Q.A forward biased diode is treated as

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set 55/2/11 markMCQ
   Q.Two statements are given – one labelled Assertion (A) and the other Reason (R). Select the correct answer from the codes below: (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false and (R) is also false. Assertion (A) : On forward biasing a p-n junction diode, the height of the barrier potential increases. Reason (R) : In forward biasing of a p-n junction diode, the direction of the applied voltage is in the same direction as the built-in potential.

   ›Reveal solutionSolution

   The assertion is false because forward biasing decreases the barrier height, not increases it. The reason is also false because the applied voltage opposes the built-in potential, not aligns with it. Hence both statements are false.

   The Concept: Barrier Potential in a p-n Junction

   A p-n junction has a depletion region at the interface, where mobile charge carriers have recombined, leaving behind fixed positive ions on the n-side and fixed negative ions on the p-side. This creates an internal electric field pointing from n to p, which gives rise to a built-in potential (or barrier potential) $V_b$. This barrier prevents further diffusion of majority carriers across the junction.

   Now, what happens when we apply an external voltage?

   Forward bias means connecting the positive terminal of the battery to the p-side and the negative terminal to the n-side. The applied voltage $V_f$ creates an electric field that opposes the built-in field. The net field across the depletion region decreases, the depletion width shrinks, and the barrier height reduces to $V_b - V_f$.

   Reverse bias does the opposite: the applied field adds to the built-in field, increasing the barrier height to $V_b + V_r$.

   So the key idea is simple: forward bias lowers the barrier; reverse bias raises it.

   ---

   Step-by-Step Analysis

   1. Examine the Assertion (A):

   "On forward biasing a p-n junction diode, the height of the barrier potential increases."

   This directly contradicts the physics we just discussed. In forward bias, the applied voltage reduces the net potential across the depletion region. The barrier height decreases, not increases. Therefore, Assertion (A) is false.

   Watch out

   A common mistake is to think that "forward" means the voltage is pushing carriers forward over the barrier, so the barrier must be higher to stop them. Actually, forward bias helps carriers cross by lowering the barrier — that's why current flows easily.

   2. Examine the Reason (R):

   "In forward biasing of a p-n junction diode, the direction of the applied voltage is in the same direction as the built-in potential."

   The built-in potential points from the n-side to the p-side (i.e., the n-side is at a higher potential relative to the p-side). In forward bias, the positive terminal is connected to p and the negative to n. So the applied voltage makes the p-side positive relative to the n-side — which is opposite to the direction of the built-in potential. Hence, Reason (R) is also false.

   3. Determine the correct code:

   Both (A) and (R) are false. This corresponds to option (D).

   Tip

   A quick memory aid: Forward = Force down (barrier decreases). Reverse = Raise (barrier increases). The applied voltage always opposes the built-in potential in forward bias.

   ✓Final answer

   The correct option is (D): both Assertion (A) and Reason (R) are false.

2. CBSE 2026Set 55/3/11 markMCQ
   Q.The process named 'minority carrier injection' in a p-n junction diode occurs during : (A) forward biasing (B) reverse biasing (C) no biasing at low temperature (D) no biasing at high temperature

   ›Reveal solutionSolution

   Minority carrier injection happens when a p-n junction is forward biased — the applied voltage reduces the barrier, allowing majority carriers from each side to cross over and become minority carriers on the other side. The correct answer is (A) forward biasing.

   The core concept: what "minority carrier injection" really means

   In a p-n junction, the p-side has holes as majority carriers and electrons as minority carriers; the n-side has electrons as majority and holes as minority. At equilibrium (no bias), the built-in potential barrier prevents net flow — only a tiny leakage current of minority carriers drifts across.

   "Minority carrier injection" is the process where majority carriers from one side are forced across the junction into the opposite side, where they become minority carriers. This is not a spontaneous event — it requires an external voltage to overcome the barrier.

