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      Class 12 PHYSICS – JEE

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      • Class 12
      • Class 12 PHYSICS – JEE
      CoursesClass 12PhysicsClass 12 PHYSICS – JEE
      • 1.Electrostatics (1)
        8
        • Lecture1.1
          Charge, Coulombs Law and Coulombs law in Vector form 41 min
        • Lecture1.2
          Electric Field; Electric Field Lines; Field lines due to multiple charges 42 min
        • Lecture1.3
          Charge Distribution; Finding Electric Field due to Different Object 01 hour
        • Lecture1.4
          Solid angle; Area Vector; Electric Flux; Flux of closed surface; Gauss Law 47 min
        • Lecture1.5
          Finding E Using Concept of Gauss law and Flux 01 hour
        • Lecture1.6
          Chapter Notes – Electrostatics (1)
        • Lecture1.7
          NCERT Solutions – Electrostatics
        • Lecture1.8
          Revision Notes Electrostatics
      • 2.Electrostatics (2)
        7
        • Lecture2.1
          Work done by Electrostatic Force; Work done by man in E-Field; Electrostatic Potential Energy 49 min
        • Lecture2.2
          Finding Electric Potential, Equipotential Surface and Motion in Electric Field 01 hour
        • Lecture2.3
          Electric Dipole and Dipole in Uniform and Non-uniform Electric field 01 hour
        • Lecture2.4
          Analysis of charge on conductors; Potential due to induced charge 58 min
        • Lecture2.5
          Conductors with cavity- Case 1: Empty cavity, Case 2: Charge Inside Cavity 41 min
        • Lecture2.6
          Connecting Two Conductors; Grounding of conductor; Electric field just outside conductor; Electrostatic pressure; Self potential Energy 54 min
        • Lecture2.7
          Chapter Notes – Electrostatics (2)
      • 3.Current Electricity (1)
        9
        • Lecture3.1
          Current, Motion of Electrons in Conductor; Temp. Dependence of Resistor 26 min
        • Lecture3.2
          Circuit Theory and Kirchoffs Laws 31 min
        • Lecture3.3
          Some Special Circuits- Series & Parallel Circuits, Open Circuit, Short Circuit 26 min
        • Lecture3.4
          Wheatstone Bridge, Current Antisymmetric 21 min
        • Lecture3.5
          Equivalent Resistance- Series and parallel, Equipotential Points, Wheatstone Bridge 25 min
        • Lecture3.6
          Current Antisymmetric, Infinite Ladder, Circuit Solving, 3D circuits 20 min
        • Lecture3.7
          Chapter Notes – Current Electricity
        • Lecture3.8
          NCERT Solutions – Current Electricity
        • Lecture3.9
          Revision Notes Current Electricity
      • 4.Current Electricity (2)
        4
        • Lecture4.1
          Heating Effect of Current; Rating of Bulb; Fuse 19 min
        • Lecture4.2
          Battery, Maximum power theorem; Ohmic and Non Ohmic Resistance; Superconductor 31 min
        • Lecture4.3
          Galvanometer; Ammeter & Voltmeter and Their Making 44 min
        • Lecture4.4
          Potentiometer and its applications ; Meter Bridge; Post Office Box; Colour Code of Resistors 32 min
      • 5.Capacitor
        6
        • Lecture5.1
          Capacitor and Capacitance; Energy in Capacitor 38 min
        • Lecture5.2
          Capacitive Circuits- Kirchoff’s Laws; Heat Production 01 hour
        • Lecture5.3
          Equivalent Capacitance; Charge on both sides of cap. Plate 52 min
        • Lecture5.4
          Dielectric Strength; Polar and Non-Polar Dielectric; Equivalent Cap. with Dielectric 01 hour
        • Lecture5.5
          Inserting and Removing Dielectric- Work (Fringing Effect), Force; Force between plates of capacitor 38 min
        • Lecture5.6
          Revision Notes Capacitor
      • 6.RC Circuits
        3
        • Lecture6.1
          Maths Needed for RC Circuits, RC circuits-Charging Circuit 19 min
        • Lecture6.2
          RC circuits-Discharging Circuit, Initial & Steady State, Final (Steady) State, Internal Resistance of Capacitor 44 min
        • Lecture6.3
          Revision Notes RC Circuits
      • 7.Magnetism and Moving Charge
        16
        • Lecture7.1
          Introduction, Vector Product, Force Applied by Magnetic Field, Lorentz Force, Velocity Selector 40 min
        • Lecture7.2
          Motion of Charged Particles in Uniform Magnetic Field 40 min
        • Lecture7.3
          Cases of Motion of Charged Particles in Uniform Magnetic Field 56 min
