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Heat of LED resistors?

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Heat of LED resistors?
Posted by Gerome on Tuesday, December 12, 2017 1:07 AM

I never noticed before, but when I was testing some DC locos with a power pack putting out 13V @ 400mA DC with a regular little LED and a 1/8 watt resistor, the resistor became too hot to hold.

Now, since my Zephyr command station is capable of 12 V @ 2.5 amps, does that mean the resistors on the LEDs from the DCC decoders are also getting hot, or do the decoders reduce this heat significantly?  I know the LEDs are better because they run cooler, but are the resistors getting hot?  I never held one under power.

Thanks

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Posted by hon30critter on Tuesday, December 12, 2017 2:28 AM

Interesting question. I have experienced hot resistors in the past. They were 1/8 watt. Since then I have used 1/4 or 1/2 watt resistors but I have never tested the heat output from the higher watt resistors. Just lazy I guess.

Dave

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Posted by faraway on Tuesday, December 12, 2017 3:14 AM

Lots of LEDs run with 20mA. A 600Ohm resistor may use 12V with 20mA. That results in 0.24W. That is to much for an 1/8W resistore. Use a 1/4W resistor or a more gentle LED.

Most LED will do fine in series with a 1.500Ohm resistor. That drops the current and 1/8W resistors are fine. That is what I install in all my engines and it works fine.

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Posted by train18393 on Tuesday, December 12, 2017 3:23 AM

The heat output of the resistor is the same independant of the wattage of the resistor, assuming they are the same ohmic value. It is a function of the dissipation of the power you are trying to loose. A 1000 ohm resistor gets rid of the same power no matter what the wattage rating. The bigger resistor feels cooler as there is more surface area to dissipate the heat, but the amount of heat generated is the same. The power used by the circuit depends upon the series resistance of the circuit not the current available from the source. This could get more complicated by figuring out the actual output of the sources and measure each with a RMS meter or oscope and use the RMS value of the voltage for comparison, but that is probably overkill. Your two power sources may not be as close as you think. 

You must be careful not to melt the plastic shell of the locomotive, as that will happen before you burn out the resistor. Use a larger wattage resistor if there is room as it will spread the heat over a larger area. If the LED is bright enough you may be able to use a larger ohm value in series as that will decrease the current, thus the heat will be reduced. 

 

Paul

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Posted by rrinker on Tuesday, December 12, 2017 6:35 AM

 If they are getting hot you are probably using too small a resistor value. White LEDs tend to be extremely bright and there's no real reason to use anything less than a 1K resistor. A 1K resistor results in less than 10ma, less than 1/10 of a watt, so a 1/8 watt resistor is fine and won't get too hot. With surface mount LEDs, and standard LEDs were there isn't a long light pipe between the LED and the front to attenuate some of the brightness, even 1K isn't enough. 1.5K, 2K, even 4.7K is not out of the question to get a pleasing light. As you go bigger on the resistor, the current decreases, thus decreasing the wattage and thus the heat generated.

 Consider this - my resistor wheels for blocl detection have a 10K resistor in them. Every axle with a resistor is the same as taking a 10K resistor and putting it directly across the rails. At a nominal 15V across the rails, that's about 2/100 of a watt. They don;t even get warm - which is a good thing, since many are metal wheelsets with plastic axles.

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Posted by gregc on Tuesday, December 12, 2017 5:31 PM

Gerome
with a regular little LED and a 1/8 watt resistor, the resistor became too hot to hold.

what size resistor?   do you know the voltage drop across the LED?

greg - Philadelphia & Reading / Reading

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Posted by BroadwayLion on Tuesday, December 12, 2017 5:39 PM

Always buy 1/4 watt resistors. They are about half the price of 1/8th watt resistors.

 

LION buys int in a box of 1000 for about $12.

 

ROAR

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Posted by richg1998 on Tuesday, December 12, 2017 5:49 PM

All I have used are 1k, 1/8 watt resistors. I have measured 9 ma many times with 20 ma LED's. No heat to worry about.

Decoder voltage was about 12.2 vdc.

Rich

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Posted by Bayfield Transfer Railway on Thursday, December 21, 2017 1:29 PM

Maybe this is a good place to get help.

I want to connect my numberboard LEDs in parallel.  This means I add the amperage.  If I have 8 LEDs, 2 per numberboard, and each uses 10 mA, what wattage does the resistor need to be?  W = V*A, and A= .010, but do I use 12 for V or some other value?

If it's 12v, then even at .01A each I'm pushing nearly a watt, and I'd be better off using 8 1/4 watt resistors located near the numberboards rather than one resistor for all.

It was my hope for convenience sake to have a bit of PC board in the engine shell with the resistors soldered on and then a plug soldered to that so that when removing the shell, one plug disconnects all the light circuits.

 

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Posted by RR_Mel on Thursday, December 21, 2017 2:04 PM

Bayfield Transfer Railway

Maybe this is a good place to get help.

I want to connect my numberboard LEDs in parallel.  This means I add the amperage.  If I have 8 LEDs, 2 per numberboard, and each uses 10 mA, what wattage does the resistor need to be?  W = V*A, and A= .010, but do I use 12 for V or some other value?

If it's 12v, then even at .01A each I'm pushing nearly a watt, and I'd be better off using 8 1/4 watt resistors located near the numberboards rather than one resistor for all.

