In a full-wave converter with mid-point (M–2) connection, each thyristor has a PIV of 400 V. For the same output voltage using a fully controlled bridge converter, what is the required PIV per thyristor?

Introduction / Context:Peak inverse voltage (PIV) rating depends on converter topology. For a given dc output, different rectifier configurations impose different maximum reverse voltages on each device. This question compares the mid-point (M–2) full-wave converter and the full bridge (B–2) converter.

Given Data / Assumptions:

• M–2 converter PIV per thyristor = 400 V.
• Same dc output voltage to be obtained using a fully controlled bridge.
• Ideal devices and transformer, standard rectifier relationships.

Concept / Approach:

For a single-phase center-tapped (M–2) converter, PIV per device ≈ 2Vm (peak of half-secondary seen reflected). For a single-phase full bridge (B–2), PIV per device ≈ Vm. Thus, for the same dc output, the B–2 requires roughly half the PIV of M–2 per device.

Step-by-Step Solution:

Verification / Alternative check:

Textbook rectifier PIV results: center-tapped diode/thyristor PIV = 2Vm; bridge PIV = Vm. The ratio is consistently 2:1 for equal dc outputs with comparable transformer secondaries.

Why Other Options Are Wrong:

• 400 V and 800 V: correspond to center-tapped values or doubled ratings; unnecessary for the bridge at the same dc output.
• 2002 V: unrealistic and not topology-derived.
• 100 V: underestimates required blocking capability.

Common Pitfalls:

• Mistaking line-to-line peak with half-winding peak.
• Ignoring that bridge shares reverse stress between legs differently than M–2.

Final Answer:

200 V