Power rating of resistors in LM317T voltage regulator circuit

I’m making a voltage regulator circuit with LM317T IC. Circuit diagram is attached.

(In the diagram 100ohm(fixed) = R1, 5.1k pot= R2)

1) I need the formula/equation to calculate power rating needed for resistors R1 and R2.

2) I want to use a 220 ohm (fixed) for R2 and 100 ohm variable resistor (flat, 1/4W probably ) as R1 to regulate voltage. Is it OK (according to formula it should work!!).

3) If I switch-on the circuit without connecting any Load, is there any problem, will it case harm to any part?

Note: My transformer provides 28.5V after rectification and smoothing.
My Load need 24 to 25V.

All looks good and sounds correct to me.

It wont hurt anything with power connected and no load.

Pete.

The power dissipation in the voltage setting resistors is but a tiny fraction of what will be dissipated in the LM317. R1 MUST be a fixed resistor, from 100 to 220Ω. The dissipation is higher at 100Ω, than it would be at 220Ω. If using 220Ω, P(R1) = E²/R1 = 1.25²/220 = 7mW.

To get ~24V out, R2 (the 5K pot) will have to be set to 4200Ω, so the power dissipated in it will be P(R2) = E²/R2 = (24-1.25)²/4200 = 123mW, so a 1/2W Pot will be fine.

What current will your load draw? Your biggest problem will be heatsinking the LM317. If your load is 1A, you will need a heatsink capable of dissipating I*E = 1* (28.5-24) = 4.5W, which will need several tens of square cm of heatsink.

polashd said:

I
Note: My transformer provides 28.5V after rectification and smoothing.
My Load need 24 to 25V.

Click to expand...

hi,
If after rect/smoothing you measure 28.5V off load this suggests that the transformer secondary you are using is 20Vrms.??

If thats the case you will not be able to get 24V regulated at 1A from the LM317, the regulator requires at least a 2V differential voltage across it in order to regulate.

I would estimate a Imax of 0.25A