LM317 keeps blowing up

[wblock]

Where is the ground connection in that picture?

[KL27x]

On second thought, I find it curious that the LM317 is dying, but the LM7815 is not. Are these the same package size?

The LM317 is dropping less than half the voltage at 13V out as the LM7815.

[rdl]

It's kind of hard to guess what's going on. I read the entire thread and still I'm not sure what the circuit is supposed to be doing. I suspect the diagram provided is not an accurate representation of what has been built. That's why I suggest getting the basic regulator circuit to work first.

[jorenca]

same package, same heatsink.

Anyway, I was recording how the constant current configuration fails, and it magically did not!
I have now connected another LM317@10mA as a load and a 20R load at 1.25V and all seems good.

I guess I somehow short out the output when I'm measuring (because I have a tiny multimeter with tiny leads that are hard to use).

[rich]

worth noting that the heatsink tab is electrically connected to Vout, so perhaps shorted Vout tab to Adj when trying to measure current in your breadboard circuit.

same package, same heatsink.

Do you mean the same type of heatsink, or are they sharing one physical heatsink?

[jorenca]

same type, heatsinks are both isolated.

[macboy]

I'm not powering the regulator with the ADJ pin floating.

I tought I should make a 10mA current sink with another LM317 to attach to my power supply output just to have the minimum load satisfied.

I saw from the LM317 datasheet that I should use 120ohms to get about 10ma current.
I'm attaching a photo of my configuration.

However, when I measured if it was working (amps from ADJ to ground) it got fried even in this set up!!

I'm desperate now.

Where is the heatsink?
Where is the input capacitor?
Where is the output capacitor?
Are you even listening?

[KL27x]

However, when I measured if it was working (amps from ADJ to ground) it got fried even in this set up!!

1. Why are you trying to measure current between ADJ and ground?
2. How are you measuring current? Do you know how to use a DMM to do this? (There are a couple ways to do this and one way to blow fuses/circuits).

[rdl]

Try building the basic circuit I posted. It's nearly fool-proof, so if it doesn't work then you've probably wired it wrong or you may have bad parts. Use 120 ohms for R1 and a 1000 ohms (1K) pot for R2. Don't worry about measuring any currents, just check the voltage between ground and +V out. You should get about 1.25 - 11.5 volts.

[ZeTeX]

Build this:

the 240ohm resistor is so that it will always draw at least 10mA, if you are going to use anything higher, use a 120R [or less] resistor between the ADJ pin to the OUT pin.
Cadj is optioal, for testing you dont need it.
the voltage divder the diode and the capacitors are a must.
keep everything as neat and clean as possible, close as possible to the LM317 and use large heatsink, if the heatsink gets over 60C put a fan on just for testing, you dont want to burn yourself.
DO NOT ground anything, only use your multimeter set on DC volt and measure directly the output, no load. see if you can get the output voltage to change by changing R2.
a picture of your setup would be nice.
good luck! 

[Audioguru]

the 240ohm resistor is so that it will always draw at least 10mA

Ohm's Law says that 1.25V/240 ohms = 5.2mA, not 10mA. The schematic is wrong and shows the 240 ohm resistor that is used for the more expensive and less current needed LM117.
Most schematics on the datasheet from the inventor (National Semi) show the LM117 and 240 ohm resistor.

You might use a low input voltage and be lucky and get an LM317 that needs a low output current. Then it will work with a 240 ohm resistor. Luck is gambling.

[jitter]

Yeah, that's why you often see R1 halved to 120 Ohms (obviously R2 must be halved too) in real world applications that draw very little current. That way there's no need for an extra load resistor on the output.

[ZeTeX]

the 240ohm resistor is so that it will always draw at least 10mA

Ohm's Law says that 1.25V/240 ohms = 5.2mA, not 10mA. The schematic is wrong and shows the 240 ohm resistor that is used for the more expensive and less current needed LM117.
Most schematics on the datasheet from the inventor (National Semi) show the LM117 and 240 ohm resistor.

You might use a low input voltage and be lucky and get an LM317 that needs a low output current. Then it will work with a 240 ohm resistor. Luck is gambling.

oh yes, my mistake  . better use 120ohm resistor.