LM317 Resistors power rating

Thread Starter

jacopo1919

Joined Apr 12, 2020
43
Hello,
I'm using a LM317 to output 5.10VDC out of 9VDC of the power supply.
The resistors would be 470R (R1) and 1k4 (R2),

I'm not sure regarding how the resistor power should be calculated (1/4W, 1/2W, 1W)? there are no informations on the datasheet and maybe you can help me to find a formula and understand why.

Thanks a lot,
J
 

dl324

Joined Mar 30, 2015
13,277
I'm not sure regarding how the resistor power should be calculated (1/4W, 1/2W, 1W)? there are no informations on the datasheet and maybe you can help me to find a formula and understand why.
The datasheet assumes you can calculate power dissipation yourself.

The voltage across the 470 ohm resistor is 1.25V.
\(P = \frac{V^2}{R}=\frac{1.25^2}{470}=3.3mW\)

The 1.4k ohm resistor will dissipate 10mW.

The set current is too low. The datasheet specifies a minimum output current of 10mA for the regulator to function, so you may need to apply a load to get the expected output voltage.
 

crutschow

Joined Mar 14, 2008
27,891
The resistors would be 470R (R1) and 1k4 (R2),
Typically R1 is set to 120 ohms to provide the minimum 10mA load current the LM317 requires to maintain regulation.
R2 would then be reduced accordingly to give your desired 5.1V output.

Based upon dl324's calculations, you should be able to calculate the power dissipated by both the new resistors.
 

Audioguru again

Joined Oct 21, 2019
3,632
Why are you using 470 ohms for R1 when the datasheet shows 240 ohms for the more expensive and lower current LM117 and assumes you can calculate using 120 ohms for an LM317?

1.25V /120 ohms= 10.4mA. Its power dissipation is 1.25V x 10.4mA= 13mW. The other resistor for a voltage across it of 5.1V - 1.25V= 3.85V will be 3.85V/10.4mA=370.2 ohms. it will dissipate 3.85V x 10.4mA= 40mW.
 
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