It was my hope for convenience sake to have a bit of PC board in the engine shell with the resistors soldered on and then a plug soldered to that so that when removing the shell, one plug disconnects all the light circuits.

 

 

I wouldn’t parallel the LEDs.  Even the same batch LEDs vary slightly in current and brightness.  I would use a separate resistor for each LED.  9 volt drop at 10ma = .09 watts.  Eight ⅛ watt resistors don’t take up much space.
 
I only use one LED per number board with a separate resistor then paralleled to a polarized micro connector.  I run my number board LEDs at 5ma, 10ma is way too bright for me.
 
 
 
Mel
 
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Posted by Bayfield Transfer Railway on Thursday, December 21, 2017 2:09 PM

Thanks.

The LEDs I have are SMT 0604s, which are pretty tiny.  Also I've noticed that on a lot of diesels at least, you can tell where the lightbulbs are behind the numberboards on the prototype and I was hoping to duplicate that effect.  But you're right about resistsors being tiny, even 1/4 w, and also very cheap.  Just heatshrink over the whole thing and there's plenty of room in an Athearn HO body.

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Posted by DigitalGriffin on Thursday, December 21, 2017 2:58 PM

Formula is V*V/R.  your input is roughly 12 volts.  Your LED light takes roughly 3.5 volts away from that leaving you 8.5Volts.  15 ma = .015 amps

V = IR
8.5 = .015R
566 = R

so lets go with common values and pull a 560Ohm resistor
V * V / R = Power
8.5 * 8.5 / 560 = .129 Watts

Barely an 1/8th watt.  So a 1/4 watt resistor is fine.

Those must be some LEDs to handle 400ma.  that's 4/10th an amp.  Most burn out over .020 amps.  (20 ma) 

 

Don - Specializing in layout DC->DCC conversions

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Posted by RR_Mel on Thursday, December 21, 2017 3:19 PM

Bayfield Transfer Railway

Thanks.

The LEDs I have are SMT 0604s, which are pretty tiny.  Also I've noticed that on a lot of diesels at least, you can tell where the lightbulbs are behind the numberboards on the prototype and I was hoping to duplicate that effect.  But you're right about resistsors being tiny, even 1/4 w, and also very cheap.  Just heatshrink over the whole thing and there's plenty of room in an Athearn HO body.

 

I use scrap Acrylic to make a light pipe type defuseser with a single 2mm LED for each number board.  I use a lot of the 0604s for other things but not in my number boards.  At 80 soldering to the 0604s isn’t an easy task without a good holder and a lot of light.  
 
 
 
Sorry for the puny pictures, many years ago with a early digital camera on it's last leg.
 
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Posted by Bayfield Transfer Railway on Friday, December 22, 2017 12:31 PM
Thanks, nice pictures. I found a place that offers 0604s with leads attached. Yes, more expensive, but at age 62 my hands aren't as steady as they used to be either.

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Michael Mornard

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Posted by Bayfield Transfer Railway on Friday, December 22, 2017 12:36 PM

DigitalGriffin

Formula is V*V/R.  your input is roughly 12 volts.  Your LED light takes roughly 3.5 volts away from that leaving you 8.5Volts.  15 ma = .015 amps

V = IR
8.5 = .015R
566 = R

so lets go with common values and pull a 560Ohm resistor
V * V / R = Power
8.5 * 8.5 / 560 = .129 Watts

Barely an 1/8th watt.  So a 1/4 watt resistor is fine.

Those must be some LEDs to handle 400ma.  that's 4/10th an amp.  Most burn out over .020 amps.  (20 ma) 

 

I'm not sure where you got 400 ma from.  Each LED is 10 mA, there are two LEDs per number board, four number boards, that's eight LEDs total.

With a 1K resistor we're talking 10 mA.  Actually, I should probably cut it down so the number boards aren't too bright.  Say a 2K resistor for roughly 6 mA.  Times eight is 48 mA, or .048A.  Times 8.5 = .408 W, so a 1/2W resistor

The missing info was how much voltage the LED uses, thanks.

 

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Michael Mornard

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Posted by DigitalGriffin on Friday, December 22, 2017 2:02 PM

I think I misread your 400mA comment as how much current your LEDs are using.

Just remember

1: if you wire your LEDs in series, the voltage drop adds up.  So two LEDs would be a 7V drop.  Three would be a 10.5V drop which would be the practical limit.  The advantage of doing it this way is your current stays the same and the power that the resistor has to disappate becomes way less because you have 5->1.5V left over after the drops.

2: If you wire your LEDs in parallel, the current adds up.  So two 10mA LEDs become 20mA.  3 would be 30mA, etc.

It sounds like you'll have a nicely lit loco when you are done.

Don - Specializing in layout DC->DCC conversions

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Posted by Bayfield Transfer Railway on Friday, December 22, 2017 2:09 PM

The information I've been missing is the voltage use of the LED.  I know that amperage in series stays constant; that is a strong argument for wiring the two LEDs of each number board in series if I decide to use two LEDs.  5 ma * 5 volts is .025W.   And I have to experiment to see if one might work.

And number boards are noticeable during the day, but just barely; they shouldn't be as bright as the headlights, which is a whole nother story.  Just like the class lights are noticeable, but not all that bright.

Disclaimer:  This post may contain humor, sarcasm, and/or flatulence.

Michael Mornard

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