Q.Explain the formation of depletion region in a p-n junction.

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.In an unbiased p-n junction, at equilibrium, which of the following statements is true? (A) Diffusion current is zero but drift current exists. (B) Diffusion current exists but drift current is zero. (C) Diffusion and drift currents are equal and opposite. (D) Both the diffusion and drift currents exist but are unequal.

   ›Reveal solutionSolution

   At equilibrium in an unbiased p-n junction, the net current is zero because the diffusion current (due to carrier concentration gradients) is exactly balanced by the drift current (due to the built-in electric field). The correct option is (C).

   The core concept: Why two currents must cancel

   A p-n junction at equilibrium means no external voltage is applied. Yet, inside the junction, two opposing processes are always active:

   1. Diffusion: Holes from the p-side (high concentration) diffuse into the n-side, and electrons from the n-side diffuse into the p-side. This creates a diffusion current in the direction of the concentration gradient.
   2. Drift: The diffusion of carriers leaves behind ionized dopant atoms (negative on the p-side, positive on the n-side), creating a built-in electric field across the depletion region. This field pushes carriers in the opposite direction — holes back to the p-side, electrons back to the n-side — producing a drift current.

   At equilibrium, these two currents must exactly cancel. If they didn’t, there would be a net flow of charge, which would change the electric field until balance is restored. This is a fundamental consequence of the condition that the Fermi level is constant throughout the junction.

   Watch out

   A common mistake is to think that at equilibrium, no currents exist at all. In reality, both diffusion and drift currents are present — they are just equal and opposite, so the net current is zero.

   Step-by-step reasoning

   1. Identify the equilibrium condition

      An unbiased p-n junction has no external voltage source. The system is in thermodynamic equilibrium, meaning the Fermi level is flat (constant) across the entire device. This implies that the net current through the junction must be zero.

   2. Recognize the two current components

      The total current $I$ across the junction is the sum of the diffusion current $I_{\text{diff}}$ and the drift current $I_{\text{drift}}$:

   $I = I_{\text{diff}} + I_{\text{drift}}$

   Both are non-zero because there is a concentration gradient (driving diffusion) and a built-in electric field (driving drift).

   3. Apply the equilibrium constraint At equilibrium, $I = 0$. Therefore:

   $I_{\text{diff}} + I_{\text{drift}} = 0 \quad \Rightarrow \quad I_{\text{diff}} = -I_{\text{drift}}$

   The negative sign indicates they flow in opposite directions. Their magnitudes are equal.

   4. Check each option
      • (A) Diffusion current is zero but drift current exists — False, because diffusion is the very reason the built-in field exists.
      • (B) Diffusion current exists but drift current is zero — False, because without drift, the field would not oppose diffusion, and carriers would keep moving until equilibrium is lost.
      • (C) Diffusion and drift currents are equal and opposite — True, as derived.
      • (D) Both exist but are unequal — False, because that would give a net current, violating equilibrium.

   Tip

   A neat way to remember this: The built-in electric field is caused by diffusion, so it always adjusts itself to exactly oppose diffusion. At equilibrium, the two are in perfect balance — like a tug-of-war where neither side moves.

   ✓Final answer

   The correct option is (C): Diffusion and drift currents are equal and opposite.

2. CBSE 2026Set DS1 markMCQ
   Q.In depletion layer of $p$-$n$ junction diode, there are:

   i) only electrons

   ii) only holes

   iii) both electrons and holes

   iv) neither electrons nor holes

   ›Reveal solutionSolution

   The depletion region is depleted of mobile charge carriers — it contains only fixed ions.

   Concept. When a $p$–$n$ junction forms, electrons from the $n$-side diffuse across and recombine with holes on the $p$-side (and vice-versa). This leaves behind, near the junction, a thin region containing only the immobile positive donor ions (on the $n$-side) and immobile negative acceptor ions (on the $p$-side). Since the free electrons and holes have recombined/been swept away, this region has practically no mobile charge carriers — hence its name, the depletion layer. It sets up the built-in potential barrier.

   ✓Final answer

   (iv) neither (mobile) electrons nor holes — only fixed ionised atoms remain.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of barrier potential for Si semiconductor is:

   (a) 0.3 V

   (b) 1.1 V

   (c) 0.7 eV

   (d) 0.7 V

   ›Reveal solutionSolution

   The depletion-layer potential barrier that must be overcome to forward-bias a silicon diode is about 0.7 volt (compare germanium's ~0.3 V).

