Energy Conversion

Thread Starter

hazcoper

Joined Oct 27, 2016
7
Hi all. Recently I had an idea. Super capacitors can store a lot of energy. And they can be charged up really quickly. The reason for that is because you can charge them with high current. Basically what I want to do is a power bank to charge my phone, I would charger them quickly, and the release the energy into my phone. I have worked out a way to charge the super capacitors, what I haven't been able is to find a way to convert the 2.7v to 5v and about 2 amperes

Can anyone help me? I need to find a dc to dc step up that can step up 2.7v to about 5v 2 amperes, to charge my phone.

Thanks for the help
José Mateus
 

tcmtech

Joined Nov 4, 2013
2,867
Super capacitors can store a lot of energy.
Compared to a common lithium battery they are very poor energy storage devices. About 1/50 the capacity of that of an equally sized lithium battery which would give you a run time of around 2 - 5 minutes between charges.

Yes, it was an idea but it was a poorly researched one. :oops:
 

wayneh

Joined Sep 9, 2010
17,496
Basically what I want to do is a power bank to charge my phone...
I recommend doing your experiment on paper first. Compare the energy needed by your phone to the stored energy in the capacitor. Don't forget conversion losses, which could be 50% or more.
 

ErnieM

Joined Apr 24, 2011
8,377
And DO NOT forget that the energy stored in a capacitor is stored in the electric field so as you extract energy the voltage decreases.

So roughly when you get have the energy out your voltage is down by 30% to 70% of the original charged voltage.

Play that against the boost switcher efficiency decreasing as the input voltage decreases.
 
Last edited:

AnalogKid

Joined Aug 1, 2013
10,987
Here is one way to work out your design on paper to see if it has merit. The total energy stored in a capacitor is 0.5 x C x V^2, one-half C V-squared. C is in farads, V is in volts, and the result is in watt-seconds.

1. Calculate the total energy in the supercap at full charge.
2. Calculate the total energy in the cap at the minimum operating voltage of the DC-DCconverter.
3. Subtract the two. The difference between the two is to total energy delivered to the phone charging system. Multiply that by 0.8, a generous estimate of the DC/DC converter efficiency.

The result is the total amount of energy available at the 5 V output to charge the phone.

4. Calculate the power needed to operate/charge the device. For example, if the phone needs 5 V at 1 A to charge a depleted battery rapidly, that is 5 W.
5. Divide the result from step 3 by the calculated phone power. The result is a first-pass approximation of how many seconds the cap can power the charger.

Example:
10 F capacitor
5 V source to charge up the cap
DC/DC converter runs down to 2 V
80% efficient DC/DC boost converter
5 W charging power into the phone

(0.5) x (10) x (5) x (5) = 125 w-s max
(0.5) x (10) x (2) x (2) = 20 w-s min
125 - 20 = 105 w-s
105 x 0.8 = 84 w-s
84 / 5 = 16.8 seconds

In this example, a fully charged 10 F supercap cap will power the phone charger for about 17 seconds before its output is too low to run the boost converter.

ak
 

GopherT

Joined Nov 23, 2012
8,009
I'm on multiple fora, and, as a group, the mods here have the lightest touch. But every now and then...

ak
Earlier this week, I pointed out that Netflix offered a neat arduino-based assemble-yourself gadget for Halloween as a promotional item and my thread was deleted because it was "too commercial." "Lightest touch" is an exaggeration - I would call it a "hair trigger".
 

Thread Starter

hazcoper

Joined Oct 27, 2016
7
View attachment 114409

Someone must have realized that my parents are Mr. and Mrs. Netflix
Here is one way to work out your design on paper to see if it has merit. The total energy stored in a capacitor is 0.5 x C x V^2, one-half C V-squared. C is in farads, V is in volts, and the result is in watt-seconds.

1. Calculate the total energy in the supercap at full charge.
2. Calculate the total energy in the cap at the minimum operating voltage of the DC-DCconverter.
3. Subtract the two. The difference between the two is to total energy delivered to the phone charging system. Multiply that by 0.8, a generous estimate of the DC/DC converter efficiency.

The result is the total amount of energy available at the 5 V output to charge the phone.

4. Calculate the power needed to operate/charge the device. For example, if the phone needs 5 V at 1 A to charge a depleted battery rapidly, that is 5 W.
5. Divide the result from step 3 by the calculated phone power. The result is a first-pass approximation of how many seconds the cap can power the charger.

Example:
10 F capacitor
5 V source to charge up the cap
DC/DC converter runs down to 2 V
80% efficient DC/DC boost converter
5 W charging power into the phone

(0.5) x (10) x (5) x (5) = 125 w-s max
(0.5) x (10) x (2) x (2) = 20 w-s min
125 - 20 = 105 w-s
105 x 0.8 = 84 w-s
84 / 5 = 16.8 seconds

In this example, a fully charged 10 F supercap cap will power the phone charger for about 17 seconds before its output is too low to run the boost converter.

ak

So, say I use 16 super capacitors, each super capacitor is 2.7v 500F. Arranging the super capacitors in series and parallel, I should be able to get a capacitor bank of 5.4V 1000F. Using your calculations:

(0.5) x (1000) x (5.4) x (5.4) = 14580 w-s max
(0.5) x (1000) x (2) x (2) = 2000 w-s min
14580 - 2000 = 12580 w-s
12580 x 0.8 = 10064 w-s
10064 / 5 = 2012.8 seconds
That would be 33 minutes

And if I arranged the super capacitors, such that I get a bank that is 10.8V 496F, technically I would about 74 minutes, but then I would need a step up step down converter, and that would change everything right?
 
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