ROUTERA


Semiconductors And Semiconductor Devices

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

How many 1s energy states are present in one mole of sodium vapor? Are they all filled in normal condition? How many 3s energy states are present in one mole of sodium vapor? Are they all filled in normal conditions?


Answer:

One mole of sodium vapor has 6.02 × 1023 atoms (Avogadro’s number). A sodium atom has atomic number 11. The electronic configuration of sodium is.


Therefore, there are 2× 6.02 × 1023 = 12.04 × 1023 (number of electrons occupying a state × Avogadro' s number) 1s energy states in one mole of sodium vapor and they are all filled in normal conditions.


Also, there are 1×6.02× 1023 = 6.02× 1023. 3s energy states in one mole. The 3s energy states is partially filled in normal conditions.



Question 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:

A single atom has discrete energy levels. Let us now take N atoms and assume that it is possible to vary the inter-atomic distance. When the inter-atomic distance (distance between the two atoms) is large, the energy levels of total N atoms coincide with those of a single atom. Let us now reduce the inter-atomic distance. This causes an atom to exert an electric force on its neighbors. The closely packed atoms have energy levels overlapped and hence closely spaced energy levels are formed so that the system now obeys Pauli Exclusion Principle. These closely spaced energy levels are discrete and called as energy bands. We don’t have continuous energy variation in a band.



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:

An insulator has a large energy band gap between the conduction band and valence band. There will be no carriers in the conduction band at 0 K as well as at 300 K.

A semiconductor has a moderate energy band gap and there will be no carriers in the conduction band at 0 K. There will be carriers in conduction band at room temperature.


A conductor has overlapped conduction and valence band (energy band gap do not exist). The conduction band is partially filled with carriers at 0 K and it will be fully filled at room temperature. Therefore, the material is a conductor in which the conduction band is partially filled at 0 K.



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:

Thermal collisions continue to generate some number of electron-hole pairs due to the jumping of electrons from valence band to conduction band while other electron-hole pairs disappear due to the recombination process. Recombination is the process where an electron moves from the conduction band to the valence band so that a mobile electron-hole pair disappears. The electron in the conduction band will lose its energy when it collides with atoms and hence it comes back to the valence band filling an empty place, thus recombination takes place. Therefore, recombination is the main reason for the conduction electrons not go on increasing with time as thermal collisions continuously generates electron-hole pair.



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 of the electrons from the valence band to the conduction band?


Answer:

Thermal collision means that the electrons have already acquired sufficient energy due to the room temperature (300 K) and it moves freely thus colliding with lattice atoms and transferring its energy to other electrons in the atom.


At room temperature, some of the electrons occupying the highest energy level in the valence band will acquire enough energy (greater than 1.1 eV) and hence they jump to the conduction band from the valence band before any collisions take place. Also some excited electrons (energy less than 1.1 eV) occupying lower energy levels in the valence band collides with other excited electrons occupying the highest energy level in the valence band. Due to this collision, there is the exchange of 0.026 eV which makes these electrons to jump from higher energy level of valence band to the conduction band. The thermal collision as well as already applied room temperature are the reason for this movement of electrons.



Question 6.

What is the resistance of an intrinsic semiconductor at 0K?


Answer:

The carriers are absent in the conduction band of an intrinsic semiconductor at 0 K. Therefore, the conductivity is zero and hence resistivity is infinity. Thus the resistance is infinite (much large) of an intrinsic semiconductor at 0 K.



Question 7.

We have valence electrons and conduction electrons in a semiconductor. Do we also have ‘valence holes’ and ‘conduction holes’?


Answer:

We have valence electrons in the outer orbit of a semiconductor. We have conduction electrons in the conduction band of semiconductor at 300 K. When an electron jumps from the valence band to the conduction band, a hole is created in the valence band at the place from where that electron jumped. There is no such concept of ‘valence holes’ and ‘conduction holes’. A place void of electron is a hole.



Question 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:

When a p-type impurity is added in a pure semiconductor, a large number of holes are created. Thus, each impurity atom is deficient of one electron and they accept the electrons which are loosely bound to other neighboring atoms. So, each impurity atom acquires negative charges. The number of negatively charged impurity atoms and the number of holes are equal in p-type semiconductor. Thus, it is electrically neutral. But when holes diffuse from the p-side to the n-side in a p-n junction or in other words we say that when an electron diffuses from the n-side to the p-side in a p-n junction, it leaves a vacant place in the n-side region and thus the n-side gets positively charged. Also, the p-side gets negatively charged due to the diffusion of electrons from the n-side.



