3-phase bridge inverter with star-connected resistive load A three-phase bridge inverter is supplied from a 400 V DC source (battery). The load is star-connected with resistance 10 Ω per phase. What is the peak value of the thyristor current under steady operation (six-step, 120° conduction)?

Introduction / Context:
Six-step three-phase bridge inverters apply stepped line-to-line voltages to a three-phase load. For a star-connected resistive load, current is instantaneously proportional to phase voltage. Knowing the peak device current is essential for device selection and thermal design.

Given Data / Assumptions:

• DC link Vdc = 400 V.
• Load: star-connected, purely resistive, R_phase = 10 Ω, no neutral.
• Ideal switches, six-step (square) operation, 120° conduction.

Concept / Approach:

In a 3-phase bridge inverter without neutral, the phase voltage waveform takes levels of ±(2/3)Vdc and ±(1/3)Vdc relative to the load neutral, derived from the instantaneous pole potentials. For a resistive load, the phase current follows the phase voltage. The peak phase voltage magnitude is 2Vdc/3, so the peak current is that voltage divided by the phase resistance.

Step-by-Step Solution:

Verification / Alternative check:

Device conduction path in each 60° interval includes one upper and one lower switch; each device current equals the phase current during its conduction interval, so peak device current equals peak phase current for a resistive load.

Why Other Options Are Wrong:

20 A and 24 A underestimate the current because they assume lower phase voltage levels; 30 A overestimates by assuming Vdc/ R_phase or ignoring the 2/3 factor.

Common Pitfalls:

Confusing line-to-line with phase voltage, and using Vdc directly instead of 2Vdc/3 for the peak phase level in a six-step inverter without a neutral connection.

Final Answer:

26.7 A