Led Resistors reach 110C!!

[dons21]

I have been wiring up some 350mA RGB (3W) lights to go behind a friends bar, but the resistors are hot! Too hot I think. I measured the resistors at 110C each with all three colours turned on.

The Leds are power by a 13.80V 20amp power supply, I have been using 33 Ohm 4 or 5 watt resistors (not sure on the wattage)

Red 2.20v @ 350mA 33 Ohm
Green 3.55v @ 350mA 33 Ohm
Blue 3.55v @ 350mA 33 Ohm

now if the calculations are correct the resistors are dissipating 3450mW if I use a ten watt or 15 watt resistor will it be cooler? I think the best solution would be to reduce the input voltage to 5v or 9v, or use individual 1-2Amp mains adapters per Light also thought about using a old pc PSU.

Would like your thoughts ideas please.


Don

 

[ericgibbs]

I have been wiring up some 350mA RGB (3W) lights to go behind a friends bar, but the resistors are hot! Too hot I think. I measured the resistors at 110C each with all three colours turned on.

The Leds are power by a 13.80V 20amp power supply, I have been using 33 Ohm 4 or 5 watt resistors (not sure on the wattage)

Red 2.20v @ 350mA 33 Ohm
Green 3.55v @ 350mA 33 Ohm
Blue 3.55v @ 350mA 33 Ohm

now if the calculations are correct the resistors are dissipating 3450mW if I use a ten watt or 15 watt resistor will it be cooler? I think the best solution would be to reduce the input voltage to 5v or 9v, or use individual 1-2Amp mains adapters per Light also thought about using a old pc PSU.

Would like your thoughts ideas please.


Don

hi,
As you calculated, about 4-5 watts/resistor!.

Are the LED's all ON at the same time or do you want to select 1 of 3??

A lower voltage supply would be a better option.

 

[dons21]

The Leds will only be on all of the time if 'white' colour is selected, the only reason at the moment for using the 13.8v supply is that i already have it.

 

[OutToLunch]

good rule of thumb is to use a resistor that is rated for twice the amount of power than the average/rms maximum power that will be seen.

still, you're going to be burning the same amount of power, so there will still be a lot of heat generated - using a resistor with higher ratings won't change that. You could split the resistance up into a few different resistors. the same power will be dissipated, but it would be spread out.

 

[dons21]

First of all thanks for your advice.

Can someone verify my calculations or have gone about this the wrong way. I am trying to estimate the heat produced from a resistor.

What I do know is that after one hour of use my resistor reaches a temp of 110 Degrees C and stays at that temp, it is a 5 watt 33 ohm dissipating 3450mW.

3450/5000*100=69% so my 5 watt resistor is operating at 69% of its capacity. At 69% it is 110C

110/69=1.594 I.e the resistor is heating up 1.594C per 1% of use.

5 watt resistor 3450/5000 *100=69.00% ; has a temp of 110.00C
10 watt resistor 3450/10000 *100=34.50% ; estimated temp 54.99C
15 watt resistor 3450/15000 *100=23.00% ; estimated temp 36.66C
20 watt resistor 3450/20000 *100=17.25% ; estimated temp 27.49C


34.50*1.594 = 54.99C
23.00*1.594 = 36.66C
17.25*1.594 = 27.49C

I know that the calculations will not be exact but can I use this method to estimate the heat produced??

Advice is welcome

Don

 

[Pommie]

I doesn't works like that. A wire wound resistor at 110°C will happily tolerate that for a very long time. Either, accept the heat generated and place the resistor in an area that can handle it, increase ventilation, or reduce the voltage as suggested earlier. If you place a 100W resistor in the same enclosure as the 5W then it will reach the same temperature, it will just take longer to get there (assuming same power dissipation).

Reducing the voltage is by far the best solution.

Mike.

 

[Roff]

Here are my thoughts, and they may be wrong.

