
Chapter 45 – Semiconductors and Semiconductor Devices
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Question 1:
How many 1s energy states are present in one mole of sodium vapour? Are they all filled in normal conditions? How many 3s energy states are present in one mole of sodium vapour? Are they all filled in normal conditions?
Answer:
For sodium, the atomic number is 11. The electronic configuration of sodium is 1s^{2} 2s^{2} 2p^{6} 3s^{1}.
One sodium atom has 11 electrons. Thus, if the sodium crystals consist of N atoms, the total number of electrons will be 11 N. We know that for each atom, there are two states in the energy level 1s. Thus, the sodium crystal will have 2 N states for 1s energy level. Similarly, the number of states in 3s energy level will also be 2 N. 1s state is filled under normal condition. But the 3s state has only one electron per sodium atom, so the 3s band will be halffilled.
Question 2:
There are energy bands in a solid. Do we have really continuous energy variation in a band ro do we have very closely spaced but still discrete energy levels?
Answer:
A solid consists of a combination of closely spaced energy levels. These energy levels are discrete but they have very small energy gap between two consecutive levels so they are reffered as band.However, the energy levels in the band are discrete.
Question 3:
The conduction band of a solid is partially filled at 0 K. Will it be a conductor, a semiconductor or an insulator?
Answer:
It will be a conductor. As the elements having partially filled conduction band belong to the category of elements whose outermost subshell consists of an odd number of electrons, they are good conductors of electricity. When an electric field is applied, electrons in the partially filled band gain energy and start drifting. So, the conductor will conduct even at 0 K. A semiconductor behaves like an insulator at 0 K and an insulator conducts poorly only at very high temperatures.
As the given material has free electrons to conduct even at 0 K, it is a conductor.
Question 4:
In semiconductors, thermal collisions are responsible for taking a valence electron to the conduction band. Why does the number of conduction electrons not go on increasing with time as thermal collisions continuously take place?
Answer:
An electron jumps from the valence band to the conduction band only when it has gained sufficient energy. The thermal collisions sometimes do not provide sufficient energy to the electron to jump. Also, energy is lost in the form of heat because of the collision of the carriers with other charge carriers and atoms. Because of all these losses, only few electrons are left with sufficient energy to jump from the valence band to the conduction band. So, the population of electron in the conduction band does not keep on increasing with time.
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Question 5:
When an electron goes from the valence band to the conduction band in silicon, its energy is increased by 1.1 eV. The average energy exchanged in a thermal collision is of the order of kT which is only 0.026 eV at room temperature. How is a thermal collision able to take some to the electrons from the valence band to the conduction band?
Answer:
Fermi level: it is the energy level occupied by the highest energy electron.
In an extrinsic semiconductor for example in ntype semiconductor, fermi level lies close to the conduction band so it needs a very small amount of energy to excite the electron from fermi level to conduction band. This energy is comparable to the thermal excitation energy. So even at room temperature,these semiconductors can conduct.For a ptype semiconductor, fermi level lies close to valence bane because here conduction takes place majorly via holes.So by the thermal excitation,a bond is broken and an electron hole pair is created.Out of this,hole comes to the valence band for conduction or equivalently an electron goes to the conduction band. In an intrinsic semiconductor, no impurity is doped so fermi level lies at the centre of band gap. Here only few electrons get sufficient energy via repeated thermal collisions to jump from the fermi level to the conduction band.Hence the conductivity of intrinsic semiconductor is less as compared to extrinsic semiconductor.
Question 6:
What is the resistance of an intrinsic semiconductor at 0 K?
Answer:
At 0 K, the valence band is full and the conduction band is empty. As no electron is available for conduction in an intrinsic semiconductor, the intrinsic semiconductor at 0 K acts as an insulator and hence offers infinite resistance.
Question 7:
We have valence electrons and conduction electrons in a semiconductor. Do we also have ‘valence holes’ and ‘conduction holes’?
Answer:
Holes do not exist in reality. They exist only virtually. When an electron jumps from the valence band to the conduction band, a vacancy is created at the place from where the electron had jumped. This vacancy is called a hole. So, a valence or conduction hole is a virtual concept only.
Question 8:
When a ptype impurity is doped in a semiconductor, a large number of holes are created, This does not make the semiconductor charged. But when holes diffuse from the pside to the nside in a pn junction, the nside gets positively charged. Explain.
Answer:
A ptype semiconductor is formed by doping a group 13 element with group 14 element (Si or Ge). As the group 13 element has only 3 electrons in its valence shell and the group 14 element has 4 electrons in its valence shell, when the group 13 element, say, Al, replaces one Si in the silicon crystal, only 3 covalent bonds are formed by it. And the fourth covalent bond is left in need of one electron. So, it creates a hole. Since the atom as a whole is electriclly neutral, the ptype semiconductor is also neutral.
In a pâ€’n junction, when the diffusion of holes takes place across the junction because of the difference in the concentration of charge carriers from p to n sides, these holes neutralise some of the electrons on the n side. So, the atom attached with that electron becomes one electron deficient and hence positively charged. This makes the n side of the pâ€’n junction positively charged and the p side of the pâ€’n junction negatively charged.
Question 9:
The drift current in a reversebiased pn junction is increased in magnitude if the temperature of the junction is increased. Explain this on the basis of creation of holeelectron pairs.
Answer:
When the temperature of a reversebiassed pâ€’n junction is increased, the breaking of bonds takes place because of the increase in the thermal energy of the charge carriers. Drift current is due to the flow of the minority carriers across the junction. So, when a pâ€’n junction is reverse biassed, the applied voltage supports the flow of minority charge carriers across the junction. Thus, the drift current increases with increase in temperature in a reversebiassed pâ€’n junction.
Question 10:
An ideal diode should pass a current freely in one direction and should stop it completely in the opposite direction. Which is closer to idealvacuum diode or a pn junction diode?
Answer:
It should be an ideal vacuum diode. When a pn junction diode is reverse biassed then a small current called reverse current flows across the diode.As the the pâ€’n junction diode allows some current in reverse biassed condition also so the given diode can not be a pn junction diode.
Question 11:
Consider an amplifier circuit using a transistor. The output power is several times greater than the input power. Where does the extra power come from?
