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The goal is to demonstrate, visually (with a light of some sort) that power is begin generated, albeit very, very small amounts of power. What kind of set up could I use to light up an LED (or any visual indicator).

My power supply is incredibly small; about 0.1 volts and I'm not sure of the current, but consider it low.

I've heard of things such as a "joule thief" that could maybe allow me to light up an LED with very low voltage, but I believe that works by sucking current, which I don't believe I have enough of to do that with this very small power supply. Another option-- if my power supply is constant, but very, very small, can I use a capacitor somehow to store the small amount of energy trickling out of my power supply to then make an LED or other indicator light up (even just flash) once every few seconds or every few minutes?

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And I assume you can't just use a voltmeter for some reason? –  Rocketmagnet Feb 28 '13 at 20:13
    
I think a small lightbuld will be better here. At 0.1 volts the LED will not show anything. –  Gustavo Litovsky Feb 28 '13 at 20:21
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Might be able to use a boost topology, to get a higher voltage, charge a capacitor over a few minutes and flash the LED using the energy stored on the cap every few minutes... –  Kvegaoro Feb 28 '13 at 20:27
    
more than 1 mili Watt or less?? Also, consider using a green LED, as it generally has the highest lumens/watt. –  hassan789 Feb 28 '13 at 22:01
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"Low" is a useless spec for current. –  Olin Lathrop Feb 28 '13 at 22:16
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4 Answers 4

This is going to be very difficult at best. Transistor circuits don't work with just 100 mV power supply voltage. If you really need to run something from this low voltage, the circuit will need some initial external power to get going. This could charge up a cap to a few volts, which could then run a boost converter which would boost the 100 mV up to the few volts to keep itself going. This is of course assuming there is enough power available to not only run the circuit at the higher voltage after conversion losses, but to have enough left over to light up a LED to a noticable level.

Let's say you get the best efficiency LED you can find and that it is bright enough a 500 µA and drops 2 V in the process. That's 1 mW. Without conversion losses and power needed by the circuit, that would mean the 100 mV supply would need to source 10 mA. If your 100 mV supply can source a few 10s of mA, maybe you have a chance, but you'd still need some initial external energy to get the bootstrap process going.

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Heh, you just wrote, "power voltage" –  gbarry Mar 1 '13 at 1:08
    
@Gbarry: Actually I said 100 mV power voltage. This makes it clear I am talking about a voltage, so "power" in this case is used to describe which voltage. How would you have said it? I suppose "power supply voltage" would be more complete, but was what I said really ambiguous? In any case, I added the extra "supply". Is that good enough for you? –  Olin Lathrop Mar 1 '13 at 13:52
    
Your technical insight is VERY appreciated, Olin. This gives me much to think about. My background is not in electricity, so forgive my ignorance with this question: If a power supply trickles out a constant amount of energy, will a capacitor eventually store up enough energy to then release it at a given amount, if only for a very small amount of time? I'm still learning, a lot. –  Zach Mar 1 '13 at 20:47
    
@Zach: Yes, a capacitor can store energy slowly and then release it quickly. However, the circuit that will occasionally pump some energy from the 100 mV to the 2 V level still has to be powered at least at its sleep current level. If there isn't enough average power to go around, this scheme isn't going to work. –  Olin Lathrop Mar 1 '13 at 21:02
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LED will be hard; instead you can use the lcd off of a small calculator. very small amount of power required to "light" them. Remove the batteries (if any) and remove the solar cell. Connect your supply to where the solar cell connects.

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I think your only way out is going to be with a galvanometer. Perhaps you can "demonstrate visually" by blocking or reflecting a light source when the galvanometer moves.

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I had not thought of a galvanometer in this way, and I will certainly consider it. Thanks for your thoughts! –  Zach Mar 1 '13 at 20:53
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If you can squeeze just a little more voltage out of your plant-based power source, the TPS61201 boost regulator can step 0.3v up to 3.3v, which would be enough to light an LED.

As other have mentioned, you may want to charge up a supercap and discharge it into the LED every few minutes. You can get 555 timers that work at 3.3v and draw 30 µa.

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You can get a microcontroller that works at 3.3 V and draws far less than 30 uA. A 555 timer probably blows the whole power budget on its own. –  Olin Lathrop Mar 4 '13 at 21:28
    
@OlinLathrop, good point. More options for control with the micro. –  tcrosley Mar 4 '13 at 21:38
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protected by W5VO Mar 4 '13 at 13:50

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