CLASS 12 PHYSICS CHAPTER 14 SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
NCERT CLASS 12 PHYSICS
CHAPTER 14 – SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
1. IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
Q1. What is semiconductor electronics?
Answer: Semiconductor electronics is the branch of electronics that deals with devices made from semiconductor materials. These devices control the flow of electric current and are used in electronic circuits.
Examples include semiconductor diodes and rectifiers.
Q2. What are semiconductors?
Answer: Semiconductors are materials whose electrical conductivity lies between that of conductors and insulators.
Their conductivity generally increases with an increase in temperature.
Examples:
Silicon (Si)
Germanium (Ge)
Q3. What are the three types of materials on the basis of electrical conductivity?
Answer:
Conductors: Materials that allow electric current to pass through them easily. Example: Copper and aluminium.
Insulators: Materials that do not allow electric current to pass through them easily. Example: Glass and rubber.
Semiconductors: Materials having conductivity between conductors and insulators. Example: Silicon and germanium.
Q4. What is an energy band?
Answer: An energy band is a range of closely spaced energy levels that electrons can occupy in a solid.
In solids, individual atomic energy levels split into a large number of closely spaced levels, forming energy bands.
Q5. What is the valence band?
Answer: The valence band is the highest energy band that is normally occupied by electrons at absolute zero temperature.
Electrons in this band are generally bound to atoms and do not contribute freely to electrical conduction.
Q6. What is the conduction band?
Answer: The conduction band is the energy band above the valence band in which electrons can move freely through the material and contribute to electric current.
Q7. What is the forbidden energy gap?
Answer: The forbidden energy gap is the energy gap between the top of the valence band and the bottom of the conduction band.
It is represented by Eg.
Eg = EC − EV
Where:
EC = Energy at the bottom of the conduction band
EV = Energy at the top of the valence band
Q8. Differentiate between conductors, semiconductors and insulators on the basis of energy bands.
Answer:
Conductors:
The valence band and conduction band overlap, or the conduction band is partially filled.
The forbidden energy gap is approximately zero.
They have high electrical conductivity.
Semiconductors:
The valence band is nearly full and the conduction band is nearly empty at low temperatures.
They have a small forbidden energy gap.
Their conductivity increases with temperature.
Insulators:
The forbidden energy gap is very large.
Electrons cannot easily move from the valence band to the conduction band.
They have very low electrical conductivity.
Q9. Why is carbon an insulator while silicon and germanium are semiconductors?
Answer: Carbon, silicon and germanium all have four valence electrons. However, their forbidden energy gaps are different.
Approximate energy gaps:
Carbon (diamond) = 5.4 eV
Silicon = 1.1 eV
Germanium = 0.7 eV
Carbon has a much larger energy gap, so ordinary thermal energy cannot easily excite electrons into its conduction band.
Silicon and germanium have smaller energy gaps, allowing some electrons to reach the conduction band at room temperature.
Therefore, carbon behaves as an insulator, whereas silicon and germanium behave as semiconductors.
Q10. What is an intrinsic semiconductor?
Answer: A pure semiconductor without any added impurity is called an intrinsic semiconductor.
Examples:
Pure silicon
Pure germanium
In an intrinsic semiconductor, the number of free electrons is equal to the number of holes.
ne = nh = ni
Q11. What are holes in a semiconductor?
Answer: When an electron gains sufficient energy and leaves a covalent bond, it creates an empty space in the bond.
This vacancy behaves like a positively charged particle and is called a hole.
Q12. How is electric current produced in an intrinsic semiconductor?
Answer: When thermal energy breaks some covalent bonds, electrons move into the conduction band and leave holes in the valence band.
Both electrons and holes participate in electrical conduction.
The movement of electrons and holes produces electric current.
Q13. What happens to an intrinsic semiconductor at absolute zero temperature?
Answer: At absolute zero temperature, all covalent bonds remain intact.
There are no free electrons in the conduction band and no holes in the valence band.
Therefore, an intrinsic semiconductor behaves like an insulator at absolute zero temperature.
