A half-wave (Greinacher) voltage doubler is assembled as follows

A half-wave (Greinacher) voltage doubler is assembled as follows. An AC source \(v_s(t)=15.0\sin(\omega t)\,\mathrm{V}\) with \(\omega=2\pi\cdot 400\,\mathrm{rad/s}\) is connected from ground to node \(P\). Capacitor \(C_1=10.0\,\mu\mathrm{F}\) joins \(P\) to node \(Q\). Diode \(D_1\) has its cathode at \(Q\) and its anode at ground. Diode \(D_2\) has its anode at \(Q\) and its cathode at node \(R\). Capacitor \(C_2=10.0\,\mu\mathrm{F}\) joins \(R\) to ground. The DC output is \(v_o=v_R\), taken with no attached load (open circuit at \(R\)). Use the constant-drop model with \(V_F=0.720\,\mathrm{V}\) for both diodes. Ignore leakage, reverse recovery, ESR, and source resistance. The circuit is in periodic steady state. Determine the DC voltage on \(C_1\) (plus at \(P\), minus at \(Q\)) and the DC voltage on \(C_2\) (plus at \(R\)). Determine the open-circuit output \(v_o\). For each diode, determine the peak inverse voltage that appears across it in the off state, with the sign convention \(v_{AK}=v_{\mathrm{anode}}-v_{\mathrm{cathode}}\), and report the most negative value of \(v_{AK}\) in volts. State whether that PIV exceeds \(2V_{s,\mathrm{peak}}\).

\(D_1\) is a negative clamp on node \(Q\); \(D_2\) then peak-rectifies the clamped waveform onto \(C_2\).