# Supercapacitor or battery with capacitor

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?

• You cannot charge a capacitor with a defined voltage .....it is a current controlled device. Jan 10, 2018 at 16:48
• The solution is likely to be rather more complicated than you think. If you connect a 2.7V capacitor to a 5V solar panel, the voltage will slowly rise from 0V. Leave it too long and it could eventually reach 5V - which would permently damage the capacitor. You also can't directly use a 2.7V capacitor to charge a phone that expects 5V (as most do) - it would need a boost converter. Jan 10, 2018 at 17:25
• @Raghuraman rajendran You will never charge a 4Ah Lithium battery from a 500F 2.7V capacitor. Assuming your lithium battery voltage is around 4V. 4A * 1 hour * 4V = 57.6kJ of energy in the battery. 0.5 * 500F * (2.7V)^2 = 1.822kJ of energy in the capacitor. So the capacitor only stores about 3% as much energy as the battery. Jan 10, 2018 at 17:51
• I'm sorry but I'm not able to follow you exactly. What i exactly want is, i want to get as much power as possible while using a hand crank generator. A battery will not be able to get all those charges. so i use a supercapacitor to get those charges as quick as possible, store it like some kind of a small buffer or something and then release it slowly towards the battery "through" a battery charger. I will not be using the supercapacitor to discharge the voltage but just for charging it up quickly....I dont know if something is wrong with the concept...if it is..please correct me Mar 24, 2018 at 11:22

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 -- >3kJ supercap vs >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.

• i can understand the fact that supercaps have low esr and batteries have high and thats the reason supercaps are able to take in much current into them. so what i ask is why cant i use this property to charge the battery using a supercapacitor instead of using the supercapacitor itself as the battery. i can eventually get a 100f supercapacitor for around 2.5 dollars or so...hence y not use it like i say? Mar 24, 2018 at 11:35
• the battery has > 50kJ and Cap of 1000F only has >3kJ how can lower capacitance charge the battery which is bigger Mar 24, 2018 at 11:56
• I think all these formulas only confuse OP. Let's put is simple: The supercapacitor can only store about 1/17 of energy required to charge the battery. So, you'd need 17 supercapacitors to charge 1 battery. That would require you to carry 3.6kg suitcase of capacitors to charge one 30g battery. and that is assuming 100% efficiency of the transfer, which is not achievable. Jan 10, 2020 at 9:04
• Yes you could that , but as Engineers we like to show others how to compute it. If he can't compare the energy values, What's the benefit? FWIW @Maple It's energy numbers that matter. Jan 10, 2020 at 16:19