Super Cap USB Charging Recharging (?)
I do not know as much as I should about "super caps". So I'll be safe here and play it dumb... I've got a chance to get a quantity of 5.5 volt rated four point four Farad super caps. I've got a ...
Forum Discussion
Here's the deal. My cell phone battery is rated at 2,800 mAh (2.8 Ah) at 3.85 V. Watt-hours equal Amp-hours x Volts, 2.8 x 3.85 = 10.78 Wh. To convert Watt-hours to Watt-seconds multiply by 3600. So 10.78 Wh = 38,808 Watt seconds, the total amount of energy stored in this fully charged battery. One Watt-second = one Joule. A Joule is the measure of energy. So the total energy stored in this battery is 38,808 Joules.
Now for the 16 farad capacitor. Total energy stored in a capacitor is total Joules = (Volts x Volts x capacitance)/2. Total Joules = (5.5V x 5.5V x 16 farads)/2, total Joules = 242 Joules stored in this 16 farad capacitor.
So the battery stores 38,808 Joules and the capacitor stores 242 Joules. Divide 38,808 Joules by 242 Joules and get 160.36.
Bottom line is that the battery stores 160.36 times more energy than the 16 farad capacitor! Of note is that you will not be able to transfer all the 242 Joules from the capacitor to the battery.
Now for the 16 farad capacitor. Total energy stored in a capacitor is total Joules = (Volts x Volts x capacitance)/2. Total Joules = (5.5V x 5.5V x 16 farads)/2, total Joules = 242 Joules stored in this 16 farad capacitor.
So the battery stores 38,808 Joules and the capacitor stores 242 Joules. Divide 38,808 Joules by 242 Joules and get 160.36.
Bottom line is that the battery stores 160.36 times more energy than the 16 farad capacitor! Of note is that you will not be able to transfer all the 242 Joules from the capacitor to the battery.