Q14. What is doping?
Answer: Doping is the process of deliberately adding a very small and controlled quantity of suitable impurity atoms to a pure semiconductor to increase its electrical conductivity.
The added impurity atoms are called dopants.
Q15. What is an extrinsic semiconductor?
Answer: A semiconductor whose conductivity is increased by adding a suitable impurity is called an extrinsic semiconductor.
There are two types:
n-type semiconductor
p-type semiconductor
Q16. What is an n-type semiconductor?
Answer: An n-type semiconductor is formed by doping a pure semiconductor such as silicon or germanium with a pentavalent impurity.
Examples of pentavalent impurities:
Phosphorus
Arsenic
Antimony
A pentavalent impurity has five valence electrons. Four electrons form covalent bonds with neighbouring semiconductor atoms, while the fifth electron is loosely bound and can become a free electron.
Thus, electrons are the majority charge carriers and holes are the minority charge carriers.
Q17. Why is an n-type semiconductor electrically neutral?
Answer: Although an n-type semiconductor contains a large number of free electrons, the donor atoms become positively charged after donating electrons.
The positive charge of the ionised donor atoms balances the negative charge of the free electrons.
Therefore, the semiconductor as a whole remains electrically neutral.
Q18. What is a p-type semiconductor?
Answer: A p-type semiconductor is formed by doping a pure semiconductor with a trivalent impurity.
Examples:
Boron
Aluminium
Indium
A trivalent impurity has three valence electrons. It forms three covalent bonds with neighbouring semiconductor atoms, leaving one incomplete bond.
This vacancy behaves like a hole.
Therefore, holes are the majority charge carriers and electrons are the minority charge carriers.
Q19. Why is a p-type semiconductor electrically neutral?
Answer: In a p-type semiconductor, the acceptor atoms become negatively charged after accepting electrons.
The positive charge of holes is balanced by the negative charge of the ionised acceptor atoms.
Therefore, the semiconductor as a whole remains electrically neutral.
Q20. Differentiate between n-type and p-type semiconductors.
Answer:
n-type semiconductor:
It is formed by doping with pentavalent impurities.
Electrons are the majority carriers.
Holes are the minority carriers.
Pentavalent impurities are called donor impurities.
Examples of dopants are phosphorus and arsenic.
p-type semiconductor:
It is formed by doping with trivalent impurities.
Holes are the majority carriers.
Electrons are the minority carriers.
Trivalent impurities are called acceptor impurities.
Examples of dopants are boron and aluminium.
Q21. What is the mass action law?
Answer: The mass action law states that, at a fixed temperature and thermal equilibrium, the product of electron concentration and hole concentration is constant for a given semiconductor.
ne × nh = ni²
Where:
ne = Electron concentration
nh = Hole concentration
ni = Intrinsic carrier concentration
Q22. What is a p-n junction?
Answer: A p-n junction is formed when p-type and n-type regions are created in the same semiconductor crystal.
The junction between these two regions is called a p-n junction.
It is the basic structure of a semiconductor diode.
Q23. What happens when a p-n junction is formed?
Answer: When p-type and n-type regions are joined:
Electrons diffuse from the n-region to the p-region.
Holes diffuse from the p-region to the n-region.
Electrons and holes recombine near the junction.
Immobile charged ions remain near the junction.
A depletion region is formed.
An internal electric field and potential barrier are established.
Q24. What is diffusion current?
Answer: Diffusion current is the current produced due to the movement of charge carriers from a region of higher concentration to a region of lower concentration.
In a p-n junction:
Holes diffuse from p-region to n-region.
Electrons diffuse from n-region to p-region.
Q25. What is drift current?
Answer: Drift current is the current produced by the movement of charge carriers under the influence of an electric field.
In a p-n junction, the electric field in the depletion region causes minority carriers to move and produces drift current.
Q26. What is a depletion region?
Answer: The depletion region is the thin region around a p-n junction that is depleted of mobile charge carriers.
