NEET Physics Class 12 Chapter 6 Solids And Semiconductors Devices Notes

Solids And Semiconductors Devices

Electronic instruments are being utilized in various fields like telecommunication, entertainment, computers, nuclear physics, and many more. Although history started with the advent of vacuum tubes, however the rapid advancement in electronics which we see today is due to the valuable contributions of semiconductor devices.

Semiconductor devices are not only small in size, consume less power, have long lifetimes, and are more efficient than vacuum tubes but also are of low cost. That is why these have replaced vacuum tubes nearly in all applications. As an example, we can consider the case of a computer.

In the early days, vacuum tube-based computers were as big as the size of a room and were capable of performing simple calculations only. At present the personal computer (PC) that you see in the laboratory or at your home is much smaller in size and capable of performing many operations. Needless to say, this is possible because of the advances in semiconductor technology. We will learn the basic concept of semiconductors.

This will enable us to understand the operation of many semiconductor devices and then we will be discussing a few semiconductor devices like diodes, and transistors along with their applications.

Energy Levels And Energy Bands Inb Solids

The electrons of an isolated atom are restricted to well-defined energy levels.

The maximum number of electrons that can be accommodated at any level is determined by the Pauli exclusion principle. The electrons belonging to the outermost energy level are called valence electrons.

For example, the electronic configuration of sodium (atomic number 11) is –1s2 2s2 2p6 3s1, here the electron belonging to the 3s level is the valence electron. Most of the solids including metals with which we are familiar occur in crystalline form.

As we know a crystal is a regular periodic arrangement of atoms separated from each other by a very small distance called lattice constant. The value of the lattice constant is different for different crystalline solids, however, it is of the order of linear dimension of atoms {~Å}.

Obviously, at such a short separation between various neighboring atoms, electrons in an atom cannot only be subjected to the Coulombic force of the nucleus of this atom but also by Coulombic forces due to nuclei and electrons of the neighboring atoms. It is this interaction that results in the bonding between various atoms which leads to the formation of crystals.

When atoms are interacting (such as in a crystal) then the energy level scheme for the individual atoms does not quite hold. The interaction between atoms markedly affects the electron energy levels, as a result there occurs a splitting of energy levels belonging to various atoms.

To understand this phenomenon in more clear terms, let us first consider the simplest case of two interacting identical atoms. Let us assume that initially they are far apart i.e. the forces of interaction between them can be neglected.

[If the distance between two atoms is much larger (~50Å) compared to their linear dimensions (~ 10Å) this assumption is reasonably correct].In such a case we may treat them as isolated with energy levels like that for the case of an isolated atom.

In crystals the number of atoms, N is very large of the order of 10 22 to 1023 per cubic centimeter, so each energy band contains as many levels as the number of atoms. The spacing between various levels within a band is therefore very small.

If for example, we assume the total width of a band of energies as 1 eV and 1022 levels are to be accommodated within this band, then the average spacing between the adjacent levels is about 10-22 eV.

For all practical purposes, therefore, energy within a band can be assumed to vary continuously. The formation of bands in a solid is shown schematically

Energy Bands:

This theory is based on the Pauli exclusion principle. In an isolated atom, the valence electrons can exist only in one of the allowed orbitals each of a sharply defined energy called energy levels. But when two atoms are brought nearer to each other, there are alterations in energy levels and they spread in the form of bands.

Energy bands are of the following types

Valence band: The energy band formed by a series of energy levels containing valence electrons is known as the valence band. At 0 K, the electrons fill the energy levels in valence band starting from the lowest one.

1. This band is always filled with electrons.
2. This is the band of maximum energy.
3. Electrons are not capable of gaining energy from external electric fields.
4. No flow of current due to electrons present in this band.
5. The highest energy level that can be occupied by an electron in valence band at 0 K is called the Fermi level.

Conduction band: The higher energy level band is called the conduction band.

1. It is also called an empty band of minimum energy.
2. This band is partially filled by the electrons.
3. In this band, the electron can gain energy from the external electric field.
4. The electrons in the conduction band are called the free electrons. They can move anywhere within the volume of the solid.
5. Current flows due to such electrons.

Forbidden energy gap (ΔEg): Energy gap between conduction band and valence band

⇒ \(\Delta E_g=(\text { C.B. })_{\min }-(\text { V.B. })_{\max }\)

Conductor, Insulator And Semiconductor:

The electrical conductivity of materials is a physical quantity that varies over a large span. On one hand, we know about metals having very large values of electrical conductivity, and on the other hand, we have insulators like quartz and mica having negligible conductivity.

Besides these there are materials having conductivity (at room temperature) much smaller, than that of metals but much larger than that of insulators these materials are called semiconductors Example Silicon and Germanium.

Not only that the conductivity of a semiconductor intermediate, but to that of metals and insulators the conductor semiconductor varies substantially with temperature. For very low temperatures (around 0K) semiconductor behaves like an insulator, however, its conductivity increases with an increase in temperature.

Conductors:

These are solids in which either the energy band containing the valence band is partially filled or the energy band containing valence electrons overlaps with the next higher band to give a new band that is partially filled too. For both these situations there are enough free levels available for electrons to which they can be excited by receiving energy from an applied electric field. Let us consider an example of sodium which is a monovalent metal.

Its band structure is such that 1s, 2s, and 2p bands are filled with electrons to their capacity however, the 3s band is only half-filled.

The reason for such a band structure is that for an isolated sodium atom in its electronic structure 1s 2, 2s2, 2p6, 3s1 the energy levels 1s, 2s, and 2p are filled while 3s contain only one electron against its capacity of accommodating two electrons.

The filled 1s, 2s, and 2p bands do not contribute to electrical conduction because an applied electric field cannot bring about intra band transitions in them.

Electrons can also not make band-to-band transitions from Is to 2s or from 2s to 2p band as for both these situations unfilled energy levels are not available.

However, electrons belonging to the 3s band can take part in intra-band transitions as half of the energy levels present in this band are available. An applied electric field can impart them an amount of energy sufficient for the transition to free energy levels, and take part in the process of conduction.

Thus the conduction properties of sodium are due to this partially filled band.. The lower half portion of this band is called the valence band and the upper half portion is called the conduction band as it is in this part when an electron reaches after receiving energy from the electric field the process of conduction starts. All monovalent metals have a half-filled conduction band like sodium.

The bivalent elements belonging to the second group of the periodic table Example magnesium, zinc, etc are also metallic. In the solid state of these materials, there is an overlapping between the highest filled band and the next higher unfilled band.

For example magnesium atom (atomic number = 12) has an electronic structure – 1s2 2s2 2p6 3s2 and in the atomic state, there is some energy gap between the filled 3s level and the next higher but unfilled 3p level.

However, during the process of crystal formation, the splitting of energy levels takes place in such a manner that the 3p band overlaps with the 3s band. In the ‘hybrid’ band’ so formed now electrons have a sufficient number of unfilled levels for transition. In such a situation if the 3s band is called the valence band then the 3p band is the conduction band and the two bands overlap.

We can conclude that for both the above metals there is no energy gap between the maximum energy of the valence band and the minimum energy of the conduction band.

The energy that an electron gains from an ordinary current source usually is 10 -4 to 10-8 eV which is sufficient to cause transition between levels inside a partially filled band. As the difference between the adjacent levels is infinitesimal, for such bands the electron can absorb infinitesimal energy in a manner like free electron.

Such electrons when reaching unfilled higher levels contribute to the process of electric conduction. In metals, both the number of free electrons and the vacant energy levels for transitions are very large which is why metals are good conductors of electricity and heat.

For metals at ordinary temperature, the electrical conductivities are in the range of 102 mho/meter to 108 mho/meter indicating this fact.

Insulator: It is a solid in which the energy band formation takes place in such a manner, that the valence band is filled while the conduction band is empty.

