# Electronic – Can a charge-pump be 100% efficient, given ideal components

dc/dc converterefficiency

A recent question about cyclically charging a capacitor reminded me of something I read once. As I remember, it demonstrated that it's impossible to construct a charge pump that is 100% efficient with ideal components, but it is possible to build a 100% efficient boost converter with an inductor if components are ideal.

Does this resonate (no pun intended) with anyone else? Any way to demonstrate or refute the truth of this?

To be clear: we are assuming we have ideal components. I realize no real circuit will be 100% efficient with real components. Diodes may have zero voltage drop. Transistors may be ideal switches that take no energy to change state. Wires may have zero resistance.

The problem with capacitors and a voltage source is like this: take a voltage source $$\V\$$ with a certain source impedance (= series resistor) $$\R\$$. Connect a capacitor $$\C\$$ to it and load it for an infinite time (any finite time will do too). The loading current will be $$I = \frac{V}{R}\exp\left(\frac{-t}{RC}\right)$$ and so the power dissipated over the resistor will be $$P = I^2 R = \frac{V^2}{R}\exp\left(\frac{-2t}{RC}\right)$$ Consequently, the total energy cost shall be $$E = \int_0^\infty P \,dt = \frac{V^2}{R} \int_0^\infty \exp\left(\frac{-2t}{RC}\right) \,dt = \frac{CV^2}{2}$$ which, as you can see, is both always positive and completely independent of $$\R\$$. Therefore, there is always a cost, and even with an ideal capacitor! Intuitively, this is because a smaller resistor causes a higher initial loading current, and hence a higher RI2 loss.