Page 47 – Very Hard Numerical Problems
Subject: Power Electronics
Level: GATE / PSU Advanced Numericals
🔹 Problem 1 – SPWM Fundamental Voltage
A 3-phase inverter operates with Vdc = 700 V and modulation index mₐ = 0.9. Find maximum fundamental line voltage.
VLL = 0.866 × mₐ × Vdc
= 0.866 × 0.9 × 700
= 0.866 × 630
= 545.6 V
Final Answer: 546 V
🔹 Problem 2 – SVPWM Switching Times
Switching period Ts = 100 μs. If T₁ = 35 μs and T₂ = 25 μs, find zero vector time.
T₀ = Ts − (T₁ + T₂)
= 100 − (35 + 25)
= 100 − 60
= 40 μs
Final Answer: 40 μs
🔹 Problem 3 – Overmodulation Region
For Vdc = 600 V and required line voltage = 580 V, determine if SPWM can achieve this.
Max SPWM line voltage = 0.866 × 600 = 519.6 V
Required = 580 V > 519.6 V
Hence SPWM cannot achieve → Overmodulation required.
Conclusion: Overmodulation region
🔹 Problem 4 – Switching Frequency Calculation
If switching period Ts = 50 μs, find switching frequency.
fs = 1 / Ts
= 1 / (50 × 10⁻⁶)
= 20 kHz
Final Answer: 20 kHz
🔹 Problem 5 – Harmonic Reduction
Switching frequency increased from 5 kHz to 15 kHz. Comment on harmonic performance.
Higher switching frequency → Harmonics shift to higher frequency → Easier filtering → Lower THD.
🔹 Problem 6 – Sector Identification
Reference angle α = 275°. Identify sector.
Each sector = 60°
240°–300° = Sector 5
275° lies in Sector 5.
Answer: Sector 5
🔹 Problem 7 – Phase Voltage
For Vdc = 500 V and mₐ = 0.8, find fundamental phase voltage.
Vphase = mₐ × Vdc / 2
= 0.8 × 500 / 2
= 400 / 2
= 200 V
Final Answer: 200 V
🔹 Problem 8 – DC Bus Utilization Comparison
Vdc = 800 V. Compare SPWM and SVPWM maximum line voltages.
SPWM max = 0.866 × 800 = 692.8 V
SVPWM max = 800 V
Improvement ≈ 107.2 V
Exam Memory Points
✔ SPWM max = 0.866 Vdc
✔ SVPWM max = Vdc
✔ T₀ = Ts − (T₁ + T₂)
✔ fs = 1/Ts
✔ Sector width = 60°
Advanced Numerical Series – Shaktimatha Learning
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