Super capacitor bank, each 2.7v 2000farad. How many times can it charge my iphone?

I'm going to buy a super capacitor bank, each capacitor are 2.7v and 2000farad. Total there are 6 of this caps connected in series. I'm going to charge this capacitor using my 80watt solar panel. Of course I will use capacitor balancing circuit and a charge controller too which is made and sold by some ebay member. My question is when this caps are fully charged, how many times it can charge my iphone from 0% to 100%?
I have calculated, the total capacitance of this bank is 333farad.
Other question is each of them are 2000farad but when connected in series the total capacitance is only 333farad, so what happened to the rest of the 1667farad?

Never use capacitor bank that contain higher voltage charge to charge a low voltage device
like iPhone. Very high risk to kill the battery or iPhone both at the same time.

Second question first. When capacitors are in series the value reduces just as when inductors or resistors are in series, the value increases. You are not losing 1667uF, except that 2000uF reduces to 333uF when so many capacitors are in series. Their behavior is like that. [a crude analogy is the way a business loses when the directors fight with each other. As against better business when they work coordinatedly.]
And now the first question: answer could be (i) once (ii) two times (iii) n times. It all depends on the charge required to charge the mobile. So learn about charges as against voltages. Q= CV. A higher capacitor can store higher charge. If mobile takes this charge, you can charge it only once. The charging of mobile is actually passing a current into the battery of the mobile and charge is not just current but charge is product, current*time. 1 amp for 1 minute means same charge as 100mA for 10 mins. The voltage across the charged capacitor will decreases as you draw current as (it/C) where i is current and t is time. So funda is missing… You need to learn that, I guess.

A one farad capacitor will drop in voltage at one volt per second at a current of one amp.
(One amp-second is one coulomb of charge).

Assuming you use a very efficient car type "12V" input phone charger that will work from 15V down to say 7V, that's an 8V discharge span.

8V * 333 Farads = 2664 Amp-Seconds.
For reference, a one amp-hour battery would be 3600 amp-seconds to discharge.

The cap bank works out at 0.74 AH.

I have no idea what capacity your phone is. Assuming it takes half an amp at 5V, so 2.5W, for four hours:

The mean cap voltage (eg. Half charge) is 11V. 2.5W at 11V is about 250mA, allowing for 90% charger efficiency.

That would give a run time of 2664 / 0.25 = 10353 seconds or 2.96 Hours.

It would not give you one full charge.

Cross check, working in Joules (J = 1/2 C V^2)
Full charge, 15V: 333/2 * 15 * 15 = 37462 Joules.
Min charge, 7V: 333/2 * 7 * 7 = 8158 Joules

Difference = 29304 Joules usable.

2.5W for four hours = 2.5 * 3600 * 4 = 36000 joules; again well over the capacitor bank storage.

An off-the-shelf USB lithium battery pack is smaller, lighter, cheaper and has far higher capacity.
Use a good car / truck USB output charger to charge that from the solar cells & you don't need the balancing circuits either…

There are masses of these little battery packs about now, even the smallest 2.6AH one is much higher usable capacity than the big capacitors; the biggest more like 40 time more, it would charge a typical phone 10 - 15 times.

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