Semiconductor Electronics Materials Devices and Simple Circuits - Class 12 Physics - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Notes - Semiconductor Electronics Materials Devices and Simple Circuits | Class 12 NCERT | Physics
Comprehensive Notes on Semiconductor Electronics: Materials, Devices, and Simple Circuits for Class 12
Introduction to Semiconductor Electronics
Semiconductor electronics play a pivotal role in modern technology, forming the backbone of numerous electronic devices like diodes, transistors, and integrated circuits (ICs). The journey from vacuum tubes to solid-state devices marked a revolution, bringing about devices that are smaller, more reliable, and energy-efficient. Understanding semiconductor electronics is crucial for grasping how today's technology works.
Classification of Materials Based on Conductivity
Metals
Metals possess very low resistivity or high conductivity. The resistivity of metals ranges from (10^{-2}) to (10^{-8} \Omega \mathrm{m}). Metals allow electric current to flow freely, making them ideal for applications requiring efficient electricity conduction.
Semiconductors
Semiconductors have intermediate resistivity and conductivity levels between metals and insulators. Their resistivity ranges from (10^{-5}) to (10^{6} \Omega \mathrm{m}). Unlike metals, semiconductors can control the flow of charge carriers, which is essential for electronic devices.
Insulators
Insulators exhibit very high resistivity, typically between (10^{11}) and (10^{19} \Omega \mathrm{m}), making them unsuitable for conducting electricity. They are used to prevent electrical currents from unintended paths.
Types of Semiconductors
Intrinsic Semiconductors
Intrinsic semiconductors are pure forms of semiconductor materials, such as silicon (Si) or germanium (Ge). Their conductivity is solely dependent on the properties of the material itself, without any added impurities.
Extrinsic Semiconductors
Extrinsic semiconductors are those to which impurities have been added to improve their conductivity. Based on the type of impurity added, they are classified into:
n-Type Semiconductors
Formed by doping silicon or germanium with pentavalent elements such as arsenic (As), which have five valence electrons. These extra electrons increase the free electron count in the material.
p-Type Semiconductors
Formed by doping silicon or germanium with trivalent elements like boron (B), which have three valence electrons. These elements create "holes" by accepting electrons, which enhances hole conductivity in the material.
Energy Bands in Solids
In solids, the energy levels of electrons form bands. The primary bands of interest are the valence band, which holds the outermost electrons, and the conduction band, which is typically empty but can conduct free electrons.
The gap between the valence band and the conduction band is known as the band gap. The size of this gap determines the electrical properties of the material:
- Insulators: Large band gap (E_g > 3 \mathrm{eV})
- Semiconductors: Moderate band gap ((0.2 \mathrm{eV} - 3 \mathrm{eV}))
- Metals: Very small or zero band gap
Semiconductor Materials
Elemental Semiconductors
Silicon (Si) and Germanium (Ge) are the most commonly used elemental semiconductors. Their atomic structures make them suitable for forming the basis of most semiconductor devices.
Compound Semiconductors
These include inorganic compounds like Gallium Arsenide (GaAs) and organic compounds like anthracene. Compound semiconductors are increasingly used in specialised applications.
Emerging Materials
After the 1990s, semiconductor devices using organic materials and semiconducting polymers started to signal a futuristic trend in polymer and molecular electronics.
Formation and Function of p-n Junction
Process of Junction Formation
A p-n junction is formed by combining p-type and n-type semiconductors. During this process, critical activities such as diffusion of electrons and holes from high concentration to low concentration regions and drift due to the electric field occur simultaneously.
Depletion Region and Its Significance
At the p-n junction, diffusion results in a region devoid of any charge carriers termed the 'depletion region’. This region develops a built-in electric field causing drift, maintaining equilibrium when diffusion and drift currents equalise.
Barrier Potential
The inherent potential barrier formed at the junction opposes further diffusion of charge carriers, thus stabilising the junction.
p-n Junction Diode Characteristics
Forward Bias Characteristics
In forward bias, the p-side of the diode is connected to the positive terminal of a battery, and the n-side to the negative terminal. This reduces the barrier potential, allowing current to flow.
