When reading about charge pumps you will encounter a phrase like "they act like resistors". The following example demonstrates why they say that.
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?
- A Charge Pump that Generates Positive and Negative High Voltages with Low Power-supply Voltage and Low Power Consumption for Non-volatile Memories
- LTSPICE: Charge Pump Voltage Doubler V3