Diode and capacitor

Thread Starter

gbox

Joined Dec 29, 2015
42
צילום מסך 2016‏.05‏.23 ב‏.18.30.18.png

I need to graph Vout.

when Vin=V C1 and C2 charge to half the voltage of Vin, both diodes are off so Vout=Vin-C1=C2.
But when Vin=-V the rightmost diode is on and Vout=-Vin+C1?
 

WBahn

Joined Mar 31, 2012
29,976
Assuming zero leakage, that the two capacitors are equal, and that they start out initially uncharged, then they will charge to the same voltage as Vin increases from zero to some positive voltage because the right diode will be reverse biased and all of the current must flow through both capacitors. In steady state, there will be zero volts across the diode (since even a small forward voltage means some current is flowing and so both capacitors will still be charging, with the voltage across them increasing, and hence the voltage across the diode decreasing.

But as the voltage is decreased, C1 will discharge but C2 will not.

This is a pretty classic flying capacitor voltage multiplier circuit.
 
Last edited:

johnmariow

Joined May 4, 2016
19
Assuming zero leakage, that the two capacitors are equal, and that they start out initially uncharged, then they will charge to the same voltage as Vin increases from zero to some positive voltage because the right diode will be reverse biased and all of the current must flow through both capacitors. In steady state, there will be zero volts across the diode (since even a small forward voltage means some current is flowing and so both capacitors will still be charging, with the voltage across them increasing, and hence the voltage across the diode decreasing.

But as the voltage is decreased, C1 will discharge but C2 will not.

This is a pretty classic flying capacitor voltage multiplier circuit.
Thanks for the correction. I didn't recognize the circuit as a multiplier.
 
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