Question about resistor heat

[Paul G]

Hello - hoping this is the best place to ask this question.

I'm creating some flood lighting using a 1/2 U channel aluminum with a few high brightness LEDs. Found a calculator and circuit to connect in series with a 12 Volt supply.

My question is not about voltages, resistor values or anything like that. However the resistor is too hot to touch when operating, although prongs, and leds and wires are cool. Just worried that it's a fire hazard.

Tech details:

12 volt supply. LEDs 2.4v @ 30mA, resistor is 150 ohm and connected in SERIES with 3 LEDs. Resistor is rated at 0.25 W so should be well under the 70mA that the calculator suggests.

Question: Is this ok for the resistor. As it's not going to be near anything flammable then I don't see it as much of a risk, but as my project is not the most fire retardant structure, I'd rather crash and burn in a real plane than a home built sim.

Thanks

Paul

[ve4anc]

Hello Paul:

Greetings from a fellow Canuck.

There is an on-line calculator at http://www.electronics2000.co.uk/calc/calcpow.php which you will find helpful in solving this problem.

Go to the bottom of the page and you can enter values. Use the drop down menus to obtain milliamps.

When I enter 60 ma and 150 ohms, it comes up with 540 milliwatts of power dissipated in the resistor ... that is twice its rated 1/4 watt value and would account for the heat you are experiencing.

You should step up to at least a 1/2 watt resistor, but I always like to be conservative by at least a factor of 2 unless space is a great consideration. Therefore a 1 watt 150 ohm resistor will serve you well and should run cool to the touch.

Regards,
Lee Smith

[Paul G]

Hi,

Thanks for the calculations. I'm trying to work out where you got 60 mA from. My calculator is based on 30. When I work out the power it comes to 70mW so a quarter watt should be enough, right?

As an aside, if I add another identical resistor in parallel to the first, will that help dissipate some of the heat? I did try this and they both get hot. I'm not trying to be a tight *** as these things are a few cents each, but I have to buy 100 of them for each value, and the store is 20 minutes drive. If I can find a one-off work around that would be better. Although I'm now starting to wonder whether I just put a resistor on each one and wire it accordingly.

Thanks again for your referral.

Paul

[AndyT]

Simple rule:
If the resistor is so hot you can get burned, then its too small for the job.

[Michael Carter]

What Andy said.

Change to a higher wattage, they'll be fine. If they're too hot, they're too small.

[P1IC]

Paul,

If your LEDs each drop 2.4V when passing a current of 30mA, and they are in series, then you must pass 30mA through the series resistor too.

Your supply is 12V, the LEDs drop a total of 3 x 2.4V= 7.2V, so your resistor needs to drop 4.8V with 30mA going through it.

So the value of the resistor should be R=V/I or 4.8/30mA, which is 160 Ohms.

And the power rating of that resistor, given by W=VxA is 4.8x30mA, or 144mW.

So with a resistor of 160 Ohms, a quarter watt rating is OK, but better with a half watt for longer life.

With a 150 Ohm resistor, the current through the circuit will be slightly higher by a couple of mA, which won't affect things significantly, so the wattage of your 150 Ohm resistor certainly ought to be enough.

If it's too hot to touch, then the assumption that your LEDs are each dropping 2.4V is wrong. Or the resistor isn't 150 Ohms! Can you measure it?

Can you measure current and voltage? If so, measure the current through the circuit, and then measure the voltage across each of the components. The current should be about 32mA and the voltages should be as above.

[ve4anc]

Since there are 3 LEDs .. each consuming 30 Ma at manufacturers max power ratings, the cumulative current would be 90 Ma plus the amount of energy dissipated in the resistor.

I had a dislexic moment and went with 2 x 30 Ma in my first post rather than current for 3 LEDs. Hence, my original use of 60 Ma in the calculator results.

Going to the calculator again, it indicates 1.215 watts.

Bottom line ... as mentioned earlier, if the resistor is running so hot that you can't touch it, it is waaaay too under rated. It should remain very close to room temperature, or just barely warm to the touch.

Generally, electronic components are rated at 25 degrees C. surface temp although internal component temps will be higher.