Answer:
The amplifier takes this energy from the power supply. Amplifiers take the energy from the power supply and control the output to match with the input signal but with greater amplitude.
Question 1:
Electric conduction in a semiconductor takes place due to
(a) electrons only
(b) holes only
(c) both electrons and holes
(d) neither electrons nor holes.
Answer:
(c) both electrons and holes
A hole is created in a semiconductor when a valence electron moves to the conduction band. When potential difference is applied across the semiconductor, the electron drifts opposite to the electric field applied, while the hole moves along the electric field. Therefore, electric conduction takes place in a semiconductor because of both electrons and holes.
Question 2:
An electric field is applied to a semiconductor. Let the number of charge carries be n and the average drift speed by v. If the temperature is increased,
(a) both n and v will increase
(b) n will increase but v will decrease
(c) v will increase but n will decrease
(d) both n and v will decrease.
Answer:
(b) n will increase but v will decrease
As we increase the temperature, additional electronâ€’hole pairs are created in a semiconductor. As a result, the number of charge carriers increases.
Now, drift velocity
vdis given by
vd=eEτmAs the temperature increases, the relaxation time of charge carriers
τdecreases. As a result, v_{d} decreases.
Question 3:
Let n_{p} and n_{e} be the number of holes and conduction electrons in an intrinsic semiconductor.
(a) n_{p} > n_{e}
(b) n_{p} = n_{e}
(c) n_{p} < n_{e}
(d) n_{p} ≠ n_{e}
Answer:
(b) n_{p} = n_{e}
As the intrinsic semiconductor is free from all impurities, the number of electrons is equal to the number of holes.
Question 4:
Let n_{p} and n_{e} be the numbers of holes and conduction electrons in an extrinsic semiconductor.
(a) n_{p} > n_{e}
(b) n_{p} = n_{e}
(c) n_{p} < n_{e}
(d) n_{p} ≠ n_{e}
Answer:
(d) n_{p} ≠ n_{e}
Extrinsic semiconductors are formed by doping either an ntype material or a ptype material with a pure semiconductor, Thus, it may have more number of holes (if the doping material is of p type) or more number of electrons (if the doping material is of n type).
Question 5:
A ptype semiconductor is
(a) positively charged
(b) negatively charged
(c) uncharged
(d) uncharged at O K but charged at higher temperatures.
Answer:
(c) uncharged
A ptype semiconductor is formed by doping a pure semiconductor with a ptype material. As impurity atoms take the position of the germanium atom in a germanium crystal, three electrons of a ptype material form covalent bonds by sharing electrons with three neighbouring germanium atoms. However, the fourth covalent bond is left incomplete, with a want of one electron. This creates a hole. As the atom as a whole is neutral, the ptype material is also neutral.
Question 6:
When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor
(a) increases
(b) decreases
(c) remains the same
(d) become zero.
Answer:
(a) increases
When an impurity (either a ptype atom or an ntype atom) is doped into an intrinsic semiconductor, it increases the number of charge carriers in the intrinsic semiconductor. As conductivity is directly related to the number of charge carriers, the conductivity of a semiconductor increases with doping.
Question 7:
If the two ends of a pn junction are joined by a wire,
(a) there will not be a steady current in the circuit
(b) there will be a steady current from the nside to the pside
(c) there will a steady current from the pside to the nside
(d) there may or may not be a current depending upon the resistance of the connecting wire.
Answer:
(a) there will not be a steady current in the circuit
In a pâ€’n junction, current flows only if it is connected to the battery. If two ends of a pâ€’n junction are joined by a wire, then there will be diffusion and drift currents in the circuit and they will cancel each other. Hence, no current will flow in the circuit.
Question 8:
The drift current in a pn junction is
(a) from the nside to the pside
(b) from the pside to the nside
(c) from the pside to the side if the junction is forwardbiased and the opposite direction if it is reversebiased
(d) from the pside to the nside if the junction is forwardbaised and in the opposite direction if it is reversebiased.
Answer:
(a) from the nside to the pside
After the diffusion of majority charge carriers across a pâ€’n junction, an electric field is set up because of the accumulation of immobile ions at the junction. These further oppose the motion of majority charge carriers across the junction. As a result, electrons from the p region start moving to the n region and holes from the n region start moving to the p region. This constitutes the drift current. As the direction of the current is opposite to the direction of the motion of the electrons, the direction of the drift current is from the n side to the p side.
In forward biasing, there is no movement of electrons from the p region to the n region and of holes from the n region to the p region. Hence, there is not drift current.
Question 9:
The diffusion current in a pn junction is
(a) from the nside ot the pside
(b) from the pside to the nside
(c) from the nside to the pside if the junction is forwardbiased and in the opposite direction if it is reversebiased
(d) from the pside to the nside if the junction is forward and in the opposite direction if it is reversebiased.
Answer:
(b) from the pside to the nside
When a pâ€’n junction is formed then because of the difference in the concentration of charge carriers in the two regions, electrons from the n region move to the p region and holes from the p region move to the n region. Since the direction of the current is always opposite to the motion of electron, the direction of the current is from the p side to the n side.
Similarly, when the junction is forward biassed, the positive terminal of the battery is connected to the p side of the pâ€’n junction and the negative terminal of the battery is connected to the n side of the pâ€’n junction. As a result, electrons in the n side of the pâ€’n junction are repelled by the negative terminal of the battery and they move to the p side, where the positive terminal of the battery attracts them. Similarly, holes from the p side of the pâ€’n junction are repelled by the positive terminal of the battery and they move to the n side, where the negative terminal of the battery attracts them. Thus, they give diffusion current from the p side to the n side across the pâ€’n junction.
In reverse biassing, there is no flow of majority carriers across the junction; hence, there is not diffusion current. Here, the flow of majority carriers is opposed by the applied voltage.
Question 10:
Diffusion current in a pn junction is greater than the drift current in magnitude
(a) if the junction is forwardbiased
(b) if the junction is reversebiased
(c) if the junction is unbiased
(d) in no case.