In addition to this, these two bands are separated by a large energy gap called the forbidden energy gap or band gap. If E c and Ev respectively denote the minimum energy in the conduction band and the maximum energy in the valence band then band gap E g is defined as E g = E c – E v

For insulators E g ~ 3 to 7 eV. As in an empty band, no electron is there to take part in the process of electric conduction, such a band does not contribute to conduction. In a filled band very large number of electrons are present but no vacant levels to which these electrons make transitions are available and hence again there will not be any conduction non such a band.

As explained earlier ordinary current sources provide only a very small energy to an electron in a solid so electrons cannot be excited from the valence band to the conduction band.

Also not only at ordinary temperatures but at elevated temperatures too, the thermal energy is much smaller than the band gap energy so electrons cannot be excited from the valence band to the conduction band by thermal means. Consequently, solids with such large band gaps are insulators.

For diamond, Eg≈6 eV hence it is an insulator. In general electrical conductivities· of insulators are in the range 10 -12 mho/metre to 10-18 mho/metre (resitivity in the range 1011 ohm-metre to 1019 ohm metre.]

Semiconductors:

In the case of semiconductors, the band structure is essentially of the same type as that for insulators with the only difference that of a relatively smaller forbidden gap. In the case of a semiconductor, this is typically of the order of 1eV.

At absolute zero temperature, the valence band is filled and the conduction band is empty, and consequently, no electrical conduction can result. This is the same behavior as observed in insulators. i.e. at absolute zero, a semiconductor behaves like an insulator.

At finite temperatures (room temperature and above) some of the electrons from near the top of the valence band acquire enough thermal energy to move into the otherwise empty conduction band.

These electrons contribute to the conduction of electricity in a semiconductor. Also, the above said transitions create some unfilled levels in the valence band and the electrons of this band can move into these levels again resulting in conduction. Thus the electrical conductivity of a semiconductor is larger than that of an insulator at room temperature.

However since the number of electrons made available to the conduction band via this process of thermal excitation is very small as compared to what is available for conduction in metals, the conductivity of semiconductors is much smaller than that of metals at a given temperature. Thus the conductivity of semiconductors lies between that of metals and insulators, that is why these are named so. The conductivity of semiconductors increases with temperature.

Semiconductors:

In the case of semiconductors, the band structure is essentially of the same type as that for insulators with the only difference being that of a relatively smaller forbidden gap. In the case of a semiconductor, this is typically of the order of 1eV.

At absolute zero temperature, the valence band is filled and the conduction band is empty, and consequently, no electrical conduction can result. This is the same behavior as observed in insulators. i.e. at absolute zero, a semiconductor behaves like an insulator.

At finite temperatures (room temperature and above) some of the electrons from near the top of the valence band acquire enough thermal energy to move into the otherwise empty conduction band. These electrons contribute to the conduction of electricity in a semiconductor.

Also, the above said transitions create some unfilled levels in the valence band and the electrons of this band can move into these levels again resulting in conduction. Thus the electrical conductivity of a semiconductor is larger than that of an insulator at room temperature.

However since the number of electrons made available to the conduction band via this process of thermal excitation is very small as compared to what is available for conduction in metals, the conductivity of semiconductors is much smaller than that of metals at a given temperature.

Thus the conductivity of semiconductors lies between that of metals and insulators, which is why these are named so. The conductivity of semiconductors increases with temperature.

Intrinsic Semiconductors:

A semiconductor free from impurities is called an intrinsic semiconductor. Ideally, an intrinsic semiconductor crystal should contain atoms of this semiconductor only but it is not possible in practice to obtain crystals with such purities.

However, if the impurity is less than 1 in 108 parts of the semiconductor it can be treated as intrinsic. For describing the properties of intrinsic semiconductors we are taking examples of silicon and germanium Both silicon and germanium are members of group IV of the periodic table of elements and are tetravalent.

Their electronic configuration is as follows:

Si(14)=1s² 2s² 2p⁶ 3s² 3p²
Ge(32)= Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p²

Both elements crystallize in such a way that each atom in the crystal is inside a tetrahedron formed by the four atoms that are closest to it. The shows one of these tetrahedral units.

Each atom shares its four valence electrons with its immediate neighbors on a one-to-one basis, so that each atom is involved in four covalent bonds. For convenience, a two-dimensional representation of the crystal structure of germanium, which can also be used for silicon.

At OK, all the valence electrons are involved in the bonding so the crystal is a perfect insulator as there are no free electrons available for conduction. At higher temperatures, however, some of the valence electrons have sufficient energy to break away from the bond and randomly move in the crystal.

Under an applied electric field these electrons drift and conduct electricity. When an electron escapes from a band it leaves behind a vacancy in the lattice. This vacancy is termed as a “hole”. The absence of an electron amounts to the presence of a positive charge of the same magnitude.

As explained later, holes also take part in conduction in semiconductors. When a covalent bond is broken, all electron-hole pair is contributed. At room temperature (300K) many electron-hole pairs are present in the crystal.

The process of electron-hole generation is explained. Let due to thermal energy an electron is set free from the covalent bond at site A whereby a hole is created at this site.

An electron from the covalent bond of a neighboring atom site B may jump to vacant site A then the bond is completed at A but a hole is created at B. In this process, a very small energy is involved compared to what is required for an electron-hole pair generation.

It is because the electron is jumping from one bond to the other and all electrons in bonding are on an average of the same energy.  when an electron jumps from C to B a hole is created at C and so on. In effect then such a vacancy or hole can be considered as mobile. Thus in a semiconductor, both electrons and holes act as charge carriers and contribute to electric conduction.

The number of electrons and holes generated by thermal means is equal for an intrinsic semiconductor. If n e and n h represent the electron and hole concentrations respectively then n i = ne = nh n e n h = ni” Here n i is intrinsic concentration.

Note:

1. A pure semiconductor is called an intrinsic semiconductor. It has thermally generated current carries.
2. They have four electrons in the outermost orbit of the atom and atoms are held together by covalent bonds.
3. Free electrons and holes both are charge carriers and n e (in C.B.) = nh (in V.B.)
4. The drift velocity of electrons (ve) is greater than that of holes (v0).
5. For them, the fermi energy level lies at the center of the C.B. and V.B.
6. In pure semiconductors, impurity must be less than 1 in 108 parts of the semiconductor.
7. In intrinsic semiconductor n e (0) = n h (0) = n i; where ne(0) = Electron density in conduction band, n h(0) = Hole density in V.B., n i = Density of intrinsic carriers.
8. The fraction of electron of the valance band present in the conduction band is given by f Ee–Eg/kT; where E g = Fermi energy or k = Boltzmann’s constant and T = Absolute temperature.
9. Because of the lower number of charge carriers at room temperature, intrinsic semiconductors have low conductivity so they have no particular use.
10. Several electrons reach from the valence band to the conduction band n = AT3/2e–Eg/2kT where A is a positive constant. (11) Net charge of a pure semiconductor is zero.

Electrical conductivity of intrinsic semiconductor:

A semiconductor, at room temperature, contains electrons in the conduction band and holes in the valence band. When an external electric field is applied, the electrons move opposite to the field and the holes move in the direction of the field, thus constituting current in the same direction. The total current is the sum of the electron and hole currents

Let us consider a semiconductor block of length l, area of cross-section A, and having electron concentration n e and hole concentration is n h, across the ends of the semiconductor creates an electric field E given by E = V/ ………

Under field E, the electrons and the holes both drift in opposite directions and constitute currents i e and I h respectively in the direction of the field.