110/69=1.594 I.e the resistor is heating up 1.594C per 1% of use.

It appears to me that you are assuming a reference temperature of zero degrees C. There is nothing magic about zero degrees, except that is is the temperature where water freezes. If your calculation method has any value(?), maybe you should use your ambient temperature (25C?) as a reference.
How did you come up with 3450mW? I couldn't get that number using any of your voltages or currents. I get 4.08W for the resistor in series with the red LED.
The temperature rise of a package for a given power dissipation is a function of surface area of the package (radiated heat), and the heat conducted through the leads, which is a complex function of lead diameter, lead length, the size and geometry of the copper that the leads are soldered to, and the type and size of the substrate that the board is made of. And maybe the phase of the moon.

 

[justDIY]

1) use more leds, so you move the energy waste from heating resistors into heating leds instead.

2) use a switchmode driver to gain more efficiency

3) use a 5v psu instead of a 14v

 

[dons21]

I didnt think it would have been that easy. Lol I think i will have to use a the 5v from a computer PSU.

 

[blueroomelectronics]

5V @ 2A switchmode wall warts are available cheap, just check ebay.
**broken link removed**

 

[dons21]

just ordered two 5v 5A power supplies from ebay 12euros each

What is the formula to work out the amount of power dissipated from a resistor?

 

[OutToLunch]

P = R*I^2
or
P = (V^2)/R

 

[dons21]

Ok thanks what is ^??

Feel stupid asking

 

[blueroomelectronics]

^ = to the power of
**broken link removed**

 

[audioguru]

Your supply voltage is way too high so instead of heating resistors why not connect some of the LEDs in series and in series with a current-limiting resistor that is small.
Five 2.2V red LEDs in series are 11V. 1V with 350mA in it is 2.85 ohms and it dissipates only 0.35W.
Maybe the 2.2V LEDs are actually 2.4V so you want only 4 in series for 9.6V. Then their resistor is 6.86 ohms. It dissipates only 0.84W.

 

[justDIY]

^ = to the power of
**broken link removed**

bill's image, incase it's not clear, is a very handy constant current source for driving leds. it is somewhat smarter than resistors in that it adapts to changes in supply voltage, as well as changes in the forward voltage of the load (Vf)

the downside is it's even less efficient than resistors alone.

 

[Roff]

bill's image, incase it's not clear, is a very handy constant current source for driving leds. it is somewhat smarter than resistors in that it adapts to changes in supply voltage, as well as changes in the forward voltage of the load (Vf)

the downside is it's even less efficient than resistors alone.

How so? Are you referring to the fact that it requires about 4 volts of headroom?

 

[justDIY]

How so? Are you referring to the fact that it requires about 4 volts of headroom?

yes, you have Vdrop * If lost in both the resistor and the regulator

basically both parts (reg and resistor) get rather hot if operating at a decent power level

 

[blueroomelectronics]

It's not my design, just something I came across.
**broken link removed**
Also this one appears interesting as you should be able to switch R1 to control the LED
**broken link removed**

 

[Blueteeth]

Just my two cents...

A constant current source using low voltage (logic level) MOSFETS is good...I believe the schematic posted by Bill above uses one. That is of course if you can choose your power supply voltage to equal the LED's Vf + headroom. But the dropout using LV mosfets is only about 0.4V ish....If you want anything less you'll have to use a super LDO reg (I've seen one thats has a dropout of 100mV @1.5A!! )

If your power supply voltage is too high (add more LEDs?) or too low...you'll have to go into 'SMPS' territory - probably easiest to make one for LED's as there are many dedicated IC's specifically for that purpose, that is driving LED's efficiently in batt powered equipment, that don't like heat. Maxim-IC is always a gem, with their samples.

Not sure if that is what you're after...but I've just finished a horrific design for a customer where power requirements were ridiculous, so my mind is swimming with 'high-efficiency' (and there for no heatsinks needed) circuits

Good luck.

Blueteeth