Day 1 ECET 2026 ECE – Diodes & Rectifiers Full Notes, Formulas & MCQs

Concept Notes

1. PN Junction Diode Basics

• A PN junction diode is formed by joining p-type and n-type semiconductors.
• It acts like a switch:
  • Forward bias → conducts current after cut-in voltage.
  • Reverse bias → ideally blocks current except small leakage current.

Cut-in voltage values:

• Silicon:
• Germanium:

Example:
If a silicon diode is connected in series with source and resistor:

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2. Half-Wave Rectifier (HWR)

• Uses one diode to convert AC into pulsating DC.
• Output is present only during one half-cycle.
• Low efficiency and high ripple.

Key facts:

• Efficiency =
• Ripple factor =
• PIV of diode =

Example:
If input peak voltage = , the PIV = .

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3. Full-Wave Rectifier (FWR)

• Uses both half-cycles of AC.
• Can be made using:
  • Center-tap transformer + 2 diodes
  • Bridge Rectifier (4 diodes)

Key facts:

• Efficiency =
• Ripple factor =
• PIV for center-tap =
• PIV for bridge =

Example:
If input peak = (with center tap), then PIV = .

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4. Bridge Rectifier

• Uses 4 diodes in bridge form.
• Does not need a center-tap transformer.
• Two diodes conduct in each half cycle.
• Better transformer utilization.

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5. Rectifier Parameters

• Average current (IDC): measures DC output.
• RMS current (IRMS): measures effective current.
• Ripple factor (γ):

Efficiency (η):

Peak Inverse Voltage (PIV): max reverse voltage diode can withstand.

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⚙️ Formulas

Diode Equation:

Thermal Voltage:

HWR Average Current:

HWR RMS Current:

FWR Average Current:

FWR RMS Current:

Ripple Factor:

• HWR:
• FWR:

Efficiency:

• HWR:
• FWR:

PIV:

• HWR:
• FWR (center-tap):
• Bridge:

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🔟 10 MCQs

Q1. The efficiency of a half-wave rectifier is approximately:
a) 20%
b) 40.6%
c) 81.2%
d) 100%

Q2. In a full-wave rectifier, the average DC current is:
a)
b)
c)
d)

Q3. The ripple factor of a half-wave rectifier is:
a) 0.48
b) 1.21
c) 0.707
d) 2.0

Q4. In a bridge rectifier, the PIV across each diode is:
a)
b)
c)
d) 0

Q5. The RMS current of a full-wave rectifier is:
a)
b)
c)
d)

Q6. A silicon diode requires minimum voltage to conduct:
a) 0.1 V
b) 0.3 V
c) 0.7 V
d) 1 V

Q7. The advantage of bridge rectifier over center-tap rectifier is:
a) Low PIV per diode
b) Fewer diodes used
c) No transformer needed
d) Reduced power factor

Q8. The ripple factor of full-wave rectifier is approximately:
a) 1.21
b) 0.48
c) 0.707
d) 0.25

Q9. In half-wave rectifier, if input peak voltage is , the PIV is:
a)
b)
c)
d) 0

Q10. The full-wave rectifier efficiency is:
a) 40.6%
b) 50%
c) 81.2%
d) 100%

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✅ Answer Key

Q No	Answer
Q1	b
Q2	c
Q3	b
Q4	a
Q5	b
Q6	c
Q7	a
Q8	b
Q9	b
Q10	c

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🧠 Explanations

• Q1: HWR efficiency = .
• Q2: FWR average current = .
• Q3: HWR ripple factor = .
• Q4: Bridge rectifier → PIV = .
• Q5: FWR RMS = .
• Q6: Cut-in voltage for Si diode = .
• Q7: Advantage = low PIV per diode → (a).
• Q8: FWR ripple factor = .
• Q9: HWR PIV = .
• Q10: FWR efficiency = .

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🎯 Motivation / Why Practice Matters

Rectifiers are the foundation of all power supply circuits.

• Without rectifiers, no electronic system can run on AC mains.
• ECET repeatedly tests efficiency, ripple factor, PIV & IDC formulas.
• Daily practice builds accuracy and speed to solve numerical questions in under 1 minute.

👉 Remember: “If you master rectifiers, you master the heart of electronics.”

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📲 CTA

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