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Can a 3000 Farad Supercap at 5V charge an iPhone? If yes how much will it charge it percent will it charge it to? Thank you

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closed as unclear what you're asking by Leon Heller, PeterJ, Null, Daniel Grillo, Nick Alexeev Sep 12 '15 at 2:56

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    \$\begingroup\$ Capacitors (even large values) lose voltage quickly when being initially discharged so the question is - how long would it take to discharge this capacitor from 5V to the minimum voltage required by the iphone charging circuit? \$\endgroup\$ – JIm Dearden Sep 10 '15 at 12:35
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    \$\begingroup\$ How much energy does an iPhone need to charge it? \$\endgroup\$ – Andy aka Sep 10 '15 at 12:36
  • \$\begingroup\$ Capacitors store energy. They're charged by the force of the voltage being applied to them, meaning an 8 V cap will only charge to 4 V if only 4 V is applied. That is why they lose voltage fast, but still have energy in them. This is just to explain that you may need a buck/boost converter to keep the voltage to the level required by the Iphone as it discharges. In terms of how much it will charge, lookup the energy capacity of the iPhone battery, then do a calculation to see how much energy can be stored in the capacitor. \$\endgroup\$ – Jarrod Christman Sep 10 '15 at 13:22
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A battery in an iPhone is a Li-Ion battery, to fully charge it needs to be at around 4.2 V, there's a charge control chip between the USB input and the battery. Let's be optimistic and assume this chip drops 100 mV or more. So for 4.2 V we would need 4.3 V at the USB input.

Lets also assume that the charging current is 0.5 A

Let's discharge the capacitor from 5 V to 4.3 V, see here for some formulas.

Discharge time = (3000 * (5 V - 4.3 V) )/ 0.5 A = 4200 seconds = 70 minutes = 1.17 hours

1.17 hours at 0.5 A is 1.17 h * 0.5 A = 0.58 Ah (Amp hours) At around 3.7 Volts that equals and energy of: 0.58 A * 3.7 V = 2.14 Wh

I see here that an iPhone 4 battery is specified to can store 5.25 Whr of energy at 3.7 V.

So our 2.14 Whr charge charges an iPhone 4 battery for 2.14/5.25 = 41 %

So no, your 5V 3000 F capacitor does not contain enough energy to fully charge an empty iPhone battery.

Possible solutions:

Charge the capacitor to 5.5 V instead of 5 V

Use more or larger capacitors

Just buy a powerbank with Li-Ion batteries inside, just like everyone else.

Disclaimer: This is a "ballpark" calculation just to quickly see what we get / how far off we are.

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    \$\begingroup\$ Another possible solution: Use a switch-mode boost converter that will supply a constant charging voltage to the iPhone from a wider range of input voltages. \$\endgroup\$ – nekomatic Sep 10 '15 at 13:35
  • \$\begingroup\$ Also the above calculation assumes that the charging process is perfectly efficient, which it isn't. \$\endgroup\$ – nekomatic Sep 10 '15 at 13:37
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As the Germans say: Jein - that yes (Ja) and no (Nein.)

There's nearly twice as much energy in a charged 3000F, 5V supercap as there is in an iPhone battery. So, yes, theoretically.

You just can't make use of all of it to charge the battery because of the way things work, so, no, not really.


An iPhone battery can store around 6 Watt Hours of energy.
Your 3000 F super cap at 5V stores around 10 Watt hours of energy.
So, there's enough in there to charge the iPhone.
BUT, you will not be able to get all of the energy from the capacitor.

As the capacitor discharges, its voltage will drop. At some point it will be too low to charge the battery. There will still be a lot of energy in the capacitor, you just won't be able to get at it.

Now, you could use a boost regulator to keep the output voltage high enough to charge the battery. I can't find one that works below 2.8V, so you will only get around 2Watt hours out of your capacitor. It will still have a lot in it, but you can't use it.

All of the above assumes no losses in the process. Since there are always losses, the picture is even worse.

I estimate around 9000F as a minimum, and you'll probably need at half as much again (14000F) to account for losses.

I see 3000F super caps at Mouser for over 50Euros each, and only 2.7 Volts.
Unless you've got a really cheap source for super caps, you'd be better off going with rechargeable batteries like everybody else.


Note:

The above values are approximate.
They wouldn't be good enough if you were to sit down and really design such a beast, but should be close enough to show the approximate size of the difficulties in making it work.


In response to stefandz's comment: A boost regulator that works down to 1.25 Volts would let you get just about 6 Watt hours out of a 3000F capacitor. So, maybe you could use 2 of your 3000F super caps to make a really expensive power bank that charges your iPhone real slow.

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  • \$\begingroup\$ Provided you don't mind charging at 100mA as the voltage droops, the MCP1640 will run down to about 1.25V in. Efficiency won't be marvellous mind you - and as you rightly point out, it's going to be a pig just managing all of this. \$\endgroup\$ – stefandz Sep 10 '15 at 15:51

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