Energy transfer from one capacitor to another
It was suggested to extract power from a generator coil into a fist cap, to disconnect this cap from the generator coil and connect it to a second cap, then disconnect the second cap from the first and pump the charge of the second cap into a load.
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We assume a voltage in
and no voltage in
at
. The switch between the two caps is closed allowing current to flow from the first to the second capacitor.
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![]() | (1) |
We first solve the homogeneous part of this differential equation with the following approach:
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This is true for all

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A solution for the homogeneous differential equation is therefore
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![]() | (3) |
We find a solution of the inhomogeneous differential equation by varying the factor.
![]() | (4) |
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We substitute this result into Eq. 4
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and thus get a particular solution for the inhomogenous differential equation Eq. 1:
![]() | (5) |
Since

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If we assume

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![]() | (6) |
We substitute

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![]() | (7) |
We get the current by taking the derivative of Eq. 7:
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![]() | (8) |
The voltage in the source cap is given by
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![]() | (9) |
We plot these three functions.
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We have
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Energy Consideration
The initial energy in the system is given by
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The energy after the charge transfer is given by
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The difference between these two energies is the heat loss.
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![]() | (10) |
The goal was to pump energy from a small HV cap into a larger LV cap (energy reservoir). This means we have






Energy transfer from one capacitor to another via a choke
We have shown in Energy transfer from ... that transfering energy from one cap to another by means of a wire with
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Once the source cap has reached a low enough potential the coil becomes an active component and converts its magnetic energy back into current and thus contributes to the filling of the target cap. The above circuit (switch closed) is described by the following differential equation.
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![]() | (11) |
We solve the homogeneous differential equation
![]() | (12) |
first using the following approach:
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This equation is true for all

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This equation is solved by
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with
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We consider

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We can rewrite this as
![]() | (13) |
![]() | (14) |
We solve the inhomogeneous differential equation by finding a particular solution

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We substitute this as

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We demand

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The special solution for our problem honoring our start conditions is therefore
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![]() | (15) |
The voltage in the target cap is given by
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The current is given by
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The voltage in the source cap is given by
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with
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The different outcome is obvious and this time highly dependent on the size of the wire resistance .
Energy consideration
The energy in

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The following figure shows the development of



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There is a point with maximal current in the choke. The voltage potential between



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Our circuit equations do not honor the diode that suppresses oscillations and prevents from being charged negatively. The energy of the coil would mainly go into
where it belongs with the diode in place. We can conclude that the choke approach transfers almost 100% of the energy in
to
with no apparent heat loss. In a real application the inductivity should be chosen as small as possible to ensure fast enough energy transfer from left to right but large enough to prevent significant voltage drops over the wire.
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Finding the optimal inductivity
We consider the inductivity to have the optimal value when the efficiency has a maximum. Handling this problem with a changing supply voltage in a bit complex. We therefore replace the energy source cap with a battery (constant voltag power supply). We then have
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![]() | (17) |
We again solve the homogenous equation first.
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This equation is solved by
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with
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We consider first.
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We solve the inhomogeneous differential equation by finding a particular solution as follows:
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We substitute this as

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The complete general solution is then
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We assume

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![]() | (18) |
We assume

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We substitute this into Eq. 18:
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