enter image description hereI have a problem:

I'm supposed to produce bicycle lamps.

The LEDs should be white, and super bright as possible, use batteries A23 12V, with a resistor (I'm using 2 MOhms,) and last for around 48 hours.

The problem is that the LEDs using this configuration, lose brightnrss after 15 seconds (maybe lose 50-60% of the brightness.)

I have tried already with 3 different kind of LEDs, always with the same result.

By the way, I used the same resistors and same batteries 2 years ago (with leds that worked,) and now I can't find LEDs that actually work.

Basically I ran out of LEDs, but now I can't find the correct leds.

-What is your recommendation? - What are the correct leds?

Characteristics of the LED I am using and fail:

Tekniske detaljer:

  • Producent (varemærke): OPTOSUPPLY
  • Type diode: LED
  • LED-diameter: 10mm
  • LED lysfarve: Kold hvid
  • Lysstyrke: 8400mcd
  • Lysvinkel: 30°
  • LED-objektiv: Diffus, hvid
  • LED strøm: 20mA
  • Driftsspænding: 2,8 - 3.6V
  • Montering: Printhul
  • Front: Konveks
  • Benafstand: 2,54mm

Any recommendation? Sorry, Im not an expert, I am a bike mechanic. I used to produce these lamps, and they used to work.

*update: I uploaded a photo of what I am trying to do. I checked that the batteries basically die after few seconds, so I realize that the consumption is extremely high. I need new leds. But I dont know which ones. As I said before, with these batteries and resistors, hundreds of lamps have been produced in 2017 and 2019. I checked an old lamp, and it has exactly the same "set of colors" resistor, and same battery of course. So the problem is the leds. (or maybe the resistors can be the same color but be different too? I dont think that).

Here is a photo of the "official" lamp that has been produced for nearly 10 years. I dont have specifications of the original leds. They are very bright, and battery last for a loooong time. Maybe the "mcd" is lower? enter image description here

  • 1
    \$\begingroup\$ Did you change how the LEDs are cooled or what kind of heatsink is attached to them? \$\endgroup\$
    – The Photon
    Mar 18, 2020 at 16:37
  • 1
    \$\begingroup\$ Schematic? If that 2M is in series with the LED, I'm surprised it works at all. If not, what's limiting the current? \$\endgroup\$ Mar 18, 2020 at 16:41
  • 1
    \$\begingroup\$ Datasheet on the battery? \$\endgroup\$
    – user16324
    Mar 18, 2020 at 16:43
  • 1
    \$\begingroup\$ A circuit diagram so we are sure what you have? \$\endgroup\$
    – Solar Mike
    Mar 18, 2020 at 16:50
  • 1
    \$\begingroup\$ "I'm using 2 MOhms" - no, it's 22 Ohms. That would make the LED draw ~300mA at 12V, way too much for your LED and the battery. Do you have another working bicycle lamp to check the resistor value? Do you know the part number or specs of the original LEDs? \$\endgroup\$ Mar 19, 2020 at 14:21

2 Answers 2


THIS A23 battery?

To get 48 hours out of it, according to the datasheet there, you'd better only be taking about 0.5 mA from it, which is 2.5% of the rated current of the LED.

I don't think it's the LED you need to worry about, I think you need to change the battery. (although for road use, that seems like a very low power LED too) and 2 Megohms doesn't sound right either, whatever circuit you used. Where did you get this design from?