Reverse Bias Characteristics
In reverse bias, the p-side is connected to the negative terminal, and the n-side to the positive terminal. This increases the barrier potential, effectively blocking any significant current flow.
V-I Characteristics
The voltage-current (V-I) characteristics of a diode display different behaviours under forward and reverse biases. In forward bias, current sharply increases after surpassing the threshold voltage. In reverse bias, a small leakage current is observed until reaching the breakdown voltage.
Applications of p-n Junction Diodes
Rectifiers
Rectifiers convert alternating current (AC) to direct current (DC). This function is crucial for powering a wide range of electronic devices.
Half-Wave Rectifier
Transforms only one half of the AC cycle into DC.
Full-Wave Rectifier
Utilises both halves of the AC cycle, providing a more efficient conversion.
Signal Diodes and Switching
Used in circuits for their ability to control current direction and switch states quickly.
Light Emitting Diodes (LEDs)
Emit light when current flows through them, widely used in displays and indicators.
Simple Circuits Using Semiconductors
Basic Circuit Components
Essential components include resistors, capacitors, inductors, and semiconductor devices.
Example Circuits
Illustrative circuits help in understanding the practical applications of semiconductor devices. Simple rectifier circuits or LED circuits are excellent examples.
Practical Applications and Experiments
Hands-on experiments and practical implementations aid in comprehending the theoretical concepts and their real-world applications.
Summary and Key Points
- Semiconductors are foundational to modern electronics.
- Materials are classified based on their conductivity: metals, semiconductors, and insulators.
- Intrinsic semiconductors are pure, while extrinsic semiconductors are doped to enhance conductivity.
- The energy band gap is crucial in determining the material’s electrical properties.
- Silicon and germanium are the primary elemental semiconductors.
- p-n junctions are fundamental components in various electronic devices.
- Diodes function by allowing current flow primarily in one direction, crucial for rectification processes.
- Rectifiers convert AC to DC, essential for various electronic applications.
Additional Resources and Further Reading
- Recommended textbooks for Class 12 Physics
- Online resources and educational videos
- Experiment kits and practical tools for exploring semiconductor electronics
Understanding semiconductor electronics involves bridging theoretical concepts with practical applications, laying the foundation for advanced studies in electronics and technology.
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NCERT Solutions - Semiconductor Electronics Materials Devices and Simple Circuits | NCERT | Physics | Class 12
In an n-type silicon, which of the following statement 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.
In an n-type silicon, the correct statement is:
(c) Holes are minority carriers and pentavalent atoms are the dopants.
Here’s why:
Electrons are the majority carriers in n-type semiconductors because pentavalent atoms (like arsenic, phosphorus, etc.) donate extra electrons.
Holes are the minority carriers because they result from the absence of electrons.
Pentavalent atoms are the dopants because they introduce extra electrons into the silicon.
Thus, (c) Holes are minority carriers and pentavalent atoms are the dopants is the correct statement.
Which of the statements given in Exercise 14.1 is true for p-type semiconductor.
The correct statement for a p-type semiconductor is:
(d) Holes are majority carriers and trivalent atoms are the dopants.
Here's why:
Holes are majority carriers: In p-type semiconductors, the doping process introduces more holes compared to electrons, making holes the primary charge carriers.
Trivalent atoms are the dopants: Trivalent impurities (elements with 3 valence electrons like Indium (In), Boron (B), or Aluminium (Al)) are used to dope the semiconductor, creating these additional holes.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated
by energy band gap respectively equal to $\left(E_{\mathrm{g}}\right)_{\mathrm{C}},\left(E_{\mathrm{g}}\right)_{\mathrm{Si}}$ and $\left(E_{\mathrm{g}}\right)_{\mathrm{Ge}}$. Which of the following statements is true?
