I've made a device that lights an LED at a very low current (<10 microamps) as a very dim indicator for use in the dark.

At its simplest, the circuit is a 3V coin battery + resistor (300K-1M ohms) + LED.

I'd like it to turn off when there's ambient light in the room and, ideally, draw no power whatsoever to save the small battery it runs on.

The darkness detectors I'm aware of, however, use a transistor and photo detector to accomplish this (see https://www.buildcircuit.com/darklight-sensor-using-transistor/), and typically draw power in ambient light when the LED is turned off. These work at when running the LED at ~10-20mA, but just tinkering, I haven't been able to make these function at the low <10uA currents I'm trying to work with.

Any ideas on a better way to switch off a low current LED circuit entirely when light is detected?

Thanks --

  • \$\begingroup\$ One possible approach would be to only turn the photo detector on for a few microseconds every few seconds. \$\endgroup\$
    – Nate S.
    Feb 7, 2019 at 17:54
  • \$\begingroup\$ @NateStrickland But then you have to worry about the power consumed by the device that does the timing and controls the photo detector... \$\endgroup\$ Feb 7, 2019 at 17:59
  • 3
    \$\begingroup\$ Add a 10uA capable solar panel and let the LED run all day \$\endgroup\$
    – sstobbe
    Feb 7, 2019 at 18:25
  • 1
    \$\begingroup\$ You might want to compare the work of hackaday.io/project/11864-tritiled , which achieves multi year battery life with a small microcontroller powering a boost controller. \$\endgroup\$
    – pjc50
    Feb 7, 2019 at 19:28
  • 1
    \$\begingroup\$ @Misunderstood The OPs question was for a photo detector that consumes little power. A solar panel is exactly that during the day it consumes no power, in fact it provides power. I don't know what the OPs widget is, so I can't speak to its utility or use case. \$\endgroup\$
    – sstobbe
    Feb 9, 2019 at 1:42

2 Answers 2


At 10 microamps, you could just add a solar cell and a (very) low leakage diode. Any reasonable amount of light would then power your circuit instead of the battery.


simulate this circuit – Schematic created using CircuitLab

The very small current back into the battery can safely be ignored, and the bas116 has nA range leakage current.


Why is this battery powered? If battery is not required then a linear 3V supply would do well.

10 µA is a bit on the high side. You should be able to get down in the sub 1 µA realm.
Depends upon whether you want to illuminate the room or the LED just be visible across the room.

There is no ROI in turning off the LED when the room is not dark. You should be able to get the LED to last years on a single CR2032. A single CR2032 should last 3 years @ 10 µA.

This project claims 5 years with a single CR2032. And it uses a cheap low efficacy LEDs.
The highest efficacy LED today (Feb 2019) is a Samsung LM301B. This LED has an efficacy of more than 150% that of the LED used in this project. The LM301B is the LED you want to use. That should extend the 5 years to more than 7.5 years.

Using a flashing LED will extend the life span. A 200 mS pulse every second will increase battery life 5X or 37.5 years when combined with the LM301B. The afore mentioned project uses a PIC µController. Is the cost and amount of work to add a µController worth 5 batteries? Don't think so. And a boost converter is not needed because a lithium battery has a very flat discharge curve. Boost comes at a cost of reduced efficiency.

A 3V lithium battery and a resistor is all you need. Let it run 24/7.

To possibly make it more efficient use a step down voltage regulator and no resistor.
For an LM301B you would need a voltage of about 2.2V


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