Question 9.

The drift current in a reverse based p-n junction increases in magnitude if the temperature of the junction is increased. Explain this on the basis of creation of hole-electron pairs.


Answer:

The electric field is directed from n-side to p-side across the p-n junction. When temperature is increased, more covalent bonds break up and hole-electron pairs are generated across the junction. The electric field attracts the electrons towards n-side while it repels the holes towards p-side. Thus the direction of the current due to this movement of holes and electrons is same as the direction of drift current flowing from the n-side to the p-side. Hence the current flowing due to the increase in temperature adds up to the drift current flowing initially in the reverse biased p-n junction diode. Therefore, the total drift current is increased.



Question 10.

An ideal diode should pass a current freely in one direction and should stop it completely I the opposite direction. Which is closer to ideal—vacuum diode or a p-n junction diode?


Answer:

Vacuum diode consists of two electrodes- a cathode and an anode. The cathode emits free electrons while the anode collects the free electrons. Thus the current only flows from the anode to the cathode. A large current flows in forward biased p-n junction diode while a small current flows in reverse-biased p-n junction diode. Therefore, vacuum diode is close to an ideal diode which passes current only in one direction.



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 output/collector current of the amplifier circuit is given by IL=IC=β×IB , where IB is the base current and . α is the fraction of emitter current received at the collector. This means that IC=α×IE and value of is about 0.95 to 0.99. Let us put in equation of and therefore . The power gain at the output is directly proportional to β2. Putting β=99 then power gain is proportional to 992=9801 Hence the output power is several times greater than the input power. The power gain or extra power is coming from the relationship between α and β parameters.




Objective I
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:

Assuming the Semiconductor is at room temperature (300°K), there are equal numbers of free electrons and holes present in the semiconductor. Hence both electrons and holes participate in the electrical conduction in a semiconductor. Therefore, the correct answer is option C.


Question 2.

An electric field is applied to a semiconductor. Let the number of charge carriers be n and the average drift speed b 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:

If the temperature is increased, there will be more breaking of the Covalent bonds and more charge carriers are generated. This leads to total increase in moving charge carriers and thus their average drift speed will now be decrease due to increase in probability of collision with the lattice atoms of semiconductor. Hence number of charge carrier n will increase but the average drift speed v will decrease. Therefore, the correct answer is option B.


Question 3.

Let np and ne be the numbers of holes and conduction electrons in an intrinsic semiconductor.
A. np > ne

B. np = ne

C. np < ne

D. np ≠ ne


Answer:

Intrinsic semiconductor is pure semiconductor which belongs to the group fourth A of the periodic table. At room temperature (300°K), some covalent bonds break and equal number of electrons and holes (the place from where the electron vacated its place) are generated. Hence np=ne. So, the correct answer is option B.


Question 4.

Let np and ne be the numbers of holes and conduction electrons in an extrinsic semiconductor.
A. np > ne

B. np = ne

C. np < ne

D. np ≠ ne


Answer:

Extrinsic semiconductor made up by adding trivalent impurities are called p-type semiconductor. In p-type semiconductor, the number of holes is greater than number of electrons (np > ne).

Extrinsic semiconductor made up by adding pentavalent impurities are called n-type semiconductor. In n-type semiconductor, the number of electrons is greater than number of holes (np < ne).


However, the question does not mention which type of extrinsic semiconductor is considered. Option B is true for intrinsic semiconductor. Hence we can say that the number of electrons and holes are not equal for an extrinsic semiconductor. Therefore, the correct answer is option D.


Question 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:

When a p-type impurity is added in a pure semiconductor, a large number of holes are created in a p-type semiconductor. Thus, each impurity atom is deficient of one electron and they accept the electrons which are loosely bound to other neighboring atoms. So, each impurity atom acquires negative charges. The number of negatively charged impurity atoms and the number of holes are equal in p-type semiconductor. Thus, it is electrically neutral. Therefore, the correct answer is option C.


Question 6.

