Page 44 – Multiple Questions & Answers (PWM & Inverters)

Subject: Power Electronics
Level: GATE / PSU Practice Set

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🔹 Q1. What is the maximum line voltage in SPWM?

V_LL,max = 0.866 V_dc

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🔹 Q2. What is the maximum line voltage in SVPWM?

V_LL,max = V_dc

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🔹 Q3. How many switching states are present in a 3-phase VSI?

8 states (6 active + 2 zero vectors)

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🔹 Q4. How many sectors are there in SVPWM?

6 sectors (each 60°)

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🔹 Q5. Define Amplitude Modulation Index.

m_a = V_m / V_carrier

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🔹 Q6. What happens when m_a > 1?

Overmodulation occurs → waveform distortion increases.

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🔹 Q7. In 180° conduction mode, how many switches conduct at a time?

Three switches conduct simultaneously.

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🔹 Q8. In 120° conduction mode, how many switches conduct at a time?

Two switches conduct simultaneously.

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🔹 Q9. What is the formula for switching time T₁ in SVPWM?

T₁ = T_s (V_ref/V_dc) sin(60° − α)

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🔹 Q10. Write the expression for zero vector time.

T₀ = T_s − (T₁ + T₂)

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🔹 Q11. What is the main advantage of SVPWM over SPWM?

15% higher DC bus utilization.

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🔹 Q12. Why must upper and lower switches of same leg never be ON together?

It causes shoot-through fault and short-circuits the DC bus.

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 Rapid Revision Summary

SPWM → 0.866 Vdc
SVPWM → Vdc
6 sectors → 6 active vectors
T₀ = Ts − (T₁ + T₂)

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Power Electronics Practice Series – Shaktimatha Learning

 

Page 43 – SVPWM Numerical Problems

Subject: Power Electronics
Topic: Space Vector PWM Numericals
Level: GATE / PSU Very Advanced

                                        

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🔹 Problem 1 – Maximum Line Voltage

Given:
DC bus voltage = 400 V
Find maximum fundamental line voltage using SVPWM.

For SVPWM:
V_LL,max = V_dc

Answer = 400 V

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🔹 Problem 2 – Compare with SPWM

Given: Vdc = 400 V

SPWM:
V_LL,max = 0.866 Vdc = 0.866 × 400 = 346.4 V

Improvement in SVPWM:
(400 − 346.4)/346.4 × 100 ≈ 15.47%

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🔹 Problem 3 – Switching Time Calculation

Given:
Vdc = 600 V
Vref = 300 V
Switching period Ts = 100 μs
α = 20°

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

Vref/Vdc = 300/600 = 0.5

T₁ = 100 × 0.5 × sin(40°)
= 100 × 0.5 × 0.643
≈ 32.15 μs

T₂ = 100 × 0.5 × sin(20°)
= 100 × 0.5 × 0.342
≈ 17.1 μs

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🔹 Problem 4 – Zero Vector Time

T₀ = Ts − (T₁ + T₂)
= 100 − (32.15 + 17.1)
≈ 50.75 μs

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🔹 GATE Concept Points

• Always identify sector first
• Use correct sine angles
• T₀ split equally at beginning & end
• Remember 15% voltage gain concept

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📌 Quick Revision

SVPWM → 6 sectors
T₁, T₂ from sine relations
T₀ = Ts − (T₁ + T₂)

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Power Electronics Advanced Series – Shaktimatha Learning

 

Page 42 – Space Vector PWM (SVPWM)

Subject: Power Electronics
Topic: Space Vector Pulse Width Modulation
Level: GATE / PSU Advanced Concept

                                           

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🔹 Why SVPWM?

SVPWM provides better DC bus utilization compared to SPWM. It increases maximum output voltage by about 15%.

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🔹 Basic Concept

• Three phase voltages represented as a single rotating vector
• Space vector rotates in α-β plane
• Inverter has 8 switching states
• 6 active vectors + 2 zero vectors

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🔹 Voltage Vector Equation

V_ref = (2/3) (V_a + aV_b + a²V_c)

Where:

• a = e^(j120°)
• Represents space vector transformation

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🔹 Maximum Output Voltage

SPWM: V_LL,max = 0.866 V_dc
SVPWM: V_LL,max = V_dc

👉 15% higher voltage utilization.

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🔹 Sector Division

• 360° divided into 6 sectors
• Each sector = 60°
• Reference vector synthesized using two adjacent active vectors

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🔹 Switching Time Calculation

T₁ = T_s (V_ref/V_dc) sin(60° − α)
T₂ = T_s (V_ref/V_dc) sin(α)

Where:

• T_s = switching period
• α = angle inside sector

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🔹 Advantages

• Better DC bus utilization
• Lower harmonic distortion
• Reduced switching losses
• Used in high-performance motor drives

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📌 Quick Revision

6 active vectors + 2 zero vectors
6 sectors (60° each)
15% higher voltage than SPWM

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Power Electronics Advanced Series – Shaktimatha Learning

 

Page 41 – Three Phase SPWM Inverter

Subject: Power Electronics
Topic: Three Phase Sinusoidal PWM
Level: GATE / PSU Advanced

                                            

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🔹 Introduction

Three phase SPWM uses three sinusoidal reference waves (120° phase shifted) compared with a high-frequency triangular carrier.

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🔹 Basic Principle

• Three sine waves: Va*, Vb*, Vc*
• Phase difference = 120°
• Compared with common triangular carrier
• Produces PWM gating signals

Switch ON when sine reference > carrier.

