I am making a device that doesn't use any batteries or outlet power. I need it to momentarily (<1 s) produce 5V DC or more after a bright light is applied to a hole with a 10 mm diameter. Due to the small diameter, 5 VDC solar cells are ruled out (they measure much more than 10 mm and don't produce 5 V if the light is applied only to an area of 10-20 mm). The receiving "sensor" can be placed anywhere from 0 to 10 mm from the hole. I have already solved the timing issue, with a 555 delay off circuit. Now I need 5.5 V.

I can't use batteries, so photodiodes are ruled out (they produce much less than 1V when they aren't powered, I would need about 15 and then they are too big to be illuminated through the 10mm hole). Solar cells are also ruled out given their inability to produce 5V with a 10mm window.

I decided to use LEDs as a voltage source using the photoelectric effect. With red LEDs, I can get spikes of up to 1.8 V per LED.

Here comes the question: If I wire four red LEDs in series, I can get spikes of 4.8-5.0 VDC. But if I wire a fifth led, the output voltage when the light goes through the window gets me spikes of only 1.5-1.8 VDC. Why?

All LEDs have been tested and they work fine. Separately, they produce 1.3-1.8 as expected depending on light angle. I have also tried to wire them in parallel and they give roughly 1.5 V as expected. I know the current is very low and it's not what I am asking. The LEDs are all wired in series. I know LEDs aren't the most 'efficient' way to do this, but the efficient ways are not usable in my application. Please stick to the question or refrain from answering if you have no idea

Why doesn't the 1.5 V of the fifth series LED get added to the previous 4.8 of the other series LEDs? All LEDs come from the same batch, same specs, same colour, they work, the wiring is OK, I have tried using different LEDs from the same batch and I always get the same result, I can add the voltage up to 4.8 V, then if I add more, the voltage drops.

Any ideas? Also, given the availability of LEDs, light sources and voltmeters, I would be very thankful it you actually test it for yourself rather than writing what you read in ChatGPT or what you just came up with. Test and then reply.

SOLVED: User @dandavis suggests it may be a case of the reverse leakage threshold being crossed.


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    \$\begingroup\$ Sounds like the 5th LED is backwards. Regardless, the current an LED can source in photovoltaic mode is very low (even compared to a photodiode which would be better), so while you can generate voltage you probably cannot extract usable power. \$\endgroup\$ Nov 18, 2023 at 15:02
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    \$\begingroup\$ Q! You say "spikes" . Why not steady DC. Please explain in question. Q2 What are you measuring with? What impedance? IF you get steady DC you can add capacitors per LED. With "spikes" timemconstats and energy may matter. || Solar cells can be cut cdimensionally small to meet your need. You seem more to be saying "With the components I am using I cannot ... " || A low voltage energy harvesting IC or circuiit will transform < 0.5V to eg 5V. \$\endgroup\$
    – Russell McMahon
    Nov 18, 2023 at 15:23
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    \$\begingroup\$ Blue LEDs should give higher V. White also. \$\endgroup\$
    – Russell McMahon
    Nov 18, 2023 at 15:24
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    \$\begingroup\$ KVL always works. If you don't get what is expected, something else has changed. \$\endgroup\$
    – RussellH
    Nov 18, 2023 at 21:00
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    \$\begingroup\$ @RussellMcMahon in my extensive testing, you'll actually get more juice from red than blue or white LEDs (especially "water clear" reds), and the most from IR. It's not the same as Vf if you're wondering (I had assumed the same), the junction converts all light with wavelengths below the LED's color into current/voltage. \$\endgroup\$
    – dandavis
    Nov 18, 2023 at 21:30

1 Answer 1


User @dandavis hits the nail:

The only other explanation is there's a reverse leakage threshold being crossed.

Voltage generation using 4 LEDs in series. Why does the voltage drop if I add a 5th LED in series?

I will also put here some of the useful contributions in case someone reading this is trying something similar:

  • Try covering the solar cell with a 1cm thick piece of styrofoam or soft plastic packing foam in front to act as a light pipe; once all the individual cells have some light, it will output its rated voltage, open circuit at least. Line the inside of the cavity with foil as well, or at least paint it bright white.
  • Red or IR leds produce the highest voltage, because the junction converts all light with wavelengths below the LED's color into current/voltage. It's not the same as Vf.
  • I have best results from slim rectangular LEDs that can be stacked. You can put some right-angle ones behind the other top-firing ones to get about 1.5 layers worth. You should be able to get 5v with 3 or 4 water clear red LEDs. Use the flat rectangular ones, about 2x5x7mm (not SMD).
  • Use the pump method to increase voltage.
  • If you can use batteries, then you can feed the output of the led stack to the base of an NPN transistor through a 100k resistor, and connect the collector to the gate of an SCR that acts as an SMD switch.
  • If you can use batteries, use photodiodes, they are much more efficient.
  • \$\begingroup\$ @dandavis makes valuable contributions and has experimented significantly so the following is NOT a slight in any way on his efforts. That comment is wrong 9it MAY be the cause),, and basing your understandings on someone's good but only perhaps-correct surmisen greatly risks trapping you in an incorrect assumption set. You have (now repeatedly) ignored other input and pursued an avenue that seems correct to you. Doing so in this case is a minor loss to you. Ifyou do the same in other areas you risk greater losses. \$\endgroup\$
    – Russell McMahon
    Nov 21, 2023 at 2:10

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