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schematic

simulate this circuit – Schematic created using CircuitLab

I don't know what I'm doing wrong. Perhaps someone can explain this. Is there any reason an ammeter would cause a voltage regulator to explode?

I'm trying to understand how to use the LM317 voltage regulator with a potentiometer to reduce 12 volts to a variable voltage between 1.25 and 10.5 volts. The 12V is generated from an ATX PC power supply, and the circuit I am using can be found here: http://www.electronics-lab.com/articles/LM317/

The only difference to the above schematic is that for R1 I am using a 680 ohm resistor, and R2 would be a 5k ohm potentiometer but I haven't even gotten that far yet. After building the circuit, I connected a multimeter to the output of the circuit and shorted what would have been the potentiometer (therefore simulating 0 ohms or full open) and read 12.5 volts. It read steady for several seconds, so I switched from volts to amps on the multimeter, and POW! The LM317 was no more.

Perhaps there is something fundamental I am missing, but what would cause this? Are my calculations (using the above website as a reference) completely wrong? If so, what would I need to do to accomplish my goal of taking a 12V supply and using a potentiometer to vary the output in as wide of a range as possible? Would there be a benefit to having a fuse (perhaps 1 amp) on the input voltage?

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You connected the meter how? Suggest you draw a schematic of your circuit, with meter included. Press ctrl-m in the edit window or click the button in the toolbar that looks like a schematic. –  Phil Frost Feb 1 at 3:58
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Fuses are always a Good Thing. –  Peter Bennett Feb 1 at 4:00
    
Thanks for the reply again Phil. I've updated my post with a schematic of the circuit I am using, and how it looks in place with the multimeter. At the point labeled "5K POT", currently this is just connected as-is with no potentiometer. As the answer below explains, it appears my silly use of the multimeter to test amperage without any other load was probably the cause, however as he also pointed out, I don't understand why I read 12.5V instead of the 1.25 V I expected to get. Thanks for the feedback. –  Mathew Tate Feb 1 at 4:12
    
I'd guess there is a good chance that you have got the LM317 pinout wrong and are eg connecting V1 to output. That would explain both high voltage out and destruction. When used correctly the LM317 is relatively hard to destroy. Shorting Vout to ground would dissipate 12+ Watts in the regulator and it would go into thermal shutdown. Check data sheet for the package you are using. Some of the small pkg versions may have differing pinouts to most. –  Russell McMahon Jun 8 at 11:56
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3 Answers 3

up vote 3 down vote accepted

If you had a short in place of the potentiometer, you should have read 1.25 volts (or so) at the output of the LM317, not 12.5 volts.

When you changed your meter to read current, presumably moving the red lead to the "10 Amp" socket, the meter would have very low resistance. When you then connected the meter between the LM317 output and ground, a very large current would flow, apparently causing the LM317 to explode.

If I recall correctly, the LM317 is spec'd to deliver 1 amp, with due regard to voltage drop, power dissipation, heatsinks and phase of the moon - that only states the current that it can safely carry under suitable conditions - it DOES NOT limit the current to that value!

The current measurement function of a multimeter should only be used with the meter connected in series with a load whose current drain you wish to measure. It must NEVER be connected directly across a power supply.

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Thank you very much. Apparently I am not well versed enough to understand how measurement of amperage works, and this would explain it. As you pointed out, I'm not sure why I didn't read 1.25V. I suppose I still do not understand this component well enough. I will continue my research, but if you have any idea why this would happen please let me know! –  Mathew Tate Feb 1 at 4:14
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An LM317 should limit the output current to no more than 3.4A, and it has thermal shutdown. Something else is going on. –  Spehro Pefhany Feb 1 at 4:16
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A short at the output can't justify damage to a regulator having internal current limiting, thermal shutdown, and safe area compensation. Also The short circuit current is limited to less than 3A, see figure 6 in datasheet. –  alexan_e Feb 1 at 9:49
    
Should go safely into shutdown. probably wrong connections. –  Russell McMahon Jun 8 at 11:57
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It's possible you connected the LM317 backwards so that it was conducting as a diode. That's why you got about 12V out rather than 1.25V. Then when you shorted the output by turning your meter to amperes, it died a horrible death.

Ammeters must only be connected in series with a load! Never parallel.

This is a view of the package from the TOP (with the part number markings visible).

enter image description here

Suggest you stay away from reading current completely, especially with that meter that has an unfused range. I hope you were wearing safety glasses.

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I wanted to follow up after some more testing. Thank you all again for the feedback, as it has helped to pinpoint the issue and I believe the circuit is working as expected now.

The first problem was, as Peter said, that I had no load on the circuit as I was trying to read amperage. I now understand that this is essentially creating a short which caused the disastrous results.

Secondly, I had a similar issue yesterday when attempting a similar circuit; in that case, as Sphero guessed, I had read the pinout incorrectly. In today's case, however, I did have it wired properly.

Third, it seems as a recurring theme, user error was to blame. When I said I believed I was reading 12.5 volts on the voltmeter during the first test, in fact I now believe I was actually misreading the decimal and was probably reading 1.25 volts instead. After reconstructing the circuit again and this time using a CPU fan as a load, testing voltage with a 0 ohm (short) in place of the potentiometer gave a reading of 1.25 volts, and replacing that short with a 4.7k ohm resistor to simulate the 5k pot I plan on using, read an output of 10+ volts and allowed the CPU fan to run properly.

Thank you all again for the input, and I hope anyone with similar issues can take away some lessons as valuable as those that I have throughout this post!

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Note that not having a load at the output may have (under certain conditions) side-effects making the device unable to maintain regulation. The regulator needs a minimum load current to maintain regulation (12mA max). Normally you should choose the output to ADJ resistance value to satisfy that requirement (as a load) so that the output voltage is constant even when the output current drops. In order to calculate the resistor use 1.25v/0.012A= 104Ohm so 100 Ohm is fine –  alexan_e Feb 1 at 10:12
    
This is interesting. If I were to, say, have an in-line voltmeter connected to the output terminals, would that satisfy such a load? Ex: amazon.com/gp/product/B00AR7ON1S/… This device states that it uses 5-15mA, and I was planning on having it wired to the output anyway. –  Mathew Tate Feb 1 at 16:22
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The min output current of a random LM317 can be anywhere from about 3.5mA to 12mA (typical and max value in the datasheet) but it is a good practice to assume the worse so you should always expect 12mA (some datasheet mention 10mA). The 680 Ohm resistor you have used sinks about 2mA so if the meter you refer to consumes another 10mA then you will be fine. I don't know the exact consumption of that meter and it should also vary somewhat. –  alexan_e Feb 1 at 16:35
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