For the below circuit according to Ohm's law the led will get 23mA.
What happens if the resistance is removed? How many amperes will the LED get? My assumption is that the wire has some kind of resistance as well...but how much ?
Thanks
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4\$\begingroup\$ You forgot the voltage drop over the diode. See Why is the diode forward voltage constant? \$\endgroup\$– CL.Commented Aug 15, 2017 at 14:23
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2\$\begingroup\$ 0Amps, because the LED will blow up. \$\endgroup\$– uhoursCommented Aug 15, 2017 at 14:28
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1\$\begingroup\$ @uhours it depends. Some LEDs can get hot but keep going being powered by a 5V source directly. It also helps when the source is only capable to give 500mA or less ;) \$\endgroup\$– Volodymyr SmoteskoCommented Aug 15, 2017 at 14:29
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6\$\begingroup\$ So it will become a SED (smoke-emitting diode) :))) \$\endgroup\$– CrisCommented Aug 15, 2017 at 14:36
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2\$\begingroup\$ It will become a DED (dark-emitting diode), though when I did this by accident a while ago, the package blew a leg off, and I was left with a DEM (dark-emitting monode) \$\endgroup\$– Neil_UKCommented Aug 15, 2017 at 15:22
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1 Answer
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Current in that case would be (5- voltage drop across LED) /(Wire Resistance + Forward Resistance of LED) and will soon probably kill LED.
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\$\begingroup\$ The LED itself also has some resistance that depends on the current through it. \$\endgroup\$– uhoursCommented Aug 15, 2017 at 14:52
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1\$\begingroup\$ It seems the output pin of the AVR on that arduino typically has 25 ohm impedance aswell. electronics.stackexchange.com/questions/139305/… \$\endgroup\$ Commented Aug 15, 2017 at 15:00
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\$\begingroup\$ Plus output impedance of 5V supply \$\endgroup\$– Trevor_GCommented Aug 15, 2017 at 15:00
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\$\begingroup\$ Plus impedance of solderd joint, plus increase in resistance cause by increasing temperature due to positive TCR. xD am I still missing anything? \$\endgroup\$– DeepCommented Aug 15, 2017 at 16:02