Voltage regulator problem

[Muhammad Nasar]

I am trying to built  a AC to DC converter, using LM317 as seen in the image but the problem is that when i ever build this circuit the Potentiometer and reference resistor burns. I don't know what is the problem plz help me out.
Plz see the image down here.
http://img534.imageshack.us/i/image0003.png/

[Psi]

LM317 are designed for up to 40V DC input

If you start with 36V AC then fullwave rectify it, you will charge up your capacitors to 50V DC

36VAC * sqrt(2) = 50VDC (peak)
This is because the 36V AC is RMS.   It will be around 50V peak to peak and that will charge up the capacitors.

The LM317 is probably dieing from too much input voltage and once it's burned out the full 50V gets onto the output which burns your 120ohm resistor and pot (if the pot is set to a low value)

[red_mamba]

you will probably have to use a SMPS chip. A lot of them have up to 60V input voltage.
Which is a better choice anyway because it won't generate as much heat as LM317

[Muhammad Nasar]

Thanks people for helping me out

[Zero999]

Use the LM317 HV which works up to 60V.

Your schematic is confusing because you've drawn the transformer s centre tapped so it appears to be running at 18V.

By the way, the LM317 only sees the difference between the input and output voltage so you can convert 50V down to any voltage above 10V. Unfortunately you loose the short circuit protection with input voltages >40V because as soon as the output is connected to 0V, the LM317 shuts down, sees the full power supply voltage and smokes.

[Muhammad Nasar]

Transformer is 36-0-36 centertaped

[Alex]

Input voltage and type of 317 used aside, the 120Ohm resistor is too low for such an output voltage (around 50V I assume). When you use the wiper of the pot and place it towads the upper half, current through the 120 Ohm resistor and the pot increases too much. Increase the upper resistor and potentiometer, ensure that the pot will dissipate significantly less power than its rating, 100mW is typical. if you want to reach the upper end of the pot but you dont want to use a high upper resistor, add a resistor in series with the pot. This will limit your adjustment range though.

[Zero999]

the 120Ohm resistor is too low for such an output voltage (around 50V I assume).

The problem is, if the resistor is much higher, the LM317 might not regulate properly with no load. The worst case minimum load current is around 10mA so by using a higher value resistor than 120R you're taking a gamble, the output voltage could rise to an unacceptable level when unloaded.

The only solution is to ensure the pot is capable of dissipating the required amount of power.

[Alex]

I have been in the situation where the load was drawing less than 12mA only once with a valve amplifier's anode supply. I added a resistor on the output to provide the minimum current for effective regulation, i did not rely on the fragile (heating) resistors that set the output voltage .


mmmns, what adjustment range do you need?

[Zero999]

I have been in the situation where the load was drawing less than 12mA only once with a valve amplifier's anode supply. I added a resistor on the output to provide the minimum current for effective regulation

That solution is only viable if he wants to adjust the voltage over a narrow range. It won't work, if he's trying to design a 1.25 to 50V PSU because the 120R required for 10.4mA at 1.25V will pull 416mA at 50V and dissipate 20.8W.

Anyway, a 1.25V to 50V linear PSU isn't viable without using a multi-tapped transformer and tap changer relays.