brightness of an LED at 10ma and again at 20ma, there is very little difference.  That being the case, why push it harder than you need to?

RJR, WHo was talking about parallel.  Regardless, if you do as Dave mentioned and use the resistor in series with the parallel LED there will be no issue.

For PS-2, MTH uses series LEDs for marker LED.  If your powering your lights from the Marker connection I would mimic the series wiring MTH does.

As far as the other bulb circuits, MTH voltage approximates 6V.  Any of the older engines with 6V CV boards used the 150 ohm resistor for the LED current limiting.

So I go with the 150.  You can certainly increase the resistance if you want.  If you go parallel, like Atlas does you need to up the resistance also.

The 1k ohm I see are when the LED are driven from TMCC AC where voltage is higher  (Atlas/Lionel).   G

GGG, my original post was written before yours, recommending LEDs in series, showed up on the forum.  Original post by Joe didn't mention series or parallel.

If wired in parallel, and voltage downstream from the resistor is near the max voltage for the LEDs, there could be a problem:  failure of one LED will cause voltage fed to the other to rise by one-half the difference between the applied voltage upstream from the resistor and the original voltage downstream; this could damage an LED running near its peak.  If one mimics the MTH wiring, which you say is in series, there should be no problem.

RJR,  Actually voltage does not rise the current rises in parallel.

For parallel LEDs fed by a series resistor, the voltage drop across the LEDs would be the same.  Let's assume 2V.  The parallel calculator will use the current you chose to use as going to both LEDs.  Let's assume 20ma.  So each LED in parallel gets 20ma, but the resistor has to source 40ma.  If one LED drops out, and the resistor is allowing 40ma, and you still have the 2V drop from the remaining LED you get the full 40ma.  Otherwise the two LED try to share the current.

The problem with Parallel is that no 2 similar LEDs are a like and due to the tolerances one may flow 25ma (path of least resistance to the current) so the other gets only 15ma.  This can cause differences in illumination, etc..

Hence why most documents recommend series wiring, besides the lower current seen by the series resistor.  The problem with series is that if one LED goes open, both go out.

I think this is why Atlas and Lionel have used parallel for diesel head lights.  If one fails you still have the headlight lite.  You just need to pick a limiting resistor value that won't burn out the second LED when the current doubles when one fails.  G

Hence my previous recommendation to run them below their ratings.

If you had two LEDs in parallel, and one fails, it would not be drawing any current?

 If you had two speakers on an amplifier and one burned out the voice coil so it's open, the amp would only see the other one still working. It would not try to drive the other.

 Here, the resistor does not burn out. So the output FET still sees it. I am confused on how it would try to drive a fail LED?? Does the LED draw on the circuit? or does it only get what's available from the resistor's draw??

I disagree as to voltage rising downstream from the resistor.  Voltage drop across a resistor is calculable by Ohm's Law.  If the current is halved, such as by one of 2 incandescent bulbs downstream from the resistor burning out, the voltage drop across the resistor will halve.  The voltage at the remaining bulb will increase as a result; this will in turn cause more current to flow through the remaining bulb.

Speakers and LEDs are different animals.  A speaker has a characteristic impedance that will work regardless of the capacity of the amplifier.  The LED presents a constant voltage load and if one burned out, the other one would see twice the current, a different situation than speakers.

If you increase the current through an LED way over it's rating, the voltage will rise very little, but soon the LED will fail from the heat.

Sorry John, I only used the speaker loading for my benefit. It doesn't apply here.

I thought that the current output would be variable dependant on the load. I didn't see why the current would double.

 I guess I have to look at it straight on (as a diode).

Please look at Kirchoff's Laws an you will see it.  A diode has a set voltage drop.  Resistors or diodes in parallel have to drop the same voltage.

If you run 10 lights off a constant voltage source and drop one out the voltage doesn't go up.  The only difference with diodes is they do not present a resistance and can try to pass unlimited current.  Hence you need the current limiting resistor.  It is sized based on the factors you use.  If you run 2 2V LED using 20ma as current and 6V as source you will get the same 100 ohm resistor for series or parallel.  The reason:  The series LED drop 2 + 2 V and the resistor has to drop 6-(2+2)=2V  so the 20ma current demands a 100 ohm resistor. V=IR so 2V divided by .02A=100 ohms.  The 20ma goes around the series loop and each component sees 20 ma, and each component drops 2V.

For the parallel set up since each LED drops the same 2V the resistor has to drop 6V-2V = 4V.  Since each LED in parallel needs 20ma that requires 20+20Ma and the current going through the resistor is 40ma.  So 4V divided by .04a gives you the 100ohm resistor again.  But this time the resistor wattage needs to be higher because the voltage drop and current is higher, even though the LEDs only see 2V drop and 20ma. 

Basically in series a single hose pumps 20gal/min to everything, while a parallel circuit needs 20gpm to each LED so for 2 the hose needs to pump 40gpm and when it gets to the LEDs the hose splits and each LED gets 20gpm.   G

No I did not say that.  In parallel they have independent current available.  The one that fails (open) would stop passing current, but the other would still work.

In series if one fails open, the continuity path to return current to the source opens so neither LED lights anymore.  G

Sorry G. Must be the painkillers!

Actually I meant to say that the spent LED doesn't draw anymore. I was thinking it was a load.

I really struggle with this stuff more than needed.

I've learned to just do what is drawn for me and not try to understand.

I'm so bad now, That I can't remember what I wired just last year in my upgrades.

I'm to the point where if I don't take notes that I can use in the future, I have to start over from scratch each time. I have an engine torn apart from last winter. I can't remember any plans I made so I'm starting fresh as if I never did it before.

 I'm soldering in so many LED's all over that it seemed a waste to have a resistor on the 10 LEDs I'm installing. If it's required, I'll just do it.

 I have my LED head, tail and ditch lights so bright on my installs, they look like real engines coming on you outside in the dark. Sad, but I can't remember what I did.

A failed diode normally acts like an open.  Nothing passes.  Sometimes they can fail shorted and than they are like a wire.  LEDs normally don't fail shorted, they just fail open.  G