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ECET 2026 ECE

Day 16 ECET 2026 ECE – SMPS Explained: Concepts

Why This Topic is Important for ECET

SMPS (Switched Mode Power Supply) is one of the most important topics in Industrial Electronics because it is widely used in computers, communication devices, embedded systems, and automation circuits. ECET frequently tests students on the working principle, efficiency, and differences between SMPS and linear power supplies. Knowing SMPS will give you an edge in both conceptual questions and numericals.


📘 Concept Notes

1. What is SMPS?

  • Switched Mode Power Supply (SMPS) is an electronic power supply that converts electrical power efficiently using switching regulators.
  • Unlike linear regulators, SMPS use high-frequency switching devices (like MOSFETs or BJTs) to regulate voltage.

2. Working Principle

  • AC is first converted into DC using a rectifier.
  • Then, DC is chopped into high-frequency AC using a high-speed switch.
  • This AC is stepped up or down using a transformer.
  • Finally, it is rectified and filtered to give a stable DC output.

3. Advantages of SMPS

  • High efficiency (70–90%).
  • Compact size (due to high-frequency operation).
  • Wide input voltage range.
  • Less heat dissipation compared to linear regulators.

4. Disadvantages of SMPS

  • More complex circuit design.
  • Generates electrical noise (EMI).
  • Requires filtering.

5. Types of SMPS Converters

  • Buck Converter → Steps down voltage.
  • Boost Converter → Steps up voltage.
  • Buck-Boost Converter → Steps up or down voltage (inverts polarity).
  • Flyback Converter → Provides multiple outputs with isolation.

Example:
If input = 12 V and buck converter duty ratio = 0.5,
Output = V_o = D \times V_{in} = 0.5 \times 12 = 6V


⚙️ Formulas

  • Output of buck converter:

 V_o = D \times V_{in}

Output of boost converter:

 V_o = \frac{V_{in}}{1 - D}

Output of buck-boost converter:

 V_o = \frac{D}{1 - D} \times V_{in}

Efficiency:

 \eta = \frac{P_{out}}{P_{in}} \times 100

Ripple voltage:

 V_{ripple} = \frac{I_{load}}{f \times C}


🔟 10 MCQs

Q1. SMPS stands for:
a) Switched Mode Power Supply
b) Standard Mode Power Supply
c) Small Mode Power Supply
d) Single Mode Power Switch

Q2. Efficiency of SMPS is generally in the range of:
a) 20–30%
b) 40–50%
c) 70–90%
d) 95–100%

Q3. A buck converter with V_{in} = 24V and duty ratio D = 0.25 gives output voltage:

Q4. In a boost converter, if V_{in} = 12V and duty ratio D = 0.5, find V_o.

Q5. Which component acts as the main switch in modern SMPS?
a) Diode
b) MOSFET
c) Resistor
d) Capacitor

Q6. The main disadvantage of SMPS compared to linear power supply is:
a) High efficiency
b) Large size
c) EMI noise generation
d) Low input range

Q7. In a buck-boost converter, if V_{in} = 10V and D = 0.4, find V_o.

Q8. Which converter can provide both step-up and step-down voltage?
a) Buck
b) Boost
c) Buck-Boost
d) None

Q9. Ripple voltage in SMPS can be reduced by:
a) Increasing resistance
b) Increasing switching frequency or capacitance
c) Decreasing duty ratio
d) Removing transformer

Q10. Which is NOT an advantage of SMPS?
a) Compact size
b) High efficiency
c) Low noise generation
d) Wide input range


✅ Answer Key

Q NoAnswer
Q1a
Q2c
Q36 V
Q424 V
Q5b
Q6c
Q7-6.67 V
Q8c
Q9b
Q10c

🧠 Explanations

  • Q1: SMPS → Switched Mode Power Supply → (a).
  • Q2: Efficiency of SMPS is typically 70–90% → (c).
  • Q3: V_o = D \times V_{in} = 0.25 \times 24 = 6V.
  • Q4: V_o = V_{in}/(1-D) = 12/(1-0.5) = 24V.
  • Q5: MOSFETs are used as high-frequency switches → (b).
  • Q6: Main drawback is EMI noise → (c).
  • Q7: V_o = (D/(1-D)) \times V_{in} = (0.4/0.6)\times 10 = 6.67V. Negative polarity → -6.67 V.
  • Q8: Buck-Boost can provide step-up and step-down → (c).
  • Q9: Increasing f or C reduces ripple → (b).
  • Q10: SMPS generates high noise, so “low noise generation” is false → (c).

🎯 Motivation / Why Practice Matters

For ECET 2026 ECE, SMPS questions test both concepts (advantages, disadvantages, components) and numericals (buck/boost/buck-boost formulas). Practicing daily:

  • Increases speed in solving formula-based numericals.
  • Improves accuracy in identifying correct converter type.
  • Helps in scoring easy but high-value marks in Industrial Electronics.

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