QUESTIONS FOR SHORT ANSWER
1. How many 1s energy states are present in one mole of sodium vapor? Are they all filled in normal conditions? How many 3s energy states are present in one mole of sodium vapor? Are they all filled in normal conditions?
ANSWER: The atomic number of sodium is 11 and there are 11 electrons in a sodium atom. The electronic configuration of electrons is -
1s²2s²2p⁶3s¹.
So there are two 1s energy states in a sodium atom wholely filled. One mole of sodium vapor will contain a 6.022x10²³ number of sodium atoms (Avogadro's number). Hence there are 2x6.022x10²³ =12.044x10²³ 1s energy states present in one mole of sodium vapor. They are all filled in normal conditions.
A sodium atom has two 3s energy states, though half-filled. Thus there will be 12.044x10²³ 3s energy states present in one mole of sodium vapor. But they are partially filled in normal conditions.
2. There are energy bands in a solid. Do we have really continuous energy variation in a band or do we have very closely spaced but still discrete energy levels?
ANSWER: There is no continuous energy variation in a band in a solid. It has very closely spaced but still discrete energy levels. These energy levels have a very small energy gap between two consecutive levels.
3. The conduction band of a solid is partially filled at 0 K. Will it be a conductor, a semiconductor, or an insulator?
ANSWER: Since the conduction band of the solid is partially filled, the electrons in this band will find empty states close to it. Thus it will be a conductor.
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: The energy gap between the valance band and the conduction band is low in semiconductors. Thermal collisions are responsible for taking a valence electron to the conduction band. But this jump of an electron leaves a vacancy in its previous position in the valence band which is commonly known as a hole. Thus a hole-electron pair is formed. The hole is sometimes filled with another covalent bond, which results in the moving of holes. But often other electrons from a hole-electron pair jump down to fill a hole losing some of their energies. This recombination makes the electron-hole pair disappear. So the forming and disappearing of electron-hole pair is a continuous process and makes a balance. That is why the number of conduction electrons does not go on increasing with time as thermal collisions continuously take place.
5. When an electron goes from the valance 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 of the electrons from the valance band to the conduction band?
ANSWER: At room temperature, some of the electrons occupying the highest energy level in the valence band already have sufficient energy that is near the required 1.1 eV energy. So when 0.026 eV of thermal collision energy is transferred to them, those electrons that get 1.1 eV or more are able to shift to the conduction band.
6. What is the resistance of an intrinsic semiconductor at 0 K?
ANSWER: At 0 K, none of the valance electrons can get the required 1.1 eV of energy to jump to the conduction band. So there are no carriers in the conduction band of an intrinsic semiconductor. Thus its conductivity is zero which means it has infinite resistance.
7. We have valence electrons and conduction electrons in a semiconductor. Do we also have "valance holes" and "conduction holes"?
ANSWER: When a valance electron leaves its place to jump to the conduction band, the vacant place on the valance band is called a hole. When an electron from the conduction band jumps down to the vacant place or hole the hole disappears, its place in the conduction band is not a vacant place or hole. So there is no concept of "valance holes" and "conduction holes", holes are always in the valance band.
8. When a p-type 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 p-side to the n-side in a p-n junction, the n-side gets positively charged. Explain.
ANSWER: In a p-type semiconductor, the majority of charge carriers are holes but it is electronically neutral as a whole because the created holes are due to the formation of covalent bonds in the trivalent impurity with the help of silicon valence electrons. So it is not charged. Similarly, an n-type semiconductor is not charged though the majority of charge carriers are electrons. When holes diffuse from the p-side to the n-side in a p-n junction, in other words, the electrons diffuse from the n-side to the p-side thus making the n-side positively charged.
9. The drift current in a reversed biased p-n junction increases in magnitude if the temperature of the junction is increased. Explain on the basis of the creation of hole-electron pairs.
ANSWER: When a hole-electron pair is created in the depletion region, the electron goes to the n-side and the hole to the p-side due to the electric field in the depletion region. This creates a current from the n-side to the p-side and is called the drift current. The reverse bias increases the electric field and the increased temperature allows the formation of more hole-electron pairs. Thus more holes are pushed to the p-side and more electrons to the n-side thus the drift current increases.
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 the ideal - a vacuum diode or a p-n junction diode?
ANSWER: A p-n junction diode allows a very small current in the opposite direction i.e. in the reverse bias. But in a vacuum diode current in the opposite direction is not possible due to its construction. Here electrons are supplied by the cathode and collected by the anode when in the forward connection resulting in a current from the anode to the cathode. But in the opposite connection, there is no electron-supplying mechanism in the anode so no current in the opposite direction. Thus a vacuum diode is closer to the ideal.
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 extra power comes from the power supply. An amplifier controls the output signal by matching it with the input signal but with greater amplitudes, in other words, an amplifier amplifies the signal with the help of power supplied to it.
OBJECTIVE - I
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).
EXPLANATION: Both electrons and holes. When a valence electron in the semiconductor moves to the conduction band, the vacancy created by it is called a hole. When an electric field is applied across the semiconductor, the conduction electrons drift opposite to the field and holes drift along the electric field. So electric conduction takes place in a semiconductor due to both electrons and holes. Option (c) is correct.
2. An electric field is applied to a semiconductor. Let the number of charge carriers be n and the average drift speed v. If the temperature is increased,
(a) both n and v will increase
(b) n will increase and v will decrease
(c) v will increase but n will decrease
(d) both n and v will decrease.
ANSWER: (b).
EXPLANATION: Due to the increase in temperature, the average energy exchanged in a collision increases which results in more valance electrons crossing the gap and the number of electron-hole pairs increases i.e. n increases.
Due to the increase in thermal collisions, the average drift speed v decreases.
Option (b) is correct.