When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor.
A. increases

B. decrease

C. remains the same

D. becomes zero


Answer:

When impurity (trivalent or pentavalent) is added into an intrinsic semiconductor, extrinsic semiconductor (p-type or n-type respectively) is obtained. This has increased holes or electrons respectively in conduction band (intrinsic carriers plus the carriers due to impurity). Hence the conductivity of the extrinsic semiconductor increases. Therefore, the correct answer is option A


Question 7.

If the two ends of a 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 a stead 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:

Let us consider there will be current (I≠0) flowing through the two ends of a P-N junction joined by a wire. This should generate heat at metal ohmic contacts (the metal used to connect p-n junction and the wire). This heat must be supplied by the p-n junction since there is no other external excitation applied. The p-n junction, therefore, would have to cool off. Clearly, under thermal equilibrium the simultaneous heating of metal ohmic contacts and cooling of the p-n junction is impossible and we conclude that I꞊0. Therefore, the correct answer is option A.


Question 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:

An open-circuited p-n junction has drift current flowing from the n-side to the p-side. The option A is true.

An open-circuited p-n junction has diffusion current flowing from the p-side to the n-side. The option B is false.


The drift current flowing from the n-side to the p-side if the junction is forward-biased, is true but the drift current flowing from the p-side to the n-side if the junction is reverse-biased, is false. Hence, the option C overall is false.


The drift current flowing from the p-side to the n-side if the junction is forward biased is false but the drift current flowing from the n-side to the p-side if the junction is reverse-biased, is true. Hence the option D overall is false.


Therefore, the correct answer is Option A.


Question 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:

An open-circuited p-n junction has drift current flowing from the n-side to the p-side. The option A is false.

An open-circuited p-n junction has diffusion current flowing from the p-side to the n-side. The option B is true.


The diffusion current flowing from the n-side to the p-side if the junction is forward-biased, is false but the diffusion current (approx zero) flowing from the p-side to the n-side if the junction is reverse-biased, is true. Hence, the option C overall is false.


The diffusion current flowing from the p-side to the n-side if the junction is forward biased, is true but the diffusion current flowing from the n-side to the p-side if the junction is reverse-biased, is false. Hence the option D overall is false.


Therefore, the correct answer is Option B.


Question 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 reverses biased

C. if the junction is unbiased

D. in no case.


Answer:

In an unbiased p-n junction, the drift current is equal to diffusion current to keep the net current equal to zero. Hence Option C is false.

The junction/barrier potential is reduced by the forward biased applied voltage and hence more holes diffuse on the n-side and more electrons diffuse on the p-side. The diffusion current is large when compared to the drift current and hence the net current is in the direction of diffusion current. Hence option A is true.


The junction/barrier potential is increased by the reverse biased applied voltage. This blocks the diffusion of holes and electrons. The drift current (in microamperes) is large when compared to the diffusion current and hence the net current is in the direction of drift current. Hence option B is false.


Therefore, the correct answer is option A.


Question 11.

Two identical p-n junctions may be connected in series with a battery in three ways figure. The potential difference across the two p-n 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


Answer:

Circuit 1: The first p-n junction is forward biased since the p-side is at greater potential than the n-side. Also, the second p-n junction is reverse biased since its p-side is at lower potential than its n-side. Therefore, the potential across the first (forward biased) and the second (reverse biased) p-n junction is not equal.

Circuit 2: The first p-n junction is forward biased since the p-side is at greater potential than the n-side. Also, the second p-n junction is forward biased since its n side is at lower potential than its p side. Therefore, the potential across the first and the second p-n junction (both forward biased) is same.


Circuit 3: The first p-n junction is reverse biased since the n-side is at greater potential than the p-side. Also, the second p-n junction is reverse biased since its p-side is at lower potential than its n-side. Therefore, the potential across the first and the second p-n junction (both reverse biased) is same.


Therefore, the correct answer is option B.


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:

The p side of the p-n junction diode is at higher potential than the n side and hence the diode is forward biased in figure (a). Assuming ideal diode case, the diode is treated as short circuit. The figure (a) network thus becomes a series source free RC circuit. The voltage equation in source free series RC circuit is given by


At t=0,


At t=RC,


Since, the charge is given by.


At t=RC, the charge is


The diode in the figure (b) is reverse biased since its p side is at low potential than its n side. Assuming ideal diode case, the diode is treated as open circuit in figure (b). Thus the charge on the capacitor at all times is


Therefore, the correct answer is option B.


Question 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 compete it.

