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I am looking for a way to power green LED (like GNL-3012GD) from MFC which provides 0.2mA @ 0.6V for about 1hr (there should be some drop off period then; 2..3hrs to restore). I understand that the amount of power is too small but what about the energy storage to drive the LED for a while (from a few seconds)? Any hints or schematics are highly welcomed.

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  • \$\begingroup\$ If you need about 2.4 V and 1 mA to power the LED, 0.2mA @ 0.6V for about 1hr will give you about 10 minutes at 50 % efficiency. But to convert only 0.6 V will be very hard. What about a MFC stack of 4 or more cells in series to get 2.4 V or more? \$\endgroup\$ – Uwe Jul 4 '18 at 20:19
  • \$\begingroup\$ @Uwe Unfortunately, I need to use only one cell, this is a restriction. \$\endgroup\$ – Mikhail Gaichenkov Jul 4 '18 at 20:22
  • \$\begingroup\$ For how long do you need to "power green LED"? What are the criteria of LED being "powered"? Is the naked eye visibility sufficient? \$\endgroup\$ – Ale..chenski Jul 4 '18 at 20:41
  • \$\begingroup\$ @Ali Chen The naked eye visibility could be sufficient. However, it would be nice to store the energy and power led on then for a few seconds. \$\endgroup\$ – Mikhail Gaichenkov Jul 4 '18 at 20:57
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    \$\begingroup\$ @Jaden mcd is a unit of measure for luminous intensity which is a better indicator of visibility than lumens which is total luminous flux. LED lumens is different than incandescent light bulb lumens. Lumens are measured isotropically where LEDs are anisotropic devices. Incandescent light bulbs are isotropic light sources. \$\endgroup\$ – Misunderstood Jul 6 '18 at 22:11
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Okay, here are some considerations.

  1. Your source has the energy storage of 0.6V* 0.2 mA * 1hr = 120 uW-h.

  2. If you get some newer LEDs, say LTST-C193TGKT-5A chip LED from Lite-ON, it has a fairly high efficiency, about 100 mcd at 5 mA., not the miserable 7 mcd @ 20 mA as for GNL-3012GD, or 50 times more efficient. I just run a test, so the light is pretty visible at 16 uA (micro Amp !!!), with forward voltage about 2.3V. This equates to power consumption at just 37 uW, while you can see the light with no mistake. If I take a proportion of 100 mcd and 16/5000, this is about 0.32 mcd of light, and there are many LEDs with 0.1 - 0.3 mcd light output. So the 0.32 mcd is pretty visible.

  3. Now, if you find some "energy-harvesting" IC operating at 500-600 mV, with output boost to 2.3-2.5 V, something like the module from Advanced Linear Devices, or LTC3108, the LTST-C193TG LED will be glowing (at 0.32 mcd) for up to 3 hours.

The hints here are (1) high-efficient LED, and (2) energy-harvesting IC. Would it fit your requirements?

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  • \$\begingroup\$ @ Ali Chen Wow! Brilliant to run the test with so efficient led! Thank you very much. The problemi is that I am asked to use GNL-3012GD only! (this is a requirement) \$\endgroup\$ – Mikhail Gaichenkov Jul 4 '18 at 21:38
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    \$\begingroup\$ The GNL-3012GD will likely emit the distinctive level of light at 1 mA (1/20 of 7 mcd = 0.35 mcd). At 2V it is 2 mW. If you manage to harvest the energy from your cell using a booster to 3 V into a 4700 uF capacitor, your LED will work for about 10 seconds. \$\endgroup\$ – Ale..chenski Jul 4 '18 at 21:52
  • \$\begingroup\$ What about the very new 18,000 mcd Brightek QBL7IG30C? \$\endgroup\$ – Misunderstood Jul 4 '18 at 21:56
  • \$\begingroup\$ @Misunderstood, the OP clearly stated that the junk GNL-3012GD is the requirement. \$\endgroup\$ – Ale..chenski Jul 4 '18 at 22:09
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    \$\begingroup\$ CORRECTION: 2 mW for 10 seconds is 0.02 J. You can harvest 0.02J from 0.00012 W source in 167 seconds, not hours. \$\endgroup\$ – Ale..chenski Jul 5 '18 at 0:47
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It may be done using no electronic DC/DC converter at all. Just use an electromechanic voltage multiplier.

You need 4 or 5 super capacitors connected in parallel by a multipole switch. Charge the capacitors for one hour. Then operate the switch by hand connecting the capacitors in series and to the LED. May be a resistor in series to the LED for current limiting.

The switch should have a pair of dual throw switches for each capacitor.

There will be no electronic losses, just the very small leakage currents of the capacitors and the switches. Electronic analog switches should not be used, their leakage current is too high.

There will be an RC time constant when charging and also when discharing the supercaps. During charge, the internal resistance of the MFC will be part of the R(C), not only the wiring resistance. The internal resistance of such a low power source may be quite high. About 3 RC time constants will be needed for a 95 % charge. If the capacity of the supercaps is too big, charging will take a very long time, but there is more energy stored to operate the LED longer.

Of course the effective capacity of 4 supercaps in parallel is 4 times the capacity of a single one. The effective capacity of 4 supercaps in series is 1/4 of a single one.