  • \$\begingroup\$ I also have doubts that one can lit leds for 48h continuously with such batteries. \$\endgroup\$
    – Fredled
    Mar 18, 2020 at 21:31
  • \$\begingroup\$ Hi Brian, thanks for your help. This is what I am doing(you can see a photo I just uploaded): I cant change the battery, it is designed with that battery (the lamps have been sold for many years and nobody complained). You are right, the batteries died after few seconds, seems that the consumption is too high. What would be the led that I can use? \$\endgroup\$
    – Pablo
    Mar 19, 2020 at 13:34
  • \$\begingroup\$ That LED. But with something more like a 22 kilohm resistor rather than 22 ohms if you really want 48 hours. Whether it's bright enough at that point is another question, More realistically, somewhere betwoeen 220 and 2kilohms to get the best compromise between battery life and brightness. Better : run 3 LEDs in series for 3x the light, only dropping about 3V across the resistor. That, with 150R (20mA) is about the brightest you can expect, and probably a few minutes battery life. Keep increasing R until it's only just bright enough and measure the life. \$\endgroup\$
    – user16324
    Mar 19, 2020 at 18:56

This is because the LEDs need exactly 12V or a little bit more than 12V to be bright. After a few seconds the voltage of the battery falls a little bit and the LEDs lose their brightness. Only one volt difference can lower the brightness by 60%. You either need a battery which gives more than 12V or Leds (or serie of Leds) which need less than 12V. And then chose to current limiting resistor accordingly. (It can't be 2MOhms. Sounds like we are not talking about the same thing) Idealy the voltage of the LED should be 2V less than that of hte battery. If the difference is too big you will waste a lot of energy through the resistor.

If you put 4 x 3V leds in serie, it won't work as soon as voltage falls bellow 12V. Moreso if the leds need a bit more than 3V. My suggestion is to mount 3 x 3.3V leds in serie (calculate the resistor so that leds use 3.3V at 11V) and the difference between the voltage of the battery and the voltage of the leds will be 2V at 12V and 1V at 11V, which is acceptable. When it falls to 11V you will still have 3.3V at the Leds. you will see a decrease in brightness, but much less than 50%. Maybe 20%. The battery can keep 11V much longer than 12V.

  • \$\begingroup\$ Hi Fredled, I uploaded a photo of what I am trying to do. I checked that the batteries basically die after few seconds, so I realize that the consumption is extremely high. I need new leds. But I dont know which ones. As I said before, with these batteries and resistors, hundreds of lamps have been produced in 2017 and 2019. \$\endgroup\$
    – Pablo
    Mar 19, 2020 at 13:42
  • \$\begingroup\$ @Pablo Is there only one led in the original lamp? (try to post a schematic of the original lamp if you can) If so, then the problem is that you connected a 3V Led where there should be a 12V led. If you put a 3V led on a 12V battery, with a resistor sized for a 12V led, you empty the battery very quickly. And it's very bad for the led too. You must find a 12v led or put 4 x 3V leds in serie, Or you must use a 3V battery (2x1.5V) with one 3V led, but then you will have 4x less light than with the 12V system, \$\endgroup\$
    – Fredled
    Mar 20, 2020 at 1:16
  • \$\begingroup\$ Hi Fredled, yes, there is one led in the original old lamps (same as the photo, 1 led, 1 resistor, 1 battery A23). Which led should I use? I see the led that I am using is 8400 mcd, is this high consumption? What mcd do you recommend for 48 hs use with a A23 battery with that resistor? Thanks! \$\endgroup\$
    – Pablo
    Mar 20, 2020 at 12:48
  • \$\begingroup\$ @Pablo You should either use a 12V LED or a battery which matches the LED voltage (or slightly more with the value of the resistor recalculated). Consumption doesn't depend directly from luminousity but from the number of amperes it draws. 20mA is very small. It won't shine a lot on 3V. but it will be much brighter on 12V system. If you put a 3V LED on a 12V battery and resistor for a 12V LED, your 3V LED was drawing a dangerouselly huge amount of current. It's even surprising the LED didn't burn. \$\endgroup\$
    – Fredled
    Mar 20, 2020 at 21:50
  • \$\begingroup\$ If the resistor is only 22 Ohm (almost nothing), then it's obviousely a 12V LED. On a 12V battery it makes sens to put a 12V LED. Thats's what you should do. \$\endgroup\$
    – Fredled
    Mar 20, 2020 at 21:55

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