        • Lecture7.4
          Force on a Current Carrying Wire on Uniform B and its Cases, Questions and Solutions 59 min
        • Lecture7.5
          Magnetic Field on Axis of Circular Loop, Magnetic field due to Moving Charge, Magnetic Field due to Current 52 min
        • Lecture7.6
          Magnetic Field due to Straight Wire, Different Methods 40 min
        • Lecture7.7
          Magnetic Field due to Rotating Ring and Spiral 41 min
        • Lecture7.8
          Force between Two Current Carrying Wires 36 min
        • Lecture7.9
          Force between Two Current Carrying Wires 58 min
        • Lecture7.10
          Miscellaneous Questions 55 min
        • Lecture7.11
          Solenoid, Toroid, Magnetic Dipole, Magnetic Dipole Momentum, Magnetic Field of Dipole 54 min
        • Lecture7.12
          Magnetic Dipole in Uniform Magnetic Field, Moving Coil Galvanometer, Torsional Pendulum 01 hour
        • Lecture7.13
          Advanced Questions, Magnetic Dipole and Angular Momentum 56 min
        • Lecture7.14
          Chapter Notes – Magnetism and Moving Charge
        • Lecture7.15
          NCERT Solutions – Magnetism and Moving Charge
        • Lecture7.16
          Revision Notes Magnetism and Moving Charge
      • 8.Magnetism and Matter
        10
        • Lecture8.1
          Magnetic Dipole, Magnetic Properties of Matter, Diamagnetism; Domain Theory of Ferro 47 min
        • Lecture8.2
          Magnetic Properties of Matter in Detail 39 min
        • Lecture8.3
          Magnetization and Magnetic Intensity, Meissner Effect, Variation of Magnetization with Temperature 55 min
        • Lecture8.4
          Hysteresis, Permanent Magnet, Properties of Ferro for Permanent Magnet, Electromagnet 31 min
        • Lecture8.5
          Magnetic Compass, Earth’s Magnetic Field 20 min
        • Lecture8.6
          Bar Magnet, Bar Magnet in Uniform Field 49 min
        • Lecture8.7
          Magnetic Poles, Magnetic Field Lines, Magnetism and Gauss’s Law 32 min
        • Lecture8.8
          Chapter Notes – Magnetism and Matter
        • Lecture8.9
          NCERT Solutions – Magnetism and Matter
        • Lecture8.10
          Revision Notes Magnetism and Matter
      • 9.Electromagnetic Induction
        14
        • Lecture9.1
          Introduction, Magnetic Flux, Motional EMF 01 min
        • Lecture9.2
          Induced Electric Field, Faraday’s Law, Comparison between Electrostatic Electric Field and Induced Electric Field 43 min
        • Lecture9.3
          Induced Current; Faraday’s Law ; Lenz’s Law 56 min
        • Lecture9.4
          Faraday’s Law and its Cases 50 min
        • Lecture9.5
          Advanced Questions on Faraday’s Law 37 min
        • Lecture9.6
          Cases of Current Electricity 59 min
        • Lecture9.7
          Lenz’s Law and Conservation of Energy, Eddy Current, AC Generator, Motor 01 hour
        • Lecture9.8
          Mutual Induction 53 min
        • Lecture9.9
          Self Inductance, Energy in an Inductor 34 min
        • Lecture9.10
          LR Circuit, Decay Circuit 01 hour
        • Lecture9.11
          Initial and Final Analysis of LR Circuit 38 min
        • Lecture9.12
          Chapter Notes – Electromagnetic Induction
        • Lecture9.13
          NCERT Solutions – Electromagnetic Induction
        • Lecture9.14
          Revision Notes Electromagnetic Induction
      • 10.Alternating Current Circuit
        8
        • Lecture10.1
          Introduction, AC/DC Sources, Basic AC Circuits, Average & RMS Value 46 min
        • Lecture10.2
          Phasor Method, Rotating Vector, Adding Phasors, RC Circuit 35 min
        • Lecture10.3
          Examples and Solutions 21 min
        • Lecture10.4
          Power in AC Circuit, Resonance Frequency, Bandwidth and Quality Factor, Transformer 51 min
        • Lecture10.5
          LC Oscillator, Question and Solutions of LC Oscillator, Damped LC Oscillator 53 min
        • Lecture10.6
          Chapter Notes – Alternating Current Circuit
        • Lecture10.7
          NCERT Solutions – Alternating Current Circuit
        • Lecture10.8
          Revision Notes Alternating Current Circuit
      • 11.Electromagnetic Waves
        4
        • Lecture11.1
          Displacement Current; Ampere Maxwell Law 14 min
        • Lecture11.2
          EM Waves; EM Spectrum; Green House Effect; Ozone Layer 36 min
        • Lecture11.3
          Chapter Notes – Electromagnetic Waves