3. Let nₚ and nₑ be the number of holes and conduction electrons in an intrinsic semiconductor.
(a) nₚ > nₑ
(b) nₚ = nₑ
(c) nₚ < nₑ
(d) nₚ ≠ nₑ.
ANSWER: (b).
EXPLANATION: In an intrinsic semiconductor, a hole is created due to a shift of a valance electron to the conduction band, they are in pairs, and the number of holes equals the number of conduction electrons.
Option (b) is correct.
4. Let nₚ and nₑ be the number of holes and conduction electrons in an extrinsic semiconductor.
(a) nₚ > nₑ
(b) nₚ = nₑ
(c) nₚ < nₑ
(d) nₚ ≠ nₑ.
ANSWER: (d).
EXPLANATION: Extrinsic semiconductors are made by doping the pure (intrinsic) semiconductors with impurities having five or three valance electrons. These impurities do not create electron-hole pairs. Either they make available extra conduction electrons or extra holes in the valance band depending upon the type of the impurity. So the number of conduction electrons and holes in an extrinsic semiconductor is not equal.
Option (d) is correct.
5. A p-type semiconductor is
(a) positively charged
(b) negatively charged
(c) uncharged
(d) uncharged at 0 K but charged at higher temperatures.
ANSWER: (c).
EXPLANATION: In a pure semiconductor such as silicon, each atom's four valance electrons make four covalent bonds with neighboring silicon atoms. When an impurity element having three valance electrons is added in a very low quantity it replaces silicon atoms at different places. Now one of the neighboring silicon atoms has four covalent bonds but the one with the impurity atom is incomplete because only silicon shares the valance electron. So a hole is created and this extrinsic semiconductor has abundant holes so it is called a p-type semiconductor. Neither the silicon nor the added impurity is electrically charged and the resulting p-type semiconductor is also uncharged.
Option (c) is correct.
6. When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor
(a) increases
(b) decreases
(c) remains the sane
(d) becomes zero.
ANSWER: (a).
EXPLANATION: When an impurity is doped into an intrinsic semiconductor, the number of charged carriers increases. Hence the conductivity of the semiconductor increases.
Option (a) is correct.
7. If the two ends of the p-n 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 n-side to the p-side
(c) there will be a steady current from the p-side to the n-side
(d) there may or may not be a current depending upon the resistance of the connecting wire.
ANSWER: (a).
EXPLANATION: When the two ends of a p-n junction are joined by a wire, there will be diffusion and drift currents that will cancel each other. So there will not be a steady current in the circuit.
Option (a) is correct.
8. The drift current in a p-n junction is
(a) from the n-side to the p-side
(b) from the p-side to the n-side
(c) from the n-side to the p-side if the junction is forward-biased and in the opposite direction if it is reverse-biased.
(d) from the p-side to the n-side if the junction is forward-biased and in the opposite direction if it is reverse-biased.
ANSWER: (a).
EXPLANATION: If an electron-hole pair is created in the depletion region, the electron is quickly pushed by the electric field toward the n-side and the hole toward the p-side. Due to this continuous process, there is a regular flow of electrons toward the n-side and the holes toward the p-side. This makes a current from the n-side to the p-side. This current is called the drift current.
Option (a) is current.
9. The diffusion current in a p-n junction is
(a) from the n-side to the p-side
(b) from the p-side to the n-side
(c) from the n-side to the p-side if the junction is forward-biased and in the opposite direction if it is reverse-biased.
(d) from the p-side to the n-side if the junction is forward-biased and in the opposite direction if it is reverse-biased.
ANSWER: (b).
EXPLANATION: Due to the presence of an electric field in the depletion region, holes in the p-side and conduction electrons in the n-side find it difficult to diffuse to the other side against the field. Only a few energetic holes and conduction electrons are able to overcome the electric field in the depletion region and the holes from the p-side crossover to the n-side while electrons from the n-side crossover to the p-side. Thus there is a diffusion current from the p-side to the n-side.
Option (b) is correct.
10. Diffusion current in a p-n junction is greater than the drift current in magnitude
(a) if the junction is forward-biased
(b) if the junction is reverse-biased
(c) if the junction is unbiased
(d) in no case.
ANSWER: (a).
EXPLANATION: When no battery is connected, the potential of the n-side is higher than the p-side. If the p-n junction is connected in forward bias, the potential of the p-side is raised and hence the height of the potential barrier decreases. The width of the depletion region is also reduced in forward bias. So more diffusion is allowed and the diffusion current increases. The drift current is unchanged because the rate of formation of new electron-hole pairs is fairly independent of the electric field. Thus, in this case, the diffusion current is more than the drift current.
If the p-n junction is unbiased, the diffusion current and drift current are in equal and opposite directions.
If the junction is reverse-biased, the diffusion current will be less than the drift current.
Hence option (a) is correct.
11. Two identical p-n junctions may be connected in series with a battery in three ways (figure 45-Q1). The potential difference across the two p-n junctions is equal in
(a) circuit 1 and Circuit 2
(b) circuit 2 and Circuit 3
(c) circuit 3 and Circuit 1
(d) circuit 1 only.
 |
| The figure for Q-11 |
ANSWER: (b).
EXPLANATION: In Circuit 1, one junction is in forward bias thus offering almost no resistance. So it can be replaced with a wire with no potential difference across it. Another junction is in reverse bias which offers maximum resistance and behaves like a cut in the circuit. So there is the maximum potential difference across it. Thus both junctions have unequal potential differences.
In Circuit 2, both junctions are in forward bias, thus offering almost no resistance. So the potential difference across each of them is equal to almost zero.
In circuit 3, both the junctions are connected in reverse bias thus offering maximum resistance and almost no current across them. Hence the potential difference across each of them is equal and maximum.
So in Circuit 2 and Circuit 3, the potential difference across the two p-n junctions are equal. Option (b) is correct.
