Hello guys
I'm studying the voltage doubler circuit as shown below and I have a confusion.In the first stage,C1 is charged to Vm (note the current direction) but in the second stage, c2 is being charged to 2Vm BUT this time the current is in opposite direction.Doesn't that mean C 1 should be discharged?

 

Yes, it does, partly loose its charge, how much is depending on the charge held in the next capacitors. But it has its charge transferred to C2 as the AC is now "pushing" it by being in series adding to the C1 voltage.
Any lost charge will be topped up on the next half cycle.

 

Hello guys
I'm studying the voltage doubler circuit as shown below and I have a confusion.In the first stage,C1 is charged to Vm (note the current direction) but in the second stage, c2 is being charged to 2Vm BUT this time the current is in opposite direction.Doesn't that mean CView attachment 136898 1 should be discharged?

Semi. If C1 and C2 are equal, C1 gives up half its charge to C2. Two capacitors in series form a voltage divider just like resistors.

ak

 

Thank you for the quick reply guys ,I attached a video which explains the voltage doubler below - It starts @02:10 .Here I see that the output voltage peak is 2Vm after several cycles of the input voltage.In all books I've read it does not mention about the cycles and discharging of C1 as seen in the video, they apply KVL directly to get Vout=2Vm (see the picture2 below).So guys, Is the explanation for the voltage doubler correct according to the book or to the video?- According to me , the real way and best way to explain the circuit is as the video , I think the book assumes the steady state behaviour of this circuit .I need your advice on this ,Am I right?

 

Hello,

The diodes act like switches. see this video for an explanation:

Bertus

 

Hello,

The diodes act like switches. see this video for an explanation:

Bertus

Hello Thank you bertus , Im a big fan of EEVblog and I just missed this part where he says ' Assume the circuit has reached a steady state' at 2:30 I can now clarify my doubt - The book also assumes the circuit reached a steady state.

 

Hello guys
I'm studying the voltage doubler circuit as shown below and I have a confusion.In the first stage,C1 is charged to Vm (note the current direction) but in the second stage, c2 is being charged to 2Vm BUT this time the current is in opposite direction.Doesn't that mean CView attachment 136898 1 should be discharged?

Using electrolytics is less common than smaller non polarised - which you can usually get away with by increasing the frequency.

The input electrolytic can be 2 in inverse series with parallel protection diodes - normally 2 identical capacitors in series gives half the capacitance, but not with the parallel diodes in place.