D. a bond is broken on the p-side and the electron freed form the bond jumps to a broken bond on the n-side to complete it.


Answer:

Diffusion is a natural phenomenon which occurs when there is concentration difference between the two regions. The covalent bond is formed between the trivalent (or pentavalent) and semiconductor atom in p-type (or n-type). Thus there are many vacant places (holes) and electrons in the p-side and n-side respectively. Thus there are less or no electrons in p-side while there are many electrons in n-side. Clearly, there is concentration difference of electrons across the junction and diffusion will take place. This forces the covalent bond to be broken on the n-side so that the electrons freed from this bond, jumps to the vacant places in the p-side. Also when the electron jumps from the n-side to the p-side, a vacant place (hole) is created in n-side which represents that a hole has been moved from the p-side to the n-side.

Therefore, the correct answer is option C.


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 concentration of impurity.


Answer:

The emitter has the greatest concentration of impurity so that the large number of carriers traverses the path to the collector via base and hence contributes to desire current. Some carriers (holes or electrons) while traversing this path get recombine with carriers (electrons or holes) present in base and thus current will decrease. If the base has impurity concentration greater than the collector, more carriers get recombine in the base and less carriers will go to the collector thus contributing current less than desired. Hence, the base should have impurity concentration less than the impurity concentration in the collector so that there will be much less recombination in base and it contributes to desire current. Therefore, the base has least impurity concentration to keep recombination much less. The collector is moderately impure. So the correct answer is option C.


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:

To make transistor work as an amplifier, the emitter-base junction must be forward biased while the collector-base junction is reverse-biased.

Therefore, the correct answer is option D.



Objective Ii
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.) Simply, when the temperature decreases the motion of electrons decreases and at 0K their motion tends to cease due. Thus, no conduction or free electrons are possible a 0K.


Therefore, option A is correct.


B.) At any temperature above 0K, in semiconductors the bonds are continuously breaking and producing electron-hole pairs. Thus, there will always be some conducting electrons present for conduction.


Therefore, option B is incorrect.


C.) As the temperature increases the heat energy tends to break the covalent bonds between the atoms which leads to the creation of free electrons. Thus, more temperature means more breaking of covalent bonds and more free electrons.


Therefore, option C is correct.


D.) The energy band gap of a conductors is very less than that of a semiconductor.


Therefore, option D is correct.


Question 2.

In a p-n junction with open ends,
A. there is not systematic motion of charge carriers

A. 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 not net charge transfer between the two sides

D. there is a constant electric field near the junction.


Answer:

For A.) and B.)


Diffusion of the electrons and holes from n and p side to p and n side, respectively, due to higher a concentration of holes on the p-side and lower at n-side and vice-versa for electrons results in systematic motion of charges take place.


Therefore, option A is incorrect and B is correct.


C.) Since, there are same number of electrons transferred as the number of holes transferred. Thus, the net charge transferred is zero.


Therefore, the option C is correct.


D.) As the During equilibrium there is no flow of charges at junction therefore the electric field there remains constant.


Therefore, the option D is correct.


Question 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:

For A.) and B.)


The formation of the new holes and conduction electrons happens continuously and throughout the materials because the heat of room temperature affects all region of the material equally. Therefore, option A is correct and B is incorrect.


For C.) and D.)


In the depletion region the electrons-hole pairs produced move away from each other due to the force of electric field. Thus, holes and conduction electrons do not recombine continuously throughout the material and option D.) is correct and C.) is wrong.


Question 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:

To make p-type semiconductor we dope Silicon with trivalent atoms, i.e., B.) Boron and D.) Aluminum, since trivalent atoms will create holes.


Question 5.

The electrical conductively 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:

For A


Increasing the temperature increases the breaking of the covalent bonds and generation of more electron-hole pairs. Thus, increasing the conductivity.


For B and C


Since doping will increase the no. of charge carriers, therefore the conductivity will increase.


For D


Irradiating UV light on germanium will provide energy enough to break the covalent bond and thus create an electron-hole pair. Thus increasing the electrical conductivity.


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 p-type semiconductor

C. an n-type semiconductor

D. a p-n junction


Answer:

For A.), B.) and C.)


In case of intrinsic semiconductor, it will always be the case that the electrons will flow to the positive terminal and holes to the negative terminal polarity because the holes and electrons are uniformly distributed. And similar would be the case for both types of semiconductor.


