Effect of source inductance (overlap) in an n-pulse rectifier An n-pulse line-commutated rectifier is fed from a source with inductance L. If the average load current is I0, what is the average output voltage drop attributable to commutation overlap?

Introduction / Context:
In line-commutated converters, finite source inductance causes a commutation overlap angle during which two valves conduct simultaneously. This overlap reduces the average DC output voltage versus the ideal value. Quantifying this drop is essential for accurate performance prediction of rectifiers, drives, and HVDC terminals.

Given Data / Assumptions:

• n-pulse rectifier (n = 6 for three-phase bridge, n = 12 for series/parallel 6-pulse pairs, etc.).
• Source inductance L represents commutating inductance per phase path.
• Average DC load current is I0 and assumed reasonably ripple-free.
• Steady operating point; small-signal effects and resistance neglected.

Concept / Approach:

During overlap, part of the line-to-line voltage is dropped across the source inductances rather than appearing at the DC output. Classical analysis integrates the line voltage lost while current transfers from the outgoing to the incoming valve. The result is a proportional reduction in average DC voltage that scales with n, ω, L, and I0.

Step-by-Step Solution:

Verification / Alternative check:

For a 6-pulse bridge, set n = 6 → ΔV = 6 ω L I0 / π, which matches standard textbook expressions. Increasing L or I0 increases the drop; higher pulse number increases commutations per cycle, thus larger total loss per electrical period.

Why Other Options Are Wrong:

(a) and (b) omit the dependence on pulse number n; (c) contains an extra 2 in the denominator leading to a lower, incorrect value; (e) has inverse dependence on n, which contradicts the physics (more commutations → more loss).

Common Pitfalls:

Confusing overlap-caused drop with resistive drop; ignoring that L here is the effective commutating inductance per path; forgetting the linear dependence on n.

Final Answer:

ΔV = (n ω L I0) / π