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 (45-Q2). The charges on the capacitors at time t =CR are respectively,
(a) VC, VC
(b) VC/e, VC
(c) VC, VC/e
(d) VC/e, VC/e.
 |
| The figure for Q-12 |
ANSWER: (b).
EXPLANATION: Before connection, the charge on the capacitor is equal to
Q =VC.
Since B is offering reverse bias to the p-n junction, there will be no current flow in the circuit and the charge on it will remain the same (=VC) even at time t =CR.
Capacitor A is offering forward bias to the p-n junction which will allow the current through it with almost no resistance. The p-n junction may be replaced as a conducting wire allowing the capacitor to start discharging. The remaining charge on the capacitor at time t is given as,
q =Qe(-t/CR)
For t =CR,
q =Qe⁻¹ =VC/e.
So option (b) is correct.
13. A hole diffuses from the p-side to the n-side in a p-n junction. This means that
(a) a bond is broken on the n-side and the electron freed from the bond jumps to the conduction band
(b) a conduction electron on the p-side jumps to a broken bond to complete it
(c) a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it
(d) a bond is broken on the p-side and the electron freed from the bond jumps to a broken bond on the n-side to complete it.
ANSWER: (c).
EXPLANATION: A hole diffuses from the p-side to the n-side in a p-n junction, which means a broken bond is created in the n-side and a broken bond disappears in the p-side by accepting a conduction electron which is a result of the broken bond on the n-side. Thus a hole in the p-side is said to be diffused to the n-side.
So option (c) is correct.
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).
EXPLANATION: The emitter is heavily doped which means that it has the highest concentration of impurity. Option (a) is incorrect.
The collector is moderately doped and the middle layer called the base is very lightly doped. This means that the base has the least concentration of impurities and the collector's impurity concentration is in between the base and emitter.
Clearly, only option (c) is correct.
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).
EXPLANATION: There is no emitter-collector junction in a transistor, hence options (a), (b), and (c) are incorrect.
A transistor has an emitter-base junction and a collector-base junction. Thus only option (d) is correct.
OBJECTIVE - II
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 the temperature
(d) the number of free electrons is less than that in a conductor.
ANSWER: (a), (c), (d).
EXPLANATION: As the temperature decreases the average energy exchanged in a collision also decreases. So fewer valance electrons cross the gap and the number of electron-hole pairs decreases. The number of such pairs is proportional to a factor T3/2 e-ΔE/2kT, where ΔE is the band gap. Clearly, at T =0 K, the number of pairs is zero. So no free electrons at 0 K. Also, the number of free electrons increases with an increase in temperature. Option (a) and (c) are correct but option (b) is incorrect.
The number of free electrons in a semiconductor is less than that in a conductor. For example, while the number of conduction electrons per m³ in copper is of the order of 10²⁸, it is of the order of 10¹⁵ in extrinsic n-type semiconductor silicon doped with phosphorus. Option (d) is correct.
2. In a p-n junction with open ends,
(a) there is no systematic motion of charge carriers
(b) holes and conduction electrons systematically go from the p-side to the n-side and from the n-side to the p-side respectively
(c) there is no net charge transfer between the two sides
(d) there is a constant electric field near the junction.
ANSWER: (b), (c), (d).
EXPLANATION: Due to the difference in concentration of holes and conduction electrons between the two sides of a p-n junction, they diffuse to the other side. But this diffusion creates a constant electric field in a region near the junction with direction from n-side to p-side. So further diffusion is difficult while any electron-hole pair created in this region drifts to the opposite direction of the diffusion. Thus a balance is reached where the diffusion and drift of charged carriers are equal. So there is a systematic motion of charged carriers between the two sides but no net charge is transferred. Thus except option (a) all other options are correct.
3. In a p-n 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), (d).
EXPLANATION: New hole and conduction electrons are continuously produced throughout the material in a p-n junction. They also continuously recombine to maintain their numbers constant at a certain temperature except in the depletion region. Due to the presence of an electric field in the depletion region, electrons of the produced pairs drift toward the n-side and holes to the p-side and they are unable to recombine. So only options (a) and (d) are correct.
4. The impurity atoms with which pure silicon may be doped to make it a p-type semiconductor are those of
(a) phosphorus
(b) boron
(c) antimony
(d) aluminum.
ANSWER: (b), (d).
EXPLANATION: A p-type semiconductor is made by doping an intrinsic semiconductor like silicon with trivalent impurity atoms. Due to this one of the covalent bonds of a silicon atom with impurity element is deficient in an electron. This results in the production of a hole there and makes the semiconductor rich in holes thus a p-type semiconductor. Out of the four options only boron and aluminum are trivalent. Hence options (b) and (d) are correct.
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: All.
EXPLANATION: With all the given methods, the number of charged carriers in germanium which is an intrinsic semiconductor increases. More the number of charged carriers more is the conductivity. Hence all of the options are true.
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 p-type semiconductor
(c) an n-type semiconductor
(d) a p-n junction.
ANSWER: (d).
EXPLANATION: An intrinsic semiconductor, a p-type semiconductor, or an n-type semiconductor allows the current to flow through it equally in both directions whatever the magnitude of the current.
Only a p-n junction allows the current to flow in one direction that is from the p-side to the n-side. Hence the given semiconducting device may be a p-n junction. Only option (d) is correct.
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: (b), (c).
EXPLANATION: By doping with a donor impurity that has five valance electrons, some of the atoms of the semiconductor are replaced by it. An impurity atom makes covalent bonds with four adjacent atoms of the semiconductor by sharing one valence electron with each of them. Thus one valence electron is unable to make a bond and is free to move. So the free electron concentration increases. Some of these electrons combine with the holes of the present electron-hole pairs making them disappear. So the hole concentration decreases. Options (b) and (c) are correct.
