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.

The horizontal resistors represent the wire resistance, the vertical resistor at the right represents the load. One capacitor would be sufficient to separate a resistive load from the gen coil. However, one might want to connect a small HV cap to the gen coil for power extraction and have a bigger LV cap (or a battery) as an energy reservoir.

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.


We first solve the homogeneous part of this differential equation with the following approach:

This is true for all only if

A solution for the homogeneous differential equation is therefore


We find a solution of the inhomogeneous differential equation by varying the factor.


We substitute this result into Eq. 4

and thus get a particular solution for the inhomogenous differential equation Eq. 1:


Since we have

If we assume

we get


We substitute into Eq. 5 and get


We get the current by taking the derivative of Eq. 7:


The voltage in the source cap is given by


We plot these three functions.

We have


Energy Consideration

The initial energy in the system is given by

The energy after the charge transfer is given by

The difference between these two energies is the heat loss.


The goal was to pump energy from a small HV cap into a larger LV cap (energy reservoir). This means we have and thus . Almost the entire energy is lost in the wire resistance as heat. For we have meaning that half the energy is lost. Note that is independent of so using thick wire wouldn't help a bit.

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 causes an unacceptable heat loss. For two identical caps half of the energy is lost. For a smaller source cap the outcome is even worse. It is therefore suggested to introduce a choke into the circuit as shown below.

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.


We solve the homogeneous differential equation


first using the following approach:

This equation is true for all if we have

This equation is solved by


We consider first.

We can rewrite this as



We solve the inhomogeneous differential equation by finding a particular solution as follows:

We substitute this as into Eq. 11.

The general solution for Eq. 11 using is then

We demand . This gets us

We further demand

The special solution for our problem honoring our start conditions is therefore


The voltage in the target cap is given by

The current is given by

The voltage in the source cap is given by



The different outcome is obvious and this time highly dependent on the size of the wire resistance .

Energy consideration

The energy in at is given by

The energy in target cap after time is given by

The following figure shows the development of (energy in ) in relation to the initial input energy .

There is a point with maximal current in the choke. The voltage potential between and plus the voltage drop over has dropped to zero and the choke starts to discharge its magentic energy into the target cap. The heat loss is minimal inspite of a significant wire resistance in this example. We create a new plot with more likely values for the components.

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.

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


We again solve the homogenous equation first.

This equation is solved by


We consider first.

We solve the inhomogeneous differential equation by finding a particular solution as follows:

We substitute this as into Eq. 17.

The complete general solution is then

We assume :


We assume :


We substitute this into Eq. 18:

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