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  • \$\begingroup\$ Thank you, Uwe! This is very clear and easy to try. I guess that the supercap should be at 3.3... 3.7V (0.6V x 5 caps). What about capacitance? The caps will be charging via the wires of MFC ( a few Ohms), so it looks like we have RC constant. \$\endgroup\$ – Mikhail Gaichenkov Jul 5 '18 at 15:57
  • \$\begingroup\$ @MikhailGaichenkov, Per my calculations and assumptions about your LED (2V 1mA visible), you need total 5-8mF to run the LED for 10 seconds (CV^2/2). You will need 4 caps (0.6*4=2.4V and ~300 Ohm resistor for LED), so four 22mF supercaps should do the job. Plus-minus. \$\endgroup\$ – Ale..chenski Jul 5 '18 at 19:11
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UPDATE 7/17/2018
Ran across a Texas Instruments Ultra Low-Power Boost Converter With Battery Management for Energy Harvester Applications. It appears to be an excellent and well documented solution for your project.

bq25504 IC datasheet

Eval board for bq25504

End of Update


1 hour run time and 2-3 hr restore time is currently not possible.

Theoretically, with 100% efficiencies, it would take at least 16 hours to harvest enough energy to light an LED for 1 hour.

A harvesting module at this input power would be 25-50% efficient best case.

It appears the MFC cannot sustain the stated 0.6V @ 200 µA and needs few hours to charge its energy capacity to deliver 1.2 mW.

It may be possible if you have a couple of days to harvest the required energy.

You need more voltage. Voltage must be more than the LED's forward voltage.
You need more power, like at least 2 mW, you have 0.12 mW. Even if you were to boost the voltage there would not be enough power.
Harvesting the power would take more than 2-3 hours. You need 16x more power than you have.
Your LED is too inefficient. A 18,000 mcd QT-BrightekQBL7IG30C would do better but still not enough voltage or power.

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  • \$\begingroup\$ I see no reference to the total mWh provided by the power source. If boosting is possible, storage of energy in a device may be possible. i.e. can you show that a very efficient power harvesting approach will not collect enough power to light a very efficient LED for 1 hour? That is the work needed to declare 'not possible.' \$\endgroup\$ – Jaden Jul 4 '18 at 21:32
  • \$\begingroup\$ @Jaden how many hours would it take to harvest a 0.12 mW source to get 2 mAh? \$\endgroup\$ – Misunderstood Jul 4 '18 at 22:15
  • \$\begingroup\$ Why are you saying "A run time of few seconds with a 1 week restore time" when you calculated the duty cycle of ~1/16 ? \$\endgroup\$ – Ben Voigt Jul 5 '18 at 0:34
  • \$\begingroup\$ @Misunderstood I commented at the declaration 'It cannot be done'. I refer you to the accepted answer. I also encourage you to accept that there are probably things you do not know and to try to do things that others say cannot be done. Or, at least, try not to discourage others from looking at the possibilities. \$\endgroup\$ – Jaden Jul 5 '18 at 4:48
  • \$\begingroup\$ @BenVoigt that was based on the answer to my question posed to Ali Chen. It sounded high but he usually gets things correct. The ALD EH4295 like it may or may not do the job with a questionable best case efficiency of 25-50%. Also based on where the OP something about the MFC not being able to sustain the output for more than an hour:--- "should be some drop off period (2..3hrs to restore). However the MFC can provide .2mA at 0.6V during one hour."--- Not enough is known about the MFC. It appears the MFC needs 2-3 hrs to charge then can provide the 200 µA for an hour. New est. 2-3 days \$\endgroup\$ – Misunderstood Jul 6 '18 at 19:58
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Not sure what you mean... that is neither enough voltage or current to directly drive ANY led. You could possibly use a boost IC like This To boost the voltage up to power the led then slowly charge a capacitor from that and switch on the led once it is charged. I don't believe that the switching process could be automated as I believe that even a 555 timer would use up more power than that MFC could supply.

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  • \$\begingroup\$ But will the LTC3105 operate with only 0.2mA @ 0.6V at input? I found no information about operation with only 0.2 mA in the datasheet. \$\endgroup\$ – Uwe Jul 4 '18 at 20:45
  • \$\begingroup\$ Thank you. Agree, we can charge a capacitor from LTC DC/DC converter for hours. but how to switch the led on then and so on? \$\endgroup\$ – Mikhail Gaichenkov Jul 4 '18 at 20:49
  • \$\begingroup\$ What about a manually operated switch to connect the LED to the charged capacitor? If no manual operation, a fully mechanic clock switch for delay. \$\endgroup\$ – Uwe Jul 4 '18 at 21:01
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    \$\begingroup\$ The charge current after conversion will be much lower than 0.2 mA. Super capacitor have the high capacity and low leakage current. \$\endgroup\$ – Uwe Jul 4 '18 at 21:33
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    \$\begingroup\$ The LTC3105 has a diagram showing no-load input current at 0.2 mA at input voltage of 0.6V, figure 3105-G16, page 5. So it looks like the entire cell power will be wasted before producing any useful output. Probably the LTC3105 isn't up to this task, it needs something better. \$\endgroup\$ – Ale..chenski Jul 4 '18 at 22:07

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