# Diode circuit with AC excitation source

1. May 23, 2016

### Gbox

1. The problem statement, all variables and given/known data
Graph $V_{out}$

2. Relevant equations

3. The attempt at a solution

When $V_{in}=V$ C1 is positive on the left and negative on the right, and C2 is negative on the the down side and positive on the upper side so there is no Vout?

2. May 23, 2016

### CWatters

That's not correct.

For the moment lets assume that V is much greater than the forward voltage of the diode...

When Vin = +V the right hand diode is reverse biased (not conducting) so that can be temporarily removed. The left hand diode is forward biased (conducting) so replace it with a wire. Redraw the circuit to show this.

Then... Are you familiar with the "Potential Divider" circuit?

3. May 23, 2016

### Gbox

Why is the left hand diode is conducting? the capacitors do not affect the voltage? do not they both charge to the same voltage and the diode is not conducting?

4. May 23, 2016

### CWatters

No they don't charge to the same voltage (unless they are same value). The two caps are in series.

5. May 24, 2016

### CWatters

This is what the circuit looks like when Vin transitions from 0 to +V (assuming +V >> Vd)....