It contains immobile positive donor ions on the n-side and immobile negative acceptor ions on the p-side.
Q27. What is potential barrier in a p-n junction?
Answer: The potential barrier is the potential difference developed across the depletion region of a p-n junction.
It opposes the further diffusion of majority charge carriers.
Its approximate value is:
Silicon: 0.7 V
Germanium: 0.2 V
Q28. What is a semiconductor diode?
Answer: A semiconductor diode is a two-terminal electronic device formed by a p-n junction.
It allows current to pass easily in one direction and offers very high resistance in the opposite direction under normal operating conditions.
Q29. What is forward biasing of a p-n junction diode?
Answer: A p-n junction diode is said to be forward biased when:
The p-side is connected to the positive terminal of a battery.
The n-side is connected to the negative terminal of the battery.
Effects of forward bias:
The potential barrier decreases.
The depletion region becomes narrower.
Majority carriers cross the junction more easily.
A large current flows through the diode after sufficient forward voltage is applied.
Q30. What is reverse biasing of a p-n junction diode?
Answer: A p-n junction diode is said to be reverse biased when:
The p-side is connected to the negative terminal of a battery.
The n-side is connected to the positive terminal of the battery.
Effects of reverse bias:
The potential barrier increases.
The depletion region becomes wider.
Majority carrier diffusion is strongly restricted.
A very small reverse current flows due to minority carriers under normal reverse-bias conditions.
Q31. Differentiate between forward bias and reverse bias.
Answer:
Forward bias:
p-side is connected to the positive terminal.
n-side is connected to the negative terminal.
Potential barrier decreases.
Depletion layer becomes narrow.
A large current can flow.
Reverse bias:
p-side is connected to the negative terminal.
n-side is connected to the positive terminal.
Potential barrier increases.
Depletion layer becomes wide.
Only a very small current flows under normal conditions.
Q32. What is the V-I characteristic of a diode?
Answer: The V-I characteristic of a diode is the graph showing the relationship between the voltage applied across the diode and the current flowing through it.
It has two regions:
Forward-bias region
Reverse-bias region
In forward bias, current increases rapidly after the cut-in voltage.
In reverse bias, a very small reverse saturation current flows until breakdown occurs.
Q33. What is cut-in voltage or threshold voltage?
Answer: Cut-in voltage is the minimum forward voltage at which the current through a diode begins to increase rapidly.
Approximate values:
Silicon diode = 0.7 V
Germanium diode = 0.2 V
Q34. What is dynamic resistance of a diode?
Answer: Dynamic resistance is the ratio of a small change in voltage across a diode to the corresponding change in current through it.
rd = ΔV/ΔI
It is measured in ohms (Ω).
Q35. What is a rectifier?
Answer: A rectifier is an electronic circuit that converts alternating current (AC) into unidirectional or pulsating direct current (DC).
A p-n junction diode is commonly used as a rectifying device.
Q36. What is half-wave rectification?
Answer: Half-wave rectification is the process in which only one half-cycle of an alternating input is allowed to pass through the load.
A single diode can be used in a half-wave rectifier.
The output is pulsating direct current.
Q37. What is full-wave rectification?
Answer: Full-wave rectification is the process in which both positive and negative half-cycles of an alternating input are converted into current flowing in the same direction through the load.
A centre-tapped circuit uses two diodes, while a bridge rectifier uses four diodes.
Q38. Differentiate between half-wave and full-wave rectifiers.
Answer:
Half-wave rectifier:
Uses one diode.
Allows only one half-cycle of AC to pass.
Output frequency is equal to input frequency.
Its maximum theoretical rectification efficiency is approximately 40.6%.
Full-wave rectifier:
Uses two diodes in a centre-tapped arrangement or four diodes in a bridge arrangement.
Uses both half-cycles of AC.
Output frequency is twice the input frequency.
Its maximum theoretical rectification efficiency is approximately 81.2%.
Q39. What is a filter circuit in a rectifier?
Answer: A filter circuit is used to reduce the fluctuations or ripples in the pulsating DC output of a rectifier.