8. Let iₑ, i₍, and iᵦ represent the emitter current, the collector current, and the base current respectively in a transistor. Then
(a) i₍ is slightly smaller than iₑ
(b) i₍ is slightly greater than iₑ
(c) iᵦ is much smaller than iₑ
(d) iᵦ is much greater than iₑ.
ANSWER: (a), (c).
EXPLANATION: The base is very lightly doped and very thin. The emitter is heavily doped and the collector is moderately doped. Since the emitter-base junction is in forward bias and the collector-base in reverse bias, most of the majority charge carriers diffuse from emitter to collector. Only a few of these charge carriers go through the base. So the base current is very low. In fact, the base current Iᵦ is only 1% to 5% of emitter current Iₑ. Hence option (c) is correct and (d) is incorrect.
Using Kirchhoff's law, we can write
Iₑ =Iᵦ +I₍
Clearly, I₍ is slightly smaller than Iᵦ. Hence option (a) is correct and (b) is incorrect.
9. In a normal operation of a transistor,
(a) the base-emitter junction is forward-biased
(b) the base-collector junction is forward-biased
(c) the base-emitter junction is reverse-biased
(d) the base-collector junction is reverse-biased.
ANSWER: (a), (d).
EXPLANATION: The base-emitter junction is always kept forward-biased. The base-collector junction is kept reverse-biased. This makes most of the majority charge carriers diffuse from the emitter to the collector. Hence only options (a) and (d) are correct.
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), (d).
EXPLANATION: NAND and NOR gates are called basic building blocks of logic circuits. Any logical gate can be constructed by repetitively using NAND or NOR gates.
An AND gate can not be prepared by repetitive use of a NOT or OR gate. Hence only options (c) and (d) are correct.
EXERCISES
1. Calculate the number of states per cubic meter of sodium in the 3s band. The density of sodium is 1013 kg/m³. How many of them are empty?
ANSWER: One sodium atom has two numbers of states in the 3s band. Its atomic number is 11 with an electronic configuration of 1s²2s²2p⁶3s¹. So out of two 3s states only one is filled in each atom.
Let us first find the number of sodium atoms in each cubic meter. The atomic mass of sodium is 23, so 23 g of sodium will have a 6.02x10²³ number of atoms. Since one cubic meter of sodium has a weight of 1013 kg, it will have a number of sodium atoms
=1013x10³x6.02x10²³/23
=2.65x10²⁸
Hence the number of states in the 3s band of one cubic meter of sodium is
=2x2.65x10²⁸
=5.3x10²⁸.
Since half of them are empty, the number of empty states in the 3s band is
=2.65x10²⁸.
2. In a pure semiconductor, the number of conduction electrons is 6x10¹⁹ per cubic meter. How many holes are there in a sample of size 1 cm x 1 cm x 1 mm?
ANSWER: In a pure or intrinsic semiconductor, the number of conduction electrons and holes are equal in number because they are in pairs. Hence the number of holes in the given sample
=0.01*0.01*0.001*6x10¹⁹
=6x10¹².
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: For the given condition,
kT =0.23 eV, where k is the Boltzmann constant.
→T =0.23 eV/k
→T =(0.23 eV)/(8.62x10⁻⁵ eV/K)
=2668 K
≈2670 K.
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-tenth of the value at 300 K? (Silicon will not retain the structure at these high temperatures.)
ANSWER: (a) The ratio of the band gap to kT at room temperature is
=1.1/(8.62x10⁻⁵*300)
≈43.
(b) Let the required temperature =T'. The ratio of the band gap to kT' is 43/10 =4.3.
So, 1.1/(kT') =4.3
→T' =1.1/(4.3k)
=1.1/(4.3*8.62x10⁻⁵)
=2967.7 K
≈3000 K.
5. When a semiconducting material is doped with an impurity, new acceptor levels are created. In a particular thermal collision, a valence electron receives 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 valance band and that the temperature T is equal to 300 K, find the energy of the acceptor levels above the valence band.
ANSWER: The electron at the top of the valance band receives energy equal to 2kT and just reaches one of the acceptor levels. Hence the energy of the acceptor levels above the valence band is =2kT.
i.e. =2*8.62x10⁻⁵*300 eV
=0.05 eV
=50 meV.
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: Maximum wavelength will be emitted when minimum energy is emitted. And it will happen when the conduction electron combines with the hole in the top level of the valence band thus releasing energy equal to the band gap. So here is the released energy,
E =3.2 eV
→hc/λ =3.2
→λ =hc/3.2
=4.14x10⁻¹⁵*3x10⁸/3.2 m
=3.88x10⁻⁷ m
=3.88x10⁻⁷*10⁹ nm
=388 nm
≈390 nm.
7. Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap?
ANSWER: The band gap will be equal to the minimum energy liberated in the recombination. Minimum energy is with the maximum wavelength released which is here 820 nm. So the band gap =E =hc/λ
=4.14x10⁻¹⁵*3x10⁸/(820x10⁻⁹) eV
=1.5 eV.
8. Find the maximum wavelength of electromagnetic radiation that can create a hole-electron pair in germanium. The band gap in germanium is 0.65 eV.
ANSWER: The minimum energy of electromagnetic radiation (i.e. having the maximum wavelength) which can create a hole-electron pair in germanium will be equal to the band gap =0.65 eV.
Hence E =hc/λ =0.65 eV
→λ =hc/0.65
=4.14x10⁻¹⁵*3x10⁸/0.65
=1.9x10⁻⁶ m.
9. In a photodiode, the conductivity 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: It is clear that the conductivity increases when the light of minimum energy with wavelength 620 nm falls on the photodiode. The valance electrons in the top levels of the valence band absorb this energy amount of energy and just reach the conduction band, thus conductivity increases. So the band gap is equal to this amount of energy.
Hence the band gap =hc/λ
=4.14x10⁻¹⁵*3x10⁸/620x10⁻⁹ eV
=2.0 eV.