Answer:
(a) if the junction is forwardbiassed
In the forward biassing of a p−n junction, the positive terminal of the battery is connected to the p side of the p−n junction and the negative terminal of the battery is connected to the n side of the p−n junction. As a result, electrons in the n side of the p−n junction are repelled by the negative terminal of the battery and move to the p side, where the positive terminal of the battery attracts the electrons. Similarly, holes from the p side of the p−n junction are repelled by the positive terminal of the battery and move to the n side, where the negative terminal of the battery attracts the holes. Thus, they give diffusion current across the p−n junction.
In case of reverse biassing, no conduction takes place across the junction because of the diffusion of majority carriers. Hence, there is no diffusion current.
If the junction is unbiased, then diffusion current is initially maximum. But at equilibrium, diffusion current becomes equal to drift current.
Question 11:
Two identical pn junction may be connected in series with a battery in three ways. The potential difference across the two pn junctions are equal in
(a) circuit 1 and circuit 2
(b) circuit 2 and circuit 3
(c) circuit 3 and circuit 1
(d) circuit 1 only.
Figure
Answer:
(b) circuit 2 and circuit 3
In circuit 1, one diode is forward biassed and the other diode is reverse biassed.
The forwardbiassed diode offers zero resistance (ideally) to the current flow, so it can be replaced by a short circuit. The voltage drop across the first diode will be zero. The second diode is reverse biassed, so it can be replaced by an open circuit; hence, the voltage drop across this diode will be maximum.
In circuit 2, both the diodes are forward biassed, so they can be replaced by short circuits; hence, the voltage drop across both of them will be minimum and equal.
In circuit 3, both the diodes are reverse biassed, so both can be replaced by open circuits; hence, the voltage drop across both of them will be maximum and equal.
Question 12:
Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t = 0 as shown in figure. The charges on the capacitors at a time t = CR are, respectively,
(a) VC, VC
(b) VC/e, VC
(c) VC, VC/e
(d) VC/e, VC/e.
Answer:
In circuit (a), the diode is forward biassed. So, it offers negligible resistance to the flow of current and can thus be replaced by a short circuit. Now, the capacitor charge will leak through the resistance and decay exponentially with time.
Capacitor charge =
VCeIn circuit (b), the diode is reverse biassed. So, it offers infinite resistance to the current flow and can thus be replaced by an open circuit. As the circuit is open now, no current can flow across the resistance. So, the charge in the capacitor cannot leak through the resistor.
Capacitor charge = VC
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Question 13:
A hole diffuses from the pside to the nside in a pn junction. This means that
(a) a bond is broken on the nside and the electron freed from the bond jumps to the conduction band
(b) a conduction electron on the pside jumps to a broken bond to complete it
(c) a bond is broken on the nside and the electron freed from the bond jumps to a broken bond on the pside to complete it
(d) a bond is broken on the pside and the electron freed from the bond jumps to a broken bond on the nside to complete it.
Answer:
(c) a bond is broken on the nside and the electron freed from the bond jumps to a broken bond on the pside to complete it
A hole diffuses from the p side to the n side in a p−n junction; that is, an electron moves from the n side to the p side. This implies that a bond is broken on the n side. As the electron travels towards the p side, which is rich in holes, it combines with a hole. A hole is created because of the deficiency of one electron. So, when an electron combines with a hole, it completes that bond.
Question 14:
In a transistor,
(a) the emitter has the least concentration of impurity
(b) the collector has the least concentration of impurity
(c) the base has the least concentration of impurity
(d) all the three regions have equal concentrations of impurity.
Answer:
(c) the base has the least concentration of impurity
When the emitter of a transistor is forward biassed, the majority carriers move from the emitter to the collector through the base. As the base is thin and lightly doped, only small amount of a combination of electrons and holes takes place, leading to weak base current. This makes the collector current equal to the emitter current.
If we make any other choice for least concentration of impurity, we will have a low value of collector current. Thus, the purpose of a transistor will not be fulfilled.
Question 15:
An incomplete sentence about transistors is given below:
The emitter−….. junction is __ and the collector−….. junction is __. The appropriate words for the dotted empty positions are, respectively,
(a) ‘collector’ and ‘base’
(b) ‘base’ and ’emitter’
(c) ‘collector’ and ’emitter’
(d) ‘base’ and ‘base’.
Answer:
(d) ‘base’ and ‘base’
In transistors, the emitter−base junction is forward biassed and the collector−base junction is reverse biassed. This is done to make majority carriers cross the emitter−base junction; this helps the electric current to flow through the transistor.
Question 1:
In a semiconductor,
(a) there are no free electrons at 0 K
(b) there are no free electrons at any temperature
(c) the number of free electrons increases with temperature
(d) the number of free electrons is less than that in a conductor.
Answer:
(a) there are no free electrons at 0 K
(c) the number of free electrons increases with temperature
(d) the number of free electrons is less than that in a conductor
In semiconductors, the valence band is full at 0 K, but the conduction band is empty. So, no free electron is available for conduction at 0 K.
As the temperature increases, covalent bonds that provide free charge carriers for conduction in a semiconductor break.
As the conduction band in metals is already partially filled at 0 K, many free electrons below the Fermi level acquire energy from an external source or temperature, jump to the conduction band and start behaving like free electrons. Hence, metals contain more free electrons than semiconductors.
Question 2:
In a pn junction with open ends,
(a) there is no systematic motion of charge carries
(b) holes and conduction electrons systematically go from the pside to nside and from the nside to pside respectively
(c) there is no net charge transfer between the two sides
(d) there is a constant electric field near the junction.
Answer:
(b) holes and conduction electrons systematically go from the pside to nside and from the nside to pside, respectively
(c) there is no net charge transfer between the two sides
(d) there is a constant electric field near the junction
Because of the difference in the concentration of charge carriers in the p−n junction, holes from the p side move to the n side and electrons from the n side move to the p side. This motion of charge carriers gives rise to diffusion current.
Because of this, a negative space charge region is formed in the p region and a positive space region is formed in the n region. This sets up an electric field across the junction. Thus, there is a constant electric field near the junction.
This electric field further opposes the diffusion of majority charge carriers across the junction. As a result, an electron from the p region starts moving to the n region and a hole from the n region starts moving to the p region. This sets up drift current. Thus, there is a systematic flow of charge carriers across the junction. Also, there is no net charge transfer between the two sides.