A capacitor connected in parallel with the load resistance can act as a filter.
Q40. How does a capacitor filter work?
Answer: A capacitor filter is connected in parallel with the load resistance.
During the rising part of the rectified voltage, the capacitor charges.
When the input voltage decreases, the capacitor discharges through the load.
This discharge maintains the output voltage between successive peaks.
Therefore, the fluctuations in the output are reduced.
A larger time constant RC generally produces better smoothing, provided the circuit is appropriately designed.
2. NCERT EXERCISE QUESTIONS AND ANSWERS
Question 14.1
In an n-type silicon, which of the following statements is true?
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
Correct option: (c) Holes are minority carriers and pentavalent atoms are the dopants.
Explanation:
An n-type semiconductor is formed by doping pure silicon with a pentavalent impurity such as phosphorus, arsenic or antimony.
A pentavalent atom has five valence electrons. Four electrons form covalent bonds with neighbouring silicon atoms, while the fifth electron becomes available for conduction.
Therefore:
Electrons are majority carriers.
Holes are minority carriers.
Pentavalent atoms are donor impurities.
Hence, option (c) is correct.
Question 14.2
Which of the statements given in Exercise 14.1 is true for p-type semiconductors?
Answer:
Correct option: (d) Holes are majority carriers and trivalent atoms are the dopants.
Explanation:
A p-type semiconductor is formed by doping pure silicon or germanium with a trivalent impurity such as boron, aluminium or indium.
A trivalent atom has three valence electrons. It forms three covalent bonds with neighbouring semiconductor atoms, leaving one incomplete bond.
This incomplete bond behaves like a hole.
Therefore:
Holes are majority carriers.
Electrons are minority carriers.
Trivalent atoms are acceptor impurities.
Hence, option (d) is correct.
Question 14.3
Carbon, silicon and germanium have four valence electrons each. These are characterised by energy band gaps (Eg)C, (Eg)Si, and (Eg)Ge. Which of the following statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
Answer:
Correct option: (c) (Eg)C > (Eg)Si > (Eg)Ge
Explanation:
The approximate forbidden energy gaps are:
Carbon (diamond) = 5.4 eV
Silicon = 1.1 eV
Germanium = 0.7 eV
Comparing the values:
5.4 eV > 1.1 eV > 0.7 eV
Therefore:
(Eg)C > (Eg)Si > (Eg)Ge
Hence, option (c) is correct.
Question 14.4
In an unbiased p-n junction, holes diffuse from the p-region to the n-region because:
(a) Free electrons in the n-region attract them.
(b) They move across the junction by the potential difference.
(c) Hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer:
Correct option: (c) Hole concentration in p-region is more as compared to n-region.
Explanation:
Diffusion takes place because of a difference in concentration.
In a p-n junction:
Hole concentration is high in the p-region.
Hole concentration is low in the n-region.
Therefore, holes naturally move from the region of higher concentration to the region of lower concentration.
This process is called diffusion.
Hence, option (c) is correct.
Question 14.5
When a forward bias is applied to a p-n junction, it:
(a) Raises the potential barrier.
(b) Reduces the majority carrier current to zero.
(c) Lowers the potential barrier.
(d) None of the above.
Answer:
Correct option: (c) Lowers the potential barrier.
Explanation:
In forward bias:
The p-side is connected to the positive terminal.
The n-side is connected to the negative terminal.
The applied voltage opposes the built-in potential barrier.
As a result:
The potential barrier decreases.
The depletion region becomes narrower.
Majority carriers can cross the junction more easily.
Forward current increases.
Hence, option (c) is correct.
Question 14.6
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?
Answer:
Given:
Input frequency:
f = 50 Hz
(a) Output frequency of a half-wave rectifier
In a half-wave rectifier, only one output pulse is obtained during one complete input cycle.
Therefore:
fout = fin
fout = 50 Hz
Output frequency = 50 Hz
(b) Output frequency of a full-wave rectifier
In a full-wave rectifier, two output pulses are obtained during one complete input cycle.
