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How does this circuit work

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jinyong

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Please help explain this voltage doubler circuit. I don't understand how point 2(the negative side of C1) can become positive all of a sudden and forward bias diode 1.

Someone tried to explain it like With V1 raising the + side of C1, node 2 will start to rise by the same amount, but is stopped short when it reaches ~0.7V and D1 turns on. Why will node 2 will start to rise by the same amount?? Please help!
 

hih dear,
check the attached file i hope it have something for you
 

for positive half cycle, the diode D1 is forward biased and D2 is reverse biased. So, the capacitor is charged with the peak value (Vpp) of the supply waveform in polarity such that current is entering at positive terminal polarity of capacitor.
So, the complete supply wave is clamped in negative direction.
For negative half cycle of the supply waveform, diode D1 is R.B. and diode D2 is F.B., so the capacitor C2 is charged with 2Vpp.
Still problem, then I may send you the scanned write up to you.
 

jinyong,
The output polarity in the diagram is reversed. This is the circuit for a negative supply. Operation is as follows:
.
On the positive half cycle, D1 is forward biased and C1 charges to the peak. Node 1 positive wrt node 2. D2 is reverse biased.
.
On the negative half cycle, VI and the voltage on C1 are in series-aiding, with a total voltage of 2Vpeak. D2 is forward biased, and C2 is charged to 2Vpeak.
.
To get a positive supply, reverse the diode directions.
Regards,
Kral
 

For the +ve cycle of source(V) diode D1 is forward fiased and the capacitor starts charging with source side +ve and the point2 is -ve ande the -ve terminal of the source is more -ve than point 2 so the diode is forward biased again when the polarity changes the diode D2 forward biases with point3 +ve so now the voltage across D2 = source voltage(V)+voltage across capacitor C1(V)i.e previously charged=2V. hence the voltage is doubled.
 

I may send you the scanned write up to you.
 

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