For D.)


In p-n junction, when the p-side is connected in forward bias, the depletion region decreases and the charge carriers are able to diffuse and a diffusion current is produced. But when the polarity is reversed, the depletion region increases, which decreases the diffusion current. Therefore, the current in p-n junction only flows in forward biasing and reduces to zero in reverse biasing.


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:

A donor will form four bonds with neighboring atoms of the semiconductors and the fifth electron of the donor atom will be responsible for conduction. Thus, the concentration of electrons tends to increase. Then some of these electrons recombine with holes which lead to the decrease in the concentration of the holes. Therefore, B and C are correct.


Question 8.

Let iE, iC and iB represent the emitter current, the collector current and the base current respectively in a transistor. Then
A. iC is slightly smaller than iE

B. iC is slightly greater than iE

C. iB is much smaller than iE

D. iB is much greater than iE.


Answer:

So, if we consider p-n-p transistor, the emitter will have holes as the majority charge carriers and they will move to the base due to forward biasing of collector-base junction. Very few holes will combine with the electrons of the base (since it is n-type) due to the thin size and light doping of the base. Because of reverse biasing at collector-base junction, majority of the holes will move to the collector side and constitute Ic which will be slightly less than Ie since no. of holes reaching the collector is less than that at emitter. Also, at the base some holes from the emitter can escape from the base terminal but due very thin size of the base a very little holes can only pass through terminal and constitute Ib and thus Ib is much smaller than Ie.


Thus, from explanation it can be seen that option A and C are correct.


Question 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:

The normal operation of a transistor is applied for amplifying signals in the circuit. Configuration A and D of biasing will allow the current to amplify according to the input current.


Therefore, option A and D are correct and B and C are incorrect as they are contrary to D and A respectively.


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:

NOR and NAND gates are the universal or the basic gates. Using these gates any other gate can be created. Following figure shows AND gate using NAND and NOR gate respectively.


AND gate using two NAND gate.




AND gate using three NOR gates



Exercises
Question 1.

Calculate the number of states per cubic metre of sodium is 3s band. The density of sodium is 1013 kg m–3. How many of them are empty?


Answer:

Electronic configuration of Sodium atom is → 1s22s22p63s1


Each 3s orbit will have two state and one of each would be occupied,


∴ in N atoms of sodium there will be 2N states possible states and N states will be empty.


Given:


mass of the sodium in 1 m3 = 1013 kgm-3


∴ the total number of atoms


=


Therefore, total number of states



And total unoccupied states are



Question 2.

In a pure semiconductor, the number of conduction electrons is 6 × 1019 per cubic metre. How many holes are there in a sample of size 1 cm × 1cm × 1 mm?


Answer:

Given: Number of conduction electrons


Since in pure semiconductor the number of holes = number of conduction electron


∴ number of holes


∴ number of holes in


∴ answer is



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:


Bandgap energy of antimonite


Now for temperature, we solve following equation,



Where,




∴ at temperature 2670 K, is equal to 0.23eV



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 tenth of the value of 300 K? (Silicon will not retain its structure at these high temperatures)


Answer:

Given:


Band gap for silicon and at room temperature


a) Ratio of band gap to


b) One-tenth of the ratio in part a)


Now,




∴ Temperature = 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 revives an energy equal to 2kT and just reaches one of the acceptor levels. Assuming that the energy of the electron was at the topo 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:

Given:


Indirectly in the problem we are given the acceptor level .


This is because the electron was at the top of the edge of valance band and reaches acceptor level on receiving the energy given.


∴ energy of the acceptor level must




Answer is 51.72meV



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:

For the electron to move from the conduction band to valance band it will have to loose energy same as the energy band gap of ZnO, i.e., 3.2.


The energy lost is in the form of electromagnetic radiation, for which the energy is given as following,



where,


,



is the wavelength





Question 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:

Minimum energy will produce maximum wavelength and ∴ we consider the energy lost by the electrons is only due to it moving from the conduction to valence band, which is the energy band gap


The energy lost is in the form of electromagnetic radiation, for which the energy is given as following,



where,


,



is the wavelength



∴ Energy band gap = 1.514eV



Question 8.

Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium. The band gap in germanium is 0.65 eV.


Answer:

Only if the electron gains energy to move to the conduction band it will lead to the creation of electron-hole pairs.



