Page 69 – Numerical Marathon (Part 2 – Ultra Hard)
Level: GATE / PSU Ultra Hard
🔹 Problem 1 – Mode Identification + Output
Vin = 24V D = 0.5 L = 20µH R = 12Ω fs = 50kHz Find mode and output voltage.Step 1: Lcritical = R(1−D)²/(2fs) = 12×0.25/(100000) = 3/100000 = 30µH Given L = 20µH < Lcritical → DCM Step 2: DCM Gain Vo/Vin = −(D²R)/(2Lfs) = −(0.25×12)/(2×20×10⁻⁶×50000) = −3/(2) = −1.5 Vo = −36V
---🔹 Problem 2 – Boost Ripple & Power
Vin=30V D=0.6 L=150µH R=15Ω fs=40kHzVo = 30/(1−0.6)=75V Io = 75/15=5A P = 75×5=375W Ripple: ΔIL=VinD/(Lfs) =30×0.6/(150×10⁻⁶×40000) =18/6 =3A
---🔹 Problem 3 – Rectifier Inversion
Vm=325V α=120°Vo=(2Vm/Ï€)cosα = (650/3.14)(−0.5) ≈ −103V Negative → Inversion Mode
---🔹 Problem 4 – Buck Control Frequency
L=100µH C=50µF Find natural frequency.ω₀=1/√(LC) =1/√(5×10⁻⁹) ≈ 14142 rad/s f₀ ≈ 2250 Hz
---🔹 Problem 5 – SEPIC Duty Range
Vin varies 18–36V Vo=24VVo/Vin=D/(1−D) Case1: 24/18=1.33 D=0.57 Case2: 24/36=0.67 D=0.4 Duty range: 0.4–0.57
---🔹 Problem 6 – Inverter THD Concept
If switching frequency increases:Harmonics shift to higher frequency THD reduces Filter size decreases
---🔹 Problem 7 – Energy Stored
L=200µH I=5AEnergy = ½LI² =0.5×200×10⁻⁶×25 =2.5mJ
---🔹 Problem 8 – MOSFET Loss
Rds(on)=0.2Ω I=4AConduction loss = I²R =16×0.2 =3.2W
---Ultra Hard Key Learning
✔ Always compare L with Lcritical ✔ DCM can produce large gain ✔ Inversion when α > 90° ✔ Switching frequency affects THD & filter size ✔ Control frequency ≠ switching frequency
Power Electronics Ultra Hard Marathon – Shaktimatha Learning
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