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This is a three phase AC voltage being rectified and boosted. I have another circuit showing the same but with only a single 'boost' (Voltage doubler).
When the switch is closed a current can flow and the voltage will drop. Play with the clock and discharge resistor to study the effects.
Notice that as the load increases (resistance decreases) to a low enough value and the button is depressed long enough, the voltage will drop so that the otherwise highest voltage will drop below the voltage of the lowest stage. despite the "boosting". Try to think about basic electricity theory and figure it out yourself before continueing to read.
Depending in the frequency of the motor, the winding resistance of the motor and the resistance of the load, the voltage may be boosted as long as sufficiënt current can be provided as opposed to the current drained through the load. If the load is too high, however, current will not be able to charge the capacitors sufficiently so that is will simply flow from ganarator to load through the diodes while being rectiefied resulting in a lower voltage at the third stage than at the fist stage due to voltage drop accross the diodes.
In this setup, when the load resistor is set to aprox 500 Ohms, depressing the switch will cause the voltages to drop. The voltage after the first boost will be higher than the starting voltage but the voltage after the second boost will be lower than the starting voltage. This is due to the diode voltage drop causing energy to be wasted (standard everycircuit set diodes).
In this example the load resistance should be set to above 1.1 kOhm for the boosting effect to remain while the switch is depressed. Any lower resistance will drain the current too fast which will cancel the effect.
While the voltage at the output of the last boost stage will be considerably higher, the current that can be drained from here will be much lower. Assuming no voltage drop accross the diodes and zero resistance conduits, if the voltage is trippled, the maximum current drain will be reduced to a third as energy needs to be conserved.
I hope you enjoy the circuit. Be carefull when attempting to reproduce this in real life since charge pumps can output huge voltage if not carefull.
More stages could be added to so that every stage adds to the starting voltage resulting in huge voltage gain at reduced current. Be mindfull that every stage will add considerable resistance and iff a current is allowed to flow it will also be dissipated through all the diodes and capacitors so that repeating this process is useless. Practical applications for multiple charge pumps are therefore more uncommon as the ammount of boosting stages increase.
Charge pumps can also be used to increase a DC voltage but require circuitry that will pulse the current. Often this is not the most efficiënt way to achieve this so that boost-converters would be used in stead. When examining a boost converter it is easy to see that it shares similarities with the charge pump.
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