The energy of electromagnetic wave,



∴ maximum wavelength of electromagnetic radiation which can create e-h pair in germanium is



Question 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:

The conduction if take place if the energy provided to the electron is more equal or more than the band gap.


Given:


Maximum wavelength at which the conduction starts


∴ this wavelength provides minimum energy to the electron to move to conduction band,


Energy of this radiation=Energy band gap



∴ energy band gap = 2.003eV ≅ 2eV



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–ΔE/2kT. Find the ratio of the concentration of conduction electrons in diamond to that in 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:

Given: energy band gap,


For silicon,


For diamond,


Also, we know that concentration of conduction electron(n) is


proportional to e–ΔE/2kT



For silicon,


For diamond,


We want,


Since, is very large, due to which the value of is in the negative power of 51, thus almost zero conduction electrons will be there in the diamond. Due to this diamond is an insulator.



Question 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:


At temperature 4K, E1= 0.74


At temperature 300K, E2=0.67


The conductivity is proportional to T3/2 e–ΔE/2kT,


Let conductivity at 4K be =σ1


and at 300k be = σ2





∴ conductivity increases by the factor of



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 × 1015 holes per cubic meter. Density of silicon is 5 × 1028 atoms per cubic meter.


Answer:

Given:


Since, the number conduction electron and holes are the same,


∴ Number of charge carriers initially


On adding the impurity to the pure silicon the conductivity becomes 100 times the initial , therefore the total charge carriers should become


(∵ σ is directly proportional to concentration of charge carriers)


Let be the holes after adding the impurity and will be the electrons


Now, the product of the total conduction electrons and holes remains nearly constant,





Solving for by discriminant method,


We get



Rejecting the second value on the basis of the fact that the initial number of holes must be smaller than the number of holes after doping.


∴ the number of holes or boron atoms added



Ratio of number of silicon atom to boron atom



∴ boron must be added in proportion of 1 boron atom in 3.59× 1010 silicon atoms.



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 × 1019 per cubic meter. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to 2 × 1023 per cubic meter. Find the concentration of the holes in the doped germanium.


Answer:

Given:


Initially,


Concentration of conduction electron


Concentration of holes


After doping,


Concentration of conduction electron


Concentration of holes


We know,


Product of concentration of conduction electron and that of holes remains constant,




∴ number of holes after doping



Question 14.

The conductivity of an intrinsic semiconductor depends on temperature as σ = σ0 e–ΔE/2kT, where σ0 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 T1=300K and T2=temperature at which the conductivity will be double of its value at T1.

And let the conductivities at temperatures T1 and T2 be σ1 and σ2 respectively.


We are given,


Band gap, ΔE= 0.650 eV



12





Taking natural log both sides,


-11.874525


On solving, we get T2=317.512K.



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. Assumed 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 0K. The concentration of the holes increase 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:

Originally the band gap was 1eV. After doping, the band gap will become =(1eV-1meV)

= (1-0.001) eV


=0.999 eV


It is given that the transition is almost forbidden at 1/50 of 0.999eV. Let the upper limit for temperature be T1 over which the transition becomes a forbidden transition.


• kT1=0.999/50


• T1=231.78K≈231.8K


Let the lower limit for temperature be T2 for which the transitions are most feasible or for which the upper levels have maximum population.


• kT2=2(0.999)


• T2=23.2K


Therefore, the required temperature range=(23.2K-231.8K)



Question 16.

In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 × 105 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 n-side to the p-side.


Answer:

Width of depletion region, d= 400 nm= 4×10-7 m

Electric Field, E= 5×105 Vm-1


(a) Let the height of potential barrier be V Volts.


We know that,


• E=V/d


• V=Ed


• V=5×105×4×10-7


• V=0.2 Volts


(b) The minimum kinetic energy required for diffusion of an electron from n-side to p-side


K.Emin= Potential barrier × Charge on an electron


= 0.2 V × 1.6 × 10-19 C


= 0.2 eV



Question 17.

The potential barrier existing across an unbiased p-n junction is 0.2 volt. 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 volt and

(c) the junction is reverse biased at 0.1 volt?