Lee

[percy]

Lee
I think you were right the first time current is the same through all components in a series ckt. What is missing is the true resistance of the leds + resistor ohms to figure a true current then you can figure Watts. But I am not sure how you put LEDS in a series CKT, I would think it would be a parallel CKT. A resistor would be in series with each LED but LEDS would be in Parallel. (I think)

percy.

[Trevor Hale]

LED's in series with one resistor in series with all, keeps all led's at the same brightness. If you do not have the resistor, the first led will be the brightest and the last led will be the dimmest.

Trev

[Paul G]

I think I'm going to change my design to a set of parallel, single LED - single resistor circuits as I've done with other parts of my sim. These LEDs run really cool and parallel means you don't have to account for how many LEDs you have.

Still confused over what was said about current. I'm pretty sure it's consistent as analogous to water flow, but voltage IS dropped per component. I'll treat advice here with more respect than the theory. Something sure is amiss but I'm not going to entertain fire risks to find out why

Thanks for all the help

Paul

[Trevor Hale]

I think I'm going to change my design to a set of parallel, single LED - single resistor circuits as I've done with other parts of my sim. These LEDs run really cool and parallel means you don't have to account for how many LEDs you have.

Still confused over what was said about current. I'm pretty sure it's consistent as analogous to water flow, but voltage IS dropped per component. I'll treat advice here with more respect than the theory. Something sure is amiss but I'm not going to entertain fire risks to find out why

Thanks for all the help

Paul

Paul please be careful though.

Resisters in parallel devide the resistance, in Series they Add.

So you can put one resistor in series with 5 LED's in Parallel, but you must still observe the curent draw of all 5 LED's.

Trev

[percy]

Paul
You are right Paul you are best off with parallel, specially if the led is rated the same as the supply voltage and the LED has an internal resistor then you do not need to mess with a resistor. I have 12v supply with several 12 volt LEDs connected in parallel with internal resistors and all works cool and fine, no resistor to mess with.

[TasKiNG]

If all the components are in series i.e.

+12V----Resistor----LED----LED----LED-----0V

Then P1IC's reply and explanation is correct. But I would not expect the resistor to get excessively hot.

Now if it is connected as follows:-

.---LED---
¦.............¦
.---LED---
¦.............¦
.---LED---
¦.............¦
R.............¦
e.............¦
s.............¦
¦.............¦
¦.............¦
.---12V----

The resistance of the 3 diodes in parallel will be 26.6 ohms.
Therefore the current (I) will be V/R = 12 / (150+26.6) = 68mA through the resistor and as Watts = I squared x R = 0.068 * 0.068 * 150.

Then Watts dissipated via the resistor would be 693mW. which would make it run very hot.

Check your circuit is correct i.e all components in series as in the top diagram.

Cheers

[Mike.Powell]

Paul,

I'd stick with the series LED arrangement. If you go with a single LED in series with a single resistor, that resistor will need to be 320 to limit the LED current to 30 ma. It'll be dropping more voltage and will dissipate 288 mW. You'll have a worse situation than you do now.

Electronic parts are rated for a certain power dissipation under specific conditions. This is typically at a standard ambient temp of 25 degrees C, but this is not the surface temp of the component. The component will be as hot as necessary to dump the waste heat. However, there is a temp limit on components, too. The power dissipation figure is valid only if the component is installed in such a fashion that it can dump the power without overheating.

The primary heat path out of a low power resistor is through its lead wires. If they are cut short and/or insulated the temp will go up until there is enough temperature differential to push the thermal energy out.

Going back to your orignal circuit, you can use multiple 150 Ohm resistors to spread the heat dissipation. Parallel two resistor and place them in series with two more paralleled resistors. You'll end up with a total of 150 Ohms with a nominal 1 watt rating.

|
|__
| |
R R
| |
---
| |
R R
| |
---
|
|

[ve4anc]

Zeesh .. that's what I get for posting before the first cup of coffee in the morning!

Multiple strikeouts.

Lee

[Paul G]

Thanks for all the ideas. i'll probably be coming back to these in the future as there's a ton of good stuff here. I went with the parallel arrangement and it solved the heat problem. Actually quite pleased with the results which I'll post when my new sim launches in a week

Paul