LM317 can someone help explain Waste heat/energy

I have known that the LM317 as a current limiting regulator gets damn hot!!!
so there is alot of waste energy

how do I minimize that?

I now want to use the LM317 to get a precise voltage. of 19.2v

can you help me know more about what waste energy would be if my device used 17v @ 300ma

would the lm317 get hot?

the input voltage to the LM317 would be from 21v-28v

The main Idea is, The special High tech PSU for LED lights has input volts from 7v-24v
I am scared to over power this module, dont want to pump 26-28v into it
so I want to drive 4x LEDs @ 300ma requires 17volt this psu if upto 78% efficient AND MY amp meter shows high then that! the efficiency is great

EDIT: Or is it possible to CLAMP a voltage at the input of my MAIN device.....?
I dont want more then 24v to go in... because of solar the voltage dips and raise

Simple DC circuits: Power dissipated, be it a resistor or an LM317 is E*I. If you are starting with 28V source, and want to drop it to 19V, there is 28-19V = 9V across the regulating element, be it a resistor or a LM317. If the current to the load is 0.3A, the power being dissipated is 0.3A*9 = 2.7W. Nothing you can do about this, short of using a Switched Mode Power Supply (Google "SMPS").

Now what you can do something about is getting rid of the heat?(Google "Heatsink"). To keep the LM317 from protecting itself and shutting down due to excessive temperature, you have to keep it's tab from exceeding about 85degC. Starting from a room temperature of 25degC, that means you can allow a temp rise of 60degC, which means the thermal resistance of the heatsink must be less than 60degC/2.7W or 22 degC/W. Look at these clip-on heatsinks] available from DigiKey.

MikeMl said:

Simple DC circuits: Power dissipated, be it a resistor or an LM317 is E*I. If you are starting with 28V source, and want to drop it to 19V, there is 28-19V = 9V across the regulating element, be it a resistor or a LM317. If the current to the load is 0.3A, the power being dissipated is 0.3A*9 = 2.7W. Nothing you can do about this, short of using a Switched Mode Power Supply (Google "SMPS").

Now what you can do something about is getting rid of the heat?(Google "Heatsink"). To keep the LM317 from protecting itself and shutting down due to excessive temperature, you have to keep it's tab from exceeding about 85degC. Starting from a room temperature of 25degC, that means you can allow a temp rise of 60degC, which means the thermal resistance of the heatsink must be less than 60degC/2.7W or 22 degC/W. Look at these clip-on heatsinks] available from DigiKey.

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truly thanks for the explaination...

heatsinks.. yeh NP... I got heaps of them at my wifes shop!!! all shapes & sizes...

unfortunatly the LED is around 3v and input is 12v so thats around 9v ont he element. & 700ma wow!

again thanks for that

I think it is quite strange that it should produce that much heat. I mean, it is supposed to produce an adjustable 1.2-28V so how should 17-20V be too hot to handle? I'm using one, but only with a 9V input.. so no heat.

Vizier87 said:

I think it is quite strange that it should produce that much heat. I mean, it is supposed to produce an adjustable 1.2-28V so how should 17-20V be too hot to handle? I'm using one, but only with a 9V input.. so no heat.

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Do the math! Then go get a 3W or 5W or 10W resistor, and power it up to the rated wattage. Now hold it your hand. You will remember what 5W feels like.

Hi,


Yes, the LM317 and similar type regulators may dissipate a lot of power simply because they are a linear device, and a linear device can not perform a true power conversion...only a switcher can do that.
For many types of loads however the current goes down when the voltage is adjusted lower so the device does not overheat as easily.
For a 23v input and 20v output at 1 amp that would mean 3v would be across the LM317, and the power therefore would be 3*1=3 watts. If it is a resistive load and we turn the output down to 10v, the current would decrease to 0.5 amps, but now we have 13 volts across the LM317 so that means we have 13*0.5=6.5 watts for the LM317 to dissipate. That's not as bad as if the current stayed the same however, because with 23v input and 10v output at 1 amp we would have 13 watts dissipated in the LM317.

A power resistor is 150 degrees C at its rated power. Don't go near one and keep it away from the pcb and plastic capacitors and wire insulation.
That is its max allowed power dissipation so use half the power rating for less heat.

Hi again,


Audioguru i am glad you brought that up as i forgot to mention something i meant to say in my previous post.

That is, sometimes a small fan helps quite a bit even with the power resistors. We cooled a 10 watt power resistor down quite
a bit with a very small fan (i think it was an old CPU fan). It was cool enough to touch easily with the fan but without it the
resistor could not be touched. It was placed about an inch away. Just have to remember if the fan dies the resistor heats
up so some method of detection might be a good idea too when using a fan like that.
It's quite amazing how much cooler it runs with a small fan.

Of course immersing it into a vat of distilled water works even better as it distributes the heat out to a much larger surface area.

MikeMl said:

Do the math! Then go get a 3W or 5W or 10W resistor, and power it up to the rated wattage. Now hold it your hand. You will remember what 5W feels like.

Click to expand...

Oww!!

It's quite amazing how much cooler it runs with a small fan.

Click to expand...

I'll keep it in mind.

A quick (novice) relative question - the LM has overheat protection. Is this temporary ie it will work again once sufficiently cooled or is it permanent like a fuse?

Tipsy said:

A quick (novice) relative question - the LM has overheat protection. Is this temporary ie it will work again once sufficiently cooled or is it permanent like a fuse?

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It protects itself to live another day. However, while in thermal shutdown, the output voltage is all over the map.

Hi again,


Yes, haha, when it goes into thermal shutdown the voltage may go to zero or close to it, but yes, it does recover once cooled down.