Answer:

Potential barrier across unbiased p-n junction = 0.2 Volts

Minimum kinetic energy required for diffusion of a hole from p-side to the n-side,


K.E.min = (Potential barrier - biasing voltage) × Charge on hole


(a) Biasing Voltage=0V


K.E.min = (0.2-0) × e


= 0.2eV


(b) Biasing Voltage for forward biased=+0.1V


K.E.min = (0.2-0.1) × e


= 0.1eV


(c) Biasing Voltage for reverse biased=-0.1


K.E.min = (0.2+0.1) × e


= 0.3eV



Question 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:

Potential barrier, d = 250 meV

Initially the kinetic energy of the hole = 300 meV


(a) When the hole approaches the junction from p-side, junction will act like forward biasing.


Therefore, the kinetic energy of the hole will decrease.


Final kinetic energy of the hole = (300-250) meV


= 50 meV


(b) When the hole approaches the junction from n-side, junction will act like reverse biasing.


Therefore, the kinetic energy of the hole in this case will increase.


Final kinetic energy of the hole = (300+250) meV


= 550 meV



Question 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 diffusion current when the diode is

A. unbiased

B. reverse biased at 200 mV and

C. forward biased at 200 mV.


Answer:

It is given that drift current is 25 μA and biasing current is 75 μA at 200 mV forward biasing.

A. We know that when the diode is unbiased,


Diffusion current = drift current = 25μA


B. When the diode is reverse biased, diffusion current = 0


C. When the diode is forward biased at 200mV,


Diffusion current - Drift current = Forward biasing current


Diffusion current – 25 μA = 75 μA


Diffusion current = (75+25) μA


Diffusion current =100 μA



Question 20.

The drift current in a p-n junction is 20.0 μA. Estimate the number of electrons crossing a cross section per second in the depletion region.


Answer:

Drift current, Id = 20μA


Where, Ne= Number of electrons crossing a cross section per second


Nh= Number of holes crossing a cross section per second


e= magnitude of charge on an electron/hole = 1.6×10-19 C


Without any biasing applied on the junction, Ne = Nh = N


Id=2N×e




N=6.25×1013



Question 21.

The current voltage characteristic of an ideal p-n junction diode is given by

i = i0 (eeV/kT – 1)

Where the drift current i0 equals 10 μA. Take the temperature T to be 300 K.

(a) Find the voltage V0 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 > V0.

(c) Find the voltage for which the dynamic resistance is 0.2 Ω


Answer:

The current voltage characteristic of ideal p-n junction diode is


(a) For large value of voltages, 1 can be neglected. So, the relation becomes



We need to find the value of V0 for which eev/kT=100


It is given that T=300K, i0=10μA and i=100


Taking natural log to both sides,




V0=0.12V


(b) Dynamic Resistance of the diode, R= Rate of change of voltage with respect to current



We know,


Differentiating both sides with respect to V,





(c) Given, R = 0.2Ω






k=8.62×10-5


T=300K


i0=10μA


R=0.2Ω


e=1.6×10-19 C


On substituting the values, we will get V=0.25 V



Question 22.

Consider a p-n junction diode having the characteristic i= i0(eeV/kT – 1) where i0 = 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:

Given, i0 =20 μA =20×10-6 A
And temperature, T=300K

a) Forward biasing voltage, V=300mV
According to the current diode equation,





b) Now the current has doubled, i.e. i = 4 A


According to the current diode equation,


• i = i0 (eeV/kT – 1)





Taking log both sides,




• V= 318mV



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.




Answer:

From the given circuit diagram, we can see that the diode is reverse biased.

• Drift current = Diffusion current = 20 μA


∴ Current through the circuit =20 μA


Voltage drop across the 20Ω resistor = 20μA×20Ω


= 4×10-4 V


Voltage through the circuit= (Voltage from the source)–(Voltage drop across the 20Ω resistor)


=5-0.0004


=4.9996 V ≈5V



Question 24.

Each of the resistances shown in figure has a value of 20 Ω. Find the equivalent resistance between A and B. Does it depend on whether the point A or B is at higher potential?





Answer:

The given figure can also be made as:

which looks like a Wheatstone bridge. Hence, current through the middle arm = 0A


∴ Current through the diode = 0A.


Equivalent circuit diagram becomes:



∴ Net resistance


• R =


• R = 20ohm



Question 25.

Find the currents through the resistances in the circuits shown in figure.




