Physics M.C.Q [1M]

2. Circuit 2 and circuit 3.

Explanation:

In circuit 1, one diode is forward biassed and the other diode is reverse biassed. The forward-biassed diode offers zero resistance (ideally) to the current flow, so it can be replaced by a short circuit. The voltage drop across the first diode will be zero. The second diode is reverse biassed, so it can be replaced by an open circuit; hence, the voltage drop across this diode will be maximum.

In circuit 2, both the diodes are forward biassed, so they can be replaced by short circuits; hence, the voltage drop across both of them will be minimum and equal.

In circuit 3, both the diodes are reverse biassed, so both can be replaced by open circuits; hence, the voltage drop across both of them will be maximum and equal.

2. Holes and conduction electrons systematically go from the p-side to the n-side and from the n side to the p-side respectively.
3. There is no net charge transfer between the two sides.
4. There is a constant electric field near the junction.

Explanation:

Because of the difference in the concentration of charge carriers in the p−n junction, holes from the p side move to the n side and electrons from the n side move to the p side. This motion of charge carriers gives rise to diffusion current.

Because of this, a negative space charge region is formed in the p region and a positive space region is formed in the n region. This sets up an electric field across the junction. Thus, there is a constant electric field near the junction.

This electric field further opposes the diffusion of majority charge carriers across the junction. As a result, an electron from the p region starts moving to the n region and a hole from the n region starts moving to the p region. This sets up drift current. Thus, there is a systematic flow of charge carriers across the junction. Also, there is no net charge transfer between the two sides.

1. There will not be a steady current in the circuit.

Explanation:

In a p‒n junction, current flows only if it is connected to the battery. If two ends of a p‒n junction are joined by a wire, then there will be diffusion and drift currents in the circuit and they will cancel each other. Hence, no current will flow in the circuit.

1. There are no free electrons at 0K.

3. The number of free electrons increases with temperature.

4. The number of free electrons is less than that in a conductor.

Explanation:

In semiconductors, the valence band is full at 0K, but the conduction band is empty. So, no free electron is available for conduction at 0K.

As the temperature increases, covalent bonds that provide free charge carriers for conduction in a semiconductor break.

As the conduction band in metals is already partially filled at 0K, many free electrons below the Fermi level acquire energy from an external source or temperature, jump to the conduction band and start behaving like free electrons. Hence, metals contain more free electrons than semiconductors.

1. The base-emitter junction is forward-biased.

4. The base-collector junction is reverse-biased.

Explanation:

In the normal operation of a transistor, the base−emitter junction is forward biassed and the base−collector junction is reverse biassed. This is done so that the conduction of majority carriers can take place across the emitter−base junction and the free electrons can reach the collector to give the output current.

1. New holes and conduction electrons are produced continuously throughout the material.

4. Holes and conduction electrons recombine continuously throughout the material except in the depletion region.

Explanation:

In a p‒n junction diode, diffusion current flows because of the diffusion of holes from the p side to the n side and of electrons from the n side to the p side. The current flowing in the diode due to the diffusion of charge carriers across the junction is called the diffusion current. The current flowing in the diode due to the movement of minority carriers across the junction due to their thermal energy is called the drift current. In an unbiased diode, the net current flowing across the junction is zero due to the cancellation of the drift current by the diffusion current. For the flow of diffusion and drift currents, holes and electrons are produced continuously throughout the material. When a hole crosses the junction, it combines with an electron on the n side. As the depletion region is devoid of free charge carriers, this recombination never takes place inside the depletion region.

1. From the n-side to the p-side.

Explanation:

After the diffusion of majority charge carriers across a p‒n junction, an electric field is set up because of the accumulation of immobile ions at the junction. These further oppose the motion of majority charge carriers across the junction. As a result, electrons from the p region start moving to the n region and holes from the n region start moving to the p region. This constitutes the drift current. As the direction of the current is opposite to the direction of the motion of the electrons, the direction of the drift current is from the n side to the p side.

In forward biasing, there is no movement of electrons from the p region to the n region and of holes from the n region to the p region. Hence, there is not drift current.

4. A p-n junction.

Explanation:

As a p−n junction allows the flow of current in forward bias and stops the current in reverse bias (almost negligible reverse leakage current flows in the reverse-biassed p−njunction), the device should be a p−n junction. Other options are examples of semiconductors that allow moderate current to flow and that do not have any effect of changing the polarity of the battery.

3. A bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it.

Explanation:

A hole diffuses from the p side to the n side in a p−n junction; that is, an electron moves from the n side to the p side. This implies that a bond is broken on the n side. As the electron travels towards the p side, which is rich in holes, it combines with a hole. A hole is created because of the deficiency of one electron. So, when an electron combines with a hole, it completes that bond.

