JEE Physics Semiconductor Electronics MCQs Set 14

Answer: (b) VC/e, VC

Explanation:
1. In circuit (1) the diode is forward biased for the discharge current, so the capacitor discharges through R: \( q=VC\,e^{-t/CR} \). At t = CR, q = VC/e.
2. In circuit (2) the diode is reverse biased, so no current flows and the capacitor keeps its charge VC.

Teacher's Note:
1. A forward biased ideal diode conducts like a wire.
2. A reverse biased diode blocks current.

Answer: (a) 1 A

Explanation:
1. The two diodes in the middle branch face each other (cathode to cathode), so whichever way the battery pushes, one of them is reverse biased and the middle branch carries no current.
2. The 10 V battery therefore drives current only through the 10 Ω resistor: \( I=\frac{10}{10}=1\ \text{A} \).

Teacher's Note:
1. A series pair of opposed diodes always blocks.
2. Then the circuit reduces to battery and resistor.

Answer: (c) the front surface becomes negatively charged

Explanation:
1. In p-type material the charge carriers are holes, which move in the direction of the conventional current (left to right).
2. Magnetic force \( \vec F=q\,\vec v\times\vec B \) with v to the right and B downward points directly away from the observer, so holes collect on the back face.
3. The back face becomes positive and the front face, left short of holes, becomes negative. The top face is not affected, as the force has no vertical component.

Teacher's Note:
1. Holes move along the current direction.
2. The Hall voltage appears across the faces perpendicular to both v and B.

Answer: (a) from the p-side to the n-side

Explanation:
1. Holes diffuse from the p-side to the n-side and electrons from the n-side to the p-side, because of the concentration differences.
2. The conventional current of both carriers is from the p-side to the n-side. Bias changes its magnitude, not its direction.

Teacher's Note:
1. Diffusion depends on the concentration gradient.
2. Drift current is the one that opposes diffusion current.

Answer: (a) \( 2\times10^{7}\ \text{V/m} \)

Explanation:
1. Energy gained over one mean free path: \( eEl=1\ \text{eV}=e\times1\ \text{V} \).
2. \( E=\frac{1}{5\times10^{-8}}=2\times10^{7}\ \text{V/m} \).

Teacher's Note:
1. Work done by the field = force × distance = eEl.
2. 1 eV is the energy for a potential difference of 1 V.

Answer: (d) Some potential difference develops across the p-n junction when it is formed.

Explanation:
1. A forward biased junction conducts, so the first statement is false.
2. A p-n junction conducts mainly in one direction and is used as a rectifier, so the second is false.
3. A reverse biased junction has very high resistance, so the third is false.
4. Diffusion of carriers sets up a barrier potential across the depletion region, so the last statement is true.

Teacher's Note:
1. Recall the barrier potential (about 0.3 V for Ge and 0.7 V for Si).
2. Forward bias conducts and reverse bias blocks.