        • Lecture11.4
          Revision Notes Electromagnetic Waves
      • 12.Photoelectric Effect
        5
        • Lecture12.1
          Recalling Basics; Photoelectric Effect 50 min
        • Lecture12.2
          Photo-electric Cell 35 min
        • Lecture12.3
          Photon Flux; Photon Density; Momentum of Photon; Radiation Pressure- Full Absorption, Full Reflection; Dual nature 52 min
        • Lecture12.4
          Chapter Notes – Photoelectric Effect
        • Lecture12.5
          Revision Notes Photoelectric Effect
      • 13.Ray Optics (Part 1)
        12
        • Lecture13.1
          Rays and Beam of Light, Reflection of Light, Angle of Deviation, Image Formation by Plane Mirror 01 hour
        • Lecture13.2
          Field of View, Numerical on Field of Line, Size of Mirror 42 min
        • Lecture13.3
          Curved Mirrors, Terms Related to Curved Mirror, Reflection of Light by Curved Mirror 40 min
        • Lecture13.4
          Image Formation by Concave Mirror, Magnification or Lateral or Transverse Magnification 01 hour
        • Lecture13.5
          Ray Diagrams for Concave Mirror 45 min
        • Lecture13.6
          Image Formation by Convex Mirror; Derivations of Various Formulae used in Concave Mirror and Convex Mirror 01 hour
        • Lecture13.7
          Advanced Optical Systems, Formation of Images with more than one Mirror 24 min
        • Lecture13.8
          Concept of Virtual Object, Formation of Image when Incident ray are Converging, Image Characteristics for Virtual Object, 55 min
        • Lecture13.9
          Newton’s Formula, Longitudinal Magnification 23 min
        • Lecture13.10
          Formation of Image when Two Plane Mirrors kept at an angle and parallel; Formation of Image by two Parallel Mirrors. 43 min
        • Lecture13.11
          Chapter Notes – Ray Optics
        • Lecture13.12
          NCERT Solutions – Ray Optics
      • 14.Ray Optics (Part 2)
        13
        • Lecture14.1
          Refractive Index, Opaque, Transparent, Speed of Light, Relative Refractive Index, Refraction and Snell’s Law, Refraction in Denser and Rarer Medium 42 min
        • Lecture14.2
          Image Formation due to Refraction; Derivation; Refraction and Image formation in Glass Slab 57 min
        • Lecture14.3
          Total Internal Reflection, Critical Angle, Principle of Reversibility 01 hour
        • Lecture14.4
          Application of Total Internal Reflection 45 min
        • Lecture14.5
          Refraction at Curved Surface, Image Formation by Curved Surface, Derivation 56 min
        • Lecture14.6
          Image Formation by Curved Surface, Snell’s Law in Vector Form 01 hour
        • Lecture14.7
          Lens, Various types of Lens, Differentiating between various Lenses; Optical Centre, Derivation of Lens Maker Formula 01 hour
        • Lecture14.8
          Lens Formula, Questions and Answers 39 min
        • Lecture14.9
          Property of Image by Convex and Concave Lens; Lens Location, Minimum Distance Between Real Image and Object 01 hour
        • Lecture14.10
          Power of Lens, Combination of Lens, Autocollimation 35 min
        • Lecture14.11
          Silvering of Lens 44 min
        • Lecture14.12
          Cutting of Lens and Mirror, Vertical Cutting, Horizontal Cutting 49 min
        • Lecture14.13
          Newton’s Law for Lens and Virtual Object 01 hour
      • 15.Ray Optics (Part 3)
        6
        • Lecture15.1
          Prism, Angle of Prism, Reversibility in Prism 51 min
        • Lecture15.2
          Deviation in Prism, Minimum and Maximum Deviation, Asymmetric, Thin Prism, Proof for formula of Thin Prism 59 min
        • Lecture15.3
          Dispersion of Light, Refractive Index, Composition of Light, Dispersion through Prism 01 hour
        • Lecture15.4
          Rainbow Formation, Scattering of Light, Tyndall Effect, Defects of Image, Spherical Defect, Chromatic Defect, Achromatism. 57 min
        • Lecture15.5
          Optical Instruments, The Human Eye, Defects of Eye and its Corrections 01 hour
        • Lecture15.6
          Microscope & Telescope 02 hour
      • 16.Wave Optics
        21
        • Lecture16.1
          Introduction to Wave Optics 11 min
        • Lecture16.2
          Huygens Wave Theory 14 min
        • Lecture16.3
          Huygens Theory of Secondary Wavelets 10 min
        • Lecture16.4
          Law of Reflection by Huygens Theory 10 min
        • Lecture16.5
