JEE Physics Semiconductor Electronics MCQs Set 05

Answer: (b) decreases

Explanation:
1. More electron-hole pairs are generated as temperature rises.
2. The number of charge carriers increases much faster than the loss of mobility.
3. So resistivity decreases.

Teacher's Note:
1. Semiconductors have a negative temperature coefficient of resistance.
2. Metals are the opposite because carrier number is fixed there.

Answer: (b) resistance of silicon will increase and that of platinum will decrease

Explanation:
1. Platinum is a metal, so its resistance falls on cooling.
2. Silicon is a semiconductor, so fewer carriers are generated on cooling and its resistance rises.

Teacher's Note:
1. Compare the signs of the temperature coefficients.
2. Metal: positive; semiconductor: negative.

Answer: (c) both electrons and holes

Explanation:
1. In a semiconductor, current is carried by conduction electrons and by holes in the valence band.
2. Both contribute to the total current.

Teacher's Note:
1. Holes are not real particles but vacancies that move opposite to electrons.
2. They are counted as positive carriers.

Answer: (d) \( n_p \neq n_e \)

Explanation:
1. An extrinsic semiconductor may be p-type (n_p greater) or n-type (n_e greater).
2. Without knowing which, the only statement always true is that they are unequal.

Teacher's Note:
1. Equality holds only for intrinsic material.
2. Doping always breaks it.

Answer: (c) uncharged

Explanation:
1. Acceptor atoms are neutral and each hole is balanced by a negative acceptor ion.
2. The crystal as a whole stays electrically neutral.

Teacher's Note:
1. Doping adds no net charge.
2. The hole is a mobile positive carrier but the lattice has an equal negative charge.

Answer: (b) the number of holes in the valence band is more than the number of electrons in the conduction band at room temperature

Explanation:
1. In p-type material holes are majority carriers.
2. So holes in the valence band outnumber conduction electrons.

Teacher's Note:
1. Majority carriers come from doping.
2. Minority carriers come from thermal generation.

Answer: (b) decreases

Explanation:
1. Extra donor electrons increase recombination with holes.
2. By the law of mass action \( n_e n_h = n_i^2 \), a larger n_e means a smaller n_h.

Teacher's Note:
1. The product of electron and hole densities stays fixed at a given temperature.
2. If one rises, the other falls.

Answer: (a) \( n_e \gg n_h \)

Explanation:
1. Phosphorus is pentavalent, so it makes an n-type semiconductor.
2. Electrons are the majority carriers, therefore \( n_e \gg n_h \).

Teacher's Note:
1. Group 15 dopant means n-type.
2. Group 13 dopant means p-type.

Answer: (d) P is a conductor and Q is a semiconductor

Explanation:
1. A rising resistance with temperature is the behaviour of a metallic conductor.
2. A falling resistance is the behaviour of a semiconductor.

Teacher's Note:
1. Use the sign of the temperature coefficient to classify the material.
2. n-type and p-type both show negative coefficients.

Answer: (a) equal

Explanation:
1. Each broken bond creates one electron and one hole together.
2. So \( n_e = n_h = n_i \).

Teacher's Note:
1. Pairs are created and destroyed together.
2. The equality is only for pure material.

Answer: (c) increases

Explanation:
1. Higher temperature gives more carriers, so the resistance of the semiconductor falls.
2. With a fixed supply voltage the current rises.

Teacher's Note:
1. Think of the resistance falling, then apply Ohm's law.
2. This is why a thermistor is used as a temperature sensor.

Answer: (c) \( 1.12 \times 10^{-19} \)

Explanation:
1. The gap of germanium is about 0.7 eV.
2. \( 0.7 \times 1.6 \times 10^{-19} = 1.12 \times 10^{-19} \) J.

Teacher's Note:
1. Convert eV to J by multiplying by \( 1.6 \times 10^{-19} \).
2. Remember Ge is about 0.7 eV and Si about 1.1 eV.

Answer: (b) holes

Explanation:
1. Holes are the majority carriers in p-type material.
2. Electrons are the minority carriers.

Teacher's Note:
1. Remember p stands for positive carriers.
2. n stands for negative carriers.

Answer: (d) copper's decrease and germanium's increase

Explanation:
1. Copper is a metal, so lattice vibrations reduce on cooling and resistance decreases.
2. Germanium loses thermally generated carriers, so its resistance increases.

Teacher's Note:
1. Opposite behaviour of conductors and semiconductors is a common question.
2. Check the carrier density for each.

Answer: (a) the resistivity of a semiconductor increases with an increase in temperature

Explanation:
1. A semiconductor's resistivity falls as temperature rises.
2. The other three statements are true.