The total current flowing through the semiconductor is i = I e + I h If v e is the drift velocity of the electrons in the conduction band and vh the drift velocity of the holes in the valence band, then we have I e = n e e A v e and I h = nh e A vh Where e is the magnitude of electron charge

⇒ \(\mathrm{i}=\mathrm{i}_e+\mathrm{i}_{\mathrm{h}}=e \mathrm{eA}\left(\mathrm{n}_e \mathrm{v}_e+\mathrm{n}_{\mathrm{h}} \mathrm{v}_{\mathrm{h}}\right)\)

If we are the drift velocity of the electrons in the conduction band and vh is the drift velocity of the holes in the valence band, then we have

Therefore i e = ne e A ve and ih = nh e A vh Where e is the magnitude of electron charge

Or \(\frac{i}{A}=e\left(n_e v_e+n_h v_h\right)\)

Let R be the resistance of the semiconductor block and  the resistivity of the block material. Then p=RA/ l.

Dividing equation 1 by equation 3, we have \(\frac{E}{\rho}=\frac{V}{R A}=\frac{i}{A},\)

Substituting in it the value of i/A from equation (2), we get

⇒ \(\begin{aligned}
& \frac{E}{\rho}=e\left(n_e v_e+n_h v_h\right) \\
& \frac{1}{\rho}=e\left(n_e \frac{v_e}{E}+n_h \frac{v_h}{E}\right) .
\end{aligned}\)

Let us now introduce a quantity, called mobility which is defined as the drift velocity per unit field and is expressed in meter2/(volt-second). Thus, the mobilities of electron and hole are given by \(\mu_e=\frac{v_e}{E} \quad \text { and } \quad \mu_n=\frac{v_h}{E}\)

The electrical conductivity s is the reciprocal of the resistively. Thus, the electrical conductivity of the semiconductor is given by

⇒ \(\begin{aligned}
& \sigma=e\left(n_e \mu_e+n_h \mu_h\right) . \\
& \sigma=e n_i\left(\mu_e+\mu_h\right)
\end{aligned}\)

since \(n_e=n_h=n_i\)

This is the required expression. It shows that the electrical conductivity of a semiconductor depends upon the electron and hole concentrations (number densities) and their mobilities. The electron mobility is higher than the hole mobility.

As temperature rises, both the concentrations n e and nh increase due to the breakage of more covalent bonds. The mobilities ne and nh, however, slightly decrease with temperature rise but this decrease is offset by the much greater increase in n e and nh. Hence, the conductivity of a semiconductor increases (or the resistivity decreases) with temperature rise.

The electrical conductivity of intrinsic (pure) semiconductors is too small to be of any practical use. If, however, a small quantity of some pentavalent or trivalent impurity is added to a pure semiconductor, the conductivity of the semiconductor is significantly increased. Such impure semiconductors are called ‘extrinsic’ or ‘impurity’ or ‘doped semiconductors.

Doping: The process of adding impurity to an intrinsic semiconductor in a controlled manner is called ‘doping’. It increases significantly the electrical conductivity of the semiconductor. The impurity atoms added are called ‘dopants’.

Extrinsic semiconductors are of two types: n-type and p-type

n-type semiconductor: When a pentavalent impurity atom (antimony, phosphorus, or arsenic) is added to a Ge(or Si) crystal, it replaces a Ge (or Si) atom in the crystal lattice. Four of the five valence electrons of the impurity atom form covalent bonds with one with each valence electron of four Ge (or Si) atoms surrounding.

Thus, by adding pentavalent impurity to pure Ge(or Si), the number of free electrons increases, that is, the conductivity of the crystal increases.

The impure Ge (or Si) crystal is called an ‘n-type’ semiconductor because it has an excess of ‘negative’ charge–carriers (electrons). The impurity atoms are called ‘donor’ atoms because they donate the conducting electrons to the crystal.

The fifth valence electrons of the impurity atoms occupy some discrete energy levels just below the condition band

These are called ‘donor levels’ and are only 0.01 eV below the conduction band in the case of Ge and 0.05 eV below in the case of Si. Therefore, at room temperature, the “fifth” electrons of almost all the donor atoms are thermally excited from the donor levels into the conduction band where they move as charge–carriers when an external electric field is applied.

At ordinary temperatures, almost all the electrons in the conduction band come from the donor levels, and only a few come from the valence band. Therefore, the main charge–carriers responsible for conduction are the electrons contributed by the donors.

Since the excitation from the valence band is small, there are very few holes in this band. The current contribution of the holes is therefore small. Thus, in an n-type semiconductor, the electrons are the ‘majority carriers’ and the holes are the ‘minority carriers.

Note: N-Type Semiconductor

These are obtained by adding a small amount of pentavalent impurity to a pure sample of semiconductor (Ge).

1. Majority charge carriers – electrons
2. Minority charge carriers – hole
3. ne > > nh ; ie > > ih
4. Conductivityσ= neµ e e
5. The donor energy level lies just below the conduction band.
6. Electrons and hole concentration: In a doped semiconductor, the electron concentration n e and the hole concentration n h are not equal (as they are in an intrinsic semiconductor). It can be shown that ne nh = ni2 where n i is the intrinsic concentration.
7. In an n-type semiconductor, the concentration of electrons in the conduction band is nearly equal to the concentration of donor atoms (N d) and very large compared to the concentration of holes in the valence band. That is n e N d > > nh.
8. Impurity atoms are called donor atoms which are elements of the V group of the periodic table. (7) The net charge on the N-type crystal is zero.
9. the mobile charge is a positive charge

p-type semiconductor: When a trivalent impurity atom (boron, aluminum, gallium, or indium) is added to a Ge (or Si) crystal, it also replaces one of the Ge (or Si) atoms in the crystal lattice. Its three valence electrons form covalent bonds with one each valence electron of these Ge (or Si) atoms surrounding it.

Thus, there remains a space, called a ‘p-type’ semiconductor because it has an excess of positive ‘acceptor’ atoms because they create holes that accept electrons.

The impurity atoms’ inductance vacant discrete levels just above the top of the valence band. These are called ‘acceptor levels’. At room temperature, electrons are easily excited from the valence band into the acceptor levels. The corresponding holes created in the valence band are the main charge–carried in the crystal when an electric field is applied.

Thus, in a p-type semiconductor the holes are the ‘majority carriers’ and the few electrons, thermally excited from the valence band into the conduction band, are ‘minority carriers’. Electron and hole concentration:

Note: P-Type Semiconductor

These are obtained by adding a small amount of trivalent impurity to a pure sample of semiconductor (Ge).

1. The majority of charge carries – holes
2. Minority charge carries – electrons
3. nh > > ne; ih > > ie
4. Conductivity  nhµ he
5. A P-type semiconductor is also electrically neutral (not positively charged)
6. Impurity is called Acceptor impurity, an element of 3 periodic table groups.
7. Acceptor energy level lies just above the valency band.
8. Electron and hole concentration: In a p-type semiconductor, the concentration of holes in the valence band is nearly equal to the concentration of acceptor atoms (N a) and very large compared to the concentration of electrons in the conduction band. That is n h = N a > > ne
9. The net charge on the p-type crystal is zero.
10. A mobile charge is a negative charge.

Distinction between intrinsic and extrinsic semiconductors

The distribution between the n-type and p-type semiconductors

Let us consider a conductor block of length l, area of cross-section A, and having electrons concentration n e and hole concentration n h. A potential difference V applied across the ends of the semiconductor creates an electric field E given by: E = V/l ……………

Under the field E, the electrons and the holes both drift in opposite directions and constitute currents i e and i h respectively in the direction of the field.

