Here https://www.maximintegrated.com/en/app-notes/index.mvp/id/725 and here https://www.analog.com/media/en/technical-documentation/data-sheets/ADP3603.pdf

The same calculation for the power dissipation given by the pump capacitor in a switching capacitor voltage regulator is made.

What I don't understand, is how a lossless component becomes a lossy component. It is said that the equivalent resistance is not just something that gives you the same RMS current, but it also models the power loss given by switching the capacitor.

It might be the power lost in the ESR or in the Ron of the switches, I thought at first, but these losses are already took into account by other factors explained in other parts of the documents.

So, what other effect does the factor \$ I_{load}*f*C_{pump} \$ really take into account?


1 Answer 1


The process of charging a capacitor via a resistor, even if the resistor is very small, produces an energy (or IIR) loss each cycle. An inductor based switching regulator doesn’t suffer from this theoretical loss because the energy that can be put into an inductor by connecting it momentarily across a supply voltage (as in the case of a boost regulator), is performed losslessly.

Going back to the charging capacitor scenario, it’s an interesting subject that has accrued several questions on this site and, it can be shown that whatever value of series charging resistor, the energy lost per cycle in the resistor is the same. Basically, the energy taken from the supply voltage to top up the capacitor is always more than what eventually gets stored in the capacitor hence, it takes on the mantle of an inefficient component even though theoretically that inefficiency is external to the capacitor.

  • \$\begingroup\$ Any reason for the downvote? \$\endgroup\$
    – Andy aka
    Sep 19, 2019 at 9:10
  • \$\begingroup\$ However, the loss that you mentioned in a simple RC circuit can be taken into account by the power loss on the resistor. In the case of the pump capacitor, the loss on all the resistors that are in the path of charge and discharge of the capacitor is already taken into account by the other factors. So it seems that IfC is taking into account a loss of a different nature. Isn't it? \$\endgroup\$
    – Tripola
    Sep 23, 2019 at 18:24
  • \$\begingroup\$ @Elia See electronics.stackexchange.com/a/55001 \$\endgroup\$
    – ManRow
    Nov 22, 2020 at 14:16

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