(a) $\left(E_{g}\right)_{\mathrm{Si}}<\left(E_{g}\right)_{\mathrm{Ge}}<\left(E_{g}\right)_{\mathrm{C}}$
(b) $\left(E_{g}\right)_{\mathrm{C}}<\left(E_{g}\right)_{\mathrm{Ge}}>\left(E_{g}\right)_{\mathrm{Si}}$
(c) $\left(E_{g}\right)_{\mathrm{C}}>\left(E_{g}\right)_{\mathrm{Si}}>\left(E_{g}\right)_{\mathrm{Ge}}$
(d) $\left(E_{g}\right)_{\mathrm{C}}=\left(E_{g}\right)_{\mathrm{Si}}=\left(E_{g}\right)_{\mathrm{Ge}}$
To determine the correct statement, let's consider the information provided in the chapter. According to the given text:
Carbon (C) has higher ionization energy because its bonding electrons lie in the second orbit, which means the energy required to take out an electron from these atoms is the highest.
Silicon (Si) and Germanium (Ge) have their bonding electrons in the third and fourth orbits, respectively. This makes it easier to ionize Si and Ge compared to C.
As a result, the energy gaps ($E_g$) between the valence band and conduction band for these materials are in the order:
$\left(E_{g}\right)_{\text{C}} > \left(E_{g}\right)_{\text{Si}} > \left(E_{g}\right)_{\text{Ge}}$
Therefore, the correct statement is:
(c) $ \left(E_{g}\right)_{\text{C}} > \left(E_{g}\right)_{\text{Si}} > \left(E_{g}\right)_{\text{Ge}}$
In an unbiased p-n junction, holes diffuse from the p-region to 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.
In an unbiased p-n junction, the reason holes diffuse from the p-region to the n-region is due to the concentration gradient between the two regions. This is described by option:
(c) hole concentration in p-region is more as compared to n-region.
This concentration gradient drives the diffusion process, leading to the movement of holes from the region where they are in higher concentration (p-region) to where they are in lower concentration (n-region).
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.
When a forward bias is applied to a p-n junction, it lowers the potential barrier. So, the correct choice is:
(c) lowers the potential barrier.
In half-wave rectification, what is the output frequency if the input frequency is $50 \mathrm{~Hz}$. What is the output frequency of a full-wave rectifier for the same input frequency.
In half-wave rectification, the output frequency is the same as the input frequency, which is $50 \mathrm{~Hz}$.
In full-wave rectification, the output frequency is twice the input frequency. Thus, for an input frequency of $50 \mathrm{~Hz}$, the output frequency will be $100 \mathrm{~Hz}$.
So,
Half-wave rectification output frequency: $50 \mathrm{~Hz}$
Full-wave rectification output frequency: $100 \mathrm{~Hz}$
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In an NPN transistor, the collector current is 24 mA. If 80% of electrons reach the collector, what is its base current in mA?
A) 36
B) 26
C) 16
D) 6
The operation of an NPN transistor involves the flow of electrons from the emitter to the collector, with a certain portion passing through the base. Given that the collector current ($I_C$) is 24 mA and that 80% of the electrons reach the collector, we can determine the base current ($I_B$).
Firstly, let's understand that the emitter current ($I_E$) is the sum of the collector current and the base current:
$$ I_E = I_C + I_B $$
Since 80% of the emitter current reaches the collector, the missing 20% is the base current. Therefore, we can calculate the emitter current as follows, knowing that $I_C$ is 80% of $I_E$:
$$ I_C = 0.8 I_E \ 24 \text{ mA} = 0.8 I_E \ I_E = \frac{24 \text{ mA}}{0.8} $$
Calculating $I_E$:
$$ I_E = 30 \text{ mA} $$
Now, we find $I_B$ by subtracting $I_C$ from $I_E$:
$$ I_B = I_E - I_C \ I_B = 30 \text{ mA} - 24 \text{ mA} $$
Calculating $I_B$:
$$ I_B = 6 \text{ mA} $$
Thus, the base current $I_B$ is 6 mA which corresponds to option D.
In the laboratory, silicon can be prepared
A) By heating carbon in an electric furnace.
B) By heating potassium with potassium dichromate.
C) By reacting silica with magnesium.
D) None of these.
The correct option is C) By reacting silica with magnesium.
In a laboratory setting, silicon can be prepared through a reaction involving silica and magnesium. Silica, which is an oxide of silicon, has the chemical formula $ \mathrm{SiO}_2 $.
In the circuit as shown in the figure, $A$ and $B$ represent two inputs, and $C$ represents the.
Options:
A) OR gate
B) NOR gate
C) AND gate
D) NAND gate
The correct option is A) OR gate.
The circuit diagram in the figure represents an OR gate. The feature of an OR gate is that the output, $C$, will be 1 if either one or both the inputs, $A$ and $B$, are at 1. Specifically, the output $C$ is 0 only when both inputs, $A$ and $B$, are 0. The Boolean expression representing this functionality is: $$ C = A + B $$ This confirms that the circuit operates as an OR gate, matching option A).
In each of the following questions, some conclusions marked 1, 2, and so on are drawn from the given statements. Select the answer choice that presents only correct conclusions.
Statements:
i. Some headphones are earphones.
ii. All earphones are telephones.
iii. No telephones are TVs.
Conclusions:
No earphones are TVs.
Some headphones are not TVs.
Some headphones are telephones.
Some telephones are not TVs.
Select the correct code:
A. (a) All follow
B. (b) Only 1, 2, and 3 follow.
C. (c) Only 2, 3, and 4 follow.
D. (d) Only 1, 3, and 4 follow.
The correct answer is A (All follow).
The conclusions can be deduced as follows from the given statements:
Conclusions 1: No earphones are TVs.
From statement ii, "All earphones are telephones," and statement iii, "No telephones are TVs," we can deduce that no earphones are TVs because the entire set of earphones is included in telephones which, in turn, are excluded from being TVs.
Conclusion 2: Some headphones are not TVs.
Since some headphones are earphones (from statement i) and these earphones cannot be TVs (from Conclusion 1), it follows logically that some headphones, specifically those that are also earphones, are not TVs.
Conclusion 3: Some headphones are telephones.
Given that some headphones are earphones (statement i) and all earphones are telephones (statement ii), this implies that some headphones must be telephones as well.
Conclusion 4: Some telephones are not TVs.
This is directly supported by statement iii that no telephones are TVs, implying some or all telephones are not TVs.
All four conclusions logically follow from the supplied premises, confirming that option A (All follow) is the correct choice.
Is $\mathrm{B}$ (boron) doped with $\mathrm{Si}$ (silicon) a p-type or an n-type semiconductor? And why?
Boron (B) has an atomic number of 5 and possesses 3 valence electrons. In the context of semiconductor doping, elements with 3 valence electrons are typically used to create p-type semiconductors. Conversely, elements with 5 valence electrons are used for making n-type semiconductors. Therefore, when Silicon (Si) is doped with Boron, it results in a p-type semiconductor due to the presence of 3 valence electrons in Boron.
Directions: Study the following information carefully to answer the question given below.
P, Q, R, S, T, V, W, and Z are eight friends studying in three different engineering colleges — A, B, and C — in three disciplines: Mechanical, Electrical, and Electronics, with not less than two and not more than three in any college. Not more than three of them study in any of the three disciplines. W studies Electrical in college B with only T, who studies Mechanical. P and Z do not study in college C and study in the same discipline, but not Electrical. R studies Mechanical in college C with V, who studies Electrical. S studies Mechanical and does not study in the same college where R studies. Q does not study Electronics.
Which of the following combinations of college-student specialization is correct?
A) C-R - Electronics
B) A-Z - Electrical
C) B-W - Electronics
D) B-W - Electrical
E) B-Z - Electronics
The correct answer is Option D: B-W-Electrical.
Based on the information provided and the analysis of it, here's how the college and discipline distribution among the friends looks:
Person | College | Discipline |
---|---|---|
P | A | Electronics |
Q | B | Electrical |
R | C | Mechanical |
S | A | Mechanical |
T | B | Mechanical |
V | C | Electrical |
W | B | Electrical |
Z | A | Electronics |
Therefore, the combination that correctly matches one of the friends with their college and discipline is:
B-W-Electrical (Option D)
This confirms that W studies Electrical engineering in college B.
"What is an organic semiconductor? Name any two semiconductors."
