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I would like to light up a 3W LED from Cree having a forward voltage of 2.9 V at 350 mA. It's maximum drive current is 1 A and gives 300 lumens/m2 output. I am considering lighting up this LED using supercapacitors.

The objective is to light the LED for 5 hours.

Based on this objective, I tried to calculate the capacitance required in reverse using the following formula:

(Energy stored in the capacitor)/(Power of the LED) = Time required for the LED to glow)

Additionally, I came to know that If a switching circuit is used, the losses for the supercapacitor charging operation would be less as a resistor dissipates 50 percent of the input energy. If the voltage drop across this circuit is 1 V, then taking this into account as well, I obtain:

\$\frac{\frac12 C V^2 - \frac12 C 1^2}{1.05(P_{led})} = 5\text{ h}\$

As the supercaps are generally rated at 2.7 V, I substitute V to be 2.7 and obtain the capacitance to be 18270 F which is extremely high. I needed to cross verify the theoretical capacitance required to get this LED to glow for 5 hours is right or not.

If I go for a 2.7 V, 500 F supercapacitor, I am obtaining the LED to glow for 30 minutes.

From what I have read about the supercaps, the charging rate is quite high for the supercaps, and assuming that I am generating electricity simultaneusly to charge the supercap and simultaneously light the LED, Would it be feasible?

I am imagining the 2.7 V 500 F supercap as a big tank with an inlet and an outlet. The inlet is quite high because of high charging rate whereas the outlet rate is slower. The inlet rate is 5 W and the outlet rate is given to the 3 W LED.

I am giving a 5 W power input using the right switching circuit to feed it into the supercapacitor. If this is the case, would I be able to instantaneously light this LED. If not how? What would be the right circuit I should implement to use it as a buffer.

shadesofpurplegreen
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    You will get more energy storage for your money by using batteries, LiPo or others. – JimmyB Jul 19 '16 at 10:33
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    Why are you surprised? 5Wh is more than most smart phone batteries have. Also, a 1W LED is pretty bright – it doesn't surprise me that you can't run one off capacitors for five hours. – Marcus Müller Jul 19 '16 at 10:35
  • But wont a supercap be best used as a buffer, rather than as an energy storage device? Since it has high charging rate, I want to use it instantaneously to drive the LED, and I am trying out some math to get the best supercap that would solve this problem for this 3 W LED. The supercap is charged instantaneously using 5 W power input. Would it be possible? – shadesofpurplegreen Jul 19 '16 at 10:36
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    Also notice that *power* and *energy* are not good bases for your calculations: The LED will not be able to consume all energy from the capacitor because the capacitor's voltage will approach 0V as it discharges while the LED will dim and probably turn off below 2V or more already. – JimmyB Jul 19 '16 at 10:36
  • But as @JimmyB said, supercap is not the energy storage of choice, here – they're used if you need to store charges for a shorter amount of time, and especially if you need discharge currents and cycle counts that chemical batteries don't stand – neither of those are your problem here, so simply choose an appropriate battery, which has a higher energy density. – Marcus Müller Jul 19 '16 at 10:37
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    You don't need a capacitor if you can only provide 5W to charge. The cap will not 'magically' be fully charged by 5W in an instant. Most LiPo's have a charging rate of `1C`, i.e. they can be fully charged from 0% to 100% within 1 hour plus some. There is no way you could provide that charge rate to a battery of 3W*5h=15Wh when you only have a 5W charger, so the battery's charging capability will be the least of your concerns. – JimmyB Jul 19 '16 at 10:39
  • exactly. Acting as if you could pull constant power from any storage device you want: If your charge rate is 5 W, and your discharge is 3W, than your charging will necessarily take \$\frac35\$ of your discharging time. – Marcus Müller Jul 19 '16 at 10:50
  • Besides, the voltage range of a (single) super cap is inappropriate for your case: It's usually charged to 2.5V - too low for your LED, and the voltage constantly drops while discharging. After only 36% of *power* consumed, you're at 2V already, 64% discharge is at 1.5V. – JimmyB Jul 19 '16 at 10:54
  • Would a supercapacitor will discharge in quite uniform way (quite like battery)? or like ordinary electrolytic capacitors, at first it will produce a huge discharge current, then gradually the current will become little? – Always Confused Jul 19 '16 at 16:38

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