Resistor heat


UPBigBoy

UPBigBoy

J&A Paint Shops
OK, here's one question I've never seen; when using a resistor to lower the voltage to an LED (3 volts operating voltage) most would use a 470 OHM resistor but if one used a 1K OHM resistor, which one would get hotter, also which one would give the longest life for both the resistor and the LED using the assumption that these where being used with DCC.
 
UPBigBoy said:
OK, here's one question I've never seen; when using a resistor to lower the voltage to an LED (3 volts operating voltage) most would use a 470 OHM resistor but if one used a 1K OHM resistor, which one would get hotter, also which one would give the longest life for both the resistor and the LED using the assumption that these where being used with DCC.
Click to expand...

The 1000 Ohm (1K Ohm) resistor would give you a dimmer LED and a longer life than the 470 Ohm resistor. It would run hotter than the 470 Ohm because it offers more resistance, but if that is an issue then just jump up to a 1k Ohm 1/4 Watt resistor instead of the smaller 1/8 Watt version. The 1/8 watt resistor will run hot and burn/blacken over time if they are overloaded.
 
Ok, some basic theory:
The resistor is not for lowering the voltage to the LED it is for limiting to current to the LED. An LED has a fixed forward voltage drop (with a given current). Taking a Red LED as an example: it has a Vforward of 1.4V and a max current of 20mA. So let's set the LED current to 10mA. So with a 12V power supply the resistor needs to drop 12- 1.4 = 10.6V at 10mA. Ohm's law states the R = E/I = 10.6/.01 = 1060 Ohms.
If you wanted the max brightness then calculate for 20mA. The resistor would then be R = 10.6/0.02 = 530 Ohm.
Note that the voltage droped across the resistor is the same for both cases but the current through the circuit (resistor and LED in series) is different.

Now we can calculate to the Heat dissipated by the resistor.
Heat is expressed in Watts and is calculated by: P = V * I. So for the two examples:
P = 10.6V * 0.01A = 0.106 Watts.
P = 10.6V * 0.02A = 0.212 Watts.


For your LED with a Vf of 3V and a 12V power supply the calculations are:
V drop = 12 - 3 = 9V
For a 470 Ohm R: I = 9/470 = 0.019Amps
Heat = 9 * 0.019 = 0.172 Watts

For a 1k Ohm R: I = 9/1000 = 0.009 Amps
Heat = 9 * .009 = 0.081 Watts

So the 470 Ohm resistor would get hotter if both resistors are the same physical size. The physical size determines how much heat it can dissipate out of the resistance element into the environment. The bigger the resistor the more heat it can get rid of.

So for your two cases I would use at least a 1/4W, 470 Ohm resistor and an 1/8W, 1k Ohm resistor. To be safe use the next higher rating resistor, 1/2W for the 470 and 1/4W for the 1k, then the resistor's temperature will stay below the operating specs.

If installing these resistors in an enclosed area then they can get hotter due to heating the air within the enclosure. When installing inside a Loco, if you can thermally connect (not electrically) the resistor to the metal frame then the heat can be conducted by the loco's frame and the resistor will stay cooler.

Hope this all makes sense and answers your question.
 
Can two resistors of the smaller rating be daisy-chained if the larger were not available? If they are linked in series, assuming it is a practical option in the context of inside an HO scale shell, you would then have the heat dissipated by two devices, thus spreading the load between them.
 
Go through the calculations and you will find that the power (heat) dissipation is the same.
From the last example in the post above:
a 1K resistor dropping 9V and passing 9mA.
You could use two 500 Ohm resistors in series to have 1k Ohm. Each resistor then drops half the voltage so the heat is: 4.5V * 0.009A = 0.0405Watts.
So you could use two 500Ohm, 1/16W resistor in series rather than on 1k, 1/8W.
The total heat dissipated is still the same but spread out to two resistors. Another advantage of this is if there is a long, narrow space where a fatter resistor will not fit but two of the smaller diameter resistors will.

The other way is to use two resistor in parallel. Two 2k Ohm in parallel = 1K and each carry half the current so the power in each is half. So again you could use two 2k, 1/16W resistors in parallel instead.

I laid out the calculations to be as simple as I can in my post above. Use these to answer these types of question you have. Setting up these calculations in an Excel spreadsheet makes it easy to play around with the 'what ifs'.
 
Last edited by a moderator:
Thanks, Walt. My question was a bit rhetorical as I already knew the answer. :eek:

I was more wanting to help point out that 'thinking outside the box' a bit can often get a modeller out of a jam. If he were tight on volume/clearance, or if he couldn't afford the costs to get the larger resistor delivered to his home, or to get one, but if he had two of the smaller ones, he could still solve his problem with a bit of creative thinking.
 
waltr said:
'CLIP'
If installing these resistors in an enclosed area then they can get hotter due to heating the air within the enclosure. When installing inside a Loco, if you can thermally connect (not electrically) the resistor to the metal frame then the heat can be conducted by the loco's frame and the resistor will stay cooler.

Hope this all makes sense and answers your question.
Click to expand...

OK, using a 1/4W 470ohm resistor, would it get hot enough to do any damage to a plastic shell and if so would a 1/2W 470 ohm resistor be better - from your info that you gave I take it that the 1/2W would be correct?

Jim
 
If you have the space for the 470 Ohm, 1/2W resistor then that is what I would use.

But, do you really need the LED to draw 19mA and be that bright? In most cases running an LED near its maximum current rating is too bright for most modeling purposes. That is why Brad suggested the 1k Ohm resistor. I have been running LED in locos at about 5mA (~2k resistor).

You should wire up the LED and 470 Ohm resistor to check out how bright it is. Then if too bright increase the resistor value. Also check how hot the resistor gets. If it too warm to comfortably hold between your fingers then it could melt the loco's shell.
I think you will find that with a 1k resistor the LED will be bright enough and the resistor will be cooler.
 
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