Answer:

(a) In fig1, both diodes D1 and D2 are forward biased. So, resistance due to diodes=0.
Net resistance = 2Ω
Net current = 2V/2Ω = 1A


(b) In fig2, diode D1 is forward biased and D2 is reverse biased. So, no current will flow through this circuit because D2 will offer infinite resistance.
Net current = 0A


(c) In fig3, diode D1 and D2, both are forward biased, So, resistance due to diodes=0.
Net resistance = 2Ω
Net current = 2V/2Ω = 2A


(d) Diode D1 is forward biased and D2 is reverse biased. So current will flow through D1 but not through D2.
Net resistance = 2Ω
Net current = 2V/2Ω = 2A



Question 26.

What are the readings of the ammeters A1 and A2 shown in figure? Neglect the resistances of the meters.




Answer:

Diode is in reverse bias.


Current i1=0A as the diode is in reverse bias.


Current i2 = 2V/10Ω = 0.2A


∴ Reading in A1=0A and reading in A2=0.2A



Question 27.

Find the current through the battery in each of the circuits shown in figure.




Answer:


fig1


In fig1, both diodes D1 and D2 have forward biasing. Net diode resistance=0.
Net resistance =(10×10)/(10+10) =5Ω
∴ Current = 5V/Net resistance =5V/5Ω =1A



fig2


In fig2, diode D1 has forward biasing and diode D2 has reverse biasing. Hence, current will flow only through diode D1.
Net resistance =10+0 =10Ω
∴ Current =5V/10Ω



Question 28.

Find the current through the resistance R in figure if (a) R = 12 Ω (b) R = 48 Ω




Answer:


Net potential difference across arm PQ is negative.


• The diode is reverse biased.


• No current flows through arm PQ


Net resistance = (R + 12) Ω


(a) R=12Ω
Net resistance =12+12 =24Ω
Current= (4V+6V)/24Ω = 0.42A


(b) R=48Ω
Net resistance=48+12 = 60Ω
Current= (4V+6V)/60Ω = 0.16A



Question 29.

Draw the current voltage characteristics for the device shown in figure between the terminals A and B.






Answer:

In the first figure it is clear that the only one diode is present between A and B. The I-V characteristics will be:


In the second figure, the diode in the upper arm in series with the 10Ω resistor has p-side connected to A and n-side connected to B.
Whereas, the other diode has exactly opposite polarities. When a potential difference is applied across terminals A and B, one of the diodes will become reverse biased and there will be no flow of current through that arm. ∴ I-V characteristics between terminals AB will be same as the previous figure.



Question 30.

Find the equivalent resistance of the network shown in figure between the points A and B.





Answer:

Case 1: Potential at terminal A > Potential at terminal B
The diode will be forward biased and it can be replaced by short circuit.


Net resistance = = 10/2 = 5Ω



Case 2: Potential at terminal A < Potential at terminal B
The diode will be reverse biased and it can be replaced by open circuit.


Net resistance=10Ω



Question 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 β.


Answer:

Current gain, β=Rate of change of collector current with respect to base current at constant output voltage

• β=


• β=0.05×103


• β=50



Question 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 β = 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 current gain β=50

Input current in a common emitter mode is the base current. Change in base current or input current = δIb=50μA


Output voltage, V0 =β×RG=50×20.5=200V


(a) Voltage Gain, VG=V0/V1


Input voltage, V1= δlb+Ri


Input resistance, Ri = 0.5kΩ = 500Ω


VG = 200/(50×106+500)


VG=8000V


(b) Change in input voltage, δV1 = δlb×Ri


= 50×106×5×102


=25×103 V


=25 mV


(c) Resistance Gain, RG= Load Resistance/Input resistance


= RL/Ri


Power gain =β2×RG


2×RL/Ri


=2500×2/0.5=2500×20.5
=2500×205=104



Question 33.

Let evaluate X for

(a) A = 1, B = 0 C = 1

(b) A = B = C = 1 and

(c) A = B = C = 0.


Answer:


(a) A=1, B=0, C=1



X = 1×1+ 0×0 + 1×0


X=1


(b) A=B=C=1



X = 1×0 + 1×0 + 1×0


X=0


(c) A=B=C=0



X = 0×1 + 0×1 + 0×1


X=0



Question 34.

Design a logical circuit using AND, OR and NOT gates to evaluate .


Answer:



Question 35.

Show that is always 1.


Answer:

We will show this using a truth table.

Also, we can let AB=X


As we know,


,