Organic semiconductors are materials where the primary components are pi-bonded molecules or polymers consisting mainly of carbon and hydrogen atoms, often including heteroatoms like nitrogen, sulfur, and oxygen. These substances are typically found in the form of molecular crystals or amorphous thin films. Generally, they act as electrical insulators, but exhibit semiconducting properties under certain conditions such as charge injection via electrodes, doping, or photoexcitation.
The band gap in organic semiconductors (meaning the energy difference between the valence band and the conduction band) ranges typically from 2.5 to 4 eV, which is larger than those found in inorganic semiconductors (usually 1 to 2 eV). This larger band gap suggests that they are more insulating unless charge carriers are present. These carriers can be generated through doping, electrode injection, or optical excitation. It is crucial to note that the primary optical excitations in these materials are neutral excitons with a Coulomb-binding energy of about 0.5 to 1.0 eV, attributed to low dielectric constants of about 3-4. This scenario complicates efficient photogenerated charge carrier production in bulk systems without binary system interactions for charge transfer. Otherwise, neutral excitons typically decay, emitting photoluminescence or dissipating non-radiatively.
The optical absorption edge of organic semiconductors is generally between 1.7 to 3 eV, corresponding to a spectral range of about 700 to 400 nm which overlaps with the visible spectrum.
Two examples of organic semiconductors are:
Doped Polyacetylene
Organic Thin Film Transistors (OTFT)
What is the branch of science that makes new substances and devices using particles of very, very small size?
A. Microscopy
B. Nanotechnology
C. Synthetics
D. Nautics
The correct option is B. Nanotechnology.
Nanotechnology is the branch of science focused on creating new substances and devices using particles that are extremely small, typically at the nanoscale level.
Among the given options, identify the chemical elements that play an important role in electronic industries.
A. Silicon
B. Germanium
C. Hydrogen
D. Oxygen
The correct options are:
A. Silicon
B. Germanium
Silicon and Germanium are crucial materials widely utilized in the electronic industries. They serve as the primary elements in the production of diodes, which are essential components in various electronic devices such as radios, televisions, and calculators.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Given values:
Collector resistance ($R_C$): $2$ kΩ
Output voltage ($V_0$): $2$ V
Base resistance ($R_B$): $1$ kΩ
Current amplification factor ($\beta$): $100$
Calculation of the base current, $I_B$:
The base current can be determined using the formula: $$ I_B = \frac{I_C}{\beta} $$ Where $I_C$ is the collector current. We can find $I_C$ by using the formula $I_C = \frac{V_0}{R_C}$: $$ I_C = \frac{2}{2000} = 1 \text{ mA} $$ Substituting $I_C$ in the base current formula: $$ I_B = \frac{1 \times 10^{-3}}{100} = 10 \mu\text{A} $$
Calculation of the input signal voltage, $V_{\text{in}}$:
The input signal voltage can be calculated using Ohm's Law applied to the base resistor: $$ V_{\text{in}} = I_B R_B $$ Substituting the values for $I_B$ and $R_B$: $$ V_{\text{in}} = 10 \times 10^{-6} \times 1000 = 10 \text{ mV} $$
Hence, the input signal voltage is 10 mV and the base current is 10 μA.
In a semiconductor, current conduction is due:
A. Only to holes.
B. Only to electrons.
C. To holes and electrons.
D. None of these.
The correct answer is C. To holes and electrons.
In a semiconductor, current conduction occurs due to the movement of both holes and electrons. In p-type materials, holes are the majority carriers as they outnumber the free electrons. Conversely, in n-type materials, free electrons are the majority carriers. Thus, both types of charge carriers contribute to current flow in semiconductors.
Photo cell is a device to
A) Store photons.
B) Measure light intensity.
C) Convert photon energy into mechanical energy.
D) Store electrical energy for replacing storage batteries.
The correct answer is B) Measure light intensity.
A photocell functions by emitting electrons when exposed to light, a phenomenon known as the photoelectric effect. The number of electrons emitted is directly proportional to the intensity of the incoming light radiation. Hence, photocells are used to measure the intensity of light based on the amount of electron emission triggered by the light.
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