Capacitor Charge Sharing

Published by Fudgy McFarlen on

When reading about charge pumps you will encounter a phrase like "they act like resistors".  The following example demonstrates why they say that.

Capacitor-Charge-Transfer-Schem

 

So each capacitor ends up with 1/2 Volt on it.  That means the system  starts with 1/2 Joule of energy and when the transfer of charge is done total energy in the system is 1/4 Joule.  

So what happened to the lost 1/4  joule?  It got dissipated in the series resistor.  What about if you do not use a resistor and just charge up C1 and connect it to C2?  

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