   Why forward bias is the only condition that does this

   1. Forward bias (p-side positive, n-side negative):

   The applied voltage opposes the built-in potential, reducing the barrier height. Majority carriers (holes from p-side, electrons from n-side) now have enough energy to diffuse across the junction. Once across, each hole finds itself surrounded by electrons on the n-side — it is now a minority carrier there. Similarly, electrons that cross become minority carriers on the p-side. This flood of excess minority carriers is precisely what "minority carrier injection" means. The injected carriers then recombine gradually, producing the forward current.

   2. Reverse bias (p-side negative, n-side positive):

   The applied voltage adds to the built-in potential, making the barrier even larger. Majority carriers cannot cross. The only current is a tiny reverse saturation current due to the existing minority carriers being swept across by the electric field — no new minority carriers are injected. In fact, any minority carriers that do appear are quickly removed, not injected.

   3. No biasing (at any temperature):

   Without an external voltage, the junction is in equilibrium. The net current is zero. There is no net injection — the small number of minority carriers that diffuse across are exactly balanced by the drift current. Temperature affects the intrinsic carrier concentration, but it does not cause injection; it only changes the equilibrium concentrations.

   Watch out

   A common mistake is to think that reverse bias also injects minority carriers because current still flows. That current comes from already present minority carriers being pulled across, not from majority carriers being forced over. Injection specifically means majority carriers become minority carriers on the other side — that only happens in forward bias.

   Step-by-step reasoning

   1. Identify what "injection" requires: For a carrier to be injected, it must cross the junction from the side where it is a majority carrier to the side where it is a minority carrier. This requires overcoming the potential barrier.

   2. Check forward bias: The applied voltage reduces the barrier. Majority carriers from each side gain enough energy to diffuse across. Holes injected into n-side become minority carriers there; electrons injected into p-side become minority carriers there. This is the textbook definition of minority carrier injection.

   3. Check reverse bias: The barrier increases. Majority carriers cannot cross. Only the small number of minority carriers already present on each side are swept across by the strong electric field — they are extracted, not injected. No new minority carriers are created.

   4. Check no bias: The barrier is at its equilibrium height. The diffusion of majority carriers is exactly balanced by the drift of minority carriers. Net injection is zero. Temperature alone cannot cause injection — it only shifts the equilibrium concentrations.

   5. Conclusion: The only condition that forces majority carriers to cross the junction and become minority carriers on the other side is forward biasing.

   Tip

   A memory aid: "Forward bias floods the other side with strangers (minority carriers); reverse bias only sweeps away the locals (existing minority carriers)."

   ✓Final answer

   The correct option is (A) forward biasing.

3. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: For germanium diode threshold voltage is nearly ______ volt.

   ›Reveal solutionSolution

   Germanium diode threshold voltage ≈ 0.3 V (silicon is ≈ 0.7 V).

   The threshold (knee or cut-in) voltage of a diode is the minimum forward voltage at which the diode starts conducting appreciably. This value depends on the semiconductor's band gap. For germanium (smaller band gap) it is about 0.3 V, whereas for silicon it is about 0.7 V.

   Hence the blank is filled by 'nearly 0.3'.

   ✓Final answer

   ≈ 0.3 V.

4. CBSE 2025Set D1 markMCQ
   Q.Reverse biased diode is (A) Zener diode (B) LED (C) Photodiode (D) both (A) and (C)

   ›Reveal solutionSolution

   A Zener diode works in its reverse breakdown region and a photodiode operates in reverse bias, so both are reverse-biased devices.

   A Zener diode is specially designed to operate in the reverse breakdown region; as a voltage regulator it is connected in reverse bias so it maintains a constant voltage.

   A photodiode is operated in reverse bias so that incident light increases the reverse (minority-carrier) current, which is then measured to sense light intensity.

   An LED, by contrast, is operated in forward bias to emit light. Hence the reverse-biased diodes are the Zener diode and the photodiode — option (D).

   ✓Final answer

   (D) both (A) and (C).

