Electrical – Supercapacitor or battery with capacitor

batteriescapacitorsupercapacitor

Maybe the question that I ask is completely dumb because I am a complete noob in this, but please help.

So first off, my question is that if I want to charge a 2.7v 500F supercapacitor, can I charge it using a 5v solar panel which practically produces less than 2.7 ( possibly around 1.5 – 2v ), or say a hand crank which too produces a very small amount of voltage but possibly some decent current?
Will I be able to charge the supercapacitor much faster by increasing the voltage or the current?

Second, if I want to use it for charging a phone (the 500F supercapacitor), how long will it take to charge my phone with, say, a 4 AH battery?

Third, the main idea I had about a supercapacitor was that it can acquire all my charges quickly and store it for a considerably longer time than a capacitor, so why not first use a capacitor which would acquire my charges very quickly and then use these capacitors to charge up my battery slowly? Is that even possible?

Best Answer

Start by reading about how to use the capacitor charge, current and energy formulae. Q=CV, Ic=CdV/dt, E= ½CV²

Then realize that Batteries have more capacitance per volume than super caps, but only above the final Vf cell voltage, so stored energy is E= ½C(Vi²-Vf²).

Current is limited by the ESR of the charging circuit and Cap/battery. Parts with lower ESR are generally rated for higher currents but this value changes with voltage and aging.

example photo not to same scale --enter image description here >3kJ supercap vs enter image description here >50kJ

This Supercap with a cost of $100 for 1000F with 10mΩ ESR has a T=RC=1second with a much faster response time than a smaller LiPo battery which has greater energy storage capacity.

This LiPo has much more Capacitance value but also much higher ESR.

E= ½C(Vi²-Vf²) for C=1000F and Vi=2.5V and Vf=0.0 (complete discharge) E= 3125 Joules

A LiPo pack with a cost of $20 is rated at 3.7V 4.4Ah for 1.0C and 0.2C to 3.0V

E= ½(Vi+Vf) *Ah 3600s/h = ½(3.7+3.0) 4.4Ah *3600 = 53,064 Joules [Watt-seconds]

You may compare the energy , cost and size with life expectancy then answer your own questions.