Average DC output of a 3-phase fully controlled bridge A 3-phase fully controlled bridge converter is fed from a 3-phase source whose phase voltage is v = Vm sin(ωt). If the firing angle is α (radians), what is the average DC output voltage Vdc (ignore overlap)?

Introduction / Context:The 3-phase fully controlled bridge (six-pulse) is a workhorse rectifier in drives and HVDC front-ends. Its average DC output depends on the firing angle α and the amplitude of the phase voltages.

Given Data / Assumptions:

• Phase voltage given as v = Vm sin(ωt), so Vm is the peak value of the phase-to-neutral sinusoid.
• Ideal commutation (no overlap), balanced source.

Concept / Approach:For a six-pulse converter expressed in terms of phase peak Vm, the average output with firing angle α is Vdc = (3 Vm / π) cos α. This follows from integrating the line-to-line segments that appear at the DC terminals during each 60° conduction interval and summing over one electrical cycle.

Step-by-Step Solution:At α = 0 (natural commutation), the ideal average is Vdo = 3 Vm / π.Introducing delay α reduces the projection by cos α: Vdc = Vdo * cos α.Therefore, Vdc = (3 Vm / π) cos α.

Verification / Alternative check:When phase voltage is specified by RMS value Vph(rms), an equivalent common form is Vdc = 2.34 Vph(rms) cos α. Rewriting with Vm = √2 Vph(rms) recovers (3 Vm / π) cos α.

Why Other Options Are Wrong:(3√2 Vm / π) cos α: Double-counts √2; Vm here is already peak.Forms with (1 ± cos α): Not applicable to this rectifier’s average expression.

Common Pitfalls:Mixing up whether Vm is phase peak, line peak, or RMS; always track the definition used in the problem statement.

Final Answer:Vdc = (3 Vm / π) cos α