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I am building a power-bank for a project, out of capacitors

  • #1

Homework Statement


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I want to build a power bank capable of holding 10,000 joules of energy, is portable (can fit in a palm and is lighter than 1kg) and safe (does not short circuit, break down or catch fire easily)

I cannot use batteries, only super capacitors

What are some problems that can arise? Please give just problems and why they would emerge, not solutions (I have to present all the research I used, it would look better for me if you don't give solutions)

Assume that the design I am using for my power bank is a basic power source, capacitor, and bulb circuit connected in parallel with nothing else

The Attempt at a Solution


Problem 1 - The circuit has a limited voltage tolerance, this is because high voltages can cause capacitors to break down and can also cause excess heat to build up

Problem 2 - The power bank can't store much energy, this is because capacitors are not as good as batteries for storing energy

Problem 3 - Capacitors release energy very quickly

Are there problems other than this ? Why do they happen?
 
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Answers and Replies

  • #2
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Start with the problem description please: What exactly do you want to achieve?
 
  • #3
gleem
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Problems in one application may be an advantage in another as for example quick discharge can be an advantage.
 
  • #4
Problems in one application may be an advantage in another as for example quick discharge can be an advantage.
Since I am making a power bank, I would want to store energy
Quick discharge would not be a problem if my laptop, or an ordinary household device with batteries that can be plugged into mains power can handle the voltage/current
I guess one problem is that I don't know how quick the capacitors would discharge
 
  • #5
gleem
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I guess one problem is that I don't know how quick the capacitors would discharg

Ah ha. That is an issue of which you should be aware. You do not seem to have any knowledge of DC circuitry with a capacitor and a resistor. Ok think about this if you connect a capacitor to a battery with just conductors the charge flows instantaneously and since there is not significant resistance the initial current is extremely large. As the capacitor charges the charges in the capacitor will inhibit further charges from being excepted and the current will decrease rapidly until the capacitor is fully charged and the current ceases. This happens is a blink of an eye.

But what if you have a resistance in series with the capacitor? The resistor will limit the maximum current flowing into the capacitor. So the capacitor will not charge as fast although as before the charge in the capacitor will inhibit subsequent charges until the capacitor is completely charges. In fact the capacitor will charge exponentially to the battery voltage. This means that for every equal time interval the capacitor will add the same fraction of of the capacity that it has left. So for example if in one second the capacitor adds charge to bring it to within 40% of maximum capacity (40% left) in the next second it will add 40% of the remaining charge capacity that is currently in the capacitor after two second the capacitor is 84% charged . In the third second it will add 40 % of the remaining capacity (16%) so the capacitor will now be 92.4% charged. This will continue until it reaches capacity. So how do you know how fast a capacitor charges with a resistance in series with it. The product of the resistance in Ohms and the capacitance in Farad gives the time in seconds for the capacitor to charge to within 36.7% of its maximum capacity.

The relationship between charge and time is:

Charge = max charge( 1 - e-t/R⋅C ) where t is time in second and R⋅C is called the time constant.

Not only does it charge this way but it also discharges in a similar manner giving up 63.2% of its remaining charge every second according to the RC value. The current changes in a similar manner decreasing has the capacitor discharges.

The relationship between discharge time and charge or current is:

present capacitor charge (current) = Max charge (current) ⋅ e-t/R⋅C

At a time equal to 4R⋅C the capacitor will be discharged or charged to about 98% of max.

Do you know the relationship between the energy stored and the capacitance or voltage or the charge stored and the voltage? These are important too.

You might be disappointing in the usable energy that a reasonable sized capacitor can store. Keep in mind that 10000 J is only 2.78 watt hours and usually only a fraction of the stored energy is easily extracted as compared to a typical laptop Li ion battery with 45 watt hours of energy.
 
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  • #6
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…the design I am using for my power bank is a basic power source, capacitor, and bulb circuit connected in parallel with nothing else
The bulb is the load?
 
  • #7
The bulb was just an example
 
  • #8
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What is the planned maximum voltage? What happens if your device needs a different voltage?
What happens to the voltage during the discharge process?

All the safety features you mentioned can be seen as potential issues as well.

Price? Mechanical robustness?
 
  • #9
The planned maximum voltage for charging the power-bank is 1.5volts (I am going to use resistors to ensure that the voltage does not go higher than this)
For discharging, The load can handle discharges of upto about 200 volts

Can you please elaborate on the safety issues part?

Price is not an issue unless the price reaches 100 dollars

Mechanical robustness is also not an issue unless my power-bank would be fragile (breaking from a 1-2m drop or not being able to carry it around)
 
  • #10
gleem
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The planned maximum voltage for charging the power-bank is 1.5volts (I am going to use resistors to ensure that the voltage does not go higher than this)
For discharging, The load can handle discharges of upto about 200 volts

Can you please elaborate on the safety issues part?

Price is not an issue unless the price reaches 100 dollars

Mechanical robustness is also not an issue unless my power-bank would be fragile (breaking from a 1-2m drop or not being able to carry it around)

The resistor will only affect the time of charging not the max voltage which is determined by the charging source.

The discharge voltage will be the charging voltage or less. If you need voltage higher than the charging voltage then you need a more complex system.

Your knowledge of DC resistor-capacitor circuits appears very weak.

If price is an issue it would be well to check out the prices of ultra capacitors in the range of 100 to 5000 Farads since you will need a few to store 10000 J and they tend not to be cheap. ( 100 -100F to 2 -5000F)
 
  • #11
CWatters
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Since I am making a power bank, I would want to store energy
Quick discharge would not be a problem if my laptop, or an ordinary household device with batteries that can be plugged into mains power can handle the voltage/current
I guess one problem is that I don't know how quick the capacitors would discharge
In most cases (but not all) the voltage is determined by the source (eg the power bank) but the current is determined by the load (the laptop).

So your capacitors will discharge at a rate determined by the laptop.

In the case of a usb charger things get a bit more complicated. As i recall all usb chargers can deliver 200mA but if both devices agree then the current can be increased eg the laptop and charger might agree a charging current of 2A.

If you are really going to build this you have a lot of research to do.
 
  • #12
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A switched voltage regulator might work. When the capacitor voltage is high, the duty cycle (ratio of on-to-off) is short.
 
  • #13
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.... In the case of a usb charger things get a bit more complicated. As i recall all usb chargers can deliver 200mA but if both devices agree then the current can be increased eg the laptop and charger might agree a charging current of 2A. ....
Just to expand on this, for anyone interested in the details - when a device is plugged into a USB port (host/computer), it is the device that is supposed to not draw more than 100 mA initially. The device can then request the higher current mode from the host, and if the host answers back that it can supply it, the device can then draw up to 500 mA if needed.

A USB 2.0 host (I'm not up on all the details of USB3.0) must provide that 100mA initially, and at least 500 mA if it acknowledges the higher current request. I've never seen any official limit on the upper current limit, but most hosts seem to limit it to ~ 1~2 amps, but 500 mA is all they are required to do to meet spec.

This all makes a lot of sense when you consider those 'self-powered' 4-port hubs that are available. They can negotiate a guaranteed 500 mA from the host, so that means they can supply 100 mA to all 4 ports, and have another 100 mA to run their internal circuits.

Things got a little fuzzier on the charger side, since many chargers are 'dumb' and can't do the negotiating thing. Apple went their own way with hanging resistors on the data lines to indicate how much current the charger can supply, The Android world worked with the USB group to define fast charge capability as the two data lines shorted to each other.

That was from memory, but I think this confirms at least most of it:

https://www.electronicdesign.com/interconnects/introduction-usb-power-delivery
 
  • #14
I see, it seems that my knowledge is quite weak

Although that is to be expected, I am still in high school and have yet to learn about capacitors in my school
At least I have a few months to build the power bank
(before you ask, if you haven't learnt about caps, then why are you building a power-bank out of them?
It's because the project to build something based on you haven't learnt about in school)
 
  • #15
unfortunately, my school is a bit weird

I have 2 months for research and 5 months for building the power-bank

and I am lazy, I wasted most of my research time

I will do more research before posting
 

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