          Deriving Laws of Refraction by Huygens Wave Theory 10 min
        • Lecture16.6
          Multiple Answer type question on Huygens Theory 41 min
        • Lecture16.7
          Conditions of Constructive and Destructive Interference 22 min
        • Lecture16.8
          Conditions of Constructive and Destructive Interference 06 min
        • Lecture16.9
          Conditions of Constructive and Destructive Interference 23 min
        • Lecture16.10
          Incoherent Sources of Light 38 min
        • Lecture16.11
          Youngs Double Slit Experiment 12 min
        • Lecture16.12
          Fringe Width Positions of Bright and Dark Fringes 15 min
        • Lecture16.13
          Numerical problems on Youngs Double Slit Experiment 11 min
        • Lecture16.14
          Numerical problems on Youngs Double Slit Experiment 19 min
        • Lecture16.15
          Displacement of Interference Pattern 19 min
        • Lecture16.16
          Numerical problems on Displacement of Interference Pattern 28 min
        • Lecture16.17
          Shapes of Fringes 37 min
        • Lecture16.18
          Colour of Thin Films 59 min
        • Lecture16.19
          Interference with White Light 32 min
        • Lecture16.20
          Chapter Notes – Wave Optics
        • Lecture16.21
          NCERT Solutions – Wave Optics
      • 17.Atomic Structure
        6
        • Lecture17.1
          Thomson and Rutherford Model of Atom; Trajectory of Alpha particle; Bohr’s Model ; Hydrogen Like Atom; Energy Levels 58 min
        • Lecture17.2
          Emission Spectra, Absorption Spectra; De Broglie Explanation of Bohr’s 2nd Postulate; Limitations of Bohr’s Model 37 min
        • Lecture17.3
          Momentum Conservation in Photon Emission, Motion of Nucleus, Atomic Collision 58 min
        • Lecture17.4
          Chapter Notes – Atomic Structure
        • Lecture17.5
          NCERT Solutions – Atomic Structure
        • Lecture17.6
          Revision Notes Atomic Structure
      • 18.Nucleus
        6
        • Lecture18.1
          Basics- Size of Nucleus, Nuclear Force, Binding Energy, Mass Defect; Radioactive Decay 01 hour
        • Lecture18.2
          Laws of Radioactive Decay 36 min
        • Lecture18.3
          Nuclear Fission; Nuclear Reactor; Nuclear Fusion- Reaction Inside Sun 30 min
        • Lecture18.4
          Chapter Notes – Nucleus
        • Lecture18.5
          NCERT Solutions – Nucleus
        • Lecture18.6
          Revision Notes Nucleus
      • 19.X-Ray
        4
        • Lecture19.1
          Electromagnetic Spectrum, Thermionic Emission; Coolidge Tube – Process 1 22 min
        • Lecture19.2
          Coolidge Tube – Process 2; Moseley’s Law; Absorption of X-rays in Heavy Metal 39 min
        • Lecture19.3
          Chapter Notes – X-Ray
        • Lecture19.4
          Revision Notes X-Ray
      • 20.Error and Measurement
        2
        • Lecture20.1
          Least Count of Instruments; Mathematical Operation on Data with Random Error 18 min
        • Lecture20.2
          Significant Digits; Significant Digits and Mathematical Operations 30 min
      • 21.Semiconductors
        9
        • Lecture21.1
          Conductor, Semiconductors and Insulators Basics Difference, Energy Band Theory, Si element 21 min
        • Lecture21.2
          Doping and PN Junction 01 hour
        • Lecture21.3
          Diode and Diode as Rectifier 01 hour
        • Lecture21.4
          Voltage Regulator and Zener Diode and Optoelectronic Jn. Devices 01 hour
        • Lecture21.5
          Transistor, pnp, npn, Modes of operation, Input and Output Characteristics, , Current Amplification Factor 01 hour
        • Lecture21.6
          Transistor as Amplifier, Transistor as Switch, Transistor as Oscillator, Digital Gates 01 hour
        • Lecture21.7
          Chapter Notes – Semiconductors
        • Lecture21.8
          NCERT Solutions – Semiconductors
        • Lecture21.9
          Revision Notes Semiconductors
      • 22.Communication Systems
        5
        • Lecture22.1
          Basic working and terms; Antenna; Modulation and Types of Modulation 32 min
        • Lecture22.2
          Amplification Modulation, Transmitter, Receiver, Modulation index 40 min
        • Lecture22.3
          Chapter Notes – Communication Systems
        • Lecture22.4
          NCERT Solutions – Communication Systems
        • Lecture22.5
          Revision Notes Communication Systems