Question 3:
In a pn junction,
(a) new holes and conduction electrons are produced continuously throughout the material
(b) new holes and conduction electrons are produced continuously throughout the material except in the depletion region
(c) holes and conduction electrons recombine continuously throughout the material
(d) holes and conduction electrons recombine continuously throughout the material except in the depletion region.
Answer:
(a) new holes and conduction electrons are produced continuously throughout the material
(d) holes and conduction electrons recombine continuously throughout the material except in the depletion region
In a pâ€’n junction diode, diffusion current flows because of the diffusion of holes from the p side to the n side and of electrons from the n side to the p side. The current flowing in the diode due to the diffusion of charge carriers across the junction is called the diffusion current. The current flowing in the diode due to the movement of minority carriers across the junction due to their thermal energy is called the drift current. In an unbiased diode, the net current flowing across the junction is zero due to the cancellation of the drift current by the diffusion current. For the flow of diffusion and drift currents, holes and electrons are produced continuously throughout the material. When a hole crosses the junction, it combines with an electron on the n side. As the depletion region is devoid of free charge carriers, this recombination never takes place inside the depletion region.
Question 4:
The impurity atoms with which pure silicon may be doped to make it a ptype semiconductor are those of
(a) phosphorus
(b) boron
(c) antimony
(d) aluminium.
Answer:
(b) boron
(d) aluminium
A ptype semiconductor is formed by doping an intrinsic semiconductor with a trivalent atom (atom having valency 3). As phosphorous and boron have three valence electrons, they can be doped with silicon to make a ptype semiconductor.
Question 5:
The electrical conductivity of pure germanium can be increased by
(a) increasing the temperature
(b) doping acceptor impurities
(c) doping donor impurities
(d) irradiating ultraviolet light on it.
Answer:
(a) increasing the temperature
(b) doping acceptor impurities
(c) doping donor impurities
(d) irradiating ultraviolet light on it
We know that the conductivity of any semiconductor can be increased by increasing the number of charge carriers. All the given methods are effective in increasing the number of free charge carriers. Hence, all options are correct.
Question 6:
A semiconducting device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. the device may be
(a) an intrinsic semiconductor
(b) a ptype semiconductor
(c) an ntype semiconductor
(d) a pn junction.
Answer:
(d) a p−n junction
As a p−n junction allows the flow of current in forward bias and stops the current in reverse bias (almost negligible reverse leakage current flows in the reversebiassed p−n junction), the device should be a p−n junction. Other options are examples of semiconductors that allow moderate current to flow and that do not have any effect of changing the polarity of the battery.
Question 7:
A semiconductor is doped with a donor impurity.
(a) The hole concentration increases.
(b) The hole concentration decreases.
(c) The electron concentration increases.
(d) The electron concentration decreases.
Answer:
(c) The electron concentration increases.
When a semiconductor is doped with a donor type such as arsenic or phosphorous, which has five valence electrons, the donor atom replaces the Si or Ge atom. As a result, four out of the five electrons of the donor atom form a covalent bond by sharing an electron with four atoms of silicon. However, the fifth electron is free to move. Also, due to the breaking up of covalent bonds at room temperature, equal number of electrons and holes are produced. Thus, the total number of holes in the ntype semiconductor is less compared to the number of free electrons.
Question 8:
Let i_{E}, i_{C} and i_{B} represent the emitter current, the collector current and the base current respectively in a transistor. Then
(a) i_{C} is slightly smaller than i_{E}
(b) i_{C} is slightly greater than i_{E}
(c) i_{B} is much smaller than i_{E}
(d) i_{B} is much greater than i_{E}.
Answer:
(a) i_{C} is slightly smaller than i_{E}
(c) i_{B} is much smaller than i_{E}
The highlighted parts could not be edited, as the meaning could not be understood. Also, please check the last line for logical accuracy. Only one option is given as correct, while the solution gives two correct options.
We know that in the transistor base is slightly doped, therefore when the majority carriers due to forward biasing of emitter base junction, feel the repulsive force from the battery and pass over to the base region. This gives the emitter current i_{E}.
As the base is thin and lightly doped, only few majority carriers of the emitter are neutralised at the base. This gives the base current. Hence, base current
iBis low.
The remaining majority carriers of the emitter pass to the collector and give collector current i_{C}.
Thus, we get the relation given below:
i_{E} = i_{B} + i_{C}
Thus, because of the base, current i_{C} is slightly smaller than i_{E} .
Hence, option (a) and (c) are correct.
Question 9:
In a normal operation of a transistor,
(a) the base−emitter junction is forwardbaised
(b) the base−collector junction is forwardbaised
(c) the base−emitter junction is reversebaised
(d) the base−collector junction is reversebaised.
Answer:
(a) the base−emitter junction is forwardbiassed
(d) the base−collector junction is reversebiassed
In the normal operation of a transistor, the base−emitter junction is forward biassed and the base−collector junction is reverse biassed. This is done so that the conduction of majority carriers can take place across the emitter−base junction and the free electrons can reach the collector to give the output current.
Question 10:
An AND gate can be prepared by repetitive use of
(a) NOT gate
(b) OR gate
(c) NAND gate
(d) NOR gate.
Answer:
(c) NAND gate
(d) NOR gate
i)
ii)
Page No 419:
Question 1:
Calculate the number of states per cubic metre of sodium in 3s band. The density of sodium is 1013 kgm^{−3}. How many of them are empty?
Answer:
Density of sodium, d = 1013
kg/m3Volume, V = 1
m3Mass of sodium, m = d
×V = 1013 × 1 = 1013 kg
Molecular mass of sodium, M = 23
We know that 23 g sodium contains 6 atoms, so the number of atoms in 1023 kg sodium will be
1013×103×6×102323=1013×623×1026=264.26×1026(a) As the number of maximum possible electrons that can occupy the 3s band is 2, the total number of states in the 3s band will be
N=2×264.26×1026=528.52×1026≈5.3×1028(b) As the atomic number of sodium is 11, its electronic configuration is
1s2,2s2,2p6,3s1.
This implies that the 3s band is halffilled in case of sodium, so the total number of unoccupied states is 2.65 × 10^{28}.
Question 2:
In a pure semiconductor, the number of conduction election 6 × 10^{19} per cubic metre. How many holes are there in a sample of size 1 cm × 1 mm?
