The Junction Rectifier — Question 8

Question 8

Peak inverse voltage is largest when source and capacitor oppose

An ideal half-wave rectifier charges a capacitor to the positive source peak $V_m=20.0\,\mathrm V$. The source then reaches $-V_m$ while capacitor discharge is negligible. The diode anode is connected to the source and cathode to the capacitor.

1. Find the diode voltage at the negative peak and the required peak inverse voltage.

2. If the capacitor has dropped by $\Delta V=1.00\,\mathrm V$ by that instant, recalculate the reverse voltage and state the conservative ideal limit.

Question 8 - Solution

Peak inverse voltage is largest when source and capacitor oppose

(a) Method and calculation

$$V_D=V_A-V_K=-V_m-(+V_m)=-2V_m.$$$$\boxed{{\mathrm{PIV}}=2V_m=40.0\,\mathrm V}.$$

(b) Interpretation and further calculation

$$V_D=-V_m-(V_m-\Delta V)\Rightarrow\boxed{|V_D|=2V_m-\Delta V=39.0\,\mathrm V}.$$$$\boxed{\max|V_D|\le2V_m}\quad\text{for the stated ideal source and capacitor conditions}.$$

Physical check

The capacitor holds its positive polarity while the source reverses. This two-peak requirement is specific to the described half-wave capacitor-input circuit.