How to Build a 5V 1A 1220mAh Supercapacitor Power Bank with AC Charging?

I am a total electronics newb and was told this is place to come for genius advice.

I need to make a contraption that charges Supercapacitors quickly from a standard wall socket (100-240VAC input) then releases the electricity back out at 5 volts 1 amp and 1220mah.

Can someone help me figure out what it would take to achieve this?

Thanks a ton!

-T

Look at the following

http://www.instructables.com/id/How-to-calculate-charging-time-energy-of-your-su/

or

http://www.illinoiscapacitor.com/pdf/papers/supercapacitors.pdf

and

http://www.maxwell.com/products/ultracapacitors/docs/an-010_charging_ultracaps.pdf

Definition of a super capacitor also might be needed

Acc most of the information I've read says 2.5 , 2v7 5.5 volts

See this link as well
http://forum.allaboutcircuits.com/showthread.php?t=47319


See also this page for Rc time constants
http://www.electronics-tutorials.ws/rc/rc_1.html

and
http://www.allaboutcircuits.com/videos/46.html

Mark's excellent references mostly cover it, but one piece is missing -- your spec of 5V at 1amp and 1220 mah.

That's a bit tricky, since the discharge curve of a capacitor (SuperCapactiors included) is not the same as a typical battery discharge curve. A battery's output voltage tends to remain fairly flat as it is discharged (there are exceptions, such as the case where the battery is being discharged at such a high rate that it's internal resistance has a more prominent effect on the discharge curve, but if we consider such cases poor design and exclude such cases, then the "fairly flat" assertion holds).

A capacitor discharge curve is reverse exponential, thus, the voltage drops quite rapidly at first and then tapers off towards a horizontal asymptote -- but it isn't really horizontal (or "flat") until it reaches zero charge, which is pretty useless.

So, to get a flat 5V you will need to convert the Capacitor voltage to 5V as it discharges. One way is to use a Switch Mode boost converter.

Check this out:

* http://dkc1.digikey.com/us/en/tod/LinearTech/...noaudio/ltc3625-supercap-charger_noaudio.html

Also, here is some crude math to get a ball park on what size super cap you will need to get 1.22A-hr:

ΔV = 5V - minimum voltage that your boost convert will work at -- let's say 1V, thus V = 5V - 1V = 4V

C = I*T/ΔV

I*T = 1.22A-hr * 3600hr/sec = 4392A-sec

C = 4392/4 = 1098F

Now, that assumes the current will be constant as the cap discharges, which won't be the case.

As the cap discharges, the boost converter will demand more current: P = I*V, so, when the cap is at 5V, a little more than 1A will be demanded (1A/Boost converter efficiency) If efficiency is 80%, then the initial current will be 1/.8 = 1.25A.

When the cap is at 1V, the boost converter will demand 5V * 1A = 5Watts ... 5W/1 volt = 5A. And lets say our boost converter is now running at 70% efficiency, so 5A/.7 = 7.1A! So, the above calculation doesn't really give a real world result.

Perhaps a better way is to use the equation for energy in a capacitor:

E = 1/2CΔV² = Pt i.e. Energy = Power * Time

C = 2*P*t*3600/ΔV² [since 't' is hours, multiply by 3600 to get seconds]

P = IV = 1A*5V but, that's for 100% efficiency, so the actual power will be more like (assuming 75% overall efficiency): 5W/.75 = 6.7W

C = 2(6.7W)(1.22hr)(3600)/4V² ≈ 3700F

BUT, this would be better solved with calculus, but it's been a while...anyone?