I have just finished making my first power supply. But when I tried to use it the potentiometer was burned. I tried different fixed resistors and all the resistors below 10k ohm gets burned.
What seems to be the problem?
If you adjust the pot. to a 0-ohm setting, your 220R resistor would get force-fed over 4W (30V/220R=~136mA x 30V = ~4.08W) of chip-frying power!
It is recommended to always use a 'minimum value' resistor in series with the pot in a voltage regulator for this reason.
In your above circuit; adding 2 parallel 10K, 0.125W resistors in series with your pot should solve the problem.
Remember that the power rating is 1/8 W for the whole resistance track. If you are only using a fraction of the track then your decrease the power rating by that fraction. For your application it makes more sense to figure out the maximum potentiometer current allowed and design to half of that.
The solution: increase the 220 he LM317 maintains 1.25 V between Vout and Vadj. The current through the 220 Ω resistor to a safe value. 1 kΩ might be good. Scale VR1 and R2 to suit.
After burning five Potentiometer and lot of resistor, finally I got the solution. Read it and don't forget to teach the stupids around the internet, who servers the circuit of LM 317 just from the data sheet and there is no knowledge of practical exercises.
Here is your solution in just one modification.
Remove the capacitor of output voltage (C3). If you need, then use a diode (IN4007) in series with output voltage and after that place the capacitor.
I am using this circuit which is variable from 2.88v to 33v. There is no burning of Potentiometer. ☺
It stops condensed voltage input to output pin of LM317.
More Information: (Optional) 1. Use resistor R1 as 1k. According to data sheet 1k is the maximum resistor value. Yes, using a 1K resistor, output voltage can be set to the least output voltage. Use formula Vout = 1.25×(1+R2/R1). Units are Ohm.