Answer:
Pure semiconductors are intrinsic semiconductors or semiconductors without any doping.
We know that for pure semiconductors, the number of conduction electrons is equal to the number of holes.
Number of electrons in volume 1 m^{3} =
6×1019Number of holes in volume 1 m^{3} =
6×1019Given volume:
V = 1 cm × 1 cm × 1 mm
⇒V = 1 × 10^{−2} × 1 × 10^{−2} × 10^{−3}
⇒V = 10^{−7} m^{3}
Now,
Number of holes in volume
107 m3:
N = 6 × 10^{19} × 10^{−7} = 6 × 10^{12}
Question 3:
Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.
Answer:
Given:
Band gap between the conduction band and the valence band, E = 0.23 eV
Boltzmann’s constant, k = 1.38 × 10^{−23} J/K
We need to find the temperature at which thermal energy kT becomes equal to the band gap of indium antimonide.
∴ kT = E
⇒1.38×1023×T=0.23×1.6×1019⇒T=0.23×1.6×10191.38×1023⇒T=0.23×1.6×1041.38⇒T=0.2666×104≈2670 K
Question 4:
The band gap for silicon is 1.1 eV. (a) Find the ratio of the band gap to kT for silicon at room temperature 300 K. (b) At what temperature does this ratio become one tents of the value at 300 K? (Silicon will not retain its structure at these high temperatures.)
Answer:
Given:
Band gap of silicon, E = 1.1 eV
Temperature, T = 300 K
Boltzmann’s constant, k =
8.62×105 eV/K
(a) We need to find out the ratio of the band gap to kT.
Ratio
=1.1kT
=1.18.62 ×105×3×102=42.53=43(b) The new ratio is
110th of the earlier ratio.
i.e. New ratio = 4.253
We know,
Ratio = (Band gap)/(kT)
⇒4.253=1.18.62×105×T⇒T=1.18.62×105×4.253⇒T=3000.4 K≈3000 K
Question 5:
When a semiconducting material is doped with an impurity, new acceptor levels are created. In a particular thermal collision, a valence electron receives an energy equal to 2kT and just reaches one of the acceptor levels. Assuming that the energy of the electron was at the top edge of the valence band and that the temperature T is equal to 300 K, find the energy of the acceptor levels above the valence band.
Answer:
Before the thermal excitation, the electron was located at the top of the valence band.
After the thermal excitation, the electron was present in the acceptor level.
This implies that the energy gap between the valence band and the acceptor level is equal to the energy absorbed by the electron in thermal excitation.
Thus,
2kT = Energy gap between the acceptor level and the valence band
⇒E=2×1.38×1023×300 ⇒E=(2×1.38×3)×1021 J⇒E=6×1.381.6×10211019 eV⇒E=6×1.381.6×102 eV⇒E=5.175×102 eV⇒E=51.75 meV
Question 6:
The band gap between the valence and the conduction bands in zinc oxide (ZnO) is 3.2 eV. Suppose an electron in the conduction band combines with a hole in the valence band and the excess energy is released in the form of electromagnetic radiation. Find the maximum wavelength that can be emitted in this process.
Answer:
Given:
Band gap = 3.2 eV
As the electron in the conduction band combines with the hole in the valence band, the minimum energy band gap (because maximum energy is released) through which the electron has to jump will be equal to the band gap of the material.
This implies that the maximum energy released in this process will be equal to the band gap of the material.
Here,E = 3.2 eVThus,⇒3.2 eV=1242 eVnmλ⇒λ=388.1 nm
Question 7:
Suppose the energy liberated in the recombination of a holeelectron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap?
Answer:
Given:
Wavelength,
λ = 820 nmThe minimum energy released in the recombination of a conduction band electron with a valence band hole is equal to the band gap of the material.
Band gap,
E=hcλ⇒E=1240820eVnmnm⇒E=1.5 eV
Question 8:
Find the maximum wavelength of electromagnetic radiation which can create a holeelectron pair in germanium. The band gap in germanium is 0.65 eV.
Answer:
Given:
Band gap of germanium, E = 0.65 eV
Wavelength of the incident radiation, λ = ?
For the electronâ€’hole pair creation, the energy of the incident radiation should be at least equal to the band gap of the material.
So,
E=hcλ⇒λ=hcE=1242 eVnm0.65 eV⇒λ=1910.7×109 m⇒λ≈1.9×106 m
Question 9:
In a photo diode, the conductive increases when the material is exposed to light. It is found that the conductivity changes only if the wavelength is less than 620 nm. What is the band gap?
Answer:
Conductivity of any material increases when the number of free charge carriers in the material increases. When a photo diode is exposed to light, additional electron hole pairs are created in the diode; thus, its conductivity increases. So to change the conductivity of a photo diode, the minimum energy of the incident radiation should be equal to the band gap of the material.
In other words,
Band gap = Energy of the incident radiation
⇒E=hcλ⇒E=1242 eVnm620 nm = 2.0 eV
Question 10:
Let ΔE denote the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to e^{−}^{Δ}^{E}^{/2kT}. Find the ratio of the concentration of conduction electrons in diamond to the in silicon at room temperature 300 K. ΔE for silicon is 1.1 eV and for diamond is 6.1 eV. How many conduction electrons are likely to be in one cubic metre of diamond?
Answer:
Given:
Number of electrons in the conduction band, n = e^{−ΔE}^{/2kT}
Band gap of diamond, ΔE_{1} = 6 eV
Band gap of silicon, ΔE_{2} = 1.1 eV
Now,
ndiamond=eΔE12kTnsilicon=eΔE22kT∴ndiamondnsilicon=e12kT(∆E1∆E2)⇒ndiamondnsilicon=e12×8.62×105×300(6.01.1)⇒ndiamondnsilicon=7.15×1042Because of more band gap, the conduction electrons per cubic metre in diamond are very less as compared to those in silicon, or we can simply say that they are almost zero.
Question 11:
The conductivity of a pure semiconductor is roughly proportional to T^{3}^{/2} e^{−}^{Δ}^{E}^{/2kT} where ΔE is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4 K to 300 K?
