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I was testing out a 5mm LED I had laying around with a 9V battery. I believe I left it connected too long and am pretty sure I burned out the LED since now it does not emit light. I forgot these little LEDs are rated for something on the order of 3 or 3.3V I believe.

I'm curious if I were to use a USB power supply of 5V would that also burn out the LED? Maybe not as fast as a 9V which seems to take them out within half a second or so I think is what I observed. How sensitive are these components to the supply voltage?

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    \$\begingroup\$ YES. What made you mistrust the specs? 20mA rated has 30mA absolute max so that's 50% tolerance. It's the effective current limiting that fuses the gold wirebond and chip. Not simply voltage. as all voltage sources have some resistance and 9V batteries are much higher R typ,. than 5V supplies, so the 5V would result in more current, like about 200mA with my rough calc. \$\endgroup\$
    – Tony Stewart EE75
    Commented Dec 19, 2020 at 6:01
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    \$\begingroup\$ LEDs die pretty much instantaneously if you don't limit the current to them with a resistor or something. Also 3.7 volts is pretty high for an LED unless it's ultraviolet. Red and yellow LEDs are around 1.7-1.8 volts, green is about 2 volts, blue is about 3 \$\endgroup\$
    – Hearth
    Commented Dec 19, 2020 at 6:01
  • \$\begingroup\$ The reason the 9V battery didn’t blow the LED immediately is probably because it is partly exhausted, increasing its internal resistance and reducing the current. If you use a 5V 1A (for example) power supply, the current will greatly exceed the rating of the LED and its life will be measured in milliseconds. \$\endgroup\$
    – DoxyLover
    Commented Dec 19, 2020 at 9:24
  • \$\begingroup\$ Next time before you do anything, research the subject a little. Learn how LEDs are used. This site has plenty of information on that. \$\endgroup\$
    – Bimpelrekkie
    Commented Dec 19, 2020 at 12:58
  • \$\begingroup\$ I actually didn't have the specs available. I was working to refurbish this old fiber optic lamp base for a Christmas Tree that no longer had its power supply around and all the LEDs appeared burned out. I replaced the LEDs and used a USB power supply last year and it worked for a while but by the next day or something the tree grew pretty dim. Then only one bulb worked faintly. I looked into replacing the LEDs last night and when I looked at the circuit I realized this thing must have been powered by some pretty low voltage enough to illuminate 3LEDs in parallel with a zener diode on the (+). \$\endgroup\$
    – jxramos
    Commented Dec 19, 2020 at 23:29

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A diode like an led is an unrestricted device and will allow as much voltage and current as put across it. If you drop 5V across an led designed for 3.3V @ 20mA, it will pull much more than 20mA and burn out pretty much immediately. Diodes have a very sharp IV curve so even a fraction of a volt above their nominal will cause it to burn out quickly. Even 0.1 volt will do that.

You almost always need a current restricting device for an led. Typically a resistor is used for low current/voltage leds.

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  • \$\begingroup\$ This is very illuminating, so indeed they are very sensitive to over voltage. I'm going to look into figuring out how to successfully drop 5V down to the LED's expected voltage. Hopefully a USB power supply will be stable enough to hold it at 3.3V with just a resistor voltage divider. \$\endgroup\$
    – jxramos
    Commented Dec 19, 2020 at 23:37
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    \$\begingroup\$ @jxramos Rather than using a voltage divider, you can use a single resistor to restrict the LED to the allowable current. If you know the LED's forward voltage and rated current you can apply the formula \$R = \frac{V_{supply}-V_{LED}}{I_{LED}}\$ (ref and online calculator). \$\endgroup\$
    – nanofarad
    Commented Dec 19, 2020 at 23:56
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    \$\begingroup\$ You DO NOT drop the voltage. Instead you limit the current with a series resistor or transistors circuit. You also DO NOT need a voltage divider, You need only one resistor in series with the LED. \$\endgroup\$
    – Audioguru
    Commented Dec 20, 2020 at 0:46
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    \$\begingroup\$ Yep. Leds are much like a resistor and will pull as much current as their quote unquote resistance allows. Basically a dead short without a current limiting device. \$\endgroup\$
    – Passerby
    Commented Dec 20, 2020 at 9:44
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    \$\begingroup\$ @nanofarad that resistor did the trick, this year I finally got around to setting up the little fiberoptic Christmas tree. The LEDs are holding steady with the 100Ohm resistor in series. \$\endgroup\$
    – jxramos
    Commented Dec 12, 2023 at 6:42

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