# Diode and capacitor

Discussion in 'Homework Help' started by gbox, May 23, 2016.

1. ### gbox Thread Starter Member

Dec 29, 2015
42
0

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?

2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
What if C1 is a thousand time larger (or smaller) than C2?

3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
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: May 23, 2016
gbox likes this.
4. ### Bordodynov Active Member

May 20, 2015
670
194
See
C2=100C1

Last edited: May 24, 2016
5. ### johnmariow New Member

May 4, 2016
20
1
Thanks for the correction. I didn't recognize the circuit as a multiplier.