   When a p-n junction forms, diffusion of majority carriers across the junction sets up a depletion region and an internal electric field, which corresponds to a potential difference — the barrier potential — that opposes further diffusion. For silicon this barrier potential is characteristically about $0.7\ \text{V}$ (for germanium it is smaller, about $0.3\ \text{V}$). This is a potential difference measured in volts, not an energy in electron-volts, so the unit-correct answer is 0.7 V rather than 0.7 eV.

   ✓Final answer

   (d) 0.7 V

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.In P-n junction diode the width of depletion layer is :

   (a) 1 m

   (b) 1 μm

   (c) 1 nm

   (d) 1 pm

   ›Reveal solutionSolution

   The depletion layer of a p-n junction is about 1 μm (≈10⁻⁶ m) wide.

   When a p-n junction is formed, electrons and holes diffuse across the junction and recombine, leaving behind immobile ionised donor and acceptor atoms. This creates a region devoid of free charge carriers — the depletion layer.

   Its width is typically of the order of a micrometre (about 10⁻⁶ m, i.e. 1 μm). Among the options, 1 μm is the correct order of magnitude.

   ✓Final answer

   (b) 1 μm.

5. CBSE 2025Set IMPROVEMENT1 markMCQ
   Q.The depletion layer of a p-n junction diode consists of:

   (a) Only holes

   (b) Only electrons

   (c) Both electrons and holes

   (d) Neither electron nor hole

   ›Reveal solutionSolution

   The depletion layer is depleted of free charge carriers; it contains only fixed (immobile) ionised donor and acceptor atoms.

   When a p-n junction is formed, free electrons from the n-side diffuse into the p-side and recombine with holes, and free holes from the p-side diffuse into the n-side and recombine with electrons, near the junction. This leaves behind immobile positively charged donor ions on the n-side and immobile negatively charged acceptor ions on the p-side, forming a narrow region called the depletion layer. This region is depleted of mobile charge carriers (free electrons and free holes) — it contains only fixed ions, which set up the barrier potential/electric field across the junction.

   ✓Final answer

   (d) Neither electron nor hole — the depletion layer contains only immobile ions.

6. CBSE 2025Set ANNUAL1 mark
   Q.A depletion layer consists of ................ . (fill in the blank)

   ›Reveal solutionSolution

   The depletion layer at a p-n junction is a thin region stripped of mobile carriers, containing only fixed ionised donor/acceptor atoms.

   When a p-n junction forms, free electrons near the junction on the n-side diffuse across and recombine with holes on the p-side (and vice-versa). This leaves behind, on either side of the junction, a narrow region of immobile ions: positively charged (uncompensated) donor ions on the n-side and negatively charged (uncompensated) acceptor ions on the p-side. This region is depleted of free (mobile) charge carriers — hence the name depletion layer — and the exposed ion charges set up a built-in electric field/potential barrier across the junction.

   ✓Final answer

   Immobile positive donor ions (n-side) and negative acceptor ions (p-side) — no free carriers.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Two statements are given below : one labelled as Assertion (A) and other labelled as Reason (R). Read the statements carefully and choose the correct option. Assertion (A) : The thickness of the depletion layer is fixed in all semi-conductor devices. Reason (R) : Free carriers are available in depletion layer.

   (a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but the Reason (R) is the incorrect explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) and Reason (R) both are false.

   ›Reveal solutionSolution

   The depletion layer's thickness is not fixed (it changes with applied bias), and it contains no free carriers at all — both statements are wrong.

   Assertion: False. The thickness of the depletion layer is not fixed — it changes with the applied voltage. Forward bias narrows the depletion layer (reduces the barrier), while reverse bias widens it (increases the barrier). It also depends on doping concentration.

   Reason: False. By definition, the depletion layer is the region devoid of free charge carriers — it contains only fixed, immobile ionised donor/acceptor atoms, not free carriers.

   Since both statements are individually incorrect, option (d) is correct.

   ✓Final answer

   (d) — both A and R are false.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Assertion: Thickness of depletion layer is fixed in all semiconductor devices. Reason: No immobile charged carrier ions are available in depletion layer.

   (a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

   (b) If both assertion and reason are true but reason is not a correct explanation of the assertion.

   (c) Assertion is true but reason is false.

