What is the maximum charge in Coulombs can be obtained in ionistor?

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SUMMARY

The maximum charge obtainable from an ionistor, which is essentially a supercapacitor, can be calculated using the formula Q=CV, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the voltage in Volts. For example, a supercapacitor rated at 12,000F and 3.3V can theoretically store approximately 39,600 Coulombs. However, practical applications must consider the limitations of supercapacitors, including maximum charge and discharge currents, to avoid damage or inefficiency.

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  • Understanding of electrical capacitance and the Farad unit
  • Familiarity with the formula Q=CV for calculating charge
  • Knowledge of supercapacitor specifications and limitations
  • Basic electrical engineering principles related to charge storage
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ivanovenkoi
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Hi
welcome to PF :smile:
ivanovenkoi said:
ionistors

what on Earth is that ?
the link you supplied goes to capacitors, and in particular, super capacitors

Capacitors have their Farad rating printed on them
 
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Ionistor is the same as capacitors. I mean capacitors exactly. I need the maximum available charge in Coulombs, not in Farads. Because Farad measures not charge but electrical capacitance.

P.S. Sorry for my poor English, I am not english speaker.

davenn said:
Hi
welcome to PF :smile:

what on Earth is that ?
the link you supplied goes to capacitors, and in particular, super capacitors

Capacitors have their Farad rating printed on them

Ionistors are capacitors in general. I need not Farad, because farad is electrical capacitance. I need charge in coulombs.
davenn said:
Hi
welcome to PF :smile:

what on Earth is that ?
the link you supplied goes to capacitors, and in particular, super capacitors

Capacitors have their Farad rating printed on them
 
ivanovenkoi said:
I need the maximum available charge in Coulombs, not in Farads.
Q=CV

So multiply the Capacitance in Farads by the Voltage in Volts, and you get the stored charge Q in Coulombs.
 
berkeman said:
Q=CV

So multiply the Capacitance in Farads by the Voltage in Volts, and you get the stored charge Q in Coulombs.
So can it be 10000F × 5V = 50000 Coulombs for ecample? Or it's too much? I need digit order. Because I read that 1 coulomb is a very big charge. So 50000 is extremely big, isn't it?
 
ivanovenkoi said:
So can it be 10000F × 5V = 50000 Coulombs for ecample? Or it's too much? I need digit order. Because I read that 1 coulomb is a very big charge. So 50000 is extremely big, isn't it?
10,000 Farads is a huge capacitor. Where are you finding something like that?
 
phyzguy said:
10,000 Farads is a huge capacitor. Where are you finding something like that?
https://en.m.wikipedia.org/wiki/SupercapacitorHere I see in the table below supercapacitors 100..12000F, 2.2..3.3V
So 12000F*3.3V is almoustly 40000 Coulombs, isn't it?
 
ivanovenkoi said:
Here I see in the table below supercapacitors 100..12000F, 2.2..3.3V
Be sure to read the limitations of supercaps. What is the application? You are not going to dump all of that charge all at once out of a supercap...
 
What berkeman said is right.
If it is for a practical application beware of the maximum charge and discharge current for the specific capacitor you buy.
 

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