2
\$\begingroup\$

I've bought a couple of candle light flashing LEDs with integrated circuits from Octosupply (Datasheet: https://www.micros.com.pl/mediaserver/info-olc.5c5800m.pdf) for a project and attached one to a LiPo pouch battery (3.7V, 1200mAh).

The LED immediately burned out, but I don't quite understand why. According to the datasheet the LED should take anything from 2.5 to 5V and the battery should provide 3.7V. I haven't measured the voltage of the battery yet.

Can someone give me a hint why this happened?

\$\endgroup\$
10
  • 1
    \$\begingroup\$ See what Spehro said. The longer lead of two should be positive. How did you connect it? EITHER the datasheet is poor or the LED was reversed or faulty. \$\endgroup\$
    – Russell McMahon
    Nov 24, 2022 at 8:12
  • \$\begingroup\$ @RussellMcMahon I hastily connected it to test the LED, but I'm not sure if I reversed the leads. I'm used to LEDs where it is irrelevant, if they're connected backwards. I learned something today. \$\endgroup\$ Nov 24, 2022 at 8:21
  • 2
    \$\begingroup\$ MANY LEDs are damaged by reverse polarity if the voltage is high enough - typically in the 5-10V range. It's always wise to be careful. \$\endgroup\$
    – Russell McMahon
    Nov 24, 2022 at 8:49
  • 1
    \$\begingroup\$ Please don't play around with litium batteries. You basically shorted the battery. An LED having a Vf range from 2.5-5 V (very poor LED) does NOT mean you can drive it with any voltage between 2.5-5 V. \$\endgroup\$
    – winny
    Nov 24, 2022 at 10:13
  • 1
    \$\begingroup\$ In case you do manage to get one of these LEDs to work before they're all broken, please tell use whether they are red or yellow. The title says Red and 625 nm should be red, but then the “symbol” column says Yellow. I'ts an amazingly bad datasheet. \$\endgroup\$
    – wrtlprnft
    Nov 24, 2022 at 20:18

4 Answers 4

6
\$\begingroup\$

Perhaps you connected it backwards. This kind of part has an internal IC that should operate from a 3.6V battery directly, at least by my reading of the current datasheet downloaded from the manufacturer.

A reverse connection could destroy the CMOS chip. The longer pin should be (+).

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ The part numbers are OSR5MS5A31A and OSR5MK5A31A of the datasheet you and OP referenced, respectively. Not sure how relevant that difference is.. \$\endgroup\$
    – Velvel
    Nov 24, 2022 at 8:55
  • \$\begingroup\$ That is quite possible. I'm afraid I haven't considered that a reversed connection could damage the LED. \$\endgroup\$ Nov 24, 2022 at 8:55
  • 4
    \$\begingroup\$ I just tested it and it works just fine, with without an additional resistor, if I actually connect it correctly! \$\endgroup\$ Nov 25, 2022 at 8:14
6
\$\begingroup\$

The burned LED is what should be the expectation by connecting it directly to battery. You seem to have a misunderstanding about the LED.

The lithium battery won't be 3.7V. It will be about 4.2V when full, and about 2.7V when empty.

For 20mA current, the LED will need some voltage between 2.5V and 5.0V. It depends on manufacturing tolerance, batch, etc.

Therefore whatever voltage the battery gives out, it will either destroy some LEDs immediately because they would pass too much current with the voltage provided by the battery, or barely even light up some LEDs because they need more voltage the battery can provide.

\$\endgroup\$
1
  • \$\begingroup\$ I just tested it and it works just fine, with without an additional resistor, if I actually connect it correctly! \$\endgroup\$ Nov 25, 2022 at 8:15
5
\$\begingroup\$

According to the datasheet the LED should take anything from 2.5 to 5V ...

Correct.

enter image description here

So, if the forward voltage of the LED is only 2.5 V then a very large current will flow and destroy the LED. A current limiting resistor will solve this.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ It's a little confusing and lazy way to write the datasheet I think. Normally a component should not fry if you stay within the "absolute maximum rating", which they specify as 5 volts. \$\endgroup\$
    – pipe
    Nov 24, 2022 at 8:01
  • \$\begingroup\$ @pipe What do you mean by lazy? If you manufacture LEDs and measure what forward voltages you get at rated 20mA, then 2.5V to 5V is the range, so how else it should be given? And thus that is also the maximum voltage you should ever give to LED in constant current mode even if you wanted to drive more than rated 20mA. \$\endgroup\$
    – Justme
    Nov 24, 2022 at 11:56
  • 3
    \$\begingroup\$ @Justme A flashing LED can be expected to have some kind of integrated circuit and it's not at all uncommon for them to operate from voltage sources rather than current sources. This particular datasheet seems quite lazy and doesn't tell you the right way to connect it, so I'd like to emphasize the datasheet is what sucks here and the only mistake the asker made was choosing a product with a poor datasheet. \$\endgroup\$
    – user253751
    Nov 24, 2022 at 12:47
  • \$\begingroup\$ I just tested it and it works just fine, with without an additional resistor, if I actually connect it correctly! \$\endgroup\$ Nov 25, 2022 at 8:15
  • 1
    \$\begingroup\$ Interestingly, the data sheet gives a reverse current with the condition of 5V reverse voltage. I would have read that as "can be connected with up to 5V reverse voltage and will pass max 10uA reverse current". But it could also be that the bottom part is actually the data sheet for the "pure" LED, after the microchip, while the top table is for the microchip. No idea though, why they would do it like that. \$\endgroup\$
    – Dakkaron
    Nov 25, 2022 at 9:12
3
\$\begingroup\$

Did you put a resistor in series to the circuit for the current limitation? A battery provide 3.7V, and without a resistor which limit the current, the current will exceed the maximum rating of the LED. The circuit will look like this (with a current of 20mA): enter image description here

I extracted the value of the resistor from this equation: Rc = (Vbatt-Vled)/If

\$\endgroup\$
3
  • \$\begingroup\$ Thank you for the feedback! I'll test the connection with another LED and see if the problem was the missing resistor or if I actually connected it backwards, which apparently can damage those LEDs. \$\endgroup\$ Nov 24, 2022 at 8:56
  • 6
    \$\begingroup\$ This is not just a LED but a LED with some kind of built-in microchip. Many of these types of LEDs are designed to be connected to voltage sources and do not need extra resistors. \$\endgroup\$
    – user253751
    Nov 24, 2022 at 12:49
  • 1
    \$\begingroup\$ I just tested it and it works just fine, with without an additional resistor, if I actually connect it correctly! \$\endgroup\$ Nov 25, 2022 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.