Question about resistor heat

[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