Slow build up, sustained release (Flashing Diode)

I am a total beginner with an idea of what I want to do but not sure how to achieve it. My problem is similar to the flashing diode scenario. Except that I cannot rely on a regular source of power for input.

Considering my input V may only be 0.1-0.2V, I need an output power of 1.5V that will last at least 5s.

So I am thinking I need a solution to take the trickle of input power and store it, until it reaches a certain threshold whereby it will be automatically discharged. Therefore lighting the diode for a few seconds before the stored charge is depleted, and then the charging cycle can commence again.

I hope that makes sense? And help would be greatly appreciated.

Ideally the charge would need to build up for not more than 6 hours before being automatically discharged over 5-10s and then repeating the cycle.

by craiggoldy
February 05, 2021

0.2V is very low, you can't really think about a circuit based on a diode (P-N junction). If your input allows it, you may be able to use a transformer, but even that is not guaranteed to raise your voltage to an acceptable level. Sure an radio antena could be boosted, but that requires an external battery too, out of question for your project. The only circuit that could be similar to yours is a RFID passive tag, but those are generally designed to a very specific frequency. So, the first problem is about boosting the voltage. But that is far from being the only problem.

Next, for the LED, you need 15 mA for 5 seconds, that makes 75 mC (milli-Coulomb) for a 100% efficient output. If you use a capacitor, that means Q = C * V = 0.075. For a final voltage at 1.6 volt ( Vthreshold for a red LED), that call for a capacitor of 0.05F. That would probably ends up by using a super cap. But the problem with those is that they are not linear neither do they allow a high voltage. Could be 2.5 volt max. So, even if we start with a fully charged cap at 2.5 V and expect to still have 1.6 volt after 5 sec, that means: 1.6 = 2.5 exp(-5/RC), assuming a linear behavior, leading us to R*C = 11.2. At 2.5V, limiting the current at around 20mA, that also implies R > 39 ohm. So, C > 0.3 F.

Now, consider the energy. Assume that we take a super cap of 1F (margin of "safety" of 3 ), at 2.5V its energy, stored, is given by 0.5 C * V^2, or 3.125. Since the efficency is, at best, 50%, we need, to raise to 2.5 volt when starting at 1.6 volt (already stored), the required energy havested each 6 hours is (2.5^- 1.6^2) = 0.2 volt * i * 21600. The 21600 is the number of seconds in 6 hours. That means that you need a constant current of at least 850uA. And if you have intermittent current, that would be more than that. Since it is a consideration of energy involved, there is no much alternative to it.

I suspect that your project if probably ... doomed to fail. Sorry to be the messenger of bad news.

by vanderghast
February 09, 2021

Thanks for the thorough answer! It has given me some food for thought. RFID passive tag would be interesting however the range of signal transmission would be another issue.

I was hoping a relay/capacitor combination would work out but I will need to do some more thinking.

by craiggoldy
February 09, 2021
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