I have a 5V schematic with 15n capacitor and I wondering if I will be able to blink an LED, powered from this capacitor. My calculations is that the capacitor will have about 0.2 uJ of energy. This I could convert to approximately 100 uSecs of 300 uA current.

So the questions are:

  1. What is the minimum energy (energy, not power, as I need to have a visible blink, not constant light) needed to make LED blink visible in ordinary office circumstances (OK, I will cover LED with my palm to avoid direct light exposure :-D)?
  2. Is it important how I will distribute this energy in time?
  3. Is there any color difference (maybe there are any human eye response difference)?
  • 1
    \$\begingroup\$ To see a led typically you need something on the order of tens of miliamps for tens of miliseconds - that's 4 orders of magnitude more then you say you can supply. \$\endgroup\$
    – jaskij
    Mar 10, 2016 at 15:15
  • 1
    \$\begingroup\$ upvoting as it's one of the few questions I've seen where the OP understands the difference between energy and power! Green is peak eye response, but not necessarily the LED with the most lumens/mW. Flicker photometers seem to suggest that all pulses < 10mS are equivalent, but I could be interpretting that wrongly, try a search for that term. \$\endgroup\$
    – Neil_UK
    Mar 10, 2016 at 15:15
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    \$\begingroup\$ Not directly related, but I saw an experimental report several years ago that the lowest-energy human visual response (in a "perfectly black" room) was achieved by a test subject who could reliably notice an emission of only (IIRC) 3 photons of red light. The applicability here, in my opinion, being that energy-for-energy, at similarl efficiency, a red LED will likely give your most reliable detection. With that said, absolute detection threshold depends on the individual human, and very much depends on the brightness, evenness, and color-balance of ambient light. \$\endgroup\$ Mar 10, 2016 at 15:34
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    \$\begingroup\$ @JanDorniak: That's a bit pessimistic. There are plenty of high-efficiency LEDs that produce a very noticeable amount of light on less than 1 mA. And a flash of 1 ms or less is still very visible. The final answer depends very much on the ambient light conditions, which the OP has acknowledged. \$\endgroup\$
    – Dave Tweed
    Mar 10, 2016 at 16:14
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    \$\begingroup\$ I got leds that are bright at 1 mA, and visible in a semi lit room at 100 micro amps. Angle is important but I'm sure I can get them down to tens of micro amps, and these are from a consumer garden led string. Nothing high grade. @jandorniak \$\endgroup\$
    – Passerby
    Mar 10, 2016 at 18:19

4 Answers 4


So, friends, I did the experiment.

The setup was two 5mm LEDs (I'm not sure what type exactly, but most probably they have 60 degs of light distribution and 40 mcd of maximum light - still didn't get the way they measure this intensity): red with 330 Ohm series resistor and green with 160. Both with 5V supply and AVR microcontroller.

With this setup I was able to see the blink as short as 1 us for green and 2 us for red LED. I should point out that I was in well lit room but I put my palms around the LEDs to make a 3 inches deep well around LEDs. I looked directly on the LEDs and I was expecting the blink. So this light is definitely not enough to notice the blink if you are not expecting one.

The current can be estimated as 3.8 Volts / 330 Ohms = 11,5 mA for red and 23 mA for green.

So the electrical power is 11,5 mAmps * 1,2 Volts = 14 mW for red and 28 mW for green.

Sequentially the blink electrical energy was as low as 28 nJ (nano Joules !!!) in both cases. Which is about ten times more than I expect to spend on a blink!

I test this on my wife and my 7-yo daughter. Same thing.

Regarding the energy distribution versus time:

Unfortunately I wasn't able to change resistors so I made just one thing: I put the LED to a constant light mode with 1% PWM. And I did not notice any difference if I change the frequency (1 us blink each 100 us is equally lit as 100 us blink each 10 ms). This is not exactly what I need but it looks like it's not a big deal how I will distribute the power in time.

Regarding the sensitivity of the different areas of an eye: I was able to see the blink only if I look exactly on the LEDs. If I shift the eye sight axis a little bit - I wasn't able to see anything. The same thing I noticed with constant lighting.

  • 1
    \$\begingroup\$ Joules are units of energy, not power. 1 Joule= 1 volt * 1 ampere * 1 second. I expect you know this since you calculated it correctly, but then your text describes it as "blink electrical power" which should be "blink electrical energy." \$\endgroup\$
    – JRE
    Mar 14, 2016 at 16:48
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    \$\begingroup\$ @JRE of course you are right! My mistype. I made the correction :) \$\endgroup\$ Mar 16, 2016 at 8:04

Wikipedia suggests something on the order of 100 photons to achieve a visible response in the most ideal conditions.

The energy in a photon can be calculated by:

$$ E = {hc \over \lambda }$$


  • h is the Planck constant, approximately 6.6×10−34 J⋅s,
  • c is the speed of light, about 3×108 m/s, and
  • λ is the wavelength of the photon.

The human eye's rod cells are most sensitive at a wavelength of 510 nm. So the energy of those photons is about 3.9×10−19 joules per photon.

Multiply that by the about 100 photons required for detection by a human eye, and you get 3.9×10−17 joules. By the law of conservation of energy, you will need at a minimum this much electrical energy to make anything visible.

Of course LEDs aren't 100% efficient. Not all colors have the same luminous efficacy, so it may be that the most efficient LED for making human-visible light isn't necessarily at the wavelength where the eye is most sensitive. I'll leave that research as an exercise, and let's just say an LED has a luminous efficacy of 25%. That increases the energy required by a factor of four, to:

1.6×10−16 joules

That is, by my rough estimation, the absolute minimum energy required to register a visual response with an LED in a human.

You have orders of magnitude more energy stored in your capacitor, so under ideal conditions, it's likely you could register a visual response even after accounting for inefficiencies in getting the power from the capacitor into the LED.

Of course in practice the room won't be perfectly dark, the viewer won't be ideally acclimated, and the LED's light won't be focused to a tiny spot. So you may require more energy. Perhaps much more.

  • \$\begingroup\$ It starts to sound interesting to wire up a relay to alternately charge a capacitor and then dump it through a typical LED, and see what the smallest capacitance that can yield a visible flash (for 5v or whatever charging) is. \$\endgroup\$ Mar 10, 2016 at 19:58
  • \$\begingroup\$ We've already established that the LED needs to be visible in "ordinary office lighting" with at most a hand providing some shade. \$\endgroup\$
    – Dave Tweed
    Mar 10, 2016 at 20:12
  • \$\begingroup\$ Is 1,5*10-17 should be 3,9 * 4 = 15.6*10-17 joules? I think that you need to divide the light energy with efficiency - not multiply. \$\endgroup\$ Mar 12, 2016 at 9:33

For some good hints about how to flash an LED on low power, look up the now-obsolete LM3909 LED flasher chip. Note how it "stacks" the voltage from the cap with the voltage from a 1.5V cell to get enough forward voltage for the LED.

One normally used a capacitor in the range of tens of µF (not tens of nF) with this chip to produces a very visible flash on a LED of only moderate efficiency. I would estimate that this supplied about 50 µJ per flash, so you're probably an order of magnitude or two short of where you need to be.


You might be able to get a visible (in office lighting, with some attention paid to optics to increase contrast) blink with 100ms of 30uA current.

There is no particular point in reducing the blink duration below about 100ms- usually LED efficiency won't be better at higher current and the eye will see the same energy as about the same brightness. A 100us pulse will appear about as bright as a 100ms pulse with 1/1000 the current, so more like 300nA-equivalant.

That might be visible with a good LED, dark-adjusted eyes and in a dark room.


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