Answer:
Here,
Conductivity at temperature
T1=
σ1Conductivity at temperature
T2=
σ2Given:
T1= 4 K
T2 = 300 K
Variation in conductivity with respect to the temperature and band gap of the material is given by
σ=T3/2eΔE/2kT∴σ2σ1=T2T13/2eΔE2/2kT2eΔE1/2kT1⇒σ2σ1=30043/2e0.67/(2×8.62×105×300)e0.74/(2×8.62×105×4)⇒σ2σ1=10463
Question 12:
Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7 × 10^{15} holes per cubic metre. Density of silicon 5 × 10^{28} atoms per cubic metre.
Answer:
Initially, the total number of charge carriers per cubic metre is given by
n_{i} = 2 × 7 × 10^{15}
⇒n_{i} = 14 × 10^{15}
Finally, the total number of charge carriers per cubic metre is given by
n_{f} = 14 × 10^{17}/m^{3}
We know that the product of the concentrations of holes and conduction electrons remains almost the same.
Let x be the number of holes.
Thus,
(7×1015)×(7×1015)=x×(14×1017x)⇒14x×1017x2=49×1030⇒x214x×101749×1030=0⇒x=14×1017±(14)2×1034+4×49×10302⇒x=14×1017±(14)2×1034+4×49×10302⇒x=28.00072×1017=14.00035×1017This is equal to the increased number of holes or the number of atoms of boron added.
Number of atoms of boron added =
(14.00035×10177×1015)=1386.035×1015Now, 1386.035 × 10^{15} atoms are added per 5 × 10^{28} atoms of Si in 1 m^{3}.
Therefore, 1 atom of boron is added per
5×10281386.035×1015atoms of Si in 1 m^{3}.
Proportion of boron impurity is
3.607×103×1013 = 3.607×1010.
Question 13:
The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is 6 × 10^{19} per cubic metref conduction electrons increases to 2 × 10^{23} per cubic metre. Find the concentration of the holes in the doped germanium.. When some phosphorus impurity is doped into a germanium sample, the concentration o
Answer:
We know that for an intrinsic semiconductor, the concentration of electrons is equal to the concentration of holes.
The product of electron hole pair concentration always remains constant.
i.e. Number of holes
×Number of conduction electrons = Constant
Initially, the number of conduction electron per cubic metre = 6 × 10^{19}
We know,
Number of holes per cubic metre = Number of electrons per cubic metre
∴ Number of holes per cubic metre = 6 ×10^{19}
After doping,
Number of conduction electrons per cubic metre = 2 × 10^{23}
Now,
Let the number of holes per cubic metre be x.
As the product of electron hole pair concentration always remains constant,
(6×1019)(6×1019)=(2×1023)x⇒x=6×6×1019+192×1023 ⇒x=18×1015=1.8×1016
Question 14:
The conductivity of an intrinsic semiconductor depends on temperature as σ = σ_{0} e^{−}^{Δ}^{E}^{/}^{2}^{kT}, where σ_{0} is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T = 300 K. Assume that the gap for germanium is 0.650 eV and remains constant as the temperature is increased.
Answer:
Let the conductivity at temperature T_{1} be
σ1and the conductivity at temperature T be
σ2.
Given:
T1 = 300 KBand gap, E = 0.650 eV
Now,
According to the question,
σ=σ0eΔE2KT
σ2 = 2σ1
⇒σ0eΔE2kT = 2×σ0eΔE2×k×T1⇒σ0eΔE2kT= 2×σ0eΔE2×k×300⇒e0.6502×8.62×105×T = 2×e0.6502×8.62×105×300⇒e0.6502×8.62×105×T = 6.96561×106On taking natural natural log on both sides, we get
0.6502×8.62×105×T = 11.874525⇒1T = 11.874525×2×8.62×1050.65⇒T = 317.51178 ≈ 318K
Question 15:
A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band. Assume that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also if kT is more than twice the gap, the upper levels have maximum population. The temperature of the semiconductor is increased from 0 K. The concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.
Answer:
Given:
Band gap = 1 eV
After doping,
Position of acceptor levels = 1 meV above the valence band
Net band gap after doping = (1 − 10^{−3}) eV = 0.999 eV
According to the question,
Any transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap.
⇒kT1=0.99950⇒T1=0.99950×8.62×105⇒T1=231.78≈232.8 K
T1is the temperature below which no transition is possible.
If kT is more than twice the gap, the upper levels have maximum population; that is, no more transitions are possible.
For the maximum limit,
KT2=2×103⇒T2=2×1038.62×105⇒T2=28.62×102=23.2 K
T2is the temperature above which no transition is possible.
∴ Temperature range = 23.2−231.8
Question 16:
In a p.n junction, the depletion region is 400 nm wide and an electric field of 5 × 10^{5} V m^{−1} exists in it. (a) Find the height of the potential barrier. (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the nside to the pside?
Answer:
Let:
Depletion region width, d = 400 nm = 4 × 10^{−7} m
Electric field, E = 5 × 10^{5} Vm^{−1}
(a) Let the potential barrier be V.
The relation between the potential and the electric field is given by
V = Ed
⇒V = E × d = 5 × d
⇒V = 5 × 10^{5} × 4 × 10^{−7} = 0.2 V
(b) To find: Kinetic energy required
Energy of any electron accelerated through a potential of V = eV
Also, the minimum energy of the electron should be equal to the band gap of the material.
∴ Potential barrier × e = 0.2 eV (e = Charge of the electron)
Question 17:
The potential barrier existing across an unbiased pn junction is 0.2 volt. What minimum kinetic energy a hole should have to diffuse from the pside to the nside if (a) the junction is unbiased, (b) the junction is forwardbiased at 0.1 volt and (c) the junction is reversebiased at 0.1 volt?
Answer:
Potential barrier = 0.2 V
(a) The minimum kinetic energy of the hole should be equal to the band gap of the material.
Band gap = eV
KE = Potential difference × e = 0.2 eV
(b) In forward biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition – Applied voltage
⇒ KE + Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 − 0.1 = 0.1 eV
(c) In reverse biassing,
Kinetic energy = Effective potential of the barrier
∴ Kinetic energy = Potential under unbiased condition + Applied voltage
⇒ KE − Ve = 0.2 eV
Here, V is the applied voltage.
⇒ KE = 0.2 + 0.1 = 0.3 eV
Question 18:
In a pn junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the pside and (b) from the nside.