        NCERT Solutions – Semiconductors

        14.1. 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.

        Answer

        The correct statement is (c).

        In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

        14.2. Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

        Answer

        The correct statement is (d).

        In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.

        14.3. 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 (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:

        The correct statement is (c).

        Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.

        The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge

        14.4. 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.

        Answer

        The correct statement is (c).

        The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

         

        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

        The correct statement is (c).

        When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.

        14.6. For transistor action, which of the following statements are correct:

        (a) Base, emitter and collector regions should have similar size and doping concentrations.

        (b) The base region must be very thin and lightly doped.

        (c) The emitter junction is forward biased and collector junction is reverse biased.

        (d) Both the emitter junction as well as the collector junction are forward biased.

        Answer

        The correct statement is (b), (c).

        For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased.

        14.7. For a transistor amplifier, the voltage gain

        (a) remains constant for all frequencies.

        (b) is high at high and low frequencies and constant in the middle frequency range.

        (c) is low at high and low frequencies and constant at mid frequencies.

        (d) None of the above.

        Answer

        The correct statement is (c).

        The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.

        14.8. 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

        Input frequency = 50 Hz

        For a half-wave rectifier, the output frequency is equal to the input frequency.

        ∴Output frequency = 50 Hz

        For a full-wave rectifier, the output frequency is twice the input frequency.

        ∴Output frequency = 2 × 50 = 100 Hz

        14.9. For a CE-transistor amplifier, the audio signal voltage across the collected 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.

        Answer

        Collector resistance, RC = 2 kΩ = 2000 Ω

        Audio signal voltage across the collector resistance, V = 2 V

        Current amplification factor of the transistor, β = 100

        Base resistance, RB = 1 kΩ = 1000 Ω

        Input signal voltage = Vi

        Base current = IB

        We have the amplification relation as:

        Voltage amplification 

        Therefore, the input signal voltage of the amplifier is 0.01 V.

        Base resistance is given by the relation:

        Therefore, the base current of the amplifier is 10 μA.

        14.10. Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.

        Answer

        Voltage gain of the first amplifier, V1 = 10

        Voltage gain of the second amplifier, V2 = 20

        Input signal voltage, Vi = 0.01 V

        Output AC signal voltage = Vo

        The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,

        V = V1 × V2

        = 10 × 20 = 200

        We have the relation:

        V0 = V × Vi

        = 200 × 0.01 = 2 V

        Therefore, the output AC signal of the given amplifier is 2 V.