5. CBSE 2025Set ANNUAL1 mark
   Q.Which type of biasing gives a semiconductor diode a very high resistance?

   ›Reveal solutionSolution

   Applying a reverse bias (p-side connected to the negative terminal, n-side to positive) pulls majority carriers further away from the junction, widening the depletion layer and hugely increasing the diode's resistance.

   A p-n junction diode can be biased two ways:

   • Forward bias (p-side to +, n-side to -): narrows the depletion region, lowers the junction's resistance, and allows a large current to flow once the barrier potential is overcome.
   • Reverse bias (p-side to -, n-side to +): pulls majority carriers away from the junction, widening the depletion region and greatly increasing the junction's resistance - only a very small reverse saturation (leakage) current flows, mainly due to minority carriers.

   So it is reverse biasing that gives the diode its very high resistance, which is exactly why diodes are used for rectification - they conduct easily one way (forward, low resistance) and block current the other way (reverse, high resistance).

   ✓Final answer

   Reverse biasing gives a semiconductor diode very high resistance.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Applying forward bias in p-n junction, the potential barrier :

   (a) decreases.

   (b) increases.

   (c) remains unchanged.

   (d) becomes zero.

   ›Reveal solutionSolution

   Forward bias applies an external voltage that opposes the built-in potential barrier at the p-n junction, reducing its effective height and allowing majority carriers to cross easily.

   At an unbiased p-n junction, diffusion of majority carriers across the junction creates a depletion region and a built-in potential barrier. When forward biased (p-side connected to positive terminal, n-side to negative), the applied electric field opposes the internal field of the junction. This narrows the depletion region and lowers the potential barrier, making it easier for majority carriers (holes from p-side, electrons from n-side) to diffuse across and constitute a large forward current.

   ✓Final answer

   In forward bias, the potential barrier decreases — option (a).

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.When a junction diode is reverse biased, the flow of current across the junction is mainly due to:

   (a) diffusion of charges

   (b) depends on the nature of material

   (c) drift of charges

   (d) both drift and diffusion of charges.

   ›Reveal solutionSolution

   Under reverse bias, the small current across a p-n junction is due to drift of minority charge carriers, not diffusion.

   In a p-n junction under forward bias, majority carriers diffuse across the junction and diffusion current dominates. Under reverse bias, the applied field opposes diffusion of majority carriers (almost completely stopped), but assists the motion of the minority carriers (electrons in the p-side, holes in the n-side) that are thermally generated near the junction; these are swept across the depletion region by the field. This gives a small, nearly voltage-independent reverse saturation current due to drift of minority carriers.

   ✓Final answer

   (c) Drift of charges.

8. CBSE 2024Set 55/2/11 markMCQ
   Q.When a p-n junction diode is subjected to reverse biasing : (A) the barrier height decreases and the depletion region widens. (B) the barrier height increases and the depletion region widens. (C) the barrier height decreases and the depletion region shrinks. (D) the barrier height increases and the depletion region shrinks.

   ›Reveal solutionSolution

   In reverse bias, the external voltage adds to the built-in potential, raising the barrier height and pushing carriers away, which widens the depletion region. The correct option is (B).

   The Concept: What Reverse Bias Does to a p-n Junction

   A p-n junction diode has a natural built-in potential (barrier) at the junction, formed by the diffusion of holes from the p-side and electrons from the n-side. This creates a depletion region — a zone depleted of free charge carriers, containing only fixed ions. The barrier height is the voltage that opposes further diffusion.

   When you apply reverse bias (positive terminal to n-side, negative to p-side), the external voltage acts in the same direction as the built-in field. It doesn't "help" current flow; instead, it reinforces the barrier. Think of it like pushing a door that's already closed — the door becomes harder to open.

   Step-by-Step Reasoning

   1. Barrier height increases

      The built-in potential $V_0$ is fixed by doping. Reverse bias adds an external voltage $V_R$ across the junction, so the total barrier becomes $V_0 + V_R$. This is larger than $V_0$, so the barrier height increases.