The total current flowing through the semiconductor is, i = i e + i h If v e, is the drift velocity of the electrons in the conduction band and vh the drift velocity of the holes in the valence band, then we have i e = n e eA v e and i h = the A vh where e is the magnitude of electron charge

⇒ \(\mathrm{i}=\mathrm{i}_e+\mathrm{i}_{\mathrm{h}}=e \mathrm{eA}\left(\mathrm{n}_e \mathrm{v}_e+\mathrm{n}_{\mathrm{h}} \mathrm{v}_{\mathrm{h}}\right)\)

Or \(\frac{i}{A}=e\left(n_e v_e+n_h v_h\right)\)

Let R be the resistance of the semiconductor block and the resistivity of the block material. Then p=RA/I

By dividing eq.1 by eq.2 we have

⇒ \(\frac{E}{\rho}=\frac{V}{R A}=\frac{i}{A},\)

Because V = iR (Ohm’s law). Substituting in it the value of i/A from eq.(ii), we get

⇒ \(\begin{aligned}
& \frac{E}{\rho}=e\left(n_e v_e+n_h v_h\right) \\
& \frac{1}{\rho}=e\left(n_e \frac{v_e}{E}+n_h \frac{v_h}{E}\right)
\end{aligned}\)

Let us introduce a quantity, called mobility which is defined as the drift velocity per unit field and is expressed in metre2 / (volt/second). Thus, the mobilities of electrons and holes are given by

⇒ \(\mu_e=\frac{v_e}{E} \quad \text { and } \mu_h=\frac{v_h}{E}\)

Introducing ue and uh in eq., we get \(\frac{1}{\rho}=e\left(n_e \mu_e+n_h \mu_h\right)\)

The electrical conductivity is the reciprocal of the resistivity. Thus, the electrical conductivity of the semiconductor is given by \(\rho=e\left(n_e \mu_e+n_n \mu_h\right)\)

This is the required expression. It shows that the electrical conductivity of a semiconductor depends upon the electron and hole concentrations (number densities) and their mobilities. The mobility of electrons is higher than the hole mobility.

As temperature rises, both the concentration n e and nh increase due to the breakage of more covalent bonds. The mobilities e and h, however, slightly decrease with temperature rise but this decrease is offset by the much greater increase in n e and n h.

Hence, the conductivity of a semiconductor increases (or the resistivity decreases) with temperature rise.

Solved Example

Example 1. The majority of charge carriers in P-type semiconductors are

1. Electrons
2. Protons
3. Holes
4. Neutrons

Solution: 3. In P-type semiconductors, holes are the majority of charge carriers

Example 2. When a semiconductor is heated, its resistance

1. Decreases
2. Increases
3. Remains unchanged
4. Nothing is definite

Answer: 1. Decreases

Example 3. Which of the following energy band diagrams shows the N-type semiconductor

Solution: 2. In an N-type semiconductor impurity energy level lies just below the conduction band.

Example 4. Which of the energy band diagrams corresponds to that of a semiconductor

Solution: 4. In semiconductors, the forbidden energy gap between the valence band and conduction band is very small, almost equal to kT. Moreover, the valence band is filled where as conduction band is empty.

Example 5. The P-N junction is-

1. An Ohmic Resistance
2. A Non-Ohmic Resistance
3. A Positive Resistance
4. A Negative Resistance

Answer: 2. An Non-Ohmic Resistance

Example 6. The mean free path of conduction electrons in copper is about 4 × 10 –8 m. For a copper block, find the electric field that can give, on average, 1eV energy to a conduction electron.
Solution: Let the electric field be E. The force on an electron is eE. As the electron moves through a distance d, the work done on it is eEd. This is equal to the energy transferred to the electron. As the electron travels an average distance of 4 × 10–8 m before a collision, the energy transferred is eE(4 × 10–8 m). To get 1 eV energy from the electric field, eE(4 × 10–8 m) = 1 eV or E = 2.5 × 10 7 V/m.

Question 7. The band gap in germanium is E = 0.68 eV. Assuming that the number of hole–electron pairs is proportional to e–E/2kT, find the percentage increase in the number of charge carriers in pure germanium as the temperature is increased from 300 K to 320 K. Solution: The number of charge carriers in an intrinsic semiconductor is double the number of hole– electron pairs. If N1 is the number of charge carriers at temperature T1 and N2 at T2, we have

⇒ \(\begin{aligned}
& \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-\Delta \mathrm{E} / 2 \mathrm{kT}_1} \\
& \mathrm{~N}_2=\mathrm{N}_0 \mathrm{e}^{-\Delta \mathrm{E} / 2 \mathrm{kT}}{ }_2
\end{aligned}\)

The percentage increase as the temperature is raised from T1 to T2 is

⇒ \(f=\frac{N_2-N_1}{N_1} \times 100=\left(\frac{N_2}{N_1}-1\right) \times 100=100\left[e^{\frac{\Delta E}{2 k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}-1\right]\)

Now, \(\frac{\Delta \mathrm{E}}{2 \mathrm{k}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right)=\frac{0.68 \mathrm{eV}}{2 \times 8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}}\left(\frac{1}{300 \mathrm{~K}}-\frac{1}{320 \mathrm{~K}}\right)=0.82\)

Thus f = 100 × [e0.82 – 1] 127.

Thus, the number of charge carriers increases by about 127%.

Example 8. A silicon specimen is made into a p-type semiconductor by doping on an average of one indium atom per 5 × 107 silicon atoms. If the number density of atoms in the silicon specimen is 5 × 1028 atoms/m3; find the number of acceptor atoms in silicon per cubic centimeter.

Solution: The doping of one indium atom in a silicon semiconductor will produce one acceptor atom in a p-type semiconductor. Since one indium atom has been dropped per 5 × 107 silicon atoms, so number density of acceptor atoms in silicon \(=\frac{5 \times 10^{28}}{5 \times 10^7}=10^{21} \text { atom } / \mathrm{m}^3=10^{15} \text { atoms } / \mathrm{cm}^3 \text {. }\)

Example 9. Pure Si at 300K has equal electron (ne) and hole (NH) concentrations of 1.5 × 1016 m–3. Dopping by indium increases nh to 3 × 1022 m–3. Calculate ne in the doped Si.
Solution: For a doped semiconductor in thermal equilibrium nenh = ni2 (Law of mass action)

⇒ \(n_e=\frac{n_i^2}{h_h}=\frac{\left(1.5 \times 10^{16}\right)^2}{3 \times 10^{22}}=7.5 \times 10^9 \mathrm{~m}^{-3}\)

Example 10. Pure Si at 300 K has equal electron (ne) and hole (NH) concentrations of 1.5 × 1016 m–3. Doping
by indium increases nh to 4.5 × 1022 m–3. Calculate ne in the doped Si
Solution: ne NH = ni2 n h = 4.5 × 1022 m–3 so, ne = 5.0 × 109 m–3

Example 11. The energy of a photon of sodium light (= 589 nm) equals the band gap of a semiconducting material. (a) Find the minimum energy E required to create a hole-electron pair. (B) Find the value of E/kT at a temperature of 300 K.
Solution: (a) The energy of the photon is E \(E=\frac{h c}{\lambda}\)

⇒ \(=\frac{1242 \mathrm{eV}-\mathrm{nm}}{589 \mathrm{~nm}}=2.1 \mathrm{eV} \text {. }\)

Thus the band gap is 2.1 eV. This is also the minimum energy E required to push an electron from the valence band into the conduction band. Hence, the minimum energy required to create a hole–electron pair is 2.1 eV.

At \(\begin{aligned}
& \mathrm{T}=300 \mathrm{~K}, \\
& \mathrm{kT}=\left(8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)(300 \mathrm{~K}) \\
& =25.86 \times 10^{-3} \mathrm{eV} . \\
& \frac{\mathrm{E}}{\mathrm{kT}}=\frac{2.1 \mathrm{eV}}{25.86 \times 10^{-3} \mathrm{eV}}=81 .
\end{aligned}\)

So it is difficult for the thermal energy to create the hole–electron pair but a photon of light can do it easily.

Junction Diode

A junction diode is a basic semiconductor device. It is a semiconductor crystal having acceptor impurities in one region (P-type crystal) and donor impurities in the other region (n-type crystal). The boundary between the two regions is called the ‘p–n junction’.

Circuit Symbol for a p-n Junction Diode:

In electronic circuits, the semiconductor devices are represented by their symbols. The symbol for the basic device is the p-n junction diode, The arrowhead represents the p -region and the bar represents the n -region of the diode. The direction of the arrow is from p to and indicates the direction of conventional current flow under forward bias. The p -side is called ‘anode’ and the n -side is called ‘cathode’.

