Page 63 – Ultra Hard Mixed DC-DC Numerical Test

Subject: Power Electronics
Level: GATE / PSU Advanced

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🔹 Problem 1 – Identify Mode + Output Voltage

Buck-Boost converter: Vin = 15 V D = 0.5 L = 50 μH fs = 40 kHz R = 5 Ω Determine:
1) CCM or DCM
2) Output voltage

Step 1 – Find Lcritical

Lcritical = R(1−D)² / (2fs) = 5(0.5)² / (2×40000) = 5(0.25) / 80000 = 1.25 / 80000 = 15.6 μH

Given L = 50 μH > Lcritical → CCM

Step 2 – Output Voltage

Vo/Vin = −D/(1−D) = −0.5/0.5 = −1 Vo = −15 V

Final Answer: CCM, Vo = −15 V

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🔹 Problem 2 – Boost Converter Ripple

Vin = 20 V D = 0.6 L = 200 μH fs = 25 kHz Find inductor ripple current.

ΔIL = Vin D / (L fs) = 20×0.6 / (200×10⁻⁶ × 25000) = 12 / 5 = 2.4 A

Answer: 2.4 A

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🔹 Problem 3 – SEPIC Duty Cycle

Vin varies 10–20 V Vo = 15 V Find duty cycle range.

Vo/Vin = D/(1−D) Case 1: Vin = 10 15/10 = 1.5 1.5 = D/(1−D) D = 0.6 Case 2: Vin = 20 15/20 = 0.75 0.75 = D/(1−D) D = 0.43

Duty cycle range: 0.43 – 0.6

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🔹 Problem 4 – Ćuk Converter Power

Vin = 12 V D = 0.5 R = 8 Ω Find output power.

Vo/Vin = −D/(1−D) = −0.5/0.5 = −1 Vo = −12 V Io = 12/8 = 1.5 A P = Vo × Io = 12 × 1.5 = 18 W

Answer: 18 W

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🔹 Problem 5 – DCM Output

Vin = 24 V D = 0.4 R = 20 Ω L = 100 μH fs = 50 kHz Find Vo (DCM).

Vo/Vin = − (D² R) / (2Lfs) = − (0.16 × 20) / (2×100×10⁻⁶ × 50000) = − (3.2) / (10) = −0.32 Vo = −0.32 × 24 = −7.68 V

Final Answer: −7.68 V

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 Master Concept Summary

✔ CCM → Gain depends only on D ✔ DCM → Gain depends on R, L, fs ✔ Always check Lcritical ✔ SEPIC = Non-inverting ✔ Ćuk = Low ripple ✔ Buck-Boost = Inverting

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Ultra Hard DC-DC Series – Shaktimatha Learning

 

Page 62 – Buck-Boost Converter in DCM (Complete Derivation)

Subject: Power Electronics
Level: GATE / PSU Conceptual + Numerical

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🔹 1️⃣ What is DCM?

DCM occurs when inductor current falls to zero before next switching cycle.
✔ Happens at light load
✔ Smaller inductance
✔ High switching frequency

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🔹 2️⃣ DCM Operation (Three Intervals)

• Interval 1 (0 → DT): Switch ON, inductor charges
• Interval 2 (DT → (D+δ)T): Diode conducts, inductor discharges
• Interval 3: Inductor current = 0

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🔹 3️⃣ Inductor Current Peak

ΔIL = (Vin × D) / (L × fs)

Since current starts from zero:

Ipeak = (Vin × D) / (L × fs)

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🔹 4️⃣ Average Output Current

Output current equals average diode current.

Io = (1/2) × Ipeak × δ

Where δ = diode conduction fraction.

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🔹 5️⃣ Final DCM Voltage Gain (Important Result)

After solving energy balance:

Vo / Vin = − (D² × R) / (2Lfs)

🔥 VERY IMPORTANT: In DCM, voltage gain depends on: ✔ Duty cycle ✔ Load resistance ✔ Inductance ✔ Switching frequency Unlike CCM, gain is NOT only function of D.

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🔹 6️⃣ Numerical Example

Vin = 20 V D = 0.4 L = 100 μH fs = 50 kHz R = 10 Ω Find output voltage (DCM).

Vo/Vin = − (D² R) / (2Lfs) = − (0.16 × 10) / (2 × 100×10⁻⁶ × 50000) = − (1.6) / (10) = −0.16

Vo = −0.16 × 20 = −3.2 V

Final Answer: −3.2 V

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🔹 7️⃣ Critical Condition (Boundary CCM/DCM)

Boundary inductance:

Lcritical = R(1−D)² / (2fs)

If L < Lcritical → DCM If L > Lcritical → CCM

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 CCM vs DCM Comparison

Feature	CCM	DCM
Inductor Current	Never zero	Falls to zero
Voltage Gain	−D/(1−D)	Depends on R, L, fs
Control	Simpler	More complex

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 GATE Strategy Tip

If question gives L, R, fs → Think DCM possibility.
If only D given → Usually CCM.

