Hi Abbas Sabti ,

your assumption is correct only in the case of charging by a DC source via a resistor.

But if you replace the resistor by an inductance, a transistor controlled by a PWM signal, and a diode (please see the attached file for a simple example), you get two advantages simultaneously:

- The final voltage at the capacitor can be higher or lower than that of the DC source.

- The efficiency can be easily about 80%, and in "benign" cases well beyond 90%.

The efficiency figure might possibly also depend on how one defines this efficiency.

https://www.purdue.edu/discoverypark/energy/assets/pdfs/SUFG/publications/SUFG%20Energy%20Storage%20Report.pdf

Efficiency is determined by the input to output energy ratio. For most energy storage technologies, energy is lost in the process of being stored and discharged. Energy may also be lost while the device is not in use, and these losses are called standby losses. Standby losses are a measure of efficiency that compares how much of energy used to charge a storage device is lost before discharge. Efficiency can be affected by ambient conditions such as temperature. Some technologies require ancillary devices, which require power, to connect them to the grid. These “parasitic” loads reduce efficiency much like standby losses.

People often play games when they write specs for their product.

I think since capacitors are not used as DC energy storage devices like batteries, It is not correct to define their efficiency from this point of view. Even if someone wants to define their efficiency in the way that you mentioned, the time in which the capacitor is going to save energy should be taken into account. Since no capacitor is ideal, when time tends to infinity, every capacitor is going to be completely depleted. So all the capacitors have got zero efficiencies when no constraint is defined for time.

On the other hand, if the capacitor is continuously connected to a DC voltage source, the efficiency is dependent on the circuit which the capacitor is supplying the power for it in parallel with the voltage source. The efficiency would be dependant on the current passing through the internal resistor of the capacitor based on "P=IR^2" and the power which is transferred to the circuit.

I'm not sure why you considered at most 50 percent efficiency for capacitors. The maximum probable power efficiency for voltage sources is 50 percent which happens when impedance matching is done and the internal impedance of the voltage source is equal to the impedance of the load circuit. But capacitors cannot be considered like voltage sources because they have not constant DC voltages or amplitudes of AC voltages when they are not connected to a voltage source.

Manufacturers have defined the efficiency of the capacitors based on their AC behavior and AC power loss due to their internal resistance when they are constantly charged and discharged and are acting as a kind of interface for transferring electrical energy usually obtaining particular signals or waveforms which is a correct criterion based on usual usages of capacitors. When the capacitors are acting as an interface for transferring energy, the comparison between their input and output power is the criterion for defining their efficiency which is completely different from the power transfer efficiency of batteries which is at most 50 percent. Capacitors do not produce any energy to be treated like batteries or energy sources when we are defining their efficiencies.