Formation of p-n Junction:

A p-n junction is not the interface between p-type and n-type semiconductor crystals pressed together. It is a single piece of semiconductor crystal having an excess of acceptor impurities on one side and donor impurities on the other. A P-type semiconductor is grown on one side of the metallic film while an N-type is grown on the other side.

PotentiaI Barrier at the Junction: Formation of Depletion Region:

A p-n junction. The p-type region has (positive) holes as majority charge carriers, and an equal number of fixed negatively-charged acceptor ions. (The material as a whole is thus neutral). Similarly, the n-type region has (negative) electrons as majority charge carriers and an equal number of fixed positively-charged donor ions.

The region on either of the junctions that become depleted (free) of the mobile charge carriers is called the ‘depletion region’. The width of the depletion region is of the order of 10 —6 m. The potential difference developed across the depletion region is called the ‘potential barrier’.

It is about 0.3 volts for Ge, p-n junction and about 0·7 volts for silicon p-n junction. It, however, depends upon the dopant concentration in the semiconductor. The magnitude of the barrier electric field for a silicon junction is

⇒ \(E_{\mathrm{i}} \approx \frac{\mathrm{V}}{\mathrm{d}} \approx \frac{0.7}{10^{-6}}=7 \times 10^5 \mathrm{Vm}^{-1}\)

Diffusion & Drift Current: Due to the concentration difference holes try to diffuse from the p side to the n side but due to the depletion layer only those holes can diffuse from the p to the n side which have high kinetic energy. Similarly, electrons of high kinetic energy also diffuse from n to p so diffusion current flows from p to n side.

Due to thermal collision or an increase in temperature, some valence electron comes in conduction band. If these occur in the depletion region then the hole moves to the p side & electron moves to the n side so a current is produced from the n to the p side it is called a drift current in the steady state both are equal & opposite.

Solved Examples

Example 12. In a p-n junction with open ends,

1. There Is No Systematic Motion Of Charge Carriers
2. Holes And Conductor Electrons Systematically Go From The P-Side And The N-Side To The P-Side Respectively
3. There Is No Net Charge Transfer Between The Two Sides
4. There Is A Constant Electric Field Near The Junction

Answer: (2,3,4)

Question 13. A potential barrier of 0.50 V exists across a P-N junction. If the depletion region is 5.0 × 10–7 m wide, the intensity of the electric field in this region is

1. 1.0 × 106 V/m
2. 1.0 × 105 V/m
3. 2.0 × 105 V/m
4. 2.0 × 106 V/m

Solution: 1. \(E=\frac{V}{d}=\frac{0.5}{5 \times 10^{-7}}=10^6 \mathrm{~V} / \mathrm{m}\)

Forward and Reverse Biasing of Junction Diode

The junction diode can be connected to an external battery in two ways, called ‘forward biasing’ and ‘reverse biasing’ of the diode. It means the way of connecting the EMF source to the P-N junction diode. It is of following two types

Forward Biasin:

A junction diode is said to be forward-biased when the positive terminal of the external battery is connected to the p -p-region and the negative terminal to the n -region of the diode

Forward-Biased Characteristics: The circuit connections. The positive terminal of the battery is connected to the p -p-region and the negative terminal to the n -n-region of the

junction diode through a potential-divider arrangement which enables to change of the applied voltage. The voltage is read by a voltmeter V and the current by a milliammeter mA.

Starting with a low value, the forward bias voltage is increased step by step and the corresponding forward current is noted. A graph is then plotted between voltage and current. The resulting curve OAB (Fig. b) is the forward characteristic of the diode. In the beginning, when the applied voltage is low, the current through the junction diode is almost zero.

It is because of the potential barrier (about 0·3 V for Ge p-n junction and about 0·7 V for Si junction) which opposes the applied voltage. With the increase in applied voltage, the current increases very slowly and non-linearly until the applied voltage exceeds the potential barrier. This is represented by the portion OA of the characteristic curve.

With further increase in applied voltage, the current increases very rapidly and almost linearly Now the diode behaves as an ordinary conductor. This is represented by the straight-line part AB of the characteristic. If this straight line is projected back. it intersects the voltage-axis at the barrier potential voltage.

Note:

1. In forward biasing width of the depletion layer decreases
2. In forward biasing resistance offered RForward≈10Ω – 25Ω
3. Forward bias opposes the potential barrier and for V > VB a forward current is set up across the junction.
4. Cut-in (Knee) voltage: The voltage at which the current starts to increase rapidly. For Ge it is 0.3 V and for Si, it is 0.7V.
5. df–diffusion dr–drift

Reverse Biasing:

A junction diode is said to be reverse-biased when the positive terminal of the external battery is connected to the n -region and the negative terminal to the p -region of the diode.

In this condition, the external field E is directed from n toward p and thus aids the internal barrier field E i Hence holes in the p-region and electrons in the n-region are both pushed away from the junction, that is, they cannot combine at the junction. Thus, there is almost no current due to the flow of majority carriers.

Reverse-Biased Characteristic: The circuit connections in which the positive terminal of the battery is connected to the n -region and the negative terminal to the p -region of the junction diode.

In a reverse-biased diode, a very small current (of the order of micro Ampere) flows across the junction due to the motion of the few thermally generated minority carriers (electrons in the p -region and holes in the n -region) whose motion is aided by the applied voltage.

The small reverse current remains almost constant over a sufficiently long range of reverse bias (applied voltage). increasing very little with increasing bias. This is represented by the part OC of the reverse characteristic curve.

Note:

1. In reverse biasing width of the depletion layer increases
2. In reverse biasing resistance offered \(R_{\text {Reverse }} \approx 10^5 \Omega\)
3. Reverse bias supports the potential barrier and no current flows across the juction due to the diffusion of the majority carriers. (A very small reverse current may exist in the circuit due to the drifting of minority carriers across the juction)
4. Break down voltage: Reverse voltage at which breakdown of semiconductor occurs. For Ge it is
   25V and for Si it is 35 V.
5. Reverse saturation current is temperature sensitive and nearly doubles for every 10ºC rise.

Avalanche Breakdown:

If the reverse bias is made very high, the minority carriers acquire kinetic energy enough to break the covalent bonds near the junction, thus liberating electron-hole pairs. These charge carriers are accelerated and produce, in the same way, other electron-hole pairs. The process is cumulative and an avalanche of electron-hole pairs is produced.

The reverse current then increases abruptly to a relatively large value (part CD of the characteristic). This is known as ‘avalanche breakdown’ and may damage the junction by the excessive heat generated.

The reverse bias voltage at which the reverse current increases abruptly is called the ‘breakdown voltage’ or ‘Zener voltage’. The numerical value of the breakdown voltage varies from tens of volts to several hundred volts depending on the number density of the impurity atoms doped into the diode.

Dynamic Resistance of a Junction Diode

The current-voltage curve of the junction diode shows that the current does not vary linearly with the voltage, that is, Ohm’s law is not obeyed. In such a situation, a quantity known as ‘dynamic resistance’ (or a.c. resistance) is defined. The dynamic resistance of a junction diode is defined as the ratio of a small change in the applied voltage (V) to the corresponding small change in current, that is \(R_{\mathrm{d}}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{i}}\)

In the forward characteristic of the p-n junction diode, beyond the turning point (knee), however, the current varies almost linearly with voltage. In this region, Rd is almost independent of V, and Ohm’s law is obeyed.

Solved Examples

Example 14. Which of the diodes is forward-biased?

1. 1, 2, 3
2. 2, 4, 5
3. 1, 3, 4
4. 2, 3, 4

Solution: 2. 2,4 and 5. P-crystals are more positive as compared to N-crystals.

