## DC voltages, someone shoot me down please!

Hi, Im trying to double DC voltage from 3v to 6v

apparently this circuit works...

charge in parallel discharge in series,

can somebody shoot me down and tell it wont work please?

Thanks

Anthony
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 Also, I know the resistors are ridiculously small, I just wanted to charge the capacitor quickly...
 Looks like a pretty standard flying capacitor based voltage doubler. It works but you don't have any power gain so while your voltage is doubled your ability to drive anything with that higher voltage is reduced. Also beware of charge sharing with whatever you connect the higher voltage to.

## DC voltages, someone shoot me down please!

well thats a relief, but why does it not give more power? i thought the capacitors could release much higher currents than batteries, so if the voltage has increased, and the current is higher, surely power will increase :(
 It doesn't give more power because if it did it would be a perpetual motion machine and violate the second law of thermodynamics. To get more power you need an active device in there somewhere. People use circuits like you made sometimes in practice. For example, in a Flash memory typically the high voltage drive used to write a bit is generated using a switched-capacitor voltage doubler similar to what you showed. Also, a capacitor stores charge. It can only release as much charge as it has stored on it. Remember current is nothing more than charge/time.
 if the battery transfers 50 coulombs of energy to a capacitor and it takes 100 seconds to do so because the current supplied from the battery is low, say 0.5A, the charge = 100 x 0.5 A = 50 Coulombs, then if this charge is released in 10 seconds instead, the current flow will be 5A... and the voltage will remain 12v (for the most part)... therefore, same charge transferred in less time = more power?
 Nice try! You came *this* close to inventing a perpetual motion machine! ;-) Power isn't charge/time, that's current. Power is voltage*current. When you charge a cap with a constant current the voltage will increase linearly since the voltage across a cap is proportional to the integral of I. In your argument, you are assuming the discharging current is constant as well. However, when you release it, the voltage is now a decaying exponential with a time constant of RC (where R is the discharging resistance). Because cap current is proportional to dV/dt, the current is also an exponential. The following link explains it pretty well: http://hyperphysics.phy-astr.gsu.edu...capdis.html#c2 You're thinking well! You'll get it! :)
 Recognitions: Gold Member Science Advisor to put it more simply.... you initially have 3V at lets say 2Amps ... that 6Watts if you double the voltage you will halve the current available so now you have 6V @ 1 Amp still = 6W you cannot generate power ( energy ) out of thin air ;) Dave

 Quote by ajd-brown if the battery transfers 50 coulombs of energy to a capacitor and it takes 100 seconds to do so because the current supplied from the battery is low, say 0.5A, the charge = 100 x 0.5 A = 50 Coulombs, then if this charge is released in 10 seconds instead, the current flow will be 5A... and the voltage will remain 12v (for the most part)... therefore, same charge transferred in less time = more power?
Yes, exactly. You get more power but for a shorter amount of time. And since energy = power * time. The energy stays the same.
Well, actually the energy is lower. The energy in a capacitor is 1/2 * charge * voltage. When charging a capacitor through a resistor half the energy gets lost in the resistance.

What carlgrace meant is the average power you get when opening and closing those switches repeatedly.