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-ΔE/2kT. Find the ratio of the concentration of conduction electrons in diamond to that silicon at room temperature 300 K. ΔE for silicon is 1.1 eV and for diamond is 6.0 eV. How many conduction electrons are likely to be in one cubic meter of diamond?
ANSWER: For silicon ΔE =1.1 eV
For diamond ΔE' =6.0 eV.
T =300 K.
From the given condition, the ratio of the concentration of the conduction electrons in diamond to that of silicon
=(e-ΔE'/2kT)/(e-ΔE/2kT)
=e(ΔE-ΔE')/2kT
=e(1.1-6.0)/2kT
=e-4.9/(2*8.62x10⁻⁵*300)
=e-94.74
=7.16x10⁻⁴².
From this ratio, we conclude that for the diamond, the population of the conduction electrons will be about 7.16x10⁻⁴² times the silicon. The number of conduction electrons per m³ for silicon at room temperature 300 K is about 7x10¹⁵. Hence the number of conduction electrons per m³ of the diamond at room temperature is
=7.16x10⁻⁴²*7x10¹⁵
=5x10⁻²⁶, which is almost zero.
11. The conductivity of a pure semiconductor is roughly proportional to T3/2 e-ΔE/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: Given, T =4 K, T' =300 K;
and ΔE =0.74 eV, ΔE' =0.67 eV.
From the given proportionality of the conductivity, the factor by which the conductivity increases is
=(T'3/2/T3/2)(e-ΔE'/2kT'/e-ΔE/2kT)
=(T'/T)3/2 e(ΔE/T -ΔE'/T')/2k
=(300/4)3/2 e(0.74/4 -0.67/300)/2k
=650*e¹⁰⁶⁰ =650*e⁵⁶⁰e⁵⁰⁰
{after putting the value of k}
=650*1.6x10²⁴³*1.4x10²¹⁷
=1.5x10⁴⁶³.
12. Estimate the proportion of the 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 7x10¹⁵ holes per cubic meter. The density of silicon is 5x10²⁸ atoms per cubic meter.
ANSWER: In pure silicon, the number of holes and the number of conduction electrons are equal. Hence the total number of charge carriers in silicon per cubic meter is,
=2*7x10¹⁵
=14x10¹⁵.
The conductivity will increase 100 times if the number of charge carriers increases 100 times. So it will be equal to 100x14x10¹⁵ =14x10¹⁷. Though a boron atom adds one hole in the semiconductor, it will also replace a silicon atom in the material. Suppose X number of boron atoms are added to one cubic meter of silicon replacing X number of silicon atoms.
Given that number of silicon atoms in a cubic meter N =5x10²⁸.
Now the number of conduction electrons per cubic meter of the doped silicon (provided by silicon atoms only)
={(N-X)/N}*7x10¹⁵
This is also the number of holes provided by silicon in the doped semiconductor.
Contribution of holes by boron =X.
Equating the total number of charge carriers in view of the increased conductivity. We get,
X+{(N-X)/N}*7x10¹⁵+{(N-X)/N}*7x10¹⁵ =14x10¹⁷
→XN+2(N-X)*7x10¹⁵ =N*14x10¹⁷
→XN +N*14x10¹⁵ +X*14x10¹⁵=N*14x10¹⁷
→X(N -14x10¹⁵) =N*14x10¹⁵*(100 -1)
Putting the value for N, we get
→X =5x10²⁸*14x10¹⁵*99/(5x10²⁸ -14x10¹⁵)
≈1.4x10¹⁸
Hence the proportion of boron impurity is
=X/N
=1.4x10¹⁸/(5x10²⁸-1.4x10¹⁸)
≈1.4x10¹⁸/5x10²⁸
≈1/3.5x10¹⁰
i.e. 1 in about 3.5x10¹⁰.
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 the conduction electrons in germanium is 6x10¹⁹ per cubic meter. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to 2x10²³ per cubic meter. Find the concentration of the holes in the doped germanium.
ANSWER: Before doping the germanium, the concentration of conduction electrons
=concentration of holes =6x10¹⁹ per m³
After doping the concentration of conduction electrons =2x10²³ per m³,
Let the concentration of holes per m³ after doping =X.
Since the product of the concentration of holes and conduction electrons are the same before and after doping, we can write
X*2x10²³ =6x10¹⁹*6x10¹⁹
→X =18x10¹⁵
=1.8x10¹⁶ per cubic meter.
14. The conductivity of an intrinsic semiconductor depends on the temperature as σ =σₒe-ΔE/2kT where σₒ is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double its value at T =300 K. Assume that the gap for germanium is 0.630 eV and remains constant as the temperature is increased.
ANSWER: Given that band gap ΔE =0.630 eV
T =300 K.
Suppose at temperature T', the conductivity is double its value at T. So,
σ' =2σ
→σₒ e-ΔE/2kT' =2σₒ e-ΔE/2kT
→e(-ΔE/2kT' +ΔE/2kT) =2
→(ΔE/2k)*(1/T -1/T') =ln 2
→(0.630/2k)(1/300 -1/T') =ln 2
→3654.3*(1/300 -1/T') =ln 2
→1/300 -1/T' =1.9x10⁻⁴
→1/T' =3.14x10⁻³
→T' =318 K.
15. A semiconducting material has a band gap of 1.0 eV. Acceptor impurities are doped into it which creates 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 the 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: There are two acceptor levels, one due to impurities which is 1 meV, and the other is the original which is 1.0 eV.
For the temperature at which hole concentration will be constant
kT =2*1 meV
→T =(2/1000)/8.62x10⁻⁵ =23 K.
Next band gap is =1.0 -1.0x10⁻³ eV
The transition in this band gap will be possible only when kT is more than 1/50 of this gap. So
kT =(1.0 -1.0x10⁻³)/50
→T =232 K.