Therefore:
fout = 2fin
fout = 2 × 50
fout = 100 Hz
Output frequency = 100 Hz
Final Answer:
Half-wave rectifier = 50 Hz
Full-wave rectifier = 100 Hz
3. IMPORTANT DERIVATIONS AND EXPLANATIONS
Q1. Explain the formation of a p-n junction.
Answer:
A p-n junction is formed by creating p-type and n-type regions in the same semiconductor crystal.
Step 1: Formation of p-type and n-type regions
The p-type region contains holes as majority carriers, whereas the n-type region contains electrons as majority carriers.
Step 2: Diffusion of charge carriers
When the junction is formed:
Electrons diffuse from the n-region to the p-region.
Holes diffuse from the p-region to the n-region.
Step 3: Recombination
Near the junction, electrons and holes recombine with each other.
Step 4: Formation of depletion region
The region around the junction becomes depleted of mobile charge carriers.
Immobile positive donor ions remain on the n-side, while immobile negative acceptor ions remain on the p-side.
Step 5: Formation of potential barrier
These fixed charges create an electric field and a potential barrier that opposes further diffusion.
At equilibrium, the diffusion current and drift current balance each other.
Therefore, the net current through an unbiased p-n junction is zero.
Q2. Explain forward biasing of a p-n junction diode with its effects.
Answer:
A p-n junction diode is forward biased when its p-side is connected to the positive terminal and its n-side is connected to the negative terminal of a battery.
Effects of forward bias:
The applied electric field opposes the built-in electric field.
The potential barrier decreases.
The depletion region becomes narrower.
Majority carriers cross the junction more easily.
A large current flows when sufficient forward voltage is applied.
For silicon, the approximate cut-in voltage is 0.7 V.
For germanium, the approximate cut-in voltage is 0.2 V.
Q3. Explain reverse biasing of a p-n junction diode with its effects.
Answer:
A p-n junction diode is reverse biased when its p-side is connected to the negative terminal and its n-side is connected to the positive terminal of a battery.
Effects of reverse bias:
The applied electric field supports the built-in electric field.
The potential barrier increases.
The depletion region becomes wider.
Majority carrier diffusion is strongly restricted.
A very small reverse saturation current flows due to minority carriers.
Under normal operating conditions, the diode offers very high resistance in reverse bias.
Q4. Explain the V-I characteristics of a p-n junction diode.
Answer:
The V-I characteristics of a p-n junction diode describe the variation of current with the voltage applied across it.
There are two operating regions.
1. Forward-bias characteristics
Initially, the current is very small.
When the applied voltage approaches the cut-in voltage, current begins to increase rapidly.
Beyond this point, a small increase in voltage can produce a large increase in current.
Approximate cut-in voltage:
Silicon = 0.7 V
Germanium = 0.2 V
2. Reverse-bias characteristics
A very small reverse current flows.
This current is called reverse saturation current.
It is mainly due to minority charge carriers.
If the reverse voltage becomes sufficiently large, breakdown can occur.
Q5. Derive the expression for dynamic resistance of a diode.
Answer:
Dynamic resistance is defined as the ratio of a small change in voltage to the corresponding change in current.
Let:
ΔV = Small change in voltage
ΔI = Corresponding change in current
According to Ohm's law for small changes:
rd = ΔV/ΔI
Therefore:
rd = ΔV/ΔI
The SI unit is ohm (Ω).
Q6. Explain the working of a half-wave rectifier.
Answer:
A half-wave rectifier converts AC into pulsating DC by using a single p-n junction diode.
Working:
During the positive half-cycle:
The diode becomes forward biased.
It allows current to flow through the load resistance.
A voltage appears across the load.
During the negative half-cycle:
The diode becomes reverse biased.
It offers very high resistance.
Almost no current flows through the load.
Therefore, only one half of the input AC waveform appears across the load.
The output is pulsating DC.
For a half-wave rectifier:
fout = fin
Its maximum theoretical rectification efficiency is approximately 40.6%.
