NEETAceToday's NEET Daily

Zener diode and voltage regulation: NEET MCQs with answers

75 questions on this topic of Electronic Devices (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Electronic Devices freeAll Electronic Devices questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. In a Zener diode voltage regulator circuit, the unregulated input voltage is 12 V and the Zener breakdown voltage is 5.0 V. If a load resistor of 1.0 kΩ is connected in parallel with the Zener diode and the series resistor is 200 Ω, what is the current passing through the Zener diode?

  1. 35 mA
  2. 25 mA
  3. 30 mA
  4. 5 mA
Show answer and explanation

Answer: (C) 30 mA

The voltage drop across the series resistor is Vs = Vin - Vz = 12 V - 5 V = 7 V. The total current from the supply is Is = 7 V / 200 Ω = 35 mA. The load current is IL = Vz / RL = 5 V / 1000 Ω = 5 mA. Therefore, the Zener current is Iz = Is - IL = 35 mA - 5 mA = 30 mA.

medium

2. A Zener diode with breakdown voltage 5 V is connected through a 500 Ω series resistor to a 20 V supply (no load). The current through the series resistor is:

  1. 10 mA
  2. 50 mA
  3. 30 mA
  4. 40 mA
Show answer and explanation

Answer: (C) 30 mA

The resistor drops Vin − Vz = 15 V, so I = 15/500 A = 30 mA.

hard

3. A Zener diode regulator has an unregulated DC input that varies from 20 V to 30 V. The breakdown voltage is 10 V, the series resistance is 500 Ω, and the load resistance is 1 kΩ. The ratio of the maximum current to the minimum current flowing through the Zener diode is:

  1. 4 : 1
  2. 2 : 1
  3. 3 : 1
  4. 5 : 2
Show answer and explanation

Answer: (C) 3 : 1

The load current is IL = Vz / RL = 10 V / 1000 Ω = 10 mA (constant). For Vin(min) = 20 V: Is(min) = (20 - 10) / 500 = 20 mA. Therefore, Iz(min) = 20 - 10 = 10 mA. For Vin(max) = 30 V: Is(max) = (30 - 10) / 500 = 40 mA. Therefore, Iz(max) = 40 - 10 = 30 mA. The ratio Iz(max) / Iz(min) = 30 mA / 10 mA = 3 : 1.

medium

4. A Zener diode with breakdown voltage 9 V is connected through a 200 Ω series resistor to a 12 V supply (no load). The current through the series resistor is:

  1. 45 mA
  2. 60 mA
  3. 15 mA
  4. 105 mA
Show answer and explanation

Answer: (C) 15 mA

The resistor drops Vin − Vz = 3 V, so I = 3/200 A = 15 mA.

easy

5. Assertion (A): A Zener diode is always operated in reverse bias breakdown region when employed as a DC voltage regulator. Reason (R): In the breakdown region, the voltage across the Zener diode remains almost constant even though the reverse current changes significantly. Which of the following is correct?

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
Show answer and explanation

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Both Assertion and Reason are true, and the Reason correctly explains the Assertion. The characteristic sharp increase in reverse current with virtually negligible change in voltage in the breakdown region allows the diode to absorb source or load fluctuations while stabilizing output voltage.

hard

6. In a Zener-regulated power supply, the unregulated input voltage increases by 2.0 V. If the series resistance is R = 250 Ω and the load resistance is kept fixed at 1 kΩ, by how much does the current through the Zener diode increase?

  1. 8 mA
  2. 2 mA
  3. 10 mA
  4. 0 mA
Show answer and explanation

Answer: (A) 8 mA

Under proper regulation, the output voltage Vz across the load remains constant, so the load current I_L = Vz / R_L does not change (ΔI_L = 0). The change in total current is ΔI = ΔVin / R = 2.0 V / 250 Ω = 8 mA. Since ΔI = ΔIz + ΔI_L, the increase in Zener current is ΔIz = 8 mA.

medium

7. A Zener diode with breakdown voltage 9 V is connected through a 200 Ω series resistor to a 20 V supply (no load). The current through the series resistor is:

  1. 145 mA
  2. 45 mA
  3. 100 mA
  4. 55 mA
Show answer and explanation

Answer: (D) 55 mA

The resistor drops Vin − Vz = 11 V, so I = 11/200 A = 55 mA.

medium

8. In a voltage regulator circuit, the unregulated input voltage is 20 V, the series resistance is Rs = 250 Ω, and the breakdown voltage of the Zener diode is 7.5 V. If the load current changes from 10 mA to 25 mA, the change in the current through the Zener diode will be:

  1. decrease by 15 mA
  2. increase by 15 mA
  3. decrease by 25 mA
  4. remain unchanged
Show answer and explanation

Answer: (A) decrease by 15 mA

Since the input voltage is fixed at 20 V and the regulated voltage is 7.5 V, the current through the series resistor is constant: Is = (20 - 7.5) / 250 = 12.5 / 250 = 50 mA. Since Is = Iz + IL, any increase in load current ΔIL = +15 mA must be balanced by an equal decrease in Zener current ΔIz = -15 mA.

medium

9. A Zener diode with breakdown voltage 9 V is connected through a 250 Ω series resistor to a 20 V supply (no load). The current through the series resistor is:

  1. 44 mA
  2. 116 mA
  3. 80 mA
  4. 36 mA
Show answer and explanation

Answer: (A) 44 mA

The resistor drops Vin − Vz = 11 V, so I = 11/250 A = 44 mA.

easy

10. If a Zener diode is connected in forward bias in a circuit with a DC power supply and a protective resistor, how does it behave?

  1. It acts like an ordinary forward-biased p-n junction diode with a knee voltage around 0.7 V.
  2. It still maintains its reverse breakdown voltage across its terminals.
  3. It offers infinite resistance and acts as an open switch for all voltages.
  4. It immediately gets damaged due to high doping levels regardless of current.
Show answer and explanation

Answer: (A) It acts like an ordinary forward-biased p-n junction diode with a knee voltage around 0.7 V.

When connected in forward bias, a Zener diode behaves like an ordinary p-n junction diode, conducting heavily once the potential barrier (knee voltage, ~0.7 V for Si) is overcome.

easy

11. A Zener diode with a breakdown voltage of 8 V is used to maintain a constant output voltage across a load resistor R_L = 4 kΩ. The DC supply voltage is 20 V, and the series resistor is R = 1.2 kΩ. The current passing through the series resistor (I), load (I_L), and the Zener diode (Iz) are respectively:

  1. I = 12 mA, I_L = 4 mA, Iz = 8 mA
  2. I = 10 mA, I_L = 2 mA, Iz = 8 mA
  3. I = 10 mA, I_L = 4 mA, Iz = 6 mA
  4. I = 8 mA, I_L = 2 mA, Iz = 6 mA
Show answer and explanation

Answer: (B) I = 10 mA, I_L = 2 mA, Iz = 8 mA

The voltage across the series resistor is 20 V - 8 V = 12 V. Total current I = 12 V / 1.2 kΩ = 10 mA. The load current is I_L = 8 V / 4 kΩ = 2 mA. Using Kirchhoff's junction rule, Iz = I - I_L = 10 mA - 2 mA = 8 mA.

medium

12. A Zener diode with breakdown voltage 9 V is connected through a 200 Ω series resistor to a 15 V supply (no load). The current through the series resistor is:

  1. 120 mA
  2. 75 mA
  3. 30 mA
  4. 45 mA
Show answer and explanation

Answer: (C) 30 mA

The resistor drops Vin − Vz = 6 V, so I = 6/200 A = 30 mA.

Keep going with Electronic Devices. All 277 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Electronic Devices