Example 15. Two identical p-n junctions may be connected in series with a battery in three ways fig. The potential difference across the two p-n junctions is equal in

1. Circuit 1 and Circuit 2
2. Circuit 2 and Circuit 3
3. Circuit 3 and Circuit 1
4. Circuit 1 only

Answer: 2. Circuit 2 and circuit 3

Example 16. Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t = 0. The charges on the capacitor at a time t = CR are, respectively,

1. VC, VC
2. VC/e, VC
3. VC, VC/e
4. VC/e, VC/e

Solution 2. VC/e, VC

Example 17. What is the current in the circuit shown below

1. 0 amp
2. 10–2 amp
3. 1 amp
4. 0.10 amp

Solution 1. The potential of the P-side is more negative than that of the N-side, hence diode is in reverse biasing. In reverse biasing it acts as an open circuit, hence no current flows.

Example 18. Assume that the junction diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7V) of its V characteristic curve. Also, assume that the voltage across the diode is independent of the current above the knee point. If VB = 5V, what should be the maximum value of R so that the voltage is above the knee joint

1. 4.3 kΩ
2. 860 kΩ
3. 4.3Ω
4. 860Ω

Answer: 1. 4.3 kΩ

Question 19. The i-V characteristic of a p-n junction diode Find the approximate dynamic resistance of the p-n junction when (a) a forward bias of 1 volt is applied, (b) a forward bias of 2 volt is applied

The current at 1 volt is 1 0 mA and at 1.2 volt it is 15 mA. The dynamic resistance in this region is

⇒ \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{i}}=\frac{0.2 \mathrm{volt}}{5 \mathrm{~mA}}=40 \Omega\)

The current at 2 volts is 400 mA and at 2.1 volts it is 800 mA. The dynamic resistance in the region is

⇒ \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{i}}=\frac{0.1 \mathrm{volt}}{400 \mathrm{~mA}}=0.25 \Omega \text {. }\)

⇒ \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{i}}=\frac{0.1 \mathrm{volt}}{400 \mathrm{~mA}}=0.25 \Omega \text {. }\)

p-n Junction Diode as a Rectifier

An electronic device that converts alternating current / voltage into direct current/voltage is called a ‘rectifier’.

A p-n junction diode offers a low resistance for the current to flow when forward-biased, but a very high resistance, when reverse-biased. It thus passes current only in one direction and acts as a rectifier.

The junction diode can be used either as a half-wave rectifier or when it allows current only during the positive half-cycles of the input a.c. supply; or as a full-wave rectifier when it allows current in the same direction for both half-cycles of the input a.c.

p-n Junction Diode as Half-Wave Rectifier: The half-wave rectifier circuit.

The a.c. input voltage is applied across the primary P₁P₂ of a transformer. S₁S₂ is the secondary coil of the same transformer. S₁ is connected to the p-type crystal of the junction diode and S₂ is connected to the n-type crystal through a load resistance RL.

During the first half-cycle of the a.c. input, when the terminal S₁ of the secondary is supposed positive and S₂ is negative, the junction diode is forward-biased. Hence it conducts and current flows through the load R L in the direction shown by arrows.

The current produces across the load an output voltage of the same shape as the half-cycle of the input voltage. During the second half-cycle of the a.c. input, the terminal S₁ is negative and S₂ is positive. The diode is now reverse-biased.

Hence there is almost zero current and zero output voltage across RL. The process is repeated. Thus, the output current is unidirectional, but intermittent and pulsating, as shown in the lower part.

Since the output current corresponds to one-half of the input voltage wave, the other half being missing, the process is called half-wave rectification.

The purpose of the transformer is to supply the necessary voltage to the rectifier. If direct current at high voltage is to be obtained from the rectifier, as is necessary for power supply, then a step-up transformer is used, In many solid-state equipment, however, a direct current of low voltage is required. In that case, a step-down transformer is used in the rectifier.

1. During the positive half-cycle
2. Diode → forward biased
3. Output signal → obtained
4. During negative half→cycle
5. Diode → reverse biased
6. Output signal → not obtained
7. Output voltage is obtained across the load resistance RL. It is not constant but pulsating
   (mixture of Ac and DC) in nature.
8. Average output in one cycle
9. ⇒ \(I_{d c}=\frac{I_0}{\pi} \text { and } V_{d c}=\frac{V_0}{\pi} ; I_0=\frac{V_0}{r_f+R_L}\)
10. (rf = forward-biased resistance)
11. r.m.s. output \(I_{r m s}=\frac{I_0}{2}, V_{r m s}=\frac{V_0}{2}\)
12. The ratio of the effective alternating component of the output voltage or current of the DC component is known as the ripple factor.

⇒ \(\mathrm{r}=\frac{\mathrm{I}_{\mathrm{ac}}}{\mathrm{I}_{\mathrm{dc}}}=\left[\left(\frac{\mathrm{I}_{\mathrm{mss}}}{\mathrm{I}_{\mathrm{dc}}}\right)^2-1\right]^{1 / 2}=1.21\)

Peak inverse voltage (PIV): The maximum reverse biased voltage that can be applied before the commencement of the Zener region is called the PIV. When the diode is not conducting PIV across it = V 0

Efficiency: It is given by % \(\eta=\frac{P_{\text {out }}}{P_{\text {in }}} \times 100=\frac{40.6}{1+\frac{r_f}{R_L}}\)

• If R L > > rf then = 40.6 % If RL = rf then = 20.3 %
• From factor \(=\frac{I_{\mathrm{rms}}}{I_{d c}}=\frac{\pi}{2}=1.57\)
• The ripple frequency (w) for a half-wave rectifier is the same as that of AC.

p-n Junction Diode as Full-Wave Rectifier: In a full-wave rectifier, a unidirectional, pulsating output current is obtained for both halves of the a.c. input voltage. Essentially, it requires two junction diodes so connected that one diode rectifies one half and the second diode rectifies the second half of the input. The circuit for a full-wave rectifier is the input and output waveform.

The a.c. input voltage is applied across the primary P₁P₂ of a transformer. The terminals S1 and S₂ of the secondary are connected to the p-type crystals of the junction diodes D₁ and D₂ whose n-type crystals are connected. A load resistance RL is connected across the n-type crystals and the central tap T of the secondary S₁S₂.

During the first half-cycle of the a.c. the input voltage, the terminal S₁ is supposed positive relative to T and S₂ is negative. In this situation, the junction diode D₁ is forward-biased and D₂ is reverse-biased.

Therefore, D₁ conducts while D₂ does not. The conventional current flows through diode D₁, load RL and the upper half of the secondary winding, as shown by solid arrows. During the second half-cycle of the input voltage, S1 is negative relative to T and S2 is positive.

Now, D1 is reverse-biased and does not conduct while D₂ is forward-biased and conducts. The current now flows through D₂, load RL and the lower half of the secondary, as shown by dotted arrows.

It may be seen that the current in the load RL flows in the same direction for both half-cycles of the a.c. input voltage. Thus, the output current is a continuous series of unidirectional pulses. However, it can be made fairly steady through smoothing filters.

During the positive half-cycle

Diode:

1. D₁ → forward biased
2. D₂ → reverse biased

Output signal → obtained due to D₁ only

During negative half-cycle

Diode:

1. D₁ → reversed biased
2. D₂ → forward biased
3. Output signal → obtained due to D₂ only
4. Fluctuating dc → Filter → constant dc.
5. Output voltage is obtained across the load resistance RL. It is not constant but pulsating in
   nature.
6. Average output \(V_{\mathrm{av}}=\frac{2 V_0}{\pi}, \mathrm{I}_{\mathrm{av}}=\frac{2 \mathrm{I}_0}{\pi}\)
7. r.m.s Output \(V_{r m s}=\frac{V_0}{\sqrt{2}}, I_{m s}=\frac{I_0}{\sqrt{2}}\)

Ripple factor: r = 0.48 = 48%

Ripple frequency: The ripple frequency of full wave rectifier = 2 × (Frequency of input ac)

Peak inverse voltage (PIV): Its value is 2V0.

