Question about charge pumps capacitors

Thread Starter


Joined Feb 1, 2013
hi all,

I know that charge pump is a DC to DC converter that uses capacitors for creating an output voltage that is either higher or lower than the input voltage .

how does it creates at the output a higher voltage than the input ?
is it lowers the current at the output in the same ratio ?

thanks !
Last edited:

Potato Pudding

Joined Jun 11, 2010
A charge pump restacks voltages.

For a voltage increase-
You charge a capacitor and then raise the negative node to a positive value, then take current off of the positive node.

With a 5 Volt supply you would likely charge the capacitor to 5 Volts then change the connections so that the capacitors negative node is raised to 5 Volts which would make the positive node 5 Volts, + 5 Volts. Diode drops will reduce that to less than 9 Volts actual output. Repeat continuously and very fast to get a reasonably steady output.

The load current must be very small or else you will end up with much less than 9 Volts out in this example.

Thread Starter


Joined Feb 1, 2013
first of all , thanks for your answer

and regard to your answer:
how do we change the connections ? with some king of switching ?
voltage increase (dc to dc) must have some king of switching in order to work ?