Between temperatures 23 K to 232 K, the hole concentration is constant. Thus the order of the temperature range in which the hole concentration is approximately constant is 20 K to 230 K.
16. In a p-n junction, the depletion region is 400 nm wide. and the electric field of 5x10⁵ V/m exists in it. (a) Find the height of the potential barrier. (b) What should be the minimum kinetic energy of a conduction electron that can diffuse from the n-side to the p-side?
ANSWER: (a) Width of the depletion region
d =400 nm =4x10⁻⁷ m
The existing electric field in this region,
E =5x10⁵ V/m
The height of the potential barrier =potential difference across the depletion region
=d*E
=4x10⁻⁷*5x10⁵ volts
=20x10⁻² volts
=0.2 volts.
(b) For diffusion, the minimum kinetic energy of a conduction electron should be equal to the work done in resisting it by the electric field. i.e.
=eE*d
= e*(0.2 volts)
=0.2 eV.
17. The potential barrier existing across an unbiased p-n junction is 0.2 volts. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if (a) the junction is unbiased. (b) the junction is forward biased at 0.1 volts and (c) the junction is reverse biased at 0.1 volts?
ANSWER: (a) The hole is assumed to have a positive charge equal to e. Suppose there is an electric field of E volts/meter across the depletion region of the unbiased p-n junction. For the diffusion of a hole from the p-side to the n-side, the minimum kinetic energy required by it is equal to the work done by the electric field in resisting the diffusion. i.e.
=eE*d, {d is the width of the depletion region}
=e(Ed)
=e*V, {Ed =V=potential barrier =0.2 volts}
=e*(0.2 volts)
=0.2 eV.
(b) When the junction is forward-biased at 0.1 volts, the potential barrier is reduced by 0.1 volts because the field provided by the battery is opposite to the field at the depletion region. Now the reduced potential barrier,
V =0.2 -0.1 =0.1 volts.
Hence the minimum kinetic energy possessed by a hole for diffusion now is,
=e*(0.1 volts)
=0.1 eV.
(c) In the reversed bias, the direction of the field at the depletion region and the direction of the field provided by the battery are the same, and hence they add up. The new potential barrier now is
=0.2 +0.1 =0.3 volts.
So the minimum kinetic energy possessed by a hole for diffusion is
=e*(0.3 volts)
=0.3 eV.
18. In a p-n 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 p-side and (b) from the n-side.
ANSWER: (a) When the hole approached the junction from the p-side, it moves across the junction against the electric field in the depletion region. The force exerted by the field is opposite to the movement. So it will lose 250 meV of its kinetic energy in crossing the junction. So the remaining kinetic energy of the hole when it crosses the junction =300 meV -250 meV
=50 meV.
(b) When the hole approached the junction from the n-side, it moves along the existing electric field in the depletion region. So the force exerted by the field is along the movement that increases its kinetic energy by the amount of the potential barrier. So the kinetic energy of the hole when it crosses the junction
=300 meV +250m meV
=550 meV.
19. When a p-n junction is reverse-biased, the current becomes almost constant at 25 µA. When it is forward-biased at 200 mV, a current of 75 µA is obtained. Find the magnitude of the diffusion current when the diode is (a) unbiased (b) reverse-biased at 200 mV and (c) forward-biased at 200 mV.
ANSWER: When a p-n junction is reverse-biased, the diffusion stops. So the diffusion current is zero. The only current is the drift current which is independent of the bias and remains constant. The 25 µA current in the reverse bias is the drift current.
(a) In the unbiased condition, the diffusion current and the drift current are equal and opposite in direction. Hence the magnitude of the diffusion current when the diode is unbiased =25 µA.
(b) Since the diffusion stops in the reverse bias, the diffusion current is zero.
(c) When the p-n junction is forward-biased, the diffusion current from the p-side to the n-side is more than the drift current from the n-side to the p-side. So
net current =diffusion current -drift current
→diffusion current =net current +drift current
→diffusion current =75 µA +25 µA
=100 µA.
20. The drift current in a p-n junction is 20.0 µA. Evaluate the number of electrons crossing a cross-section per second in the depletion region.
ANSWER: In a drift current in the depletion region, holes from newly created hole-electron pairs move toward the p-side, and the electrons toward the n-side. If in a drift current, the number of holes and electrons crossing a cross-section per second are n each then the total charge crossing per second =2ne. Hence the drift current is
i =2ne =20.0 µA
→2ne =20.0x10⁻⁶
Charge on an electron e =1.602x10⁻¹⁹ C
So the number of electrons crossing per second in the depletion region
n =20.0x10⁻⁶/2e
=20.0x10⁻⁶/(2*1.602x10⁻¹⁹)
=6.2x10¹³.
21. The current-voltage characteristic of an ideal p-n junction diode is given by
i =iₒ (eeV/kT -1)
where the drift current iₒ equals 10 µA. Take the temperature T to be 300 K.
(a) Find the voltage Vₒ 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 > Vₒ.
(c) Find the voltage for which the dynamic resistance is 0.2 Ω.
ANSWER: (a) From the given condition we neglect the term 1 and equate,
eeVₒ/kT =100
→eVₒ/kT =ln 100
→Vₒ =(kT*ln 100)/e
=8.62x10⁻⁵*300*ln 100
=(0.12 eV)/e
=0.12 V.
(b) The dynamic resistance is given as dV/di. We take the differential of both sides of the characteristic expression,
di =iₒ*(eeV/kT)*(edV/kT)
→(dV/di)*(iₒe/kT)*(eeV/kT) =1
→dV/di =kTe-eV/kT/eiₒ.
(c) Given the dynamic resistance dV/ di =0.2 Ω, iₒ =10 µA, T =300 K. Putting the values in the above expression,
0.2 =8.62x10⁻⁵*300*(e-eV/kT)/(e10x10⁻⁶)
→(e-eV/kT) =0.2e/2586
(Since the unit of k is eV/K, e from the denominator on the RHS will be eliminated for uniformity of units)
→eeV/kT =2586/0.2 =12930
→eV/kT =ln (12930) =9.5
→eV =8.62x10⁻⁵*300*9.5 =0.25 eV
→V =0.25 V.