Answer:
Given:
Potential barrier, d = 250 meV
Initially,
Kinetic energy of a hole = 330 meV
We know that the kinetic energy of a hole decreases when the junction is forward biassed (because of the energy loss in crossing the junction).
Also, the kinetic energy of a hole increases when the junction is reverse biassed (because reverse bias voltage pushes the hole on the nside towards the junction) in the given the pn junction diode.
(a) The kinetic energy of a hole decreases under forward bias.
∴ Final kinetic energy = (300 − 250) meV
= 50 meV
(b) The kinetic energy of a hole increases under reverse bias.
∴ Final kinetic energy = (300 + 250) meV
= 550 meV
Question 19:
When a pn junction is reversebiased, the current becomes almost constant at 25 µA. When it is forwardbiased at 200 mV, a current of 75 µA is obtained. Find the magnitude of diffusion current when the diode is
(a) unbiased,
(b) reversebiased at 200 mV and
(c) forwardbiased at 200 mV.
Answer:
Given:
Drift current (current under reverse bias), i_{1} = 25 µA
Forward bias voltage, V = 200 mV
Net current under forward bias, i_{2} = 75 µA
(a) When the pâ€’n junction is in unbiased condition, no net current flows across the junction.
i.e. Drift current = Diffusion current
∴ Diffusion current = 25 µA
(b) Under reverse bias, the built in the potential and applied voltage opposes the motion of the majority carriers across the junction.
Thus, the diffusion current becomes zero.
(c) Under forward bias, the voltage supports the motion of majority carriers across the junction.
Let the actual current be x.
So,
(x − Drift current) = Forwardbiassed current
⇒x25 μA=75 μA⇒x=(75+25) μA⇒x=100 μA
Page No 420:
Question 20:
The drift current in a pn junction is 20.0 µA. Estimate the number of electrons crossing a cross section per second in the depletion region.
Answer:
Given:
Drift current, i_{d} = 20 µA = 20 × 10^{−6} A
Both holes and electrons are moving and contributing to the current flow.
We know that current is the rate of the flow of charge.
Thus, we need to find the number of electrons crossing unit area per second.
Now,
t = 1 s
id =QT∵T=1 s∴id=Q=ne⇒n=ideSo, the total number of charge carriers crossing the depletion region is given by
n=20×1062×1.6×1019⇒n=6.25×1013Also, the number of electrons crossing the depletion region is given by
ne=n2=6.25×10132⇒ne=3.1×1013
Question 21:
The current−voltage characteristic of an ideal pn junction diode is given by
i=i0(eeV/KT1)where, the drift current i_{0} equals 10 µA. Take the temperature T to be 300 K. (a) Find the voltage V_{0} for which
eeV/kT=100.One can neglect the term 1 for voltages greater than this value. (b) Find an expression for the dynamic resistance of the diode as a function of V for V > V_{0}. (c) Find the voltage for which the dynamic resistance is 0.2 Ω.
Answer:
(a) The currentâ€’voltage relationship of a diode is given by
i=i0(eeV/kT1)For a large value of voltage, 1 can be neglected.
i≈i0eeV/kTAgain, we need to find the voltage at which
eeV/kT=100
⇒eVkT=ln 100⇒V=ln 100×kTe⇒V=2.303×log 100×8.62×105×300e⇒V=0.12 V(b) Given:
i=i0(eeV/kT1) …(1)
We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.
i.e.
R = dVdiAs the exponential factor dominates the factor of 1, we can neglect this factor.
Now, on differentiating eq. (1) w.r.t. V, we get
didV=i0ekTeeV/kT⇒1R=ei0kTeeV/kT⇒R=kTei0eeV/kT …(2)(c) Given:
R = 2 â„¦
On substituting this value in eq. (2), we get
2=8.62×105×300e×10×106eeV/8.62×105×300⇒V=0.25 V
Question 22:
Consider a pn junction diode having the characteristic
ii0(eeV/kT1) where i0=20μA. The diode is operated at T = 300 K. (a) Find the current through the diode when a voltage of 300 mV is applied across it in forward bias. (b) At what voltage does the current double?
Answer:
(a) Given:
Drift current, i_{0} = 20 × 10^{−6} A
Temperature, T = 300 K
Applied voltage, V = 300 mV
The variation in the current with respect to the voltage is given by
i=i0eeVKT1⇒i=20×106e0.38.62×300×1051⇒i=20×105e1008.611⇒i=2.18 A≈2 A(b) We need to find the voltage at which the current doubles so that the new value of the current becomes 4 A.
⇒4=20×106eeV8.62×3×1031⇒eV×1038.62×31=4×10620⇒eV×1038.62×3=200001⇒V×1038.62×3=12.2060⇒V=12.206×8.63×3103⇒V=318 mV
Question 23:
Calculate the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is 20 µA.
Figure
Answer:
(a) From the circuit diagram, it can be said that the diode is reverse biassed, with applied voltage of 5.0 V.
Under reverse bias condition,
Current in the circuit = Drift current
So, the current in the circuit is 20 µA.
(b) Voltage across the diode will be equal to the voltage of the battery minus the voltage drop across the 20 ohm resistor.
⇒V=5iR⇒V=5(20×20×106)⇒V=5(4×104)⇒V=104(500004)⇒V=49996×104⇒V=4.9996 V≅5 V
Question 24:
Each of the resistance shown in figure has a value of 20 â„¦. Find the equivalent resistance between A and B. Does it depend on whether the point A or B is at higher potential?
Figure
Answer:
According to the Wheatstone bridge principle, if the bridge is balanced, then the current flow across the central resistor is zero. So, to simplify the circuit, we can remove the central resistor.
The given circuit is also balanced, so there is no current through the diode.
Hence, the net resistance of the circuit is given by
RP=R1R2R1+R2For R1=R2=R,RP=R2=402⇒RP=20 Ω
Question 25:
Find the currents through the resistance in the circuits shown in figure.
Figure
Answer:
We know that under forward bias, an ideal diode acts as a short circuit; and under reverse bias, an ideal diode acts as an open circuit.
(a) Since both the diodes are forward biassed, they can be replaced by short circuits.