Efficiency: \(n_{\%}=\frac{81.2}{1+\frac{r_1}{R_L}} \text { for } r_f<<R_L, \eta=81.2 \%\)

Full wave bridge rectifier: Four diodes D₁, D₂, D_3, and D₄ are used in the circuit. During the positive half cycle, D₁ and D₃ are forward biased and D₂ and D₄ are reverse biased. During the negative half cycle, D₂ and D₄ are forward-biased biased and D₁ and D₃ are reverse-biased

Different Types of Junction Diodes

The junction diodes are of many types. The important types are Zener diode, photodiode, light-emitting diode (LED) and solar cell.

Zener Diode: It is a voltage-regulating device based upon the phenomenon of avalanche breakdown in a junction diode. When the reverse bias applied to a junction diode is increased, there is an abrupt rise in the (reverse) current when the bias reaches a certain value, known as ‘breakdown voltage’ or ‘Zener voltage’.

Thus, in this region of the reverse characteristic curve, the voltage across the diode remains almost constant for a large range of currents. Hence the diode may be used to stabilize voltage at a predetermined value. It is then known as the ‘Zener diode’. It can be designed, by properly controlled doping of the diode, to stabilize the voltage at any desired value between 4 –100 volts.

Parallel with the load RL. The Zener diode is selected with a Zener voltage VZ equal to the voltage desired across the load. The fluctuating d.c. the input voltage may be the d.c. the output of a rectifier. Whenever the input voltage increases, the excess voltage is dropped across the resistance R.

This causes an increase in the input current i. This increase is conducted by the Zener diode, while the current through the load and hence the voltage across it remains constant at VZ.

Likewise, a decrease in the input voltage causes a decrease in the input current i. The current through the diode decreases correspondingly, again maintaining the current through the load constant.

Since the resistance R absorbs the input voltage fluctuations to give a constant output voltage VZ, the circuit cannot work if the input voltage falls below VZ..

Photodiode: A photodiode is a reverse-biased p-n junction made from a photosensitive semiconductor. The junction is embedded in clear plastic.

The upper surface across the junction is open to light, while the remaining sides of the plastic are painted black or enclosed in a metallic case. The entire unit is extremely small, of the order of a 0·1 inch size.

The circuit is When no light is falling on the junction and the reverse bias is of the order of a few tenths of a volt, an almost constant small current (A) is obtained. This “dark” current is the reverse saturation current due to the thermally generated minority carriers (electrons in the -region and holes in the n -region).

When light of appropriate frequency is made incident on the junction, additional electron-hole pairs are created near the junction (due to breaking of covalent bonds).

These light-generated minority carriers cross the· (reverse-biased) junction and contribute to the (reverse) current due to thermally-generated carriers. Therefore, the current in the circuit increases (a fraction of a mA).

This, so-called ‘photoconductive’ current varies almost linearly with the incident light flux. The p–n photodiodes can operate at frequencies of the order of 1 MHz. Hence they are used in highspeed reading of computer punched cards, light-detection systems, light-operated switches, electronic counters, etc.

Light-Emitting Diode (LED): When a p–n junction diode is forward-biased. both the electron and the holes move towards the junction. As they cross the junction, the electrons fall into the holes (recombine). Hence, energy is released at the junction (because the electrons fall from a higher to a lower energy level). In the case of Ge and Si diodes, the energy released is infrared radiation. If, however, the diode is made of gallium arsenide or indium phosphide, the energy released is visible light. The diode is then called a ‘light-emitting diode’ (LED).

LEDs have replaced incandescent lamps in many applications because of their low input power, long life, and fast on-off switching. LEDs that can emit red, yellow, orange, green, and blue light are commercially available.

The semiconductor used for the fabrication of visible LEDs must at least have a band gap of 1.8 eV (the spectral range of visible light is from about 0.4 m to 0.7 m, i.e., from about 3 eV to 1.8 eV).

The compound semiconductor Gallium Arsenide – Phosphide (GaAs1 –xPx) is used for making LEDs of different colors. GaAs 0.6 P0.4 (Eg ~ 1.9 eV) is used for red LED. GaAs (Eg ~ 1.4 eV) are used for making infrared LED.

These LEDs find extensive use in remote controls, burglar alarm systems, optical communication, etc. Extensive research is being done for developing white LEDs which can replace incandescent lamps.

They are extensively used in fancy electronic devices like calculators, etc.

Solar Cell; A solar cell is a p-n junction that generates emf when solar radiation falls on the p-n junction. It works on the same principle (photovoltaic effect) as the photodiode, except that no external bias is applied and the junction area is kept much larger for solar radiation to be incident because we are interested in more power.

A p-Si wafer of about 300 m is taken over and a thin layer (~0.3 m) of n-Si is grown on one side by a diffusion process. The other side of p-Si is coated with a metal (back contact). On the top of the n-Si layer, a metal finger electrode (or metallic grid) is deposited.

This acts as a front contact. The metallic grid occupies only a very small fraction of the cell area (<15%) so that light can be incident on the cell from the top.

The generation of emf by a solar cell, when light falls on, is due to the following three basic processes: generation, separation, and collection—

Generation of e-h pairs due to light (with h > E g ) close to the junction; Separation of electrons and holes due to electric field of the depletion region.

Electrons are swept to the n-side and holes to the p-side; The electrons reaching the n-side are collected by the front contact and holes reaching the p-side are collected by the back contact.

Thus p-side becomes positive and the n-side becomes negative giving rise to photovoltage. When an external load is connected a photocurrent I L flows through the load. A typical I-V characteristics of a solar cell is shown. Note that the I – V characteristics of solar cells are drawn in the fourth quadrant of the coordinate axes.

This is because a solar cell does not draw current but supplies the same to the load. Semiconductors with band gap close to 1.5 eV are ideal materials for solar cell fabrication. Solar cells are made with semiconductors like Si (E g = 1.1 eV), GaAs (E g = 1.43 eV), CdTe (E g = 1.45 eV), CuInSe2 (E g = 1.04 eV), etc.

The important criteria for the selection of a material for solar cell fabrication are band gap (~1.0 to 1.8 eV), high optical absorption (~104 cm–1), electrical conductivity, availability of the raw material, and cost. Note that sunlight is not always required for a solar cell. Any light with photon energies greater than the bandgap will do.

Solar cells are used to power electronic devices in satellites and space vehicles and also as a power supply to some calculators. Production of low-cost photovoltaic cells for large-scale solar energy is a topic for research.

Solved Example

Example 20. A zener diode of voltage VZ (= 6) is used to maintain a constant voltage across a load resistance R L (=1000) by using a series resistance RS (=100). If the e.m.f. or source is E (= 9V), calculate the value of current through series resistance, Zener diode, and load resistance. What is the power being dissipated in the Zener diode?
Solution: Here, E = 9V; VZ = 6; RL = 1000Ω and Rs = 100Ω

Potential drop across series resistor V = E – VZ = 9 –6 = 3 V

Current through series resistance RS is \(I=\frac{V}{R}=\frac{3}{100}=0.03 \mathrm{~A}\)

Current through load resistance RL is IL \(=\frac{V_{\mathrm{Z}}}{R_L}=\frac{6}{1000}=0.006 \mathrm{~A}\)

Current through Zener diode is PZ = V – IL = 0.03 – 0.006 = 0.024 A

Power dissipated in Zener diode is PZ = VZ IZ = 6 × 0.024 = 0.144 Watt

Example 21. An A.C. of 200 rms voltage is applied to the circuit containing the diode and the capacitor and it is being rectified. The potential across the capacitor C in volts will be

1. 1500
2. 200
3. 283
4. 141

Answer: 3. 283

Question 22. Input is applied across A and C and output is taken across B and D, then the output is

1. Zero
2. Same as input
3. Full wave rectified
4. Half wave rectified

Answer: 3. Full wave rectified

Question 23. Two junction diodes one of germanium (Ge) and the other of silicon (Si) are connected to a battery of EMF 12 V and a load resistance of 10. The germanium diode conducts at 0.3 V and the silicon diode at 0.7 V. When a current flows in the circuit, the potential of terminal Y will be

1. 12V
2. 11V
3. 11.3V
4. 11.7V

Answer: 4. 11.7V

Question 24. Potential barrier developed in a junction diode opposes-

1. Minority carriers in both regions only
2. Majority carriers
3. Electrons in N-region
4. Holes in P-region

Answer: 2. Majority carriers

Question 25. Avalanche breakdown in a semiconductor diode occurs when-

1. Forward current exceeds a certain value
2. Reverse bias exceeds a certain value
3. Forward bias exceeds a certain value
4. The potential barrier is reduced to zero

Answer: 2. Reverse bias exceeds a certain value

Example 26. A potential barrier of 0.50 V exists across a p-n junction.

