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I was trying to make 1A dummy load using lm317 (didn't intend to use it,just trying!). So, I connected the circuit as shown :

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I put 1 ohm resistor as a load and another 1 ohm resistor instead of R1, when I connected the circuit with 5V, the voltage across the load resistance was about 1.6V (which also means 1.6A). I wanted to try a 24V supply as the input but when i did, it exploded (The meter read about 6 amps before the explosion). what could possibly be the reason? here's a photo of the poor IC after the accident..

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It was tied to a large heat sink though...

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  • \$\begingroup\$ LM317 is voltage regulator, not current regulator. And you said it is 1A (datasheet says max 1.5 A) and you wonder why it exploded at 6 A... it simply heated so quickly from within the heat sink was unable to get the heat within such short timeframe. Device exploded from within because you exceeded its abilities (specifications). \$\endgroup\$ – Anonymous Oct 27 '16 at 21:26
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    \$\begingroup\$ @Anonymous According to this datasheet onsemi.com/pub_link/Collateral/LM317-D.PDF page 9, figure 26, what the OP has should be a current limiter. which should have limited the circuit to ~1 Amp (Iout = 1.25/R1) \$\endgroup\$ – ambitiose_sed_ineptum Oct 27 '16 at 21:33
  • \$\begingroup\$ But for some reason whole device is called "1.5 A Adjustable Output, Positive Voltage Regulator". While it says that it has "internal current limiting, thermal shutdown" and "Internal Short Circuit Current Limiting Constant with Temperature", there're always gray areas not covered or missed by the developers, when device still thinks it is not short circuit and its current limiting does not work leading to almost immediate damage. \$\endgroup\$ – Anonymous Oct 27 '16 at 21:41
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    \$\begingroup\$ How large was the heat sink, and how well the IC was attached to it (thermally)? \$\endgroup\$ – Ale..chenski Oct 28 '16 at 2:28
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    \$\begingroup\$ @anonymous. A LM317 in this configuration becomes a constant current regulator, whatever the name they gave to the device. And the overcurrent/short circuit protection mechanism has nothing to do with it. Ask a new question here, we'll explain why... \$\endgroup\$ – dim Oct 28 '16 at 3:14
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It's fairly likely you mixed the leads up, or it wasn't an LM317 to begin with. Note that the pinout is different from that of, say, an LM340 type of regulator.

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At 24V the device should have been dissipating around 24W which is a lot of power, but might be okay with a large heat sink.

6A indicates that it was not working correctly, and the power dissipation would have been very high.

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  • \$\begingroup\$ I know that and i was paying attention (I really hated that pinout), but I think you're right... \$\endgroup\$ – iMohaned Oct 27 '16 at 21:56
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Constant current supply is one thing but it should not be mistaken to be short circuit protection. The device under the circuit condition was working as a constant current source with 3.4 Votls circuit, under constant current of 1 ampear and within the power limits of the device. When you apply 24 volts, it had to do double function of dropping the regulator voltage to 3.4 volts dissipating 20.6 volts and also drawing additional 1 amplears. In this case the load impedence is not matched with the source impedence and hence a short circuit occured and all protection function of the device collapsed, letting ohms law to take it's toll !

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