   (d) Both assertion and reason are false.

   ›Reveal solutionSolution

   The depletion layer's thickness is NOT fixed — it changes with applied bias — and the depletion layer DOES contain immobile ionised dopant ions, so both statements are false.

   Assertion check: In a p-n junction, the depletion-region width depends on the applied voltage. Under forward bias, the width of the depletion layer decreases; under reverse bias, it increases; it also depends on doping concentration. So the assertion, that the thickness is 'fixed in all semiconductor devices', is FALSE.

   Reason check: The depletion layer is called so because it is depleted of MOBILE charge carriers (free electrons and holes), but it is NOT devoid of charge altogether — it contains immobile ionised donor atoms (positive ions) on the n-side and immobile ionised acceptor atoms (negative ions) on the p-side. These immobile ions are exactly what create the internal electric field/potential barrier at the junction. So the statement 'no immobile charged carrier ions are available' is also FALSE.

   Since both the assertion and the reason are false statements, the correct option is (d).

   ✓Final answer

   (d) Both assertion and reason are false.

9. CBSE 2025Set ANNUAL1 mark
   Q.What is the effect on the width of a depletion region of a p-n junction diode if the doping concentration is increased?

   ›Reveal solutionSolution

   More dopant ions per unit volume means equilibrium (same barrier potential) is reached over a narrower depletion width.

   The depletion region forms when majority carriers diffuse across the junction and leave behind fixed ionized dopant atoms. As doping concentration increases, there are more ionized dopant atoms per unit volume available near the junction, so the same total (equilibrium) amount of fixed charge — and hence the same barrier potential — is reached over a narrower region. So the width of the depletion region decreases as doping concentration increases (and, correspondingly, the junction capacitance increases, since capacitance is inversely related to depletion width).

   ✓Final answer

   Increasing the doping concentration decreases (narrows) the width of the depletion region.

10. CBSE 2024Set IMPROVEMENT1 mark
    Q.Why do holes diffuse from p-region to n-region in an unbiased p-n junction?

    ›Reveal solutionSolution

    Diffusion always moves majority carriers from a region of high concentration to low concentration.

    In the p-region, holes are the majority carriers and are present in high concentration; in the n-region, holes are minority carriers, present in very low concentration. Whenever a p-n junction is formed, this large concentration gradient of holes across the junction drives them to diffuse from the p-region (high hole concentration) into the n-region (low hole concentration), exactly as any species diffuses from higher to lower concentration. (Similarly, electrons diffuse from the n-region into the p-region.) This diffusion happens even without any external bias, purely due to the concentration difference.

    ✓Final answer

    Because of the concentration gradient — holes are far more concentrated in the p-region than in the n-region, so they diffuse from p to n.

11. CBSE 2024Set A1 markMCQ
    Q.In depletion layer of p-n junction diode, there are -

    (a) Electron

    (b) Protons

    (c) Mobile ions

    (d) Immobile ions

    ›Reveal solutionSolution

    The depletion region is depleted of free charge carriers and contains only fixed ionised dopant atoms.

    When a p-n junction forms, mobile electrons from the n-side diffuse into the p-side and mobile holes from the p-side diffuse into the n-side; these recombine near the junction. This leaves behind a thin region on either side of the junction that has no free (mobile) charge carriers, only the immobile ionised dopant atoms — positive donor ions on the n-side and negative acceptor ions on the p-side, fixed in the crystal lattice. This region is called the depletion layer (or depletion region), and it is these fixed/immobile ions that set up the barrier potential.

    ✓Final answer

    (d) Immobile ions.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The potential barrier in the depletion layer is due to -

    (a) ions

    (b) holes

    (c) electrons

    (d) forbidden gap

    ›Reveal solutionSolution

    The depletion layer's potential barrier arises from the fixed, immobile ions left behind after diffusion of mobile carriers, not from the carriers themselves.

    At a p-n junction, electrons diffuse from the n-side to the p-side and holes diffuse from the p-side to the n-side. This leaves behind fixed ionized donor atoms (positive) on the n-side and fixed ionized acceptor atoms (negative) on the p-side, near the junction — the depletion region has no mobile carriers, only these immobile ions. The resulting space-charge region of positive and negative ions sets up an internal electric field, which is the potential barrier that opposes further diffusion.

    ✓Final answer

    The potential barrier is due to the (immobile) ions in the depletion layer — option (a).