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p-n junction forward and reverse characteristic curves: NEET MCQs with answers

50 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. While plotting the reverse characteristic of a silicon p-n junction diode, a student observes that increasing the reverse voltage from 2 V to 10 V causes negligible change in reverse current. What primarily limits this reverse current below breakdown?

  1. The rate of thermal generation of minority carriers in and near the depletion layer
  2. The high contact resistance between the semiconductor and metal leads
  3. The magnitude of the barrier potential which prevents minority carrier flow
  4. The resistance of the current-limiting resistor connected in the circuit
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Answer: (A) The rate of thermal generation of minority carriers in and near the depletion layer

In reverse bias, the electric field sweeps minority carriers across the junction. Because all thermally generated minority carriers within a diffusion length reach the junction, the current is limited purely by their generation rate, leading to reverse saturation current.

medium

2. Consider the following statements regarding the experimental determination of diode characteristics: Statement A: In forward bias, the static resistance (V/I) decreases as the forward bias voltage increases beyond the threshold. Statement B: Dynamic resistance of a diode is strictly equal to static resistance at all operating points. Choose the correct option:

  1. Both Statement A and Statement B are true
  2. Statement A is true but Statement B is false
  3. Statement A is false but Statement B is true
  4. Both Statement A and Statement B are false
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Answer: (B) Statement A is true but Statement B is false

Statement A is true because current increases exponentially while voltage increases slowly past threshold, causing V/I to decrease. Statement B is false because static resistance R_s = V/I while dynamic resistance r_d = dV/dI; for a non-linear device, these two values are distinct.

hard

3. In a diode experiment, forward current I is measured as a function of forward voltage V. If the diode follows the relation I = I₀ [exp(e V / k_B T) - 1], what is the theoretical dynamic resistance r_d of the diode when biased well above the threshold such that exp(e V / k_B T) >> 1?

  1. r_d = V / I
  2. r_d = e · I / (k_B T)
  3. r_d = (k_B T / e) · I
  4. r_d = k_B T / (e · I)
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Answer: (D) r_d = k_B T / (e · I)

For I ≈ I₀ exp(e V / k_B T), differentiating with respect to V gives dI/dV = (e / k_B T) · I. Therefore, the dynamic resistance r_d = dV/dI = k_B T / (e · I). At room temperature, this evaluates to approximately 26 mV / I.

medium

4. The forward characteristic of a germanium diode differs from that of a silicon diode in having a:

  1. Lower threshold voltage (about 0.3 V)
  2. Higher threshold voltage (about 1.5 V)
  3. No threshold
  4. Negative threshold
Show answer and explanation

Answer: (A) Lower threshold voltage (about 0.3 V)

Germanium has a smaller band gap, so its knee voltage (about 0.3 V) is lower than that of silicon (about 0.7 V).

NCERT Class 11

easy

5. A student records the following data while plotting the forward characteristic curve of a silicon p-n junction diode: at a voltage of 0.72 V, the current is 15 mA, and at 0.76 V, the current rises to 35 mA. The dynamic resistance of the diode in this operating region is:

  1. 1.5 Ω
  2. 4.0 Ω
  3. 5.0 Ω
  4. 2.0 Ω
Show answer and explanation

Answer: (D) 2.0 Ω

Dynamic resistance r_d is given by ΔV / ΔI. Here, ΔV = 0.76 V - 0.72 V = 0.04 V, and ΔI = 35 mA - 15 mA = 20 mA = 20 × 10⁻³ A. Therefore, r_d = 0.04 / (20 × 10⁻³) = 2.0 Ω.

hard

6. At an operating point on the reverse-bias curve of a diode, a student measures the reverse voltage to be 10 V and the reverse current to be 2.0 µA. When the voltage is increased to 15 V, the current becomes 2.1 µA. What are the reverse static resistance at 10 V and the reverse dynamic resistance in this interval?

  1. 2.0 × 10⁶ Ω and 1.0 × 10⁷ Ω
  2. 5.0 × 10⁶ Ω and 5.0 × 10⁷ Ω
  3. 5.0 × 10⁶ Ω and 5.0 × 10⁶ Ω
  4. 1.0 × 10⁷ Ω and 2.5 × 10⁶ Ω
Show answer and explanation

Answer: (B) 5.0 × 10⁶ Ω and 5.0 × 10⁷ Ω

Static resistance R_static = V / I = 10 V / (2.0 × 10⁻⁶ A) = 5.0 × 10⁶ Ω (5.0 MΩ). Dynamic resistance r_dynamic = ΔV / ΔI = (15 - 10) V / [(2.1 - 2.0) × 10⁻⁶ A] = 5 V / (0.1 × 10⁻⁶ A) = 5.0 × 10⁷ Ω (50 MΩ).

medium

7. Consider the following two statements regarding the experimental determination of p-n junction diode characteristics: Statement I: In the forward bias experiment, the voltage step size chosen near and above the knee voltage should be smaller than that chosen well below the knee voltage. Statement II: Above the knee voltage, current increases exponentially with voltage, so large voltage steps can cause rapid current surges exceeding the meter's range.

  1. Both Statement I and Statement II are correct, and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct, but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct, and Statement II is the correct explanation of Statement I

Below the knee voltage, current is negligibly small, so larger voltage increments (e.g., 0.1 V) are acceptable. Above the knee voltage, current rises steeply, so fine increments (e.g., 0.02 V) must be taken to accurately map the curve and prevent current from abruptly exceeding the milliammeter range.

hard

8. A circuit designed to plot the forward characteristic of a diode contains a DC power supply V_s, a series resistor R, and the diode. The load line equation for this circuit on the I-V plane (where I is vertical axis and V is horizontal axis) is:

  1. I = V/R - V_s/R
  2. I = -V/R + V_s/R
  3. I = -V/R - V_s/R
  4. I = V_s/R
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Answer: (B) I = -V/R + V_s/R

Applying Kirchhoff's loop rule: V_s = I·R + V, where V is the voltage across the diode and I is the circuit current. Rearranging for current I gives I = -(1/R)·V + (V_s/R). This represents a straight line with slope -1/R and vertical intercept V_s/R.

medium

9. While plotting the forward characteristic curve of a silicon diode, a student accidentally connects the voltmeter in series with the diode and the milliammeter in parallel across the diode. What will the student observe upon gradually increasing the supply voltage?

  1. The diode burns out immediately due to excessive power dissipation
  2. The milliammeter burns out due to excessive current while the voltmeter reads zero
  3. Both meters show normal forward-bias characteristic readings
  4. The voltmeter reads almost the entire supply voltage while the milliammeter reads near zero current
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Answer: (D) The voltmeter reads almost the entire supply voltage while the milliammeter reads near zero current

An ideal or practical voltmeter has a very high internal resistance. Connected in series, it restricts the total circuit current to an extremely negligible value, so the milliammeter (in parallel with the diode) reads near zero. Almost the entire supply voltage drops across the high-resistance voltmeter.

easy

10. Consider the following two statements regarding the circuit setup used to study the forward and reverse characteristics of a p-n junction diode: Statement I: A milliammeter is chosen for the forward-bias circuit, whereas a microammeter is chosen for the reverse-bias circuit. Statement II: The reverse saturation current is primarily due to minority charge carriers whose concentration is very low compared to majority carriers. In light of the above statements, choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Forward current is carried by majority carriers and is typically on the order of milliamperes, requiring a milliammeter. Reverse current below breakdown is carried by thermally generated minority carriers and is typically on the order of microamperes, requiring a microammeter. Thus, both statements are correct.

medium

11. A student sets up the circuit to study diode characteristics. The DC supply provides a constant 5.0 V, connected in series with a 200 Ω resistor and a forward-biased silicon diode. If the voltage drop across the diode under these conditions is 0.7 V, the power dissipated in the series resistor is:

  1. 92.45 mW
  2. 15.05 mW
  3. 107.5 mW
  4. 43.0 mW
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Answer: (A) 92.45 mW

The voltage drop across the series resistor is V_R = V_supply - V_diode = 5.0 V - 0.7 V = 4.3 V. The power dissipated in the resistor is P = (V_R)² / R = (4.3)² / 200 = 18.49 / 200 = 0.09245 W = 92.45 mW.

easy

12. While plotting the I-V characteristics of a p-n junction diode, which of the following represents the correct meter combination used in the circuit?

  1. Forward bias: Microammeter; Reverse bias: Milliammeter
  2. Forward bias: Milliammeter; Reverse bias: Microammeter
  3. Forward bias: Ammeter (0-5 A); Reverse bias: Milliammeter
  4. Forward bias: Microammeter; Reverse bias: Microammeter
Show answer and explanation

Answer: (B) Forward bias: Milliammeter; Reverse bias: Microammeter

In forward bias, current is carried by majority carriers and is typically of the order of milliamperes (mA). In reverse bias, minority carrier drift produces a tiny current of the order of microamperes (μA) for Ge or nanoamperes for Si, requiring a microammeter.

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