# Capacitor inductive charging circuit

Thread Starter

#### Mark Flint

Joined Jun 11, 2017
97
I have 2 x 12.5v capacitor arrays made from 2.5v 1F caps.

After an operation in my circuit I have 6v in one array and 10v in the other.
I want to move 2.5v from the 6v up into the 10v array in the most efficient way. So the ideal result would be 3.5v in one array, and 12.5v in the other.
I have been reading an article on inductive charging by @crutschow which look to be what I need (https://bit.ly/2lJlB16)
I think I need a custom version of this circuit to match my components.
I'm not good with Spice and I can't understand the schematic on the page, but I would like to learn what's happening in that schematic, and build it at http://www.falstad.com as I have a bit of experience with that simulator.

The screenshot of the circuit published by crutschow: Last edited by a moderator:

#### crutschow

Joined Mar 14, 2008
27,184
I want to move 2.5v from the 6v up into the 10v array in the most efficient way. So the ideal result would be 3.5v in one array, and 12.5v in the other.
I have been reading an article on inductive charging by @crutschow which look to be what I need
Unfortunately that simple circuit can only go from an initial higher voltage to an initial lower voltage and, with two capacitors, won't give a higher voltage then either one had at the beginning.
For that you would need to use a switching boost converter.

Also, for the beginning and ending voltages you stated, the total energy of your two capacitors would be greater at the end then at the start, which is not possible due to the conservation of energy law (capacitor energy is proportional the the square of their voltage [E=1/2 CV²]).

Thread Starter

#### Mark Flint

Joined Jun 11, 2017
97
Unfortunately that simple circuit can only go from an initial higher voltage to an initial lower voltage and, with two capacitors, won't give a higher voltage then either one had at the beginning.
For that you would need to use a switching boost converter.
Thank you. Yes. In that case I can change my circuit so that I have a 25v cap (C2) which can be discharged into C1, also at 25v. At that point in time C1 will be at about 10v. So I would like to calculate components for discharge to the lowest voltage I can discharge C2 (presuming it is at 24v at the start of transfer) into C1 by using the inductor--diode method.

I have a copy of LT Spice. Learning to use it is another thing, but I'll make a start with a tutorial. It seems like this is going to be an essential part of my learning process.

#### crutschow

Joined Mar 14, 2008
27,184
Theoretically, the best you could do with 100% efficiency and two equal size caps, one at 25V and one at 10V, is end up with one cap at 26.9V,
This is because of the capacitor energy being equal to voltage squared, i.e. √(10²+25²) = 26.9.
So you see, you gain very little voltage by trying to transfer the energy between a low voltage and a high voltage capacitor, especially when you include the conversion process inefficiency.

LTspice does have a somewhat steep learning curve, but I think you will find it worthwhile if you want to learn something more about electronics. It allows you to easily experiment with various circuits and see the voltages and currents anywhere in the circuit.
Read the tutorials and play with some of the test circuits to help the process.

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Thread Starter

#### Mark Flint

Joined Jun 11, 2017
97
Also, for the beginning and ending voltages you stated, the total energy of your two capacitors would be greater at the end then at the start, which is not possible due to the conservation of energy law (capacitor energy is proportional the the square of their voltage [E=1/2 CV²]).
Yes, conservation of energy is pretty solid But the capacitor energy equations involving CV² only applies to resistive charging, right?

#### WBahn

Joined Mar 31, 2012
26,398
Yes, conservation of energy is pretty solid But the capacitor energy equations involving CV² only applies to resistive charging, right?
What does the type of charging have to do with it?

At any given time, the energy stored on a linear capacitor is ½CV². Unless you are bringing in energy from somewhere else, the sum of the energies on all caps at the end of the process can be no greater than the sum of the energies on all caps at the start.

Thread Starter

#### Mark Flint

Joined Jun 11, 2017
97
What does the type of charging have to do with it?

At any given time, the energy stored on a linear capacitor is ½CV². Unless you are bringing in energy from somewhere else, the sum of the energies on all caps at the end of the process can be no greater than the sum of the energies on all caps at the start.
Yes - perhaps I am confusing terms (charging vs. storage) - I'm not saying there can be "more" in an isolated system - or that the amount of energy stored is not ½CV². I'm was trying to clarify that there are methods of charging a capacitor that do not result in 50% loss. For example, it should be possible to charge a capacitor with 80% efficiency or better.

#### crutschow

Joined Mar 14, 2008
27,184
But the capacitor energy equations involving CV² only applies to resistive charging, right?
No.
I'm was trying to clarify that there are methods of charging a capacitor that do not result in 50% loss. For example, it should be possible to charge a capacitor with 80% efficiency or better.
Yes.
But the equation for energy stored in a capacitor is not affected by the efficiency of the charging or how it is charged, resistor involved or not. It only determines how much of the total energy ends up stored in the capacitor.
Even with 100% charging efficiency, the energy stored in the capacitor is always still E=1/2 CV² where V is the ending capacitor voltage.

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