I accidently applied around 5V to the output pin of the 555 timer.The output is zero when I ground the trigger, and an LED connected between 5V and pin 7 will not respond when 5V is applied to the threshold. Pin 8 is also connected to 5V. Is it damaged? Could it be something else?
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asked May 7, 2015 at 16:35
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1 Answer
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4
Sounds dead.
If you have another working chip you can set up a circuit with that chip (e.g. an LED blinker). If the circuit behaves as expected with that chip, then swap in the one you think is damaged and see if the circuit still behaves. If it doesn't then the chip is dead.
answered May 7, 2015 at 16:44
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\$\begingroup\$ Yeah the problem is that I don't have another one. And the circuit was working before on the breadboard, it's just after soldering it I made a mistake and connected the output pin to the voltage source and now it's not responding. \$\endgroup\$ Commented May 7, 2015 at 16:46
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\$\begingroup\$ That's why I need a definite answer whether shorting 5V to pin 3 will kill it or not. \$\endgroup\$ Commented May 7, 2015 at 16:48
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\$\begingroup\$ If it was working and isn't anymore in the same circuit then it's dead. It's entirely possible to fry the output if you connected 5v while it was trying to drive the out to ground. \$\endgroup\$ Commented May 7, 2015 at 17:04
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\$\begingroup\$ Is there any damage at all on the device case? Many devices these days are designed to take into account what could (within reason) go wrong, so they will gracefully fail rather than go up in a puff of smoke or worse. But you might see a bulge or crack or surface discoloration. You could also check pin-to-pin resistance to get a rough idea, but you still need a good device for comparison. But if pin 3 to both 5v pin or ground is open that would not be good. \$\endgroup\$– user3169Commented May 7, 2015 at 22:21