# Variable output LM317 power supply circuit - potentiometer burning

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?     • Transformer used backwards, doing 10X step-up? :) – Kaz Feb 18 '16 at 22:35
• I think you might have a transformer problem. Measure the AC voltage out of with nothing connected and see if its much higher than expected ... – brhans Feb 18 '16 at 22:39
• The transformer is 220 to 24 and gives 24.2 ACV – user90575 Feb 18 '16 at 22:42
• What's your output voltage and the voltage on the Adjust pin if you remove VR1 completely and rely on R2 to set the output voltage? – pipe Feb 18 '16 at 23:07
• OFF TOPIC: fill empty spaces on the PCB with copper areas when making the PCB. Then much less copper needs to be etched away: just the gap between the tracks and copper area. It is faster and requires less chemical. – Kaz Feb 19 '16 at 1:51

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.

• Then, with their 5K resistance when the pot is shorted, the pair of 10K resistors would each get ~0.09W.
• By time the pot then reached a high enough setting (5K) to start passing them for Vdrop (and thus power dissipated), the circuit resistance (from the 220R resistor to GND through the pot) would be just over 10K, leaving only 3mA of current @ 30V, for a max load (neglecting any flow through the parallel 12K) of 0.09w; low enough to not fry your components anymore.
• The 220 Ω resistor won't have 30 V across it; the LM317 maintains 1.25 V between Vout and Vadj. The current through the 220 R will be $I = \frac {V}{R} = \frac {1.25}{220} = 5.7 \ \mathrm {mA}$. – Transistor Jan 17 '18 at 19:31
• Your potentiometer probably has a power rating of 1/8 W or so. For a 10 kΩ pot this gives a maximum current of $I = \sqrt { \frac {P}{R} } = \sqrt { \frac {0.125}{220} } = 3.5 \ \mathrm {mA}$.
• The LM317 maintains 1.25 V between Vout and Vadj. The current through the 220 R will be $I = \frac {V}{R} = \frac {1.25}{220} = 5.7 \ \mathrm {mA}$.
• When your pot is set to maximum resistance the 5.7 mA is divided almost equally between VR1 and R2 and the pot will be running close to its rated power dissipation.
• As you reduce the pot resistance it will pass more and more current and at 10% resistance it will be passing > 5 mA so the resistance track will overheat.

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.

1. Use a Diode within input and output voltage to protect reverse voltage.

Enjoy.

• Your suggested diode on the output destroys the voltage regulation as the voltage drop across the diode will vary with current. C3 will not cause any problems in the OP's design as D5 protects the LM317. Be careful about referring to "the stupids"; you may find yourself included. Welcome to EE.SE. – Transistor Jan 17 '18 at 19:28