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🔹 Modulation Index

m_a = V_m / V_carrier

• m_a ≤ 1 → Linear region
• m_a > 1 → Overmodulation

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🔹 Line Voltage Fundamental

V_LL1 = (√3 / 2) × m_a × V_dc

Important for GATE numericals.

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🔹 Advantages

• Lower harmonic distortion
• Better voltage control
• Suitable for motor drives
• Improved power quality

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🔹 Numerical Example

Given:
V_dc = 600 V
m_a = 0.9

V_LL1 = (√3 / 2) × 0.9 × 600

= 0.866 × 0.9 × 600
≈ 468 V (Fundamental Line Voltage)

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🔹 GATE Important Points

• Line vs Phase voltage difference
• Linear modulation region
• Carrier frequency selection
• THD comparison with square wave

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Quick Revision

Three sine waves (120° shifted)
Common triangular carrier
V_LL1 = (√3 / 2) m_a V_dc

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Power Electronics Series – Shaktimatha Learning

 

Page 40 – Three Phase Voltage Source Inverter (VSI)

Subject: Power Electronics
Topic: Three Phase Bridge Inverter
Level: GATE / PSU Core Concept

                                          

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🔹 Introduction

A three-phase VSI converts DC voltage into three-phase AC output. It is widely used in motor drives, EVs, UPS and industrial applications.

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🔹 Circuit Configuration

• Six switches (S1–S6)
• Three legs (Phase A, B, C)
• Each leg has two switches
• Upper & lower switches are complementary

Important: Never turn ON both switches of same leg (shoot-through fault).

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🔹 180° Conduction Mode

Each switch conducts for 180°. At any instant, three switches are ON.

• Produces quasi-square output
• Lower harmonic performance than PWM

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🔹 120° Conduction Mode

Each switch conducts for 120°. Only two switches conduct at a time.

• Reduced switching overlap
• Different line voltage waveform

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🔹 Line Voltage Expression

V_LL = √3 × V_ph

For square wave inverter:

Fundamental Line Voltage: V_LL1 = (√6 / π) × V_dc

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🔹 Advantages

• High efficiency
• Suitable for high power applications
• Used in induction motor drives
• Basis for PWM inverters

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🔹 GATE Important Points

• Difference between 120° & 180° conduction
• Line vs phase voltage relation
• Switching sequence table
• Fundamental RMS calculation

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 Quick Revision

Three Phase VSI → 6 switches
180° mode → 3 ON
120° mode → 2 ON
V_LL1 = (√6 / π) V_dc

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Power Electronics Series – Shaktimatha Learning

 

Page 39 – Sinusoidal PWM (SPWM) Inverter

Subject: Power Electronics
Topic: Sinusoidal Pulse Width Modulation
Level: GATE / PSU Very Important

                                        

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🔹 Why PWM is Needed?

• Square wave inverter has high THD
• PWM reduces lower order harmonics
• Improves output waveform quality
• Better control of output voltage

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🔹 Basic Principle

Compare:

• High frequency triangular carrier wave
• Low frequency sinusoidal reference wave

Switch ON when sine > carrier.

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🔹 Modulation Index

m_a = V_m / V_carrier

Where:

• m_a = Amplitude modulation index
• m_a ≤ 1 → Linear region
• m_a > 1 → Overmodulation

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🔹 Output Fundamental Voltage

V₁ = m_a × V_dc / 2

(For full bridge SPWM inverter)

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🔹 Frequency Modulation Index

m_f = f_carrier / f_reference

Higher m_f → Better harmonic performance.

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🔹 Numerical Example

Given:
V_dc = 400 V
m_a = 0.8

V₁ = 0.8 × 400 / 2

= 160 V (fundamental peak)

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🔹 GATE Important Points

• SPWM reduces lower order harmonics
• Linear control when m_a ≤ 1
• Overmodulation distorts waveform
• Used in motor drives & UPS

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Quick Revision

SPWM → Compare sine & triangle
V₁ = m_a V_dc/2

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PWM Series – Shaktimatha Learning

 

Page 38 – Single Phase Voltage Source Inverter (VSI)

Subject: Power Electronics
Topic: Single Phase VSI (Full Bridge)
Level: GATE / PSU Core

                                              

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🔹 What is an Inverter?

An inverter converts DC power into AC power.

Applications:

• UPS systems
• Solar inverters
• Motor drives
• HVDC systems

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🔹 Single Phase Full Bridge VSI

• Uses 4 switches (IGBT/MOSFET)
• DC input voltage = V_dc
• Output = Square AC waveform

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🔹 Output Voltage

V_o = +V_dc (for 0 to π)
V_o = −V_dc (for π to 2π)

Square wave AC output

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🔹 RMS Output Voltage

V_rms = V_dc

(for square wave inverter)

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🔹 Fourier Series (Important for GATE)

V_o(t) = (4V_dc/π) [ sin ωt + 1/3 sin 3ωt + 1/5 sin 5ωt + ... ]

Only odd harmonics present.

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🔹 Numerical Example

Given: V_dc = 200 V

Fundamental peak voltage:

V_1peak = 4V_dc/π

= 4 × 200 / 3.14
= 254.8 V

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🔹 GATE Important Points

• Only odd harmonics exist
• THD is high for square wave
• Filters required
• RMS = V_dc

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 Quick Revision

Square Wave Inverter → Odd Harmonics
Fundamental = 4V_dc/π

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Inverter Series – Shaktimatha Learning