Net resistance = 2 â„¦
i=2 V2 Ω=1 A(b) One diode is forward biassed and the other is reverse biassed; thus, the resistance of one becomes ∞. The circuit becomes open at the position of the second diode, so no current flows through this arm.
∴i=22+∞=0 A(c) Both of them are forward biassed. Thus, the resistance is zero.
We can replace both the resistors by short circuits.
Net resistance = 2 â„¦
∴i=22=1 A(d) One of them is forward biassed and the other is reverse biassed.
Thus, the current passes through the forwardbiassed diode and the other acts as open circuit.
∴ i=22=1 A
Question 26:
What are the readings of the ammeters A_{1} and A_{2} shown in figure. Neglect the resistance of the meters.
Figure
Answer:
From the circuit diagram, we can see that one diode (the upper one) is reverse biassed.
We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.
Thus,
In the given circuit, one diode is reverse biassed. So, it can be replaced by an open circuit. Hence, the current in this branch will be zero.
So, the current through A_{1} is zero.
For A_{2},
Current =
210= 0.2 A
Question 27:
Find the current through the battery in each of the circuits shown in figure.
Figure
Answer:
We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.
(a) In the given circuit diagram, both diodes are forward biassed. So, the resistance of both of them is zero. Thus, the diode resistance is zero.
i=510×1010+10=55=1 A(b) One diode is forward biassed and the other is reverse biassed. The reversebiassed diode is replaced by an open circuit, so no current flows through this branch.
The current passes through the forwardbiassed diode only.
i=VRnet = 510=0.5 A
Question 28:
Find the current through the resistance R in figure if (a) R = 12â„¦ (b) R = 48â„¦.
Figure
Answer:
(a) When R = 12 â„¦:
The 4 V battery is forward biassing the diode and the 6 V battery is reverse biassing the diode, so the diode is effectively reverse biassed. It acts like an open circuit so that no current flows through this branch. Hence, to simplify the circuit, this branch can be removed.
The current through R on applying the KVL in the circuit is given by
i=1024=0.4166=0.42 A(b) Similarly, for R = 48 â„¦,
i=1048+12=1060=0.16 A
Question 29:
Draw the currentvoltage characteristics for the device show in figure between the terminals A and B.
Figure
Answer:
(a) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the diode will get forward biassed by the applied voltage. So, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forwardbiassed diode.
(b) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the upper diode will get forward biassed and the lower diode will get reverse biassed by the applied voltage. So, this lower branch can be replaced by an open circuit; hence, the current flow through this branch will be zero. The current flows only through the upper diode, so the circuit on simplification will become identical to the circuit in part (a). Hence, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forwardbiassed diode.
Question 30:
Find the equivalent resistance of the network shown in figure between the points A and B.
Figure
Answer:
Let the potentials at A and B be V_{A} and V_{B}, respectively.
(i) When V_{A} > V_{B}, that is, a battery is connected between points A and B, with its positive terminal connected to point A and its negative terminal connected to point B:
As the potential on the pside of the diode is greater than the potential on the nside of the diode, the diode is forward biased; thus, it can be replaced by a short circuit.
As the two resistances are connected in parallel, the effective resistance becomes
Equivalent resistance
=102=5 Ω(ii) When V_{A} < V_{B}:
The diode is reverse biased, so it is replaced by an open circuit. Thus, no current flows through this branch.
∴ Equivalent resistance = 10 â„¦
Question 31:
When the base current in a transistor is changed from 30µA to 80µA, the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain β.
Answer:
Given:
Change in the base current,
δIb=(8030) μAChange in the collector current,
δIc=(3.51) mABase current gain = Rate of change of collector current with respect to base current
Thus,
β=δlcδlb at constant Vcc⇒β=2.5×10350×106⇒β=25050=50∴ Current gain = 50
Question 32:
A load resistor of 2kâ„¦ is connected in the collector branch of an amplifier circuit using a transistor in commonemitter mode. The current gain β = 50. The input resistance of the transistor is 0.50 kâ„¦. If the input current is changed by 50µA. (a) by what amount does the output voltage change, (b) by what amount does the input voltage change and (c) what is the power gain?
Answer:
Given:
Base current gain,
β=50Change in base current,
δIb=50 μALoad resistance,
RL= 2 kâ„¦
Input resistance,
Ri= 0.50 kâ„¦
(a) The change in output voltage is given by
V0=Ic×RL∵Ic=β×Ib∴V0=β×Ib×RL⇒V0=50×50 μA×2 kΩ⇒V0=5 V(b) The change in input voltage is given by
δVi=δlb×Ri⇒δVi=50×106×5×102⇒δVi=25×103⇒δVi=25 mV(c) Power gain is given by
β2×RLRi⇒2500×20.5⇒2500×205=104
Question 33:
Let
X=ABC+BCA+CAB.. Evaluate X for
(a) A = 1, B = 0, C = 1,
(b) A = B = C = 1, and
(c) A = B = C = 0.
Answer:
Given:
Output
X=ABC+BCA+CAB(a) A = 1, B = 0, C = 1
X=1.(0.1)¯+0.(1.1)¯+1.(1.0)¯ =1.0¯+0.1¯+1.0¯ =1.1+0.0+1.1 =1+0+1 =1+1 =1(b) A = B = C = 1
X=1.(1.1)¯+1.(1.1)¯+1.(1.1)¯ =1.1¯+1.1¯+1.1¯ =1.0+1.0+1.0 =0+0+0 =0(c) A = B = C = 0
X=0.(0.0)¯+0(0.0)¯+0.(0.0)¯ =0.0¯+0.0¯+0.0¯ =0.1+0.1+0.1 =0+0+0 =0
Page No 421:
Question 34:
Design a logical circuit using AND, OR and NOT gates to evaluate
ABC+BCA.
Answer:
X = ABC¯+BCA¯ =A(B¯+C¯)+B(C¯+A¯) =AB¯+AC¯+BC¯+BA¯ =AB¯+C¯(A+B)+BA¯
Question 35:
Show that
AB+ABis always 1.
Answer:
Given:
X=AB+ABLet:
AB = y
∴
X=y+yFor y = 0,
X = 0 + 1
= 1
For y = 1,
X = 1 + 0
= 1
Thus, X = 1 for all values of y = AB.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity