DIY power bank with supercapacitors

Supercapacitors have a colossal capacity compared to conventional capacitors. They also have a number of advantages over lithium-ion batteries, such as: they are not afraid of low temperatures and are not afraid of complete discharge. All this forced me to make a power bank using supercapacitors.
Ordinary power banks, if left idle, will discharge over time, since the elements have a self-discharge. And one fine day, when you need to take the power bank, say, on a hike, it will be “dead” and show no signs of life.
The same model, made on ionistors, will always be ready for work if it is first charged.

Will need

The controller performs several functions at once: it controls the charging and discharging of the battery, protects against short circuits, and displays the capacity of the entire battery on the display.
Micro USB is used to connect the power supply for charging. The remaining two USB are outputs for connecting a load.

Manufacturing a power bank using ionistors

We solder two supercapacitors in series."Plus" to "minus".

We also solder the next two. And now we solder these two pairs in parallel to each other, “plus” to “plus”, “minus” to “minus”.

The result was a battery made from ionistors with a maximum voltage of 5.4 V and a capacity of 1000 Farads.
Solder the controller.

We insert the USB cable and charge our new power bank.

As soon as it’s charged, connect the phone and check if it’s charging.

Yes, everything works fine - the mobile phone is charging.
To insulate all contacts and make some semblance of a case, we will put on shrink film and blow everything with a hot air gun.

Let's cut out a window for the display using a utility knife.

Side view, USB ports open.

Conclusion

In conclusion, I would like to add a fly in the ointment: ionistors have a high self-discharge compared to lithium-ion batteries, therefore, after a full charge, the time of its use is more limited than that of commercially produced devices.

Watch the video

come back

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Comments (14)

#1 Guest Alexander Guests 15 September 2019 19:51
9

I would like to add a couple more fly in the ointment.

2.The total capacity will not be 1000 farads, but 500. Explanation in the 8th grade physics textbook.

3. You have two capacitors of the same capacity, say each 1000 µF/10 V. One is fully charged (Uc1 = 10 V), the second is completely discharged (0 V). Find the total energy of both capacitors. Then, we connect the capacitors in parallel. As a result, the first one is discharged, the second one is charged. Theoretically, the voltage on both capacitors will be equal to 5 V (in practice, a little less). Once again, find the total energy of the two capacitors, which are now only half charged. Now explain why in the first case the total energy turned out to be twice as large as in the second case. What happened? Where did the energy go when it was redistributed between capacitors?

Conduct an experiment and make sure that a power bank is “faith in a miracle” without “knowledge of physics”.
- Answer
1. #2 Guest Yuri Guests 16 September 2019 21:37
  1
  
  One is charged, the second is discharged (empty), there is no energy in it. What can be the total energy if it is not in the same circuit? Parallel conductors charge and discharge simultaneously.
  
  Answer
2. #3 Guest Yuri Guests 16 September 2019 22:04
  2
  
  I have a charged capacitor, and in the store nearby there is a whole box of discharged ones, let’s look for their total power? Maybe we’ll connect it in a circuit and wonder why there is a charge on one parallel capacitor and zero on the others?
  
  Answer
3. #4 Slavs Guests 29 October 2019 19:09
  1
  
  The total capacity will actually be 250 farads when two identical 500 F conductors are connected in series.
  
  Answer
#5 Guest Sergey Guests 16 September 2019 18:27
3

On the author's soap.... read...
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#6 Alexander Vishnevetsky Guests 17 September 2019 23:16
4

The capacity for a series connection is 1/(1/500+1/500), i.e. 250. When parallel, the sum is 250 + 250, a total of 500, not 1000
And the controller from a lithium power bank will charge up to 4.2 instead of the full 5.4V
And I don’t think that this garbage will charge the phone by any significant percentage...
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#7 Ilya Guests September 18, 2019 10:13
3

And this fool has an energy capacity 2 times less than a small compact LiPoly battery with a capacity of 1000 milliamp-hours
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#8 Alex Guests 18 September 2019 14:22
9

"CONVENTIONAL POWER BANKS, if left idle, discharge over time, since the elements are SELF-CHARGING. And one fine day, when you need to take a power bank, say, on a hike, it will be “dead” and will not show signs of life.
The same model, made on ionistors, will ALWAYS be ready for work if it is first charged.
...
In conclusion, I would like to add a fly in the ointment: ionistors HAVE A HIGH SELF-CHARGE compared to lithium-ion batteries, therefore, after a full charge, the time of its use is more limited than that of commercially produced devices."
As if the beginning was written by 1 person, but the result is different? I remembered a joke - ... Thus, the more I drink, the less I drink...
- Answer
1. #9 Well Guests 18 September 2019 14:31
  4
  
  You are simply ZERO at electronics! Everything is written correctly: if the battery is discharged to zero, then you can’t charge it - the thing is damaged. A ionistors They can lie discharged for years! Their disadvantage is that they quickly discharge themselves, faster than batteries....
  Well, what’s not clear??? If you don’t understand, then don’t consider others stupid!
  
  Answer
  1. #10 Alex Guests 13 November 2020 08:43
    2
    
    Yes, you have “a lot of knowledge”
    Lithium batteries cannot be discharged to zero because they are protected from this and the cells themselves lose only 10% of their charge per year... of course, if you go camping once every 10 years, then yes, they will discharge to zero
    
    Answer
#11 Glory Guests 21 September 2019 10:38
2

Author, did you manage to charge the phone? How long did this “power bank” produce current? Or, as in the video, 44% of the charge, as it was and remains?
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#12 Guest Anonymous Guests 25 September 2019 16:01
5

Some of the bugs have already been described, so I’ll add that the standard board of the power bank has protection against deep discharge (2.5-3.2V) so as not to kill lithium, because its operating voltage range is 3.2-4.2 V.... . Hence the logic ionistors (supercapacitors) 1. They will charge not to the nominal 5.4, but only to 4.2 V (i.e. less than 80%), and besides, when they are discharged, they will discharge to only 3.2 V (i.e., roughly speaking, only 20% or up to 60%)... Therefore, we will get some 20% of their capacity...
If we charge with “standard means”, then the power bank board simply will not be able to provide full current for very fast charging, and they will charge a little faster than lithium. They will discharge just as slowly, because the phone and the board limit the current... oh yes, they already talked about the wild self-discharge that lasts for hours, and not for months, for lithium, right? Those. By the time you need to charge your phone from them, there will already be something in the ionistors... and there will be nothing.

taking into account their price combined with efficiency, this is one of the stupidest applications ionistors. If an instantaneous charge from an external high-precision source had been used, everything would have gone well. Add a boost that could suck ionistors up to 0.5-1V would have smoothed out the story a little... and so... the "Oskolkovo bonus" for budget development.
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#13 Guest Alexander Guests September 27, 2019 11:53
0

Supercapacitors?, well, a very interesting power bank.
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#14 Novel Guests 30 September 2019 23:57
0

Discarding the theory, in the first photo it is soldered with an untinned tip. This is the big fat point: he just needs to do writing and not practical work.
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