The following circuit can be used for doubling any DC source voltage (up to 15 V DC). The presented design will double any voltage between 4 to 15 V DC and will be able to operate loads at current not more then 30 mA.As can be seen in the diagram, this DC voltage doubler circuit employs just a single IC 4093 for achieving the proposed result.
The IC 4093 has six gates in all which are all effectively for generating the discussed voltage doubling actions.
Two of the gates out of the six are configured as an oscillator. The extreme left of the diagram shows the oscillator section.
The 100 K resistor and the 0.01 capacitor form the basic frequency determining components.
A frequency is imperatively required if a voltage stepping actions needs to be implemented, therefore here too the involvement of an oscillator becomes necessary.
These oscillation become useful for initialing the charging and discharging a set of capacitors at the output which amounts to the multiplying of the voltage across the set of capacitors in a such a way that the result becomes twice the applied supply voltage.
However the voltage from the oscillator cannot be preferably applied directly to the capacitors, rather its done through a group of gates of the IC arranged in a parallel way.
These parallel gates together produce a good buffering to the applied frequency from the generator gates so that the resultant frequency is stronger with respect to current and does not falter with relatively higher loads at the outputs.
But still keeping the specifications of a CMOS IC in mind the output current handling capacity cannot be expected to be larger than 40 mA.
Higher loads than this will result in the deterioration of the voltage level toward the supply level.
The output capacitor values can be increased to 100uF for getting reasonably higher efficiency levels from the circuit.
With 12 volts as the supply input to the IC, an output of around 22 volts may be acquired.