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Read the following… (Class 12 Physics SQP 2023-24, Q29)

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PhysicsCBSE Class 12 · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 12 Physics sample paper 2023-24: question 29

From the official CBSE sample question paper for 2023-24, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedSemiconductor Electronics: Materials, Devices and Simple Circuits

Read the following paragraph and answer the questions that follow.
A semiconductor diode is basically a pn junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device. When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal, it is said to be forward biased. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. An ideal diode is one whose resistance in forward biasing is zero and the resistance is infinite in reverse biasing. When the diode is forward biased, it is found that beyond forward voltage called knee voltage, the conductivity is very high. When the biasing voltage is more than the knee voltage the potential barrier is overcome and the current increases rapidly with increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.

  1. i.

    In the given figure, a diode D is connected to an external resistance R = 100 and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:
    (a) 40 mA (b) 20 mA (c) 35 mA (d) 30 mA

    1. (A)40 mA
    2. (B)20 mA
    3. (C)35 mA
    4. (D)30 mA
    1 mark
  2. ii.

    In which of the following figures, the pn diode is reverse biased?

    1. (A)Diode with p-side at -12 V and resistor R end at -5 V
    2. (B)Diode p-side earthed, resistor R end at -10 V
    3. (C)Diode with resistor R end at +5 V and other end earthed
    4. (D)Diode p-side at +5 V, resistor R end at +10 V
    1 mark
  3. iii.

    Based on the V-I characteristics of the diode, we can classify diode as

    1. (A)bilateral device
    2. (B)ohmic device
    3. (C)non-ohmic device
    4. (D)passive element
    1 mark
  4. iv.

    Two identical PN junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be
    (1) P N N P, (2) P N P N, (3) N P N P, each connected across a battery with + and − terminals.

    1. (A)in the circuits (1) and (2)
    2. (B)in the circuits (2) and (3)
    3. (C)in the circuits (1) and (3)
    4. (D)only in the circuit (1)
    1 mark

  5. OR

    iv.

    The V-I characteristic of a diode is shown in the figure. The ratio of the resistance of the diode at I = 15 mA to the resistance at V = -10 V is

    1. (A)100
    2. (B)
    3. (C)10
    4. (D)
    1 mark

Figure in the paper: Circuit: diode D in series with R = 100 Ω and a 3.5 V cell. Part ii figures: four diode-resistor circuits with the given potentials. Part iv (OR) figure: three series PN junction connections with battery. Final V-I graph for silicon diode: forward current I (mA) axis marked 10, 15, 20, 30; voltage axis marked 0.5, 0.7, 0.8 V; steep rise near 0.7 V; reverse side at -10 V shows current about 1 µA. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Subtract the barrier potential from the emf before using Ohm's law.

Hint 2

Reverse bias means n-side at higher potential; equal junction voltages need both junctions facing the same way.

Hint 3

For resistance from a graph, use in forward bias and in reverse bias.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) Voltage across R V, so A mA. Option (d).

    1 mark

  2. (ii) A diode is reverse biased when its n-side is at a higher potential than its p-side. In circuit (c) the n-side is held at +5 V through R while the p-side is earthed (0 V), so it is reverse biased. Option (c).

    1 mark

  3. (iii) The V-I graph of a diode is not a straight line (current does not vary linearly with voltage), so the diode is a non-ohmic device. Option (c).

    1 mark

  4. (iv)(a) For the same potential difference across both junctions, both must be biased the same way, i.e. connected in the same orientation. In (2) PNPN and (3) NPNP both junctions face the same way; in (1) PNNP one is forward and one reverse biased. Option (b).

    1 mark

  5. (iv)(b) From the graph, near mA: . At V, A: . Ratio . Option (d).

Answer: (i) (d) 30 mA; (ii) (c); (iii) (c) non-ohmic; (iv) (b) circuits (2) and (3) OR (d)

Where the marks are

  • (i) 30 mA, option (d) (1 mark)
  • (ii) option (c), n-side at higher potential (1 mark)
  • (iii) non-ohmic device, option (c) (1 mark)
  • (iv) circuits (2) and (3), option (b); OR ratio , option (d) (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Semiconductor Electronics: Materials, Devices and Simple Circuits, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.