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What temperature can 3w resistors safely run at?

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bigal_scorpio

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Hi to all,

I have just built a project that uses 1W LEDs which are run from 13v or so (car) and worked out that they needed 30r to give the 240mA which they take, bear in mind that they are run through ULN2803s. This is all ok when they are running on PWM at about 50% duty when they are running in sequence but they must all be on solid whenever the car is moving so I don't break the law.

Thats when the problems start! On full the resistors quickly heat up and reach 180c in just a few minutes, where they seem to stabilise. Now I'm no expert on cooking but I know that the wife puts things in the oven at 180 and sometimes less and they always come out cooked!

The resistors in question are axial cylindrical at around 12mm length and 4.5mm in diameter. They are actually 33r as I erred on the side of caution with the current. What is the max temp they can run at? All the info on the net seems to be about the cement type and as they are a completely different construction I thought it better to ask for expert advice rather than my sons car being a fire brigade statistic.

Al
 
I'd suggest buying a high-power resistor with a built-in heat sink. Generally those are required for automotive applications such as this, because they can dissipate a lot of heat. Try a google search for "high power resistor" and see what you can find :)
 
13V-3V for the LED = 10V across the resistor.

10V/33ohms = 300mA

0.3A * 10V = 3W

You are using 3 watts in your 3 watt resistors. At 3W you are at the burn out temperature of the resistor.

Maybe you have two LEDs in series. ????
 
The ULN2803 that drives the LEDs and resistors has a saturation voltage drop of about 1.1V to 1.4V.
Then the resistors have 270mA and dissipate 2.4W.
But maybe the battery is charging at 14.4V, not 13V.
And maybe the "3V" LEDs are actually only 2.8V.

Don't operate a resistor at its maximum power rating because it will burn plastic, burn you and maybe delaminate a pcb.
If the resistor is enclosed then it will get hotter than if it is in free air.
 
Hi DerStrom8,

If only I could use the bigger cement type, but I need 16 of them and I would need a shoe box for the circuit. ;)

Hi Ron,

I did mention that I am switching them through a ULN2803 which drops the 13 volts by 1.4 or so and the LEDs are 3.6v, leaving 8v by 33r =242mA and amps at 1.96 I think. But the problem is that I need to run 16 of them too, all separately for patterns so I can't series them. :(

I am thinking at the moment about a switching supply maybe of about 4v at 6A. It would be more efficient but it would also be my umpteenth attempt at a SMPS with all the others being failures.

Any thoughts? Al
 
Hi Audioguru,

The LEDs are definately 3.6 to 4.2 drop from Cree D/S. At the moment the supply is my bench PSU as I'm still in the build and test phase and the voltage is set at 13v. They are drawing a measured 240mA.

Maybe the 3W resistors are less than 3W? I can only go on the vendors values. They certainly are the weak link in the otherwise sound circuit. See above post for reason I'm not using cement types.

Al
 
The LEDs drop 3.6 to 4.2 volts.
Put the LEDs in pairs with one resistor for 2 LEDs. Now the pair will drop 7.2 to 8.4 volts. The resistor value will need to be changed. There will be 1/2 as many resistors in the box. (or use two 3W resistors so they will be at 1/2 rating)

7.2 to 8.4V for the LEDs + 1.4V for the ULN2803 so 8.6 to 9.8V across the silicon.

Much less resistor heat loss.
 
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I would need a shoe box for the circuit.
Better a shoe-box than a fire-brigade statistic :). If you can't connect the LEDs in series and don't have a SMPS then you have to dissipate a lot of heat somewhere.
 
Hi DerStrom8,

If only I could use the bigger cement type, but I need 16 of them and I would need a shoe box for the circuit. ;)

Hi Ron,

I did mention that I am switching them through a ULN2803 which drops the 13 volts by 1.4 or so and the LEDs are 3.6v, leaving 8v by 33r =242mA and amps at 1.96 I think. But the problem is that I need to run 16 of them too, all separately for patterns so I can't series them. :(

I am thinking at the moment about a switching supply maybe of about 4v at 6A. It would be more efficient but it would also be my umpteenth attempt at a SMPS with all the others being failures.

Any thoughts? Al

https://www.electro-tech-online.com/custompdfs/2012/08/lm2576hv.pdf 2 or 3 of these would do the job using smps I used a couple of these and they are incredibly simple. If you connect it up right, it can be used for a constant current source, but you're doing series, so I guess you wouldn't need that.
Here's my method of making it a constant current source.
 
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I would not run the LEDs strait off a car battery anyway. I would use a 5 or 6 volt regulater.
 
I would not run the LEDs strait off a car battery anyway. I would use a 5 or 6 volt regulater.

Car battery voltages can vary quite a bit, but if there is only one LED and a resistor, the current will not vary too much. A 5 or 6 V regulator will be more sensitive to voltage spikes, and a linear regulator with a smaller resistor will dissipate the same amount of power as a larger resistor if the LED current is the same.
 
and a linear regulator with a smaller resistor will dissipate the same amount of power as a larger resistor if the LED current is the same.
Yes but you will only need one regulator! and then you can use smaller resisters.
 
Linear regulator????

There are 16 LEDs at 240mA each.
3.84 amps
14 volts - 6 volts = 8 volts.
8V * 3.84 A = 30 watts.
 
Linear regulator????

There are 16 LEDs at 240mA each.
3.84 amps
14 volts - 6 volts = 8 volts.
8V * 3.84 A = 30 watts.
Ooops. That would be one big regulater. You could use 4 TO220 regulators or......
So I gess we are back to a SMPS?
 
Would this possibly be a way to do your project with minimal wasted heat, if I have the uln connected correctly, just do it twice
 
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I have fire hose and turnouts in my car, give me a call when it smokes :) I would go with a switcher.
 
Maybe the 3W resistors are less than 3W? I can only go on the vendors values. They certainly are the weak link in the otherwise sound circuit. See above post for reason I'm not using cement types.

They probably are - if they were smaller than that they would already have burnt-out - as others have suggested, you need heatsinked ceramic resistors.
 
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