        14.11. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

        Answer

        Energy band gap of the given photodiode, Eg = 2.8 eV

        Wavelength, λ = 6000 nm = 6000 × 10−9 m

        The energy of a signal is given by the relation:

        E = 

        Where,

        h = Planck’s constant

        = 6.626 × 10−34 Js

        c = Speed of light

        = 3 × 108 m/s

        E 

        = 3.313 × 10−20 J

        But 1.6 × 10−19 J = 1 eV

        ∴E = 3.313 × 10−20 J

        The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.

        14.12. The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni= 1.5 × 1016 m−3. Is the material n-type or p-type?

        Answer

        Number of silicon atoms, N = 5 × 1028 atoms/m3

        Number of arsenic atoms, nAs = 5 × 1022 atoms/m3

        Number of indium atoms, nIn = 5 × 1020 atoms/m3

        Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3

        Number of electrons, ne = 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022

        Number of holes = nh

        In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:

        nenh = ni2

        Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

        14.13. In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by

        where n0 is a constant.

        Answer

        Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV

        The temperature dependence of the intrinsic carrier-concentration is written as:

        Where,

        kB = Boltzmann constant = 8.62 × 10−5 eV/K

        T = Temperature

        n0 = Constant

        Initial temperature, T1 = 300 K

        The intrinsic carrier-concentration at this temperature can be written as:

         … (1)

        Final temperature, T2 = 600 K

        The intrinsic carrier-concentration at this temperature can be written as:

         … (2)

        The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

        Therefore, the ratio between the conductivities is 1.09 × 105.

         

        14.14. In a p-n junction diode, the current I can be expressed as

        where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kBis the Boltzmann constant (8.6×10−5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10−12 A and T = 300 K, then

        (a) What will be the forward current at a forward voltage of 0.6 V?

        (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

        (c) What is the dynamic resistance?

        (d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

        Answer

        In a p-n junction diode, the expression for current is given as:

        Where,

        I0 = Reverse saturation current = 5 × 10−12 A

        T = Absolute temperature = 300 K

        kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1

        V = Voltage across the diode

        (a) Forward voltage, V = 0.6 V

        ∴Current, I 

        Therefore, the forward current is about 0.0256 A.

        (b) For forward voltage, V’ = 0.7 V, we can write:

        Hence, the increase in current, ΔI = I‘ − I

        = 1.257 − 0.0256 = 1.23 A

        (c) Dynamic resistance 

        (d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

        14.15. You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

        Answer

        (a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.

        Hence, the output of the NOR Gate =

        This will be the input for the NOT Gate. Its output will be = A + B

        ∴Y = A + B

        Hence, this circuit functions as an OR Gate.

        (b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.

        Hence, the output of the given circuit can be written as:

        Hence, this circuit functions as an AND Gate.

        14.16. Write the truth table for a NAND gate connected as given in Fig. 14.45.

        Hence identify the exact logic operation carried out by this circuit.

        Answer

        A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

        Hence, the output can be written as:

        The truth table for equation (i) can be drawn as:

        A

        Y

        0

        1

        1

        0

        This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

        14.17. You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

        Answer

        In both the given circuits, A and B are the inputs and Y is the output.

        (a) The output of the left NAND gate will be, as shown in the following figure.

        Hence, the output of the combination of the two NAND gates is given as:

        Hence, this circuit functions as an AND gate.

        (b) is the output of the upper left of the NAND gate and is the output of the lower half of the NAND gate, as shown in the following figure.

        Hence, the output of the combination of the NAND gates will be given as:

        Hence, this circuit functions as an OR gate.

        14.18. Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

        (Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

        Answer

        A and B are the inputs of the given circuit. The output of the first NOR gate is. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.

        Hence, the output of the combination is given as:

        The truth table for this operation is given as:

        A

        B

        Y (=A + B)

        0

        0

        0

        0

        1

        1

        1

        0

        1

        1

        1

        1

        This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

         

        14.19. Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

        Answer

        (a) A acts as the two inputs of the NOR gate and Y is the output, as shown in the following figure. Hence, the output of the circuit is.

        The truth table for the same is given as:

        A

        Y

        0

        1

        1

        0

        This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

        (b) A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates areas shown in the following figure.

        are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:

        The truth table for the same can be written as:

        A

        B

        Y (=A⋅B)

        0

        0

        0

        0

        1

        0

        1

        0

        0

        1

        1

        1

        This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

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