   2. Depletion region widens

      The stronger electric field pulls majority carriers (holes on p-side, electrons on n-side) further away from the junction. More fixed ions are uncovered, so the depletion region expands on both sides. The width $W$ is proportional to $\sqrt{V_0 + V_R}$, so it widens with increasing reverse bias.

   3. What about current?

      Reverse bias doesn't cause significant current (except a tiny leakage) because the barrier prevents majority carriers from crossing. This is consistent with the widening depletion region — fewer carriers are available near the junction.

   Watch out

   A common mistake is to think reverse bias "reduces" the barrier because it opposes forward current. Actually, reverse bias adds to the barrier — only forward bias reduces it.

   Tip

   Remember the mnemonic: Reverse = Raise barrier, Widen depletion. Forward = Flatten barrier, Narrow depletion.

   Final Answer

   ✓Final answer

   The correct option is (B): the barrier height increases and the depletion region widens.

9. CBSE 2024Set ANNUAL1 mark
   Q.Which type of biasing gives a semiconductor diode very high resistance?

   ›Reveal solutionSolution

   Reverse bias widens the depletion region, giving very high resistance to current flow.

   When a p-n junction diode is reverse biased, the applied voltage widens the depletion region and increases the potential barrier at the junction, sweeping the majority carriers further away from the junction. This makes the junction offer an extremely high resistance to the flow of (majority-carrier) current, allowing only a very small, nearly-constant reverse saturation current (due to minority carriers) to flow. In forward bias, by contrast, the depletion region narrows and the junction offers a very low resistance.

   ✓Final answer

   Reverse biasing gives a semiconductor (p-n junction) diode very high resistance.

10. CBSE 2023Set 55/3/11 markMCQ
    Q.If a p-n junction diode is reverse biased,

    (a) the potential barrier is lowered.

    (b) the potential barrier remains unaffected.

    (c) the potential barrier is raised.

    (d) the current is mainly due to majority carriers.

    ›Reveal solutionSolution

    Reverse bias raises the potential barrier across a p-n junction by widening the depletion region, making it harder for majority carriers to cross — so the correct option is (C).

    Concept and Intuition

    A p-n junction diode has a potential barrier (also called the built-in potential) at the junction. This barrier exists because, at equilibrium, diffusion of majority carriers (holes from p-side, electrons from n-side) creates a depletion region with an internal electric field that opposes further diffusion. The height of this barrier is determined by the doping concentrations and temperature — it’s fixed for a given diode at equilibrium.

    Now, what happens when you apply an external voltage? The key idea is that the external bias either opposes or aids the internal electric field. In reverse bias, the positive terminal of the battery is connected to the n-side and the negative terminal to the p-side. This external field points in the same direction as the internal field — from n to p. So the net field across the junction increases, pulling more majority carriers away from the junction. This widens the depletion region and raises the potential barrier.

    In contrast, forward bias reduces the barrier. And current in reverse bias is due to minority carriers (not majority), which is very small (leakage current).

    Watch out

    Common Mistake

    Many students think reverse bias "lowers" the barrier because they confuse it with forward bias. Remember: reverse bias = barrier raised; forward bias = barrier lowered.

    Step-by-Step Reasoning

    1. Recall the equilibrium condition: At zero bias, the p-n junction has a built-in potential barrier $V_0$ (typically ~0.7 V for silicon). This barrier prevents net current flow.

    2. Apply reverse bias: Connect the positive terminal of a battery to the n-side and negative to the p-side. The external voltage $V_R$ creates an electric field from n to p — same direction as the internal field.

    3. Effect on the depletion region: The external field adds to the internal field, pulling holes (on p-side) further away from the junction and electrons (on n-side) further away. This widens the depletion region.

    4. Effect on the potential barrier: The total potential across the junction becomes $V_0 + V_R$ (since the external voltage adds to the built-in potential). So the barrier height increases — it is raised.

    5. Effect on current: Majority carriers (holes on p-side, electrons on n-side) now face an even larger barrier, so diffusion current drops to nearly zero. The only current is a tiny reverse saturation current due to minority carriers drifting across — so option (d) is false.

