Heat Dissipation in a Constant Current Source

Discussion in 'Electronics Resources' started by sailmike, Sep 15, 2014.

Nov 11, 2013
147
3
I need to figure out whether the LM317 I choose for my project can handle the voltage and current. The LM317 has a junction to ambient of 70 C/W and a junction to case of 5 C/W. This is for the surface mount version. The pad for pin 4 can be made large to help with the heat dissipation and I have the room for that. The supply voltage will be a battery pack of 9 volts and the LM317 will be powering an LED that uses about 3.1 volts and 350 mA.

I tested the voltage drop across the through hole version of the LM317 when it was powering a LED of 6 volts and 100 mA and it was 4.5 volts. This was with a 12 volt source.

I just need to figure out whether the LM317 can handle the power and how to calculate it.

Thanks,
Mike

2. crutschow Expert

Mar 14, 2008
16,576
4,478
The dissipated power is simple to calculate. It's the input voltage (Vin) minus the LED voltage (Vled) times the LED current (Iled) or (Vin - Vled) * Iled.

Thus for a 9V supply, 3.1V LED, and 350mA current the power dissipated in the LM317 wuld be (9V - 3.1V) * .35A = 2.07W. This gives a junction temperature rise for the surface mount device with no heat-sink of 2.07 * 70C = 145C. For a 25C ambient the junction temperature would thus be 170C. Since that is well over the maximum allowed junction temperature you should add a heat-sink to lower the thermal resistance to air to less than half that or <35C/W (30C/W for the heat sink alone) or use a TO-220 version of the device attached to a small heat sink.

Nov 11, 2013
147
3
Thank you for the quick reply and clear explanation. Looks like another option is to lower the source voltage. I'm going to be using C or D batteries, so four batteries gives me 6 volts and a power of 1.015W to dissipate. This leads me to another question, what's the minimum supply voltage required for the LM317 to maintain 350 mA? There's also going to be a 7805 to supply 5V to various IC's, so I'll need to figure out the minimum to this too.

Thanks,
Mike

4. crutschow Expert

Mar 14, 2008
16,576
4,478
You can find that in the data sheets. I think it's around 2 to 3V drop across the regulator depending upon the current. For operation with a lower battery voltage you can use an LDO type of regulator which can work with a volt or so drop across them.