Maharashtra Board Class 11 Physics Chapter 14 Semiconductors Solutions

Choose the Correct Option

Question 1. Electric conduction through a semiconductor is due to:
(a) Electrons
(b) Holes
(c) None of the options
(d) Both electrons and holes
Answer: (d) Both electrons and holes
In simple words: In semiconductors, electric current is carried by both free electrons moving in one direction and empty spaces called holes moving in the opposite direction.

🎯 Exam Tip: Always remember that unlike metals which only use electrons, semiconductors rely on both electrons and holes for conduction.

Question 2. The energy levels of holes are:
(a) In the valence band
(b) In the conduction band
(c) In the band gap but close to valence band
(d) In the band gap but close to conduction band
Answer: (c) In the band gap but close to valence band
In simple words: Holes are empty spaces left by electrons, and their energy levels sit just above the valence band in the small gap between the bands.

🎯 Exam Tip: Remember that acceptor impurities create energy levels close to the valence band, which is where holes are easily formed.

Question 3. Current through a reverse biased p-n junction, increases abruptly at:
(a) Breakdown voltage
(b) 0.0 V
(c) 0.3 V
(d) 0.7 V
Answer: (a) Breakdown voltage
In simple words: When a diode is connected backward, it blocks current until the voltage gets too high (breakdown voltage), causing a sudden flood of current.

🎯 Exam Tip: Do not confuse breakdown voltage with knee voltage (0.7 V for Silicon, 0.3 V for Germanium), which applies to forward bias.

Question 4. A reverse biased diode, is equivalent to:
(a) an off switch
(b) an on switch
(c) a low resistance
(d) None of the options
Answer: (a) an off switch
In simple words: When a diode is reverse biased, it blocks the flow of electric current, acting just like an open or "off" switch in a circuit.

🎯 Exam Tip: Remember that in ideal conditions, a forward-biased diode acts as a closed (on) switch, while a reverse-biased diode acts as an open (off) switch.

Question 5. The potential barrier in p-n diode is due to:
(a) depletion of positive charges near the junction
(b) accumulation of positive charges near the junction
(c) depletion of negative charges near the junction
(d) accumulation of positive and negative charges near the junction
Answer: (d) accumulation of positive and negative charges near the junction
In simple words: Near the junction, positive ions on the n-side and negative ions on the p-side collect to form a barrier that stops more charges from crossing over.

🎯 Exam Tip: The potential barrier is created by immobile donor and acceptor ions left behind in the depletion region near the junction.

Answer the Following Questions

Question 1. What is the importance of energy gap in a semiconductor?
Answer:
1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
2. This band gap is present only in semiconductors and insulators.
3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
4. Band gap in semiconductors is of the order of 1 eV.
5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction. This energy gap essentially determines how easily a material can conduct electricity under different conditions.
In simple words: The energy gap is the small hurdle electrons must jump over to conduct electricity. In semiconductors, this hurdle is small enough that heat or voltage can easily help electrons cross it and start conducting.

🎯 Exam Tip: Clearly define the energy gap first, mention its typical value for semiconductors (~1 eV), and explain how crossing it enables electrical conduction.

Question 2. Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer: To make germanium an n-type semiconductor, a pentavalent impurity element such as arsenic (As), phosphorus (P), or antimony (Sb) should be used. Adding these elements introduces extra free electrons into the crystal lattice, which significantly increases its electrical conductivity.
In simple words: To make an n-type semiconductor, we add elements with five outer electrons, like phosphorus or arsenic, which leave behind extra free electrons to carry the electric current.

🎯 Exam Tip: Always specify that pentavalent impurities (like phosphorus or arsenic) are used for n-type semiconductors, while trivalent impurities are used for p-type.

Question 3. What causes a larger current through a p-n junction diode when forward biased?
Answer: In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction. This reduction in barrier height facilitates the easy flow of majority carriers.
In simple words: In forward bias, the barrier inside the diode becomes very thin, making it very easy for electric current to flow through it.

🎯 Exam Tip: Mention the decrease in depletion region width and the resulting low resistance path to secure full marks.

Question 4. On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer: For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor. This concentration of carriers increases significantly with any rise in temperature.
In simple words: In a pure semiconductor, electricity flows because of both free electrons and holes. Its ability to conduct electricity depends on how many of these charge carriers are available.

🎯 Exam Tip: Remember that for pure (intrinsic) semiconductors, the number of electrons equals the number of holes, and conductivity depends directly on this carrier concentration.

