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This question already has an answer here:

I'm working on an RGB LED project. I'm using this as the power supply: https://www.sparkfun.com/products/114 It uses an LM317 which according to the data sheet can supply up to 1.5A. The LED's I'm using are these: http://www.adafruit.com/products/314

And the mains power supply I'm using is a 12v DC 2 AMP one.

Here's the problem. If I have just all of the red parts of the LED's lit up (I have 8 LED's), within about a minute the voltage regulator is far too hot to touch and I disconnected it when it started to smell funny. I attached a coin to the heatsink part of the voltage regulator (Not great I know, but I'm still waiting for some proper heat sinks I ordered to arrive) and within about a minute that too was far too hot to touch.

How can this be? Each red part of the LED uses 20mA, so multiplied by 8 that is 160mA. So the circuit isn't drawing too much current, is it? I'm using 150 Ohm resistors for each LED.

Why is it heating up so much, and what am I doing wrong?

Thanks

(Also, if I powered up all red, green, and blue parts of all 8 LED's, the voltage regulator gets blisteringly hot)

(I've just being reading about "switching" regulators being used as they produce less heat, would this be a better idea?)

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marked as duplicate by The Photon, user17592, Dave Tweed, Olin Lathrop, W5VO May 5 '13 at 17:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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A linear regulator dissipates heat proportional to the amount of voltage it must drop, and the amount of current flowing through it.

Input supply = 12 Volts

Option A:

Output Vout = 3.3 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R = 7.5 mA per LED = 60 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 522 mW

Over half a watt is being dissipated by the linear regulator in this case. Add the green and blue channels, and dissipation will tend to thrice that figure.

Option B:

Output Vout = 5 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R = 19.33 mA per LED = 154.67 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 1082.67 mW

Over 1 watt is being dissipated by the linear regulator in this case. 1 Watt is a substantial amount of power being emitted as heat. The regulator getting too hot to touch is not unexpected in this case. Adding the green and blue channels will make things much worse.

Recommendations:

  1. Use the breadboard power supply in 3.3 Volt mode until you have a good heat-sink attached, even 7.5 mA per LED channel should provide a reasonable amount of illumination from the LEDs.
  2. Switch over to a DC-DC switching regulator or buck regulator, those waste much less power as heat.
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Problem:
You power the LM317 with 12V and it generates an output of 5V @160 mA.
So 12V - 5V = 7V @ 160 mA that the lm317 needs to dissipate (heat).

P(watt) = I * U
= 0.160 A * 7V
= 1.12 W of heat

The datasheet of the LM317 says:

The Thermal resistance, junction to case is 5 °C/W
The Thermal resistance, junction to ambient is 80 °C/W (no heatsink)
so total thermal resistance is: 85 °C/W

Temperature rise is: 1.12 W * 85 °C/W = 95.2 °C
When the room temperature is 21 °C the case temperature will be 116.2 °C
The LM317 has protection so it starts to shut off when it gets to hot.

Recommendations:

You really need a better heatsink OR lower current usage OR a lower input voltage OR ...

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  • 1
    \$\begingroup\$ What does 12 - 5 equal? \$\endgroup\$ – Andy aka May 5 '13 at 0:36
  • \$\begingroup\$ Math at 3 in the morning =/ . \$\endgroup\$ – Spikee May 5 '13 at 10:19
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Yes, the LM317 can handle up to 1.5A, but when dropping 12V down to 5V, it needs to get rid of 10.5 watts worth of heat in the process! There's no way it can do this without a substantial heatsink attached.

Even at the 160 mA your red LEDs are drawing, the regulator is dissipating well over 1 watt. Its temperature will rise until the amount of heat being transferred to the air equals the amount of heat being generated. If there isn't much surface area over which that can happen, the temperature will quickly rise above the maximum rating of the device. The coin adds a little area (and a little thermal mass), but not nearly enough.

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  • \$\begingroup\$ Thanks for the reply, it's helped me understand what's going wrong. What would you suggest I do? I wanted to make a fairly small project, and having a large heatsink would make it bigger. \$\endgroup\$ – Lloyd May 4 '13 at 21:50
  • \$\begingroup\$ Two quick suggestions: 1) If your project has a metal case, you could try bolting your regulator directly to it (use short wires to connect it to the PCB if necessary). 2) Use a 3-terminal switching regulator, which will be much more efficient to begin with. \$\endgroup\$ – Dave Tweed May 4 '13 at 21:54
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The quoted maximum current the LM317 can supply requires that you provide a way for the chip to get rid of the heat it must produce to 'boil away' the voltage difference. You don't say what the output voltage of the LM317 is, I assume 5V. That means that each part LED that you turn on caused the LM317 to dissipate 0.02 * (12 - 5 ) = 0.14 Watt. A TO220 package without any additional heatsink can dissipate ~ 1 W (but it will get hot!). For 24 LED parts that is 3.4 Watt. That requires a decent heatsink for the LM317.

A heatsink can be calculated much like a resistor. You want to dissipate 4 Watt, over a temperature difference of let's say 40 degrees C. Hence you need a heatsink of (at most) 40 degrees / 4 Watt = 10C/W. That is not a very big one.

You did not give details about your circuit, but other options could be

  • using a power supply that outputs 9V instead of 12V, or even one that outputs 5V

  • if you use 2803-style open-collector drivers: feed the LEds directly from the 12V. This will not reduce the total power dissipation, but it will distribute it over all resistors.

  • use a small switching converter to produce the 5V

(I removed the suggestion to put the same color LEDs in series, because that is not possible with RGB LEDs, and Loyd seems to want to control the LEDs individually.)

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  • \$\begingroup\$ Thanks for the reply. I'm actually using a ULN2003 chip. Would you recommend I buy a 5V mains supply or use a switching converter? \$\endgroup\$ – Lloyd May 4 '13 at 22:02
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    \$\begingroup\$ A 5V power supply is likely more convenient, and cheaper. Any plug-in supply that hooks up to a USB plug (ie most phone chargers) puts out 5V, so you can generally find one for free. \$\endgroup\$ – Eric Gunnerson May 5 '13 at 2:47
  • \$\begingroup\$ @lloyd: I don't know your situation, so that's for you to decide. If you stick with 12V I suggest you power the LEDs directly from the 12V. If 5V is possible, go for that (smallest chance). \$\endgroup\$ – Wouter van Ooijen May 5 '13 at 6:08
  • \$\begingroup\$ Funny, I got two downvotes!? Please comment and state why! \$\endgroup\$ – Wouter van Ooijen May 5 '13 at 6:11
  • \$\begingroup\$ I think I will use a 5V power supply. I don't know why you was downvoted, I havent down voted any replies on this thread. \$\endgroup\$ – Lloyd May 5 '13 at 12:35

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