1. Increases..

Explanation:

When an impurity (either a p-type atom or an n-type atom) is doped into an intrinsic semiconductor, it increases the number of charge carriers in the intrinsic semiconductor. As conductivity is directly related to the number of charge carriers, the conductivity of a semiconductor increases with doping.

1. i_C is slightly smaller than i_E.

3. i_B is much smaller than i_E.

Explanation:

The highlighted parts could not be edited, as the meaning could not be understood. Also, please check the last line for logical accuracy. Only one option is given as correct, while the solution gives two correct options.

We know that in the transistor base is slightly doped, therefore when the majority carriers due to forward biasing of emitter base junction, feel the repulsive force from the battery and pass over to the base region. This gives the emitter current i_E.

As the base is thin and lightly doped, only few majority carriers of the emitter are neutralised at the base. This gives the base current.

Hence, base current (i_B) is low.

The remaining majority carriers of the emitter pass to the collector and give collector current i_C.

Thus, we get the relation given below:

i_E = i_B + i_C

Thus, because of the base, current i_C is slightly smaller than i_E.

2. Boron.

4. Aluminium.

Explanation:

A p-type semiconductor is formed by doping an intrinsic semiconductor with a trivalent atom (atom having valency 3). As phosphorous and boron have three valence electrons, they can be doped with silicon to make a p-type semiconductor.

1. If the junction is forward-biased.

Explanation:

In the forward biassing of a p−n junction, the positive terminal of the battery is connected to the p side of the p−n junction and the negative terminal of the battery is connected to the n side of the p−n junction. As a result, electrons in the n side of the p−n junction are repelled by the negative terminal of the battery and move to the p side, where the positive terminal of the battery attracts the electrons. Similarly, holes from the p side of the p−n junction are repelled by the positive terminal of the battery and move to the nside, where the negative terminal of the battery attracts the holes. Thus, they give diffusion current across the p−n junction.

In case of reverse biassing, no conduction takes place across the junction because of the diffusion of majority carriers. Hence, there is no diffusion current.

If the junction is unbiased, then diffusion current is initially maximum. But at equilibrium, diffusion current becomes equal to drift current.

3. NAND gate.
4. NOR gate.

3. The electron concentration increases.

Explanation:

When a semiconductor is doped with a donor type such as arsenic or phosphorous, which has five valence electrons, the donor atom replaces the Si or Ge atom. As a result, four out of the five electrons of the donor atom form a covalent bond by sharing an electron with four atoms of silicon. However, the fifth electron is free to move. Also, due to the breaking up of covalent bonds at room temperature, equal number of electrons and holes are produced. Thus, the total number of holes in the n-type semiconductor is less compared to the number of free electrons.

3. Both electrons and holes.

Explanation:

A hole is created in a semiconductor when a valence electron moves to the conduction band. When potential difference is applied across the semiconductor, the electron drifts opposite to the electric field applied, while the hole moves along the electric field. Therefore, electric conduction takes place in a semiconductor because of both electrons and holes.

2. From the p-side to the n-side.

Explanation:

When a p‒n junction is formed then because of the difference in the concentration of charge carriers in the two regions, electrons from the n region move to the p region and holes from the p region move to the n region. Since the direction of the current is always opposite to the motion of electron, the direction of the current is from the p side to the n side.

Similarly, when the junction is forward biassed, the positive terminal of the battery is connected to the pside of the p‒n junction and the negative terminal of the battery is connected to the n side of the p‒njunction. As a result, electrons in the n side of the p‒n junction are repelled by the negative terminal of the battery and they move to the p side, where the positive terminal of the battery attracts them. Similarly, holes from the p side of the p‒n junction are repelled by the positive terminal of the battery and they move to the n side, where the negative terminal of the battery attracts them. Thus, they give diffusion current from the p side to the n side across the p‒n junction.

In reverse biassing, there is no flow of majority carriers across the junction; hence, there is not diffusion current. Here, the flow of majority carriers is opposed by the applied voltage.

3. Uncharged.

Explanation:

A p-type semiconductor is formed by doping a pure semiconductor with a p-type material. As impurity atoms take the position of the germanium atom in a germanium crystal, three electrons of a p-type material form covalent bonds by sharing electrons with three neighbouring germanium atoms. However, the fourth covalent bond is left incomplete, with a want of one electron. This creates a hole. As the atom as a whole is neutral, the p-type material is also neutral.

1. Increasing the temperature.
2. Doping acceptor impurities.
3. Doping donor impurities.
4. Irradiating ultraviolet light on it.

Explanation:

We know that the conductivity of any semiconductor can be increased by increasing the number of charge carriers. All the given methods are effective in increasing the number of free charge carriers. Hence, all options are correct.