Page 60 – Ćuk Converter (Detailed Derivation + Numericals)
Subject: Power Electronics
Level: GATE / PSU Advanced
🔹 1️⃣ Circuit Overview
Ćuk converter uses:
✔ Two inductors (L1, L2)
✔ Energy transfer capacitor (C1)
✔ Output capacitor (C2)
✔ Switch + Diode
Special Feature:
- Continuous input current
- Continuous output current
- Low ripple
🔹 2️⃣ Derivation of Voltage Gain (CCM)
Mode 1 (Switch ON):
- L1 charges from source
- Diode OFF
- Capacitor C1 discharges to L2
Mode 2 (Switch OFF):
- L1 transfers energy to C1
- Diode ON
- L2 supplies load
Apply Volt-Second Balance on L1:
Vin × D + (Vin − Vc1)(1 − D) = 0
Apply Volt-Second Balance on L2:
(−Vc1 − Vo)D + (−Vo)(1 − D) = 0
Solving both equations gives:
Vo / Vin = − D / (1 − D)
Important: Same gain as Buck-Boost but with smoother current.
🔹 3️⃣ Numerical Problem 1 – Output Voltage
Vin = 20 V D = 0.6 Find output voltage.
Vo = − (0.6 / 0.4) × 20 = −1.5 × 20 = −30 V
Answer: −30 V
🔹 4️⃣ Numerical Problem 2 – Inductor Current
Vin = 24 V Vo = −48 V R = 12 Ω Find output current and average L2 current.
Io = 48 / 12 = 4 A |Vo/Vin| = 48/24 = 2 2 = D/(1−D) D = 0.667 IL2(avg) = Io = 4 A
Note: In Ćuk converter, output inductor current equals load current.
🔹 5️⃣ Numerical Problem 3 – Ripple Current
Vin = 30 V D = 0.4 L1 = 150 μH fs = 40 kHz Find ripple current in L1.
ΔIL1 = (Vin × D) / (L1 × fs) = (30 × 0.4) / (150×10⁻⁶ × 40000) = 12 / 6 = 2 A
Answer: 2 A
🔹 6️⃣ Critical Inductance Condition
To maintain CCM: L > (R (1−D)²) / (2 fs)
Same expression form as buck-boost.
Exam Comparison – Buck-Boost vs Ćuk
| Parameter | Buck-Boost | Ćuk |
|---|---|---|
| Inductors | 1 | 2 |
| Ripple | Medium | Low |
| Gain | −D/(1−D) | −D/(1−D) |
| Exam Weight | High | Moderate-High |
Advanced Converter Series – Shaktimatha Learning
.png)
No comments:
Post a Comment