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I have a 5v 2amp external usb battery pack that dis/charges 4 18650 vape style batteries. Each is 3.7v 2500mah.

My Question is, what size supercapacitor would replace the batteries, preferably more than the 10000 mah of the combined batteries? Big enough to hold and discharge over at least 2 hours. Then use solar panel to recharge the supercap. Solar panel also has a usb port but it's not always sunny in Oregon. Should I do away with the battery pack and charge directly from supercap? Phone and tablet are 3000mah and 4000mah respectively. Both charge at 5v 2amp. I'm an ac man learning to live in a dc world. Trying to cheat on Thomas Edison with Nicola Tesla.

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    \$\begingroup\$ Work it out yourself from Q=CV. (Coulombs=Farads * Volts.) 1 Coulomb = 1 amp second, so you can convert from amp-hours to coulombs. \$\endgroup\$ – Brian Drummond May 20 '15 at 20:21
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    \$\begingroup\$ You would need about 8.8 Wh. I have a 350F D Cell supercapacitor here that's 0.4Wh. A minimum of 22 of them. And these things are 1.3" diameter cylinders that are about 2.5" high. And I'm not taking into account the fact that you would need switching regulators to keep a steady voltage. You would need a lot of very large capacitors. \$\endgroup\$ – scld May 20 '15 at 20:21
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    \$\begingroup\$ How many horses would it take to pull my carriage as fast as a ferrari \$\endgroup\$ – Andy aka May 20 '15 at 20:42
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    \$\begingroup\$ What size capacitor, you wonder? About 2 gallons, I would guess. \$\endgroup\$ – Nick Alexeev May 20 '15 at 20:55
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    \$\begingroup\$ -1 for thinking anyone at Radio Shack is a tech geek! \$\endgroup\$ – DoxyLover May 20 '15 at 20:58
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If you're trying to power a circuit off of capacitors instead of a battery for a long while it will not work. The two devices have different physics in terms of their energy storage. A capacitor's voltage does not change instantaneously, but it can supply a large amount of current instantaneously and discharge rapidly.

What are you trying to power, and for how long? -Ryan

Here you go, just as Adam said, this will be a very large capacitor. My solution only brings you to a half charge because I assumed 500mA.

enter image description here

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  • \$\begingroup\$ Its an external usb battery back that uses li-on batts. Looking for super caps to replace the lions. I read its got braking circuit built in or something to slow the discharge. Still kinda new tech. Oldest reference i found for battery to capacitor replacement is 3 years ago. When big caps were 10farad. Lol \$\endgroup\$ – Nobody Atall May 20 '15 at 21:02
  • \$\begingroup\$ lol, alright what voltage will the capacitor bank begin at and where will you declare the device can no longer operate... that is what is the max and min voltage your device will accept from the battery pack? And what discharge current is apparent on the array (Load current). \$\endgroup\$ – Rwsselby May 20 '15 at 21:04
  • \$\begingroup\$ Phone input 5v roughly 1amp (2400mah battery) 2 hours to charge normally. Tablets are 5v 2amp chargers (4000mah battery), 2-3 hours to charge. This is what i use the battery pack to charge. Voltage in of battpack is up to 2amps. Output is variable .5, 1, or 2 amps at 5volts. \$\endgroup\$ – Nobody Atall May 20 '15 at 21:14
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    \$\begingroup\$ @NobodyAtall - That's not what he asked. Let's say you've got a capacitor which starts at 5 volts. What is the minimum voltage you're willing to allow before you declare the cap useless? 4.9 volts? 4.5 volts? 4 volts? What voltage exactly? \$\endgroup\$ – WhatRoughBeast May 20 '15 at 22:19
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We can calculate a very rough estimate without too much trouble. I'll assume the battery voltage drops linearly from 3.7V to 3.2V as it discharges. That gives a rough total energy of:

$$E_{batt} = 4 \times 3.45\ \mathrm V \times 2700\ \mathrm{mA} \times 3600\ \mathrm s \approx 119.2\ \mathrm{kJ}$$

To make a capacitor provide the same energy over the same voltage range, we need:

$$\Delta E_{cap} = E_{batt} = \frac 12 CV_i^2 - \frac 12 CV_f^2$$

$$119.2\ \mathrm{kJ} = \frac 12 C((3.7\ \mathrm{V})^2 - (3.2 \mathrm{V})^2)$$

$$C \approx 69100\ \mathrm{F}$$

The biggest supercapacitor on DigiKey is 5000 F, costs $200, is only rated for 2.7 V, and is almost twice the size of a D battery. So I don't think you're going to get equivalent performance from a standalone capacitor anytime soon. A switching regulator might improve your performance by an order of magnitude (or even two), but will also drive up the cost, size, and complexity of your battery replacement.

If you're willing to accept a shorter use time, you can drop the capacitance proportionately. If you only want 10.2 mAh (which is not very much at all):

$$C \approx 69100\ \mathrm F \times \frac {10.2\ \mathrm{mAh}} {2700\ \mathrm{mAh}} \approx 261\ \mathrm{F}$$

I see a 330 F supercapacitor rated for 5V that's a third of an inch tall and costs a few dollars. That might do the job, at least for a little while.

EDIT: Oops, I was looking at a 330 mF (millifarad) capacitor. ~300 farad capacitors are much larger and more expensive. So even the shorter use time is not practical.

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  • \$\begingroup\$ Thanks for the math. Ive forgotten all equations. Ok this sounds like what im looking for, could i put 3-4 in sequence? I picture electrons as water, capacitors as buckets, when bucket is full it dumps to the next bucket. But not till its full. So still a bit fuddled on this bit. \$\endgroup\$ – Nobody Atall May 20 '15 at 21:45
  • \$\begingroup\$ Putting capacitors in series increases the combined voltage rating but reduces the combined capacitance. You want them in parallel for extra capacitance. \$\endgroup\$ – Adam Haun May 20 '15 at 21:51
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    \$\begingroup\$ I would recommend strongly that you don't do any of this. \$\endgroup\$ – scld May 21 '15 at 0:18
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    \$\begingroup\$ If I were to actually attempt to replace a battery bank with a capacitor bank, which I wouldn't, I would probably design a buck/boost voltage regulator with 2.5A current limiting and short circuit protection. Keep the output voltage at 5V, charge the supercap stack to something like 7.5V (I would put 3 in series and then put the trio in parallel with other trios, regulated by a supercap charging IC). There are all in one circuits that do the buck/boost, balanced charging,short circuit protection. LT makes a good one. \$\endgroup\$ – scld May 21 '15 at 14:11
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    \$\begingroup\$ After days of searching for alternative power, I found 2.7v 500f edlc capacitor with 5v step up board suits my needs perfectly. \$\endgroup\$ – Nobody Atall Sep 18 '15 at 8:35

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