Q7. Explain the working of a full-wave rectifier.
Answer:
A full-wave rectifier converts both half-cycles of AC into pulsating DC.
It can be made using:
Two diodes and a centre-tapped transformer, or
Four diodes in a bridge arrangement.
Working of a centre-tapped full-wave rectifier:
During the positive half-cycle:
One diode becomes forward biased.
The other diode becomes reverse biased.
Current flows through the load in one direction.
During the negative half-cycle:
The second diode becomes forward biased.
The first diode becomes reverse biased.
Current again flows through the load in the same direction.
Thus, both half-cycles contribute to the output.
For a full-wave rectifier:
fout = 2fin
Its maximum theoretical rectification efficiency is approximately 81.2%.
Q8. Explain the working of a capacitor filter in a rectifier.
Answer:
A capacitor filter is used to reduce the ripples present in the pulsating DC output of a rectifier.
Working:
The capacitor is connected in parallel with the load resistance.
When the rectified voltage increases, the capacitor charges.
When the input voltage decreases, the capacitor discharges through the load resistance.
The discharge supplies current to the load between successive peaks.
As a result, the output voltage becomes smoother.
A capacitor filter reduces AC fluctuations and produces a more nearly steady DC output.
4. IMPORTANT EXAMINATION QUESTIONS
Q1. Why does the conductivity of a semiconductor increase with temperature?
Answer: When temperature increases, more covalent bonds break due to thermal energy. This produces more free electrons and holes. The increase in charge carriers increases the electrical conductivity of the semiconductor.
Q2. Why are holes considered positive charge carriers?
Answer: A hole represents the absence of an electron in a covalent bond. When neighbouring electrons move to fill this vacancy, the hole appears to move in the opposite direction. Its behaviour is equivalent to that of a positive charge.
Q3. Why is doping done in semiconductors?
Answer: Doping is done to increase and control the electrical conductivity of a pure semiconductor by introducing suitable impurity atoms.
Q4. Why are pentavalent impurities called donor impurities?
Answer: Pentavalent impurities have five valence electrons. Four electrons form covalent bonds, while the fifth electron can become available for conduction. Therefore, these impurities donate free electrons and are called donor impurities.
Q5. Why are trivalent impurities called acceptor impurities?
Answer: Trivalent impurities have only three valence electrons. They can accept an electron from a neighbouring covalent bond to complete their bonding, leaving a hole behind. Therefore, they are called acceptor impurities.
Q6. Why does a p-n junction develop a potential barrier?
Answer: Electrons and holes diffuse across the junction and recombine near it. This leaves fixed positive and negative ions on opposite sides of the junction. These ions create an electric field and a potential barrier that opposes further diffusion.
Q7. Why is the depletion region called a depletion layer?
Answer: It is called a depletion layer because it is depleted of mobile charge carriers such as free electrons and holes.
Q8. Why does the depletion layer become narrow during forward bias?
Answer: During forward bias, the applied voltage opposes the built-in electric field. This reduces the potential barrier and allows majority carriers to move towards the junction, making the depletion layer narrower.
Q9. Why does the depletion layer become wider during reverse bias?
Answer: During reverse bias, the applied voltage supports the built-in electric field. This increases the potential barrier and pulls majority carriers away from the junction, making the depletion layer wider.
Q10. Why does a diode conduct in forward bias but not easily in reverse bias?
Answer: In forward bias, the potential barrier decreases and majority carriers cross the junction easily. In reverse bias, the potential barrier increases and the depletion layer becomes wider, restricting majority carrier flow.
Q11. Why is the reverse current in a diode very small?
Answer: Reverse current is mainly produced by minority charge carriers. Their concentration is very low compared with majority carriers, so the reverse current is very small under normal conditions.
Q12. Why is a diode used as a rectifier?
Answer: A diode allows current to flow easily in forward bias but offers very high resistance in reverse bias. This one-way conduction property allows it to convert AC into pulsating DC.
Q13. Why is the output of a full-wave rectifier better than that of a half-wave rectifier?
