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?

  • \$\begingroup\$ You cannot charge a capacitor with a defined voltage .....it is a current controlled device. \$\endgroup\$ Jan 10, 2018 at 16:48
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    \$\begingroup\$ 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. \$\endgroup\$
    – Simon B
    Jan 10, 2018 at 17:25
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    \$\begingroup\$ @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. \$\endgroup\$
    – user4574
    Jan 10, 2018 at 17:51
  • \$\begingroup\$ 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 \$\endgroup\$ Mar 24, 2018 at 11:22

1 Answer 1


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.

  • \$\begingroup\$ 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? \$\endgroup\$ Mar 24, 2018 at 11:35
  • \$\begingroup\$ the battery has > 50kJ and Cap of 1000F only has >3kJ how can lower capacitance charge the battery which is bigger \$\endgroup\$ Mar 24, 2018 at 11:56
  • \$\begingroup\$ 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. \$\endgroup\$
    – Maple
    Jan 10, 2020 at 9:04
  • \$\begingroup\$ 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. \$\endgroup\$ Jan 10, 2020 at 16:19

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