Resistor Power Rating

[ibwev]

I am attempting to calculate the power rating of R1 and R2 on page 1 of the datasheet of the LM317 voltage regulator (https://www.onsemi.com/pub_link/Collateral/LM317-D.PDF). I will use 240Ω for R1 as suggested by the datasheet and 2.05KΩ for R2 to yield 12 volts.

The datasheet on page 7 reads "
the LM317 develops and maintains a nominal
1.25 V reference (Vref) between its output and adjustment
terminals. This reference voltage is converted to a
programming current (IPROG) by R1 (see Figure 17), and this
constant current flows through R2 to ground."

1) Would the power rating of R2 be as follows:
P=(Vref)(minimum load current)
= 1.25 X 0.01
= 0.0125
So use a resistor with a power rating greater than or equal to 0.025W (double calculated power rating or 0.0125+0.0125)

2) How do I calculate the power rating of R1?

 

[JimB]

Refering to page 7 and figure 19 of the datasheet:

For R1 Power = Vref^2 /R1 = 1.25 x 1.25 / R1

For R2 Power = (Vout - Vref)^2 / R2

JimB

 

[DerStrom8]

in order to determine the power rating you'll need for a resistor, look at the absolute maximum voltage and current that will be going through it. Then it is usually a good idea to choose a resistor rated for at least twice this value, which will handle fluctuations--not to mention the fact that the voltage is probably measured in RMS, so you will have peaks to worry about.

The power rating of the resistor is not critical provided it is high enough to handle the maximum power that it might have to dissipate. When in doubt, go for a higher wattage.

 

[chemelec]

For Just that Voltage Regulator on Page 1, I use 1/4 watt resistors, But even 1/8 watt resistors will work Ok.

If your making the Constant Current circuit, than Resistor Wattage is Important to determine.