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Advanced DC-DC Control Series – Shaktimatha Learning

 

Page 61 – SEPIC Converter (Design Example + Component Selection)

Subject: Power Electronics
Level: GATE / PSU + Practical Design

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🔹 1️⃣ What is SEPIC?

SEPIC = Single-Ended Primary Inductor Converter ✔ Non-inverting output ✔ Can step-up or step-down ✔ Continuous input current ✔ Widely used in battery-powered systems

Voltage Gain (CCM):

Vo / Vin = D / (1 − D)

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🔹 2️⃣ Design Problem

Design a SEPIC converter with:
Vin = 12 V
Required Vo = 24 V
Load Power = 48 W
Switching Frequency fs = 50 kHz
Allow ripple current = 30% of average current

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🔹 3️⃣ Step 1 – Duty Cycle

Vo/Vin = D/(1−D) 24/12 = D/(1−D) 2 = D/(1−D) 2(1−D) = D 2 − 2D = D 2 = 3D D = 0.667

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🔹 4️⃣ Step 2 – Output Current

Io = P / Vo = 48 / 24 = 2 A

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🔹 5️⃣ Step 3 – Inductor Design

Average input current:

Pin ≈ Pout Iin = P / Vin = 48 / 12 = 4 A

Allow 30% ripple:

ΔIL = 0.3 × 4 = 1.2 A

Inductor formula:

L = (Vin × D) / (ΔIL × fs)

L = (12 × 0.667) / (1.2 × 50000) = 8 / 60000 = 133 μH

Choose: L1 = L2 ≈ 150 μH (standard value)

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🔹 6️⃣ Step 4 – Coupling Capacitor (C1)

Approximation formula:

C1 ≥ Io × D / (ΔVc × fs)

Assume allowed ripple ΔVc = 1 V

C1 = (2 × 0.667) / (1 × 50000) = 1.334 / 50000 = 26.6 μF

Select: 33 μF

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🔹 7️⃣ Step 5 – Output Capacitor

Co ≥ Io × D / (ΔVo × fs)

Assume ripple 0.5 V

Co = (2 × 0.667) / (0.5 × 50000) = 1.334 / 25000 = 53 μF

Select: 68 μF

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🔹 8️⃣ Final Design Summary

Parameter	Value
Duty Cycle	0.667
Inductors	150 μH
C1	33 μF
Co	68 μF

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 Exam Memory Capsule

✔ SEPIC gain = D/(1−D) ✔ Non-inverting output ✔ Two inductors ✔ Design based on ripple percentage ✔ Very common in battery & automotive systems

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Advanced DC-DC Design Series – Shaktimatha Learning

 

Page 60 – Ćuk Converter (Detailed Derivation + Numericals)

Subject: Power Electronics
Level: GATE / PSU Advanced

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🔹 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

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🔹 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.

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🔹 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

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🔹 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.

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🔹 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

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🔹 6️⃣ Critical Inductance Condition

To maintain CCM: L > (R (1−D)²) / (2 fs)

Same expression form as buck-boost.

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 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

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Advanced Converter Series – Shaktimatha Learning

 

Page 59 – Buck-Boost Converter Numerical Problems (CCM)

Subject: Power Electronics
Level: GATE / PSU Numerical Practice

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🔹 Problem 1 – Output Voltage Calculation

A buck-boost converter operates in CCM with:
Vin = 24 V
Duty cycle D = 0.6
Find output voltage.

Vo / Vin = − D / (1 − D)

Vo = − (0.6 / 0.4) × 24 = −1.5 × 24 = −36 V

Final Answer: −36 V

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🔹 Problem 2 – Duty Cycle Required

Vin = 48 V Required Vo = −24 V Find duty cycle.

|Vo/Vin| = D / (1 − D) 24/48 = D/(1−D) 0.5 = D/(1−D) 0.5(1−D) = D 0.5 − 0.5D = D 0.5 = 1.5D D = 0.333

Final Answer: D = 0.333 (33.3%)

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🔹 Problem 3 – Inductor Current

Vin = 20 V Vo = −40 V Load resistance R = 10 Ω Find output current and average inductor current.

Io = |Vo| / R = 40 / 10 = 4 A For buck-boost in CCM: I_L(avg) = Io / (1 − D) First find D: |Vo/Vin| = D/(1−D) 40/20 = D/(1−D) 2 = D/(1−D) 2(1−D) = D 2 − 2D = D 2 = 3D D = 0.667 Now, IL(avg) = 4 / (1 − 0.667) = 4 / 0.333 ≈ 12 A

Final Answer: Io = 4 A IL(avg) ≈ 12 A

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🔹 Problem 4 – Critical Inductance

Vin = 30 V Vo = −60 V R = 12 Ω Switching frequency fs = 25 kHz Find critical inductance.

