Half-wave rectifier with capacitor-input filter: If the input to the ideal diode rectifier is v(t) = Vm sin(ωt), what is the required Peak Inverse Voltage (PIV) rating of the diode?

Difficulty: Easy

Correct Answer: 2 Vm

Explanation:


Introduction / Context:
PIV (Peak Inverse Voltage) is the maximum reverse voltage that a rectifier diode must withstand. Choosing an adequate PIV is crucial for reliability, especially in capacitor-input filters where the reservoir capacitor holds a voltage between conduction intervals.


Given Data / Assumptions:

  • Half-wave rectifier.
  • Capacitor-input (reservoir) filter holds approximately Vm at the output between peaks.
  • Ideal diode and negligible transformer resistance for peak estimation.


Concept / Approach:

At the instant just after the capacitor has charged to Vm, the input becomes negative. The diode is then reverse-biased by the instantaneous source voltage in series with the stored capacitor voltage. With the source at approximately −Vm and the capacitor still ~ +Vm, the diode sees a reverse voltage close to Vm + Vm = 2 Vm.


Step-by-Step Solution:

At the negative peak: vs ≈ −Vm.Capacitor voltage: vc ≈ +Vm (little discharge between peaks under light load).Reverse voltage across diode: VR,max ≈ vc − vs ≈ Vm − (−Vm) = 2 Vm.


Verification / Alternative check:

Classic rectifier design tables list: Half-wave with capacitor-input → PIV ≈ 2 Vm; Full-wave bridge with capacitor-input → PIV ≈ Vm per diode, confirming the result.


Why Other Options Are Wrong:

  • Vm: valid for some bridge configurations, not for half-wave with a reservoir capacitor.
  • 3 Vm or 4 Vm: overly conservative for the idealised case; increases cost without benefit.
  • Vm/2: far too low and unsafe.


Common Pitfalls:

  • Ignoring the stored capacitor voltage when computing reverse stress.


Final Answer:

2 Vm.

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