Semi conductor devices practice · Medium

Option A (The donor level lies closely below the bottom of the conduction band) is correct because in an n-type semiconductor, donor impurities create energy levels just below the conduction band, allowing electrons to easily move into the conduction band.

Option B (Both Statements are false) is correct because Carbon is an insulator, Silicon and Germanium are semiconductors, and the ionization energy is least for Ge, followed by Si and then C.

Increases the potential barrier is correct because reverse biasing widens the depletion region and increases the potential barrier.

Option B (Phosphorus and boron respectively) is correct because Phosphorus is a pentavalent impurity used for creating n-type semiconductors, while Boron is a trivalent impurity used for creating p-type semiconductors.

Option D (100) is correct because the current gain is the ratio of the change in collector current to the change in base current.

Option C (12) is correct because in hexagonal close packing (hcp), each atom is surrounded by 12 nearest neighbors, giving a coordination number of 12.

Option C (Three) is correct because an OR gate can be constructed using three NAND gates.

Both statements are false. The resistivity of a semiconductor decreases with increasing temperature due to increased charge carriers.

Option A is correct because a wider base region increases electron-hole recombination, decreasing collector current, and the emitter current is the sum of base and collector currents.

Option A (to enable the collector to collect 95% of the holes or electrons coming from the emitter side) is correct because a thin, lightly doped base allows most carriers from the emitter to reach the collector.

Option C is correct because metals are better conductors, and the energy gap between valence and conduction bands is small or non-existent in metals.

Option B (Decreases) is correct because increasing the forward voltage reduces the width of the depletion region in a diode.

Option B (Covalent) is correct because semiconductors like silicon and germanium form covalent bonds with their neighboring atoms.

Option A (Zener diode) is correct because Zener diodes are designed to operate in the reverse breakdown region and maintain a relatively constant voltage, making them suitable for voltage regulation.

The correct answer is A (6) because in a simple cubic crystal, each atom is directly adjacent to six other atoms.

Option A (Zero) is correct because in Boolean algebra, the AND operation between a variable and its complement always results in zero.

Option D ($p$-type semiconductor) is correct because boron, being a trivalent impurity, creates holes in the silicon lattice, leading to $p$-type conductivity.

Option C is correct because antimony (Sb) is a pentavalent impurity, adding it to germanium creates an N-type semiconductor with more electrons than holes.

Option B (Pentavalent elements) is correct because pentavalent impurities donate extra electrons, making them donor-type impurities in semiconductors.

Option C is correct because a semiconductor has an energy gap of approximately 0.7 eV between the conduction and valence bands.

Option A (Wave nature of electron) is correct because the formation of covalent bonds is explained by the overlapping of atomic orbitals, which is a consequence of the wave nature of electrons.

Option A (${\mu _e} > {\mu _h}$) is correct because electrons generally have higher mobility than holes in semiconductors due to their smaller effective mass and less frequent scattering.

Option B is correct because increasing the doping in the base region of an NPN transistor increases recombination, thus reducing collector current.

Option C (${V_I} > {V_z}$) is correct because the input voltage must be greater than the Zener voltage for the diode to operate in the breakdown region and regulate the voltage.

Option B (Doped impurity is high) is correct because Zener breakdown occurs in heavily doped p-n junctions due to strong electric fields.

Option A ((A) and (B) both are incorrect) is correct because a Zener diode is used in reverse bias as a voltage regulator, and the potential barrier for a p-n junction is typically around 0.3V for Germanium and 0.7V for Silicon.

Option D is correct because an AND gate can be created by using NAND gates or NOR gates repetitively.

Option A is correct because both statements are correct applications of De Morgan's laws and double negation.

Option C (It's emitter junction is in forward bias and collector junction is in reverse bias) is correct because this configuration places the transistor in the active region, which is necessary for amplification.

Option B (Reverse bias voltage applied must be greater than zener breakdown voltage) is correct because a Zener diode operates in the reverse breakdown region to maintain a constant voltage.