22. Consider a p-n junction diode having the characteristic i =iₒ (eeV/kT -1) where iₒ =20 µA. 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) V =300 mV =0.3 V,
iₒ =20 µA =2x10⁻⁵ A, T =300 K.
The current through the diode is,
i =2x10⁻⁵(e0.3e/8.62x10⁻⁵*300 -1)
=2.1 A
≈2 A.
(b) Suppose at voltage V' the current doubles. So, i' =2i
→iₒ(eeV'/kT -1) =2iₒ(eeV/kT -1)
→eeV'/kT -1 =2eeV/kT -2
(Numerical terms -1 and -2 are negligible in comparison to the other terms on each side, so neglect.)
→e(eV'/kT -eV/kT) =2
→e(V'-V)e/kT =2
→(V' -V)e/kT =ln 2
→V' -V =(kT/e)*ln 2
→V' =V +(8.62x10⁻⁵*300)*ln 2
=0.3 +0.018 volts
=0.318 volts =318 mV.
23. Calculate the current through the circuit and the potential difference across the diode shown in the figure (45-E1). The drift current for the diode is 20 µA.
 |
| The figure for Q-23 |
ANSWER: (a) Since the diode is in reverse bias, it will not allow batteries to send current through it. But the drift current is independent of the bias and its direction is from n-type to p-type semiconductor which is in the direction of the potential difference in the given circuit. So the only current in the circuit will be the drift current =20 µA.
(b) Suppose the potential difference across the diode is V. Then,
iR +V =5 volts
→V =5 -20x10⁻⁶*20 volts
=5 -0.0004 volts
=4.9996 volts
≈5 V.
24. Each of the resistances shown in the figure (45-E2) has a value of 20 Ω. Find the equivalent resistance between A and B. Does it depend on whether point A or B is at higher potential?
 |
| The figure for Q-24 |
ANSWER: Since all four resistances have equal value =20 Ω, the given circuit forms a Wheatstone bridge. In a balanced Wheatstone bridge, the middle wire has zero current because the potential difference across it is zero. Thus we can remove the middle wire of the given circuit containing the diode. So the remaining circuit has two parallel resistances, each having a resistance of R =20 Ω +20 Ω =40 Ω. Let the equivalent resistance = R', Then,
1/R' =1/40 +1/40 =2/40 =1/20
→R' =20 Ω.
The equivalent resistance R' will not depend on whether point A or B is at higher potential.
-----------------------------
In problems, 25 to 30, assume that the resistance of each diode is zero in forward bias and is infinity in reverse bias.
25. Find the currents through the resistances in the circuits shown in Figure (45-E3).
 |
| The Figure for Q-25 |
ANSWER: (a) E =2 V, R =2 Ω, Both the diodes are in forward bias and hence will have zero resistance. So from Ohm's law, the current through the resistance will be
i =E/R
=2/2 A
=1 A.
(b) Here one diode is in forward bias giving zero resistance while another is in reverse bias giving infinite resistance. Since they are connected in series, the total resistance in the circuit will be infinite and no current will flow.
i =E/∞ =zero.
(c) Both of the parallel connected diodes are in forward bias offering no resistance. Hence the circuit will be similar to the first circuit. Current through the resistance
i =2/2 =1 A.
(d) In this circuit one of the two parallel connected diodes is in forward bias resulting in zero resistance. Another one is in reverse bias giving infinite resistance. So all the current will pass through the first diode without resistance. The equivalent resistance is zero. We also can see it by calculating two resistances zero and ∞ in parallel.
1/R =1/0 +1/∞ =∞ +0 =∞
→R =1/∞ =0.
So only resistance in the circuit is 2 Ω. The current through the resistance is,
i =2/2 =1 A.
26. What are the readings of the ammeters A₁ and A₂ shown in figure (45-E4). Neglect the resistances of the meters.
 |
| The figure for Q-26 |
ANSWER: The diode is in reverse bias giving infinite resistance, so the current in the upper loop will be zero. The reading of ammeter A₁ will be zero.
Since the resistance of the diode is infinite, the effect will be as if no upper loop but only a lower loop. In this loop, E =2 V, R =10 Ω. So the current in the lower loop
i =V/R =2/10 =0.2 A.
Hence the reading of ammeter A₂ =0.2 A.
27. Find the current through the battery in each of the circuits shown in Figure (45-E5).
 |
| The figure for Q-27 |
ANSWER: In the first figure, both of the diodes are in forward bias and hence no resistance by them. The equivalent resistance of the circuit will be for two 10 Ω parallel resistances. Hence,
1/R =1/10 +1/10 =1/5
→R =5 Ω.
E =5 V.
Hence the current through the battery
i =E/R
=5/5 =1 A.
In the second figure, the lower diode is in reverse bias, thus having infinite resistance. This diode will act as a break in the wire. Since the upper diode is in the forward bias with zero resistance, the effective simple circuit will have a 5 V battery and a 10 Ω resistance. The current through the resistance and the battery will be
i =5/10 =0.5 A.
28. Find the current through the resistance R in figure (45-E6) if (a) R =12 Ω (b) R =48 Ω.
 |
| The figure for Q-28 |
ANSWER: (a) When R =12 Ω, the net potential difference between the two ends will be applying a reverse bias on the diode. Thus no current through the diode and it will act as a break in the wire. The remaining circuit has emf =4+6 =10 V and resistance =12+12 =24 Ω. Hence current through R is
i =10/24 =0.42 A.