    6. Evaluate the options:

       • (a) "potential barrier is lowered" — false, that’s forward bias.
       • (b) "potential barrier remains unaffected" — false, it changes with bias.
       • (c) "potential barrier is raised" — true.
       • (d) "current is mainly due to majority carriers" — false, it’s due to minority carriers.

    Tip

    Quick Mnemonic

    Reverse bias = Raised barrier. Forward bias = Flattened (lowered) barrier. The first letters match.

    ✓Final answer

    The correct option is (C) — the potential barrier is raised when a p-n junction diode is reverse biased.

11. CBSE 2023Set 55/4/11 markMCQ
    Q.The threshold voltage for a p-n junction diode used in the circuit is $0.7\ \text{V}$. The type of biasing and current in the circuit are : (A) Forward biasing, $0\ \text{A}$ (B) Reverse biasing, $0\ \text{A}$ (C) Forward biasing, $5\ \text{mA}$ (D) Reverse biasing, $2\ \text{mA}$

    ›Reveal solutionSolution

    The diode is forward-biased because the battery’s positive terminal connects to the p-side and negative to the n-side. However, the applied voltage (0.5 V) is less than the threshold voltage (0.7 V), so the diode remains in the “off” state and no current flows. The correct choice is (A).

    

    ---

    The concept: Why a diode doesn’t conduct below its threshold

    A p-n junction diode is not a perfect switch that turns on the instant you apply any forward voltage. In forward bias, the external voltage must overcome the built-in potential barrier (about 0.7 V for silicon) before significant current can flow. Below that threshold, the diode’s resistance is extremely high — effectively an open circuit. This is why the circuit current is zero when the battery supplies only 0.5 V.

    Watch out

    A common mistake is to assume that any forward bias automatically produces current. In reality, the diode conducts only when the applied voltage exceeds the threshold (knee) voltage — typically 0.7 V for silicon.

    ---

    Step-by-step reasoning

    1. Identify the biasing from the circuit diagram

       The diode symbol shows the p-side (anode) connected to the positive terminal of the 0.5 V cell, and the n-side (cathode) connected to the negative terminal. This is the standard forward-bias configuration: the external field opposes the built-in field, reducing the barrier.

    2. Check the applied voltage against the threshold

       The battery supplies only 0.5 V. The diode’s threshold voltage is given as 0.7 V. Since $0.5 < 0.7$, the applied voltage is insufficient to overcome the barrier.

    3. Determine the diode’s state

       In this sub-threshold region, the diode behaves like a very large resistor — practically an open circuit. No charge carriers can cross the junction in significant numbers.

    4. Calculate the current

       With the diode effectively open, the series circuit is broken. By Ohm’s law, $I = \frac{V}{R}$, but here the diode’s infinite resistance means $I = 0\ \text{A}$. The 100 Ω resistor plays no role because no current flows through it.

    Tip

    You can think of the diode as a “voltage-controlled switch”: it stays off until the forward voltage reaches the threshold, then turns on abruptly. Below threshold, treat it as an open circuit.

    ---

    Final answer

    ✓Final answer

    The circuit is forward-biased, but the current is $0\ \text{A}$. The correct option is (A).

12. CBSE 2023Set ANNUAL1 mark
    Q.Is the Junction diode D forward or reverse biased in the given diagram?

    ›Reveal solutionSolution

    The anode (p-side) of D is connected (through R) to the higher potential (+5 V) terminal and the cathode (n-side) to the lower potential (+2 V) terminal, so D is forward biased.

    A junction diode conducts easily (is forward biased) when its p-side (anode) is at a higher potential than its n-side (cathode). Here the anode of D is joined through the resistor R to the +5 V terminal, while the cathode is joined to the +2 V terminal. Since +5 V > +2 V, the anode side is at the higher potential, which drives conventional current from the +5V terminal through R and D to the +2 V terminal — i.e., D is forward biased.

    ✓Final answer

    The junction diode D is forward biased.