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Max power output of an LM317 driver?

Flugufrelsarinn

Flugufrelsarinn

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I am looking at doing a 12X build. The only issue is I am not sure if I will be able to build my own driver that will work.

Would it be possible to build a 480mA driver with an LM317 if I use a combination of 2x8.2, 47, 9.1 and 100[ohm] parallel resistors as the resistance to set the LM317?

I am not sure how to calculate the power dissipated by the load resistors. Do I use the 1.25[V] V_ref value, or the 9[V] input voltage?


Sorry for my grammar, I typed this on my phone.
 





djQUAN

djQUAN

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for 480mA you'd need 2.604ohms

if you parallel all those resistors, you'd get 2.5969ohms for about 481mA but that's a lot of resistors to connect in parallel which won't look neat in a driver.

It would be simpler to use 2.7ohms and 68ohms or 3.3ohms and 12ohms in parallel. Both would be under 1% off of the desired value.

Also, 3.9ohms and 8.2ohms or 4.7ohms and 5.6ohms in parallel is still under 2% error from the ideal.

For the power calculation, use 1.25V as it will be the voltage across the resistors.
 
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The Lightning Stalker

The Lightning Stalker

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The voltage drop will be the same, but the current through each resistor will be different. To
get the current, you take 1.25 / the resistor's value in ohms = I. Then you take V * I to get
the wattage dissipated by the resistor. I always derate the resistor by 1/3 when using
through-hole unless they are on a heatsink. Otherwise they tend to leave nasty brown marks
on the board. Let's say you went with the 2Ω7 and 68Ω combination. For the 2Ω7, the
current through it would be 1.25 / 2.7 = 463mA. 463 * 1.25 = 579mW. 579 * 3 = 1737mW.
So this one should be at least 2W. Do the same thing for the other resistor to determine its
value.
 
The Lightning Stalker

The Lightning Stalker

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The Lightning Stalker said:
The voltage drop will be the same, but the current through each resistor will be different. To
get the current, you take 1.25 / the resistor's value in ohms = I. Then you take V * I to get
the wattage dissipated by the resistor. I always derate the resistor by 1/3 when using
through-hole unless they are on a heatsink. Otherwise they tend to leave nasty brown marks
on the board. Let's say you went with the 2Ω7 and 68Ω combination. For the 2Ω7, the
current through it would be 1.25 / 2.7 = 463mA. 463 * 1.25 = 579mW. 579 * 3 = 1737mW.
So this one should be at least 2W. Do the same thing for the other resistor to determine its
value.
Click to expand...

attachment.php
 
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