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I was thinking about how you can juice the maximum out of an ultracapacitor, because they usually work at no more than 2.5V, and once discharged below 2V they won't be able to sustain an LED anymore. This means you're effectively using only 20% of what the capacitor has to offer. If an ultracap can sustain normally an LED for 5mins, with this variation you'll be able to sustain it for at least 20mins. The principle behind it is probably known to most of us, just a joule thief, but the trick is using a germanium transistor instead of a silicon one, because they can work at a lower base voltages of around 0.2-0.3V. This means the joule thief can discharge the cap to 0.3-0.4V before it completely stops working. In contrast a silicon transistor would stop working at 0.7-0.8V, which is kind of wasteful - especially keeping in mind that a joule thief increases efficiency with lower voltages. The last 0.4V a germanium transistor can give, will probably extend the working of the circuit by a few minutes. Also, since a wide variety of supply voltages are used from 2.5V down to 0.4V, something has to be put in place to protect the LED from overcurrent. A simple CCS should do the job, limiting the current to ≈20mA. A 1F capacitor might give you an hour or so of sustained operation. I just need to get ahold of supee ultracaps and I'll give this a go. You can use 20F capacitors to sustain this thing for 20 hours or so. A solar LED lamp might be a good application. Only problem I have to solve is the extremely high voltages this thing can generate at 2.5V supply voltage. I guess high voltage transistors could handle the spikes, but I don't like seeing 300V+ in such a low powered circuit. On the other hand any attempt to suppress them would mean dissipating them, thus discharging the cap faster. One way to do it, is with a floating regulator, like the one on the right, but the circuit will perceive it as a load, and fasten the cap discharge.
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