Answer: A full-wave rectifier uses both half-cycles of the input AC, whereas a half-wave rectifier uses only one half-cycle. Therefore, a full-wave rectifier provides a higher average output and a higher ripple frequency, making filtering easier.
Q14. Why is a capacitor connected in parallel with the load resistance in a rectifier?
Answer: A capacitor connected in parallel with the load charges when the output voltage increases and discharges when the output voltage decreases. This reduces the fluctuations in the rectified output and makes it smoother.
Q15. What happens to the output frequency when a full-wave rectifier is used?
Answer: The output frequency becomes twice the input frequency because two output pulses are produced during each complete input cycle.
fout = 2fin
5. IMPORTANT FORMULAS
Forbidden energy gap:
Eg = EC − EV
Intrinsic semiconductor:
ne = nh = ni
Mass action law:
ne × nh = ni²
Dynamic resistance:
rd = ΔV/ΔI
Half-wave rectifier output frequency:
fout = fin
Full-wave rectifier output frequency:
fout = 2fin
Time period and frequency:
T = 1/f
Ohm's law:
V = IR
Electric power:
P = VI
Energy consumed:
E = Pt
Resistance:
R = V/I
Capacitor charge:
Q = CV
Time constant of an RC circuit:
τ = RC
Rectification efficiency:
η = (DC output power / AC input power) × 100%
Maximum theoretical efficiency of half-wave rectifier:
η = 40.6%
Maximum theoretical efficiency of full-wave rectifier:
η = 81.2%
6. IMPORTANT VALUES AND CONSTANTS
| Quantity | Approximate Value |
|---|---|
| Energy gap of carbon (diamond) | 5.4 eV |
| Energy gap of silicon | 1.1 eV |
| Energy gap of germanium | 0.7 eV |
| Cut-in voltage of silicon diode | 0.7 V |
| Cut-in voltage of germanium diode | 0.2 V |
| Half-wave rectifier efficiency | 40.6% |
| Full-wave rectifier efficiency | 81.2% |
| Electronic charge | 1.6 × 10⁻¹⁹ C |
| Planck's constant | 6.63 × 10⁻³⁴ J s |
7. IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Electric current | Ampere (A) |
| Electric potential | Volt (V) |
| Resistance | Ohm (Ω) |
| Dynamic resistance | Ohm (Ω) |
| Energy | Joule (J) |
| Power | Watt (W) |
| Frequency | Hertz (Hz) |
| Time | Second (s) |
| Capacitance | Farad (F) |
| Electric charge | Coulomb (C) |
| Energy gap | Joule (J) or electron volt (eV) |
8. QUICK REVISION POINTS
Semiconductors have electrical conductivity between conductors and insulators.
Silicon and germanium are common semiconductor materials.
The valence band is the energy band containing bound electrons.
The conduction band contains electrons that can contribute to electrical conduction.
The forbidden energy gap is the energy difference between the conduction band and valence band.
The energy gap of silicon is approximately 1.1 eV.
The energy gap of germanium is approximately 0.7 eV.
A pure semiconductor is called an intrinsic semiconductor.
An impurity-doped semiconductor is called an extrinsic semiconductor.
Pentavalent impurities produce n-type semiconductors.
Trivalent impurities produce p-type semiconductors.
Electrons are the majority carriers in n-type semiconductors.
Holes are the majority carriers in p-type semiconductors.
The mass action law is ne × nh = ni².
A p-n junction consists of p-type and n-type regions in the same semiconductor crystal.
The depletion region is depleted of mobile charge carriers.
Forward bias reduces the potential barrier.
Reverse bias increases the potential barrier.
A diode conducts easily in forward bias.
A rectifier converts AC into pulsating DC.
A half-wave rectifier uses one half-cycle of AC.
A full-wave rectifier uses both half-cycles of AC.
Half-wave output frequency is equal to input frequency.
Full-wave output frequency is twice the input frequency.
A capacitor filter reduces ripples in rectifier output.
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