Friday, 27 February 2026

Page 48 – Ultra Hard Multi-Step Problems

Subject: Power Electronics
Level: GATE / PSU Ultra Advanced

🔹 Problem 1 – Combined SPWM + Frequency + Voltage

A 3-phase inverter has Vdc = 750 V. Modulation index mₐ = 0.85. Switching frequency = 12 kHz. Fundamental frequency = 50 Hz. Find: 1) Maximum line voltage 2) Switching period 3) Frequency ratio mf

Step 1: Line voltage (SPWM) VLL = 0.866 × mₐ × Vdc = 0.866 × 0.85 × 750 = 552 V

Step 2: Switching period Ts = 1 / fs = 1 / 12000 = 83.3 μs

Step 3: Frequency ratio mf = fs / f = 12000 / 50 = 240

Final Answers: VLL ≈ 552 V Ts ≈ 83 μs mf = 240

🔹 Problem 2 – SVPWM Sector + Switching Times

Vdc = 600 V Ts = 100 μs Reference voltage = 400 V Angle α = 35° Find sector and T₁, T₂ (approx).

Sector width = 60° 35° lies in Sector 1

T₁ = Ts (Vref/Vdc) sin(60° − α) T₂ = Ts (Vref/Vdc) sin(α)

Vref/Vdc = 400/600 = 0.667 T₁ ≈ 100 × 0.667 × sin(25°) ≈ 100 × 0.667 × 0.422 ≈ 28 μs T₂ ≈ 100 × 0.667 × sin(35°) ≈ 100 × 0.667 × 0.574 ≈ 38 μs

T₀ = 100 − (28 + 38) = 34 μs

Final: Sector 1, T₁ ≈ 28 μs, T₂ ≈ 38 μs, T₀ ≈ 34 μs

🔹 Problem 3 – Overmodulation Decision + THD Concept

Vdc = 650 V Required line voltage = 600 V Check if SPWM works.

SPWM max = 0.866 × 650 = 563 V Required = 600 V > 563 V → SPWM cannot achieve → Overmodulation needed → THD increases

Conclusion: Operates in overmodulation region.

🔹 Problem 4 – Multi-Concept Trap (Phase + Line)

Vdc = 500 V mₐ = 0.9 Find: 1) Phase voltage 2) Line voltage

Phase voltage Vphase = mₐ × Vdc / 2 = 0.9 × 500 / 2 = 225 V

Line voltage VLL = √3 × Vphase = 1.732 × 225 ≈ 390 V

Final: Vphase = 225 V, VLL ≈ 390 V

Ultra Exam Memory Zone

✔ SPWM → 0.866 Vdc ✔ SVPWM → Vdc ✔ mf = fs / f ✔ T₀ = Ts − (T₁ + T₂) ✔ Overmodulation → Higher THD ✔ Sector width = 60°

Ultra Advanced Practice Series – Shaktimatha Learning

Shaktimatha 369 Technologies

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