In the circuit shown below, the input voltage is sinusoidal and \(2.5~\text{V}\) peak-to-peak. The capacitors are \(20~\mu\text{F}\) each. Assume ideal diodes (zero knee voltage) and \(R_L\) is very large (almost infinite). Which one of the following options is \emph{closest to the peak-to-peak voltage across \(R_L\), after a large number of cycles?}
Show Hint
- A Greinacher/Villard stage ideally doubles the peak (and hence doubles the peak-to-peak) of the AC source at the high-impedance output.
- With ideal diodes and large \(R_L\), the stored charge is retained, giving a nearly constant doubled level.
Step 1: The network \((C_1,D_1,D_2,C_2)\) is a (Greinacher/Villard) \emph{voltage doubler}. After many cycles and with an almost open‐circuit load, \(C_1\) charges to the source peak \(V_p\), and the subsequent half–cycle stacks the source peak with the charge on \(C_1\) through \(D_2\). Hence \(C_2\) charges to approximately \(2V_p\). Step 2: The output across \(R_L\) (in steady state) therefore has a peak value \(\approx 2V_p\). Consequently, the peak–to–peak value at the output is about \(2\times(2V_p)=4V_p\). For the given input of \(2.5~\text{V}\) peak–to–peak (i.e., \(V_{pp,in}=2V_p=2.5~\text{V}\Rightarrow V_p=1.25~\text{V}\)), the doubler produces an output close to
\[
V_{pp,out}\approx 2\,V_{pp,in}\approx 2\times 2.5=5.0~\text{V}.
\]
(With ideal diodes and very large \(R_L\), ripple is negligible, so the obtained value is the closest among the choices.)