Teacher's Note:
1. Be careful with words like false.
2. Eliminate the three true statements first.

Answer: (a) \( \mu_e > \mu_h \)

Explanation:
1. Electrons move freely in the conduction band, while holes move by hopping among bonds.
2. Hence electrons are more mobile.

Teacher's Note:
1. Mobility is always taken as positive.
2. For Si, mobility is about 1350 for electrons and 480 for holes in cm²/Vs.

Answer: (b) just below the conduction band

Explanation:
1. Donor atoms give a level slightly below the conduction band.
2. Electrons are easily excited from it into the conduction band.

Teacher's Note:
1. Donor level is near the conduction band.
2. Acceptor level is near the valence band.

Answer: (d) 1.1 eV

Explanation:
1. Silicon has an energy gap of about 1.1 eV at room temperature.
2. Germanium is 0.7 eV and diamond is 5.4 eV.

Teacher's Note:
1. Memorise these three standard values.

Answer: (c) positive and negative ions

Explanation:
1. Donor ions (positive) are exposed on the n side.
2. Acceptor ions (negative) are exposed on the p side.

Teacher's Note:
1. Ions are fixed in the crystal and cannot move.
2. They produce the barrier electric field.

Answer: (b) diffusion of free carriers

Explanation:
1. Carriers diffuse from high to low concentration across the junction.
2. They recombine and leave uncovered ions, which form the depletion layer.

Teacher's Note:
1. Diffusion creates the layer; drift then opposes diffusion.
2. At equilibrium the two currents cancel.

Answer: (c) there is an electric field at the junction directed from the n-type side to the p-type side

Explanation:
1. The n side loses electrons and becomes positive, while the p side becomes negative.
2. So the barrier field points from n to p, and the n side is at higher potential.

Teacher's Note:
1. The field opposes further diffusion of majority carriers.
2. It helps minority carriers drift across.

Answer: (a) if the junction is forward biased

Explanation:
1. Forward bias lowers the barrier, so more majority carriers cross.
2. Diffusion current then exceeds the drift current.

Teacher's Note:
1. Unbiased: they are equal and opposite.
2. Reverse biased: drift dominates.

Answer: (b) reverse

Explanation:
1. Under reverse bias, the diffusion of majority carriers stops.
2. Minority holes of the n region are swept to the p region, giving a small net flow.

Teacher's Note:
1. This small leakage is the reverse saturation current.
2. It depends on temperature.

Answer: (c) 300 \( \Omega \)

Explanation:
1. Voltage across R = 2 − 0.5 = 1.5 V.
2. \( R = 1.5 / (5 \times 10^{-3}) = 300 \ \Omega \).

Teacher's Note:
1. Subtract the diode drop first.
2. Then apply Ohm's law to the resistor.

Answer: (c) \( 10^5 \)

Explanation:
1. \( E = V/d = 0.1 / 10^{-6} = 10^5 \) V/m.

Teacher's Note:
1. Average field is potential divided by thickness.
2. Note how large the field is even for a small voltage.

Answer: (a) high potential is at the n side and low potential at the p side

Explanation:
1. The n side is left with positive ions, so it is at the higher potential.
2. This difference is the barrier potential.

Teacher's Note:
1. The barrier potential is about 0.7 V for Si and 0.3 V for Ge.
2. It cannot be measured with a voltmeter.

Answer: (d) suddenly increases

Explanation:
1. At a large enough reverse voltage, breakdown (Zener or avalanche) occurs.
2. The current then rises suddenly.

Teacher's Note:
1. Below breakdown the reverse current is almost constant.
2. The breakdown voltage is the basis of Zener diodes.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true, as the resistance of a semiconductor changes with current because heating changes the carrier number.
2. The reason is a general definition of current and does not explain why Ohm's law fails.

Teacher's Note:
1. Ohm's law needs a constant resistance.
2. In semiconductors, the carrier number varies with temperature and field.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In metals, more collisions raise resistance.
2. In semiconductors, more carriers are freed and resistance falls. The reason correctly explains the assertion.

Teacher's Note:
1. Compare what changes with temperature in each material.
2. Number of carriers dominates in semiconductors.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Because the gap is small, thermal energy can excite many electrons across it.
2. More carriers mean lower resistance, so R explains A.

Teacher's Note:
1. Small gap means easy thermal excitation.

Answer: (c) A is true, but R is false.

Explanation:
1. Doping adds majority carriers, so conductivity increases and A is true.
2. Doping does not change the temperature, so R is false.

Teacher's Note:
1. Doping adds carriers, not heat.