For buck-boost: L_critical = (R (1−D)²) / (2 fs)

First find D: |Vo/Vin| = 60/30 = 2 2 = D/(1−D) D = 0.667 Now, L = (12 × (0.333)²) / (2 × 25000) = (12 × 0.111) / 50000 = 1.332 / 50000 = 26.6 μH

Final Answer: 26.6 μH

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🔹 Problem 5 – Ripple Current

Vin = 24 V D = 0.5 L = 200 μH fs = 50 kHz Find inductor ripple current.

ΔIL = (Vin × D) / (L × fs) = (24 × 0.5) / (200×10⁻⁶ × 50000) = 12 / 10 = 1.2 A

Final Answer: 1.2 A

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 Exam Memory Capsule

✔ Vo/Vin = −D/(1−D) ✔ Output is negative ✔ IL(avg) = Io/(1−D) ✔ Lcritical = R(1−D)² / (2fs) ✔ ΔIL = VinD / (Lfs)

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Buck-Boost Numerical Series – Shaktimatha Learning

 

Page 58 – Advanced DC-DC Converters (Buck-Boost, Ćuk, SEPIC)

Subject: Power Electronics
Level: GATE / PSU Advanced

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🔹 1️⃣ Inverting Buck-Boost Converter

Output voltage is inverted and can be higher or lower than input.

Voltage Gain (CCM):

Vo / Vin = − D / (1 − D)

• D → Duty cycle
• Output polarity → Negative

Key Points:

• Simple structure
• Single inductor
• Higher ripple compared to Cuk

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🔹 2️⃣ Ćuk Converter

Uses capacitor for energy transfer and two inductors for smooth current.

Voltage Gain (CCM):

Vo / Vin = − D / (1 − D)

Special Feature:

• Continuous input current
• Continuous output current
• Low ripple

Applications:

• Precision power supplies
• Low ripple DC systems

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🔹 3️⃣ SEPIC Converter

Single-Ended Primary Inductor Converter Non-inverting output, can step up or step down.

Voltage Gain (CCM):

Vo / Vin = D / (1 − D)

Features:

• Non-inverting output
• Two inductors
• Used in battery systems

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🔹 4️⃣ Comparison Table

Converter	Polarity	Voltage Gain	Ripple
Buck-Boost	Inverting	−D/(1−D)	Medium
Ćuk	Inverting	−D/(1−D)	Low
SEPIC	Non-Inverting	D/(1−D)	Low

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 Exam Memory Capsule

✔ Buck-Boost → Simple, inverting ✔ Ćuk → Smooth current, low ripple ✔ SEPIC → Non-inverting ✔ Gain depends only on Duty Cycle ✔ CCM analysis important for GATE

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Advanced DC-DC Converter Series – Shaktimatha Learning

 

Page 57 – Ultra Hard Mixed Test (Multilevel + PWM)

Subject: Power Electronics
Level: GATE / PSU Ultra Advanced

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🔹 Q1.

A CHB inverter has 4 H-bridges per phase. Find number of output levels.

Levels = 2n + 1 = 2(4) + 1 = 9

Answer: 9 Levels

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🔹 Q2.

For a 5-level NPC inverter, find number of clamping diodes per phase.

Diodes = (m−1)(m−2) = (5−1)(5−2) = 4×3 = 12

Answer: 12 diodes

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🔹 Q3.

For a 7-level FC inverter, find number of flying capacitors per phase.

Capacitors = (m−1)(m−2)/2 = (7−1)(7−2)/2 = (6×5)/2 = 15

Answer: 15 capacitors

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🔹 Q4.

A 3-level NPC inverter operates with Vdc = 900 V. Find switch voltage stress.

Stress = Vdc/(m−1) = 900/(3−1) = 450 V

Answer: 450 V

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🔹 Q5.

In Phase-Shifted PWM for 7-level CHB, find carrier phase shift.

Number of carriers = m−1 = 6 Phase shift = 360°/6 = 60°

Answer: 60°

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🔹 Q6.

Increasing number of levels mainly reduces: A) Switching loss B) Voltage rating C) THD D) DC source

Answer: C

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🔹 Q7.

Which inverter requires multiple isolated DC sources? A) NPC B) CHB C) FC D) 2-Level

Answer: B (CHB)

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🔹 Q8.

Which PWM technique is best for CHB inverter? A) Level-shifted B) Phase-shifted C) Sinusoidal only D) Random PWM

Answer: B

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🔹 Q9.

For 9-level NPC inverter, find switch stress if Vdc = 1200 V.

Stress = 1200/(9−1) = 1200/8 = 150 V

Answer: 150 V

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🔹 Q10.

More flying capacitors lead to: A) Easier balancing B) Complex control C) Lower switching frequency D) Lower voltage stress

Answer: B

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 Ultra Memory Capsule

✔ CHB → Multiple DC sources ✔ NPC → Many diodes ✔ FC → Many capacitors ✔ Stress = Vdc/(m−1) ✔ Levels = 2n + 1 ✔ PS-PWM shift = 360°/(m−1)

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Ultra Advanced Mixed Test – Shaktimatha Learning