(b) When R =48 Ω. Suppose the current through 4 V battery = i, through 6 V battery =i' and through 1 Ω resistance =i". From Kirchoff's loop law in the upper loop,
12i -4 -6 +48i' =0
→12i +48i' =10 ---- (i)
For the lower loop,
-48i' +6 +1i" =0
→-48i' +i" =-6
→i" =48i' -6 ----- (ii)
At the left junction,
i =i' +i"
From (i)
12i' +12i" +48i' =10
→60i' +12i" =10
→30i' +6i" =5
→30i' +6(48i' -6) =5
→318i' -36 =5
→i' =41/318 =0.13 A.
29. Draw the current-voltage characteristics for the device shown in Figure (45-E7) between terminals A and B.
 |
| The figure for Q-29 |
ANSWER: From problems 25 to 30, given that the resistance of each diode is zero in forward bias and infinity in reverse bias. So the current will be zero if B is at a higher potential than A. When A is at a higher potential than B, then diode resistance is zero due to forward bias and the current will be i =V/R =V/10. The current-voltage characteristic will be as follows.
 |
| Figure - 1 |
In the second figure, when A is at a higher potential than B, the lower branch is ineffective due to reverse bias. The upper diode is in forward bias, so it has zero resistance. The current will be through 10 Ω resistance and its value,
i =V/10.
We plot this potential difference and current on the positive X-axis and Y-axis respectively.
When B is at a higher potential than A, the upper diode is reverse biased and the lower one is forward biased. So the current will flow only through the lower diode and resistance. The current through the device is,
i =V/20.
We plot this potential difference and the current on the negative X-axis and Y-axis respectively. The current-voltage characteristic drawn for the device is as below.
 |
| Figure - 2 |
30. Find the equivalent resistance of the network shown in Figure (45-E8) between points A and B.
 |
| The figure for Q-30 |
ANSWER: We have to take the resistance of a forward-biased diode as zero and reverse-biased as infinite.
In the given figure, the diode is in the forward bias if point A is at a higher potential than B. So it will offer no resistance and the network will act as two 10 Ω resistances connected in parallel. Its equivalent resistance R is given as,
1/R =1/10 +1/10 =1/5
→R =5 Ω.
When point B is at a higher potential than A, the diode is in the reverse bias offering infinite resistance and no current will flow through that branch. So only remaining resistance is 10 Ω in the lower wire. So the equivalent resistance of the network in this condition is R =10 Ω.
31. When the base current in a transistor is changed from 30 µA to 80 µA, the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain ß.
ANSWER: The current gain is given as,
ß =ΔIC/ΔIB
Given that change in the base current
ΔIB =80 -30 =50 µA =50x10⁻³ mA
Change in collector-current
ΔIC =3.5 -1.0 =2.5 mA.
Hence,
ß =(2.5 mA)/(50x10⁻³ mA)
=50.
32. A load resistor of 2 kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain ß =50. The input resistance of the transistor is 0.50 kΩ. If the input current is changed by 50 µA, (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: The input current is the base current. Here change in input current
ΔIB =50 µA. Current gain ß =50. Hence the change in collector current,
ΔIC =ßΔIB
=50*50µA
=2500 µA
The connected load resistance in the collector branch =2 kΩ.
(a) The change in output voltage,
ΔVC =R*ΔIC
=2x10³*2500x10⁻⁶ V
=5.0 V.
(b) The input resistance of the transistor
r =0.5 kΩ =500 Ω
The change in input current ΔIB =50 µA
Hence the change in input voltage,
ΔVB =r*ΔIB
=500*50x10⁻⁶ V
=0.025 V
=25 mV.
(c) Voltage gain =ΔVC/ΔVB
=5.0/(25x10⁻³) =200.
Given that current gain ß =50.
Hence, power gain =voltage gain*current gain
=200*50
=10⁴.
33. Let X =AB̅C̅ +BC̅A̅ +CA̅B̅. Evaluate X for
(a) A =1, B =0, C =1,
(b) A =B =C =1 and
(c) A =B =C =0.
ANSWER: The given function may be written as,
X =A AND NOT(B AND C) +B AND NOT(C AND A) +C AND NOT(A AND B)
(a) If A =1, B =0,C =1 then,
B AND C =0, NOT(B AND C) =1,
So, A AND NOT(B AND C) =1
for the next term,
C AND A =1, NOT (C AND A) =0,
So, B AND NOT(C AND A) =0
For the last term,
A AND B =0, NOT (A AND B) =1,
So, C AND NOT(A AND B) =1
The value of X =1 +0 +1 =1
Because the + symbol represents OR. If any of the variables on RHS is 1 in the OR function, the result is 1.
(b) A =B =C =1
For the first term,
B AND C =1, NOT(B AND C) =0,
A AND NOT(B AND C) =0
Similarly, the other two terms will also be zero. Now the function X is
X =0 +0 +0
=0 OR 0 OR 0
=0.
(c) A =B =C =0
For the first term,
B AND C =0, NOT(B AND C) =1
A AND NOT(B AND C) =0 AND 1 =0.
Similarly, the other two terms will also be 0.
Hence, X =0 +0 +0
→X =0 OR 0 OR 0
=0.
34. Design a logical circuit using AND, OR, and NOT gates to evaluate AB̅C̅ +BC̅A̅.
ANSWER: The given expression may be written as,
A AND NOT(B AND C) +B AND NOT(C AND A)
The logical circuit to evaluate it may be understood through the following figure.
 |
| Design of logical circuit for Q-34 |
35. Show that AB +A̅B̅ is always 1.
ANSWER: AB +A̅B̅ means,
(A AND B) OR NOT(A AND B).
A or B may take a value of either 1 or 0. Depending upon the values of A and B, the value of AB will be either 1 or 0. So the value of "NOT(AB)" will be 0 and 1 respectively. Thus the given expression is either,
1 OR 0 =1
or
0 OR 1 =1.
So the given expression is always zero.
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