Zener Diode

Answer: D. NCERT Class 12 Physics Chapter 14, page 336, states that a Zener diode is fabricated by heavily doping both the p- and n-sides, giving a depletion region less than 10⁻⁶ m thick.

Why A is wrong: A is wrong because the heavy doping is applied to both sides; asymmetric doping is not what defines a Zener diode.

Why B is wrong: B is wrong because a Zener is a plain p-n junction. An insulating layer between the regions would block the junction current the device depends on.

Why C is wrong: C is wrong because light doping widens the depletion region and lowers the junction field — the opposite of what the Zener design needs.

Answer: C. NCERT Class 12 Physics Chapter 14, page 336, describes the Zener diode as always used in reverse bias in the breakdown region, where the voltage across it stays close to V_z.

Why A is wrong: A is wrong because in forward bias the device behaves like any ordinary silicon diode and its Zener rating plays no role — this is the documented forward-bias distractor for this exam pattern.

Why B is wrong: B is wrong because below breakdown only the tiny reverse saturation current flows and the voltage across the diode simply follows the supply; nothing is stabilised.

Why D is wrong: D is wrong because below the cut-in voltage the junction is essentially off, so it neither conducts usefully nor holds a fixed voltage.

Answer: A. NCERT Class 12 Physics Chapter 14, page 336, gives the depletion region of a Zener diode as very thin, less than 10⁻⁶ m, because of the heavy doping on both sides.

Why B is wrong: B is wrong because 10⁻⁴ m is about a hundred times too wide; a junction that thick would need hundreds of volts to reach the breakdown field.

Why C is wrong: C is wrong because 10⁻² m is a centimetre — larger than most diode packages, let alone the junction inside one.

Why D is wrong: D is wrong because 10⁻¹ m is a macroscopic length and would give a junction field of only tens of volts per metre at normal reverse bias.

Answer: B. E = V/d = 5.0 V ÷ (1.0 × 10⁻⁶ m) = 5.0 × 10⁶ V/m, the same order of magnitude quoted in NCERT Class 12 Physics Chapter 14, page 336, as the field that triggers Zener breakdown.

Why A is wrong: A is wrong because it uses a depletion width of 10⁻³ m instead of 10⁻⁶ m — three orders of magnitude too thick for a heavily doped Zener junction.

Why C is wrong: C is wrong because it multiplies the voltage by the thickness instead of dividing, which also yields volt-metres rather than volts per metre.

Why D is wrong: D is wrong because it is a thousand times too large; it corresponds to a depletion width of 10⁻⁹ m, thinner than a few atomic spacings.

Answer: A. The applied 4.0 V is below V_z, so the diode stays in the low reverse-current region described in NCERT Class 12 Physics Chapter 14, page 336; the current is negligible, the drop across the resistor is negligible, and essentially the whole 4.0 V appears across the diode.

Why B is wrong: B is wrong because a Zener cannot raise the voltage above the supply. It clamps at 6.0 V only when the source can push it past breakdown, and 4.0 V cannot.

Why C is wrong: C is wrong because 2.0 V comes from subtracting the supply from the rating (6.0 − 4.0), an operation with no circuit meaning here.

Why D is wrong: D is wrong because zero across the diode would require it to act as a short circuit; a reverse-biased Zener below V_z is closer to an open circuit.

Answer: D. The Zener rating describes the reverse breakdown point only. In forward bias the device is an ordinary silicon junction, so it conducts at the usual forward drop — the behaviour NCERT Class 12 Physics Chapter 14, page 336, contrasts with the reverse breakdown use.

Why A is wrong: A is wrong because V_z is a reverse-bias property. Nothing in forward conduction holds the voltage at 8.0 V.

Why B is wrong: B is wrong because it adds two drops that never occur together; the forward and reverse characteristics are separate branches of the same curve.

Why C is wrong: C is wrong because a conducting silicon junction always sustains a finite forward drop of roughly 0.7 V, not zero.

Answer: B. NCERT Class 12 Physics Chapter 14, page 336, attributes Zener breakdown to the very high junction field — of order 5 × 10⁶ V/m at only a few volts of reverse bias — pulling electrons out of the host atoms, an internal field-emission process.

Why A is wrong: A is wrong because that describes avalanche breakdown, a collision-multiplication process. It is the standard alternative offered alongside the Zener mechanism, and the heavy doping here makes the depletion layer too thin for carriers to gain that much energy between collisions.

Why C is wrong: C is wrong because reverse bias widens the depletion region rather than removing it; the junction never becomes forward biased.

Why D is wrong: D is wrong because Zener breakdown is non-destructive and reversible when the current is limited. Melting would permanently destroy the device.

Answer: C. In breakdown each diode holds its own V_z, so the pair holds 5.0 + 3.0 = 8.0 V; the resistor takes 12.0 − 8.0 = 4.0 V, giving I = 4.0 V ÷ (1.0 × 10³ Ω) = 4.0 mA. The clamping behaviour is the reverse-breakdown property described in NCERT Class 12 Physics Chapter 14, page 336.

Why A is wrong: A is wrong because it ignores the diodes entirely and puts the full 12.0 V across the resistor.

Why B is wrong: B is wrong because it subtracts only the 3.0 V diode; series elements in breakdown each hold their own V_z, so both must be subtracted.

Why D is wrong: D is wrong because it subtracts only the 5.0 V diode and omits the 3.0 V one.