Question 5. Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:
1. In a p-type semiconductor, holes are majority charge carriers.
2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
3. During transportation of hole, there is an indirect movement of electrons.
4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping. This difference in mobility is primarily due to the band structure of the semiconductor material.
In simple words: Electrons move much faster and more easily than holes. Since n-type semiconductors rely on fast-moving electrons while p-type rely on slower holes, the n-type conducts electricity better.

🎯 Exam Tip: Clearly state that the mobility of electrons (majority carriers in n-type) is higher than the mobility of holes (majority carriers in p-type) to get full marks.

Answer in Detail

Question 1. Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.
ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.
iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.
Energy Gap Diagram:
• Conduction band
• Energy gap
• Valence band
iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.
Metal Band Overlap Diagram:
• Conduction Band
• Energy bands overlap
• Valence Band
v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in electrical conduction quite easily. This classification helps us understand the electrical conductivity of different materials under various physical conditions.
In simple words: Solids are grouped into conductors, insulators, and semiconductors based on how easily their electrons can move. Conductors have overlapping energy bands so electrons flow easily, while insulators have a large gap that blocks electron movement, and semiconductors have a tiny gap.

🎯 Exam Tip: Draw neat, labeled diagrams of the energy bands for conductors, insulators, and semiconductors to secure full marks in this descriptive question.

Question 2. Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

In simple words: An intrinsic semiconductor is completely pure and does not conduct electricity very well. An extrinsic semiconductor has added impurities which greatly increase its ability to conduct electricity even at room temperature.

🎯 Exam Tip: When writing comparison tables, ensure that each point on the left directly corresponds to the contrasting point on the right for maximum clarity.

Question 3. Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region. This region acts as a barrier for further charge flow.
ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.
iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.
iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.
v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.
vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.
vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.
In simple words: The depletion region is like a security gate at the border between the p-type and n-type sides. It stops too many charges from crossing over continuously, keeping the diode stable and working properly.

🎯 Exam Tip: Clearly mention how the potential barrier stops the continuous diffusion of majority carriers to achieve electrostatic equilibrium.

Question 4. Explain the I-V characteristic of a forward biased junction diode.
Answer: In a forward biased p-n junction diode, the p-region is connected to the positive terminal and the n-region to the negative terminal of the external voltage source. As the applied voltage increases, the depletion region width decreases and the barrier potential is overcome. Once the voltage exceeds the knee voltage (cut-in voltage), the forward current increases exponentially with a small increase in voltage.
In simple words: When we connect the diode in forward bias, it allows current to flow easily once the voltage is high enough to push past the internal barrier.

🎯 Exam Tip: Always draw the I-V characteristic curve showing the knee voltage clearly to secure full marks in this descriptive question.

Question 4. Explain the I-V characteristics of a forward biased diode.
Answer:
Visual Representation of the I-V Characteristic Curve:

• Y-axis: Forward current (mA) ranging from 0 to 10 mA
• X-axis: Forward voltage (V) ranging from 0 to 0.7 V
• Key Features: Shows a flat curve initially, followed by a sharp rise at the 'Knee' point (0.7 V for Silicon) in the forward bias region.

1. Figure given below shows the I-V characteristic of a forward biased diode.
2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
6. To limit current flowing through the diode, resistors are used in series with the diode.
7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage. This characteristic behavior is fundamental to the operation of modern semiconductor devices.
In simple words: When a diode is connected in forward bias, it doesn't conduct much electricity at first. Once the voltage reaches a certain level (called the knee voltage), the diode suddenly lets a large amount of current flow through it easily.

🎯 Exam Tip: Remember to mention the specific knee voltage values for both silicon (0.7 V) and germanium (0.3 V) to secure full marks.

Question 5. Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:
1. A p-n junction can be connected to an external voltage supply in two possible ways.
2. A p-n junction is said to be connected in a forward bias when the p-region is connected to the positive terminal and the n-region is connected to the negative terminal of the external voltage source. This external bias alters the potential barrier and directly controls the flow of charge carriers across the junction.
In simple words: An external voltage can either shrink or widen the middle barrier (depletion region) of a p-n junction. Connecting it in forward bias pushes the charges together, making this barrier thinner and allowing current to flow.

🎯 Exam Tip: Clearly distinguish between forward bias (which decreases the depletion width) and reverse bias (which increases it) in your explanation.