1. If the depletion region is 5.0 × 10–7 m wide, what is the intensity of the electric field in this region?
2. An electron with a speed of 5.0 × 105 m/s approaches the p-n junction from the n-side. With what speed will it enter the p-side?

Solution: The electric field is E = V/d \(=\frac{0.50 \mathrm{~V}}{5.0 \times 10^{-7} \mathrm{~m}}=1.0 \times 10^6 \mathrm{~V} / \mathrm{m} \text {. }\)

Suppose the electron has a speed of v1 when it enters the depletion layer and v2 when it comes As the potential energy increases by e × 0.50 V, from the principle of conservation of energy,

Or \(\frac{1}{2} \mathrm{mu}_1^2=\mathrm{e} \times 0.50 \mathrm{~V}+\frac{1}{2} \mathrm{mu}_2^2\)

Or, \(\frac{1}{2} \times\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(5.0 \times 10^5 \mathrm{~m} / \mathrm{s}\right)^2=1.6 \times 10^{-19} \times 0.5 \mathrm{~J}+\left(9.1 \times \frac{1}{2} 10^{-31} \mathrm{~kg}\right) \mathrm{v}_2^2\)

Solving this, 2 = 2.7 × 105 m/s

Example 27. A 2 V battery may be connected across points A and B. Assume that the resistance of each diode is zero in forward bias and infinity in reverse bias. Find the current supplied by the battery if the positive terminal of the battery is connected to point A the point B.

Solution: When the positive terminal of the battery is connected to point A, diode D₁ is forward-biased and D₂ is reverse-biased. The resistance of the diode D₁ is zero, and it can be replaced by a resistance-less wire.

Similarly, the resistance of the diode D₂ is infinite, and it can be replaced by a broken wire. The equivalent circuit. The current supplied by the battery is 2 V/10  = 0.2 A.

When the positive terminal of the battery is connected to point B, the diode D 2 is forward-biased and D1 is reverse-biased. The equivalent circuit. The through the battery is 2 V/20Ω = 0.1 A.

Example 28. What is the reading of the ammeters A1 and A2? Neglect the resistance of the meters.

Answer: zero 0.2 A

Example 29. Calculate the value of V0 and if the Si diode and the Ge diode start conducting at 0.7 V and 0.3 V respectively, in the given circuit. If the Ge diode connection is reversed, what will be the new values of V 0 and I?
Solution: The effective forward voltage across the Ge diode is 12 V – 0.3 = 11.7. This will appear as the output voltage across the lad, that is, V0 = 11.7 V The current in the load is

⇒ \(i=\frac{V_0}{R_L}=\frac{11.7}{5 K \Omega}=2.34 \mathrm{~mA} .\)

On reversing the connections of Ge diode, it will be reverse-biased and conduct no current. Only the Si diode will conduct. The effective forward voltage across the Si diode is 12 V – 0.7V = 11.3 V. This will appear as output, that is V0 = 11.3 V.

The current in the load \(\mathrm{i}=\frac{\mathrm{V}_0}{\mathrm{R}_{\mathrm{L}}}=\frac{11.3}{5 \mathrm{k} \Omega}=2.26 \mathrm{~mA} .\)

Junction Transistor:

Transistor structure and action:

A transistor has three doped regions forming two p–n junctions between them. There are two types of transistors.

n–p–n transistor: Here two segments of n-type semiconductor (emitter and collector) are separated by a segment of p-type semiconductor (base).

p–n–p transistor: Here two segments of a p-type semiconductor (termed as emitter and collector) are separated by a segment of an n-type semiconductor (termed as a base).

The schematic representations of an n–p–n and a p–n–p configuration. All three segments of a transistor have different thickness and their doping levels are also different. In the schematic symbols used for representing p–n–p and n–p–n transistors, the arrowhead shows the direction of conventional current in the transistor.

A brief description of the three segments of a transistor is given below:

Emitter: This is the segment on one side of the transistor. It is of moderate size and heavily doped. It supplies a large number of majority carriers for the current flow through the transistor.

Base: This is the central segment. It is very thin and lightly doped.

Collector: This segment collects a major portion of the majority of carries supplied by the emitter. The collector side is moderately doped and larger as compared to the emitter.

In the case of a p–n junction, there is a formation of depletion region across the junction. In the case of a transistor, depletion regions are formed at the emitter-base–junction and the base-collector junction.

The transistor works as an amplifier, with its emitter-base junction forward-biased and the base-collector junction reverse-biased. This situation, is where VCC and VEE are used for creating the respective biasing. When the transistor is biased in this way it is said to be in active state.

We represent the voltage between the emitter and base as VEB and that between the collector and base as V CB., the base is a common terminal for the two power supplies whose other terminals are connected to the emitter and collector, respectively.

So, the two power supplies are represented as VEE and V CC’ respectively. In circuits, where the emitter is the common terminal, the power supply between the base and emitter is represented as VBB and that between the collector and emitter as VCC’.

The heavily doped emitter has a high concentration of majority carriers, which will be holes in a p–n–p transistor and electrons in an n–p–n transistor. The majority of carriers enter the base region in large numbers. The base is thin and lightly doped. So, for the majority of carriers, there would be few.

In a p–n–p transistor the majority of carriers in the base are electrons since the base is of n-type semiconductor. The large number of holes entering the base from the emitter swamps the small number of electrons there. As the base-collector–junction is reverse biased, these holes, which appear as minority carriers at the junction, can easily cross the junction and enter the collector.

The holes in the base could move either towards the base terminal to combine with the electrons entering from outside or cross the junction to enter into the collector and reach the collector terminal. The base is made thin so that most of the holes find themselves near the reverse-biased base-collector junction and so cross the junction instead of moving to the base terminal.

Note: Due to forward bias a large current enters the emitter-base junction, but most of it is diverted to the adjacent reverse-biased base-collector junction and the current coming out of the base becomes a very small fraction of the current that entered the junction.

If we represent the hole current and the electron current crossing the forward-biased junction by the sum Ih + Ie. We see that the emitter current IE = Ih + Ie but the base current IB << Ih + Ie because a major part of IE goes to the collector instead of coming out of the base terminal.

The base current is thus a small fraction of the emitter current. It is obvious from the above description and also from a straightforward application of Kirchoff’s law. that the emitter current is the sum of the collector current and base current:

IE = IC + IB We also see that IC < IE’ Our description of the direction of motion of the holes is identical to the direction of the conventional current. But the direction of motion of electrons is just opposite to that of the current.

Thus in a p–n–p transistor the current enters from the emitter into the base whereas in an n–p–n transistor it enters from the base into the emitter. The arrowhead in the emitter shows the direction of the conventional current.

We can conclude that in the active state of the transistor, the emitter-base junction acts as a low resistance while the base-collector acts as a high resistance. In a transistor, only three terminals are available viz emitter (E), base (B), and collector (C).

Therefore in a circuit, the input/output connections have to be such that one of these (E, B, or C) is common to both the input and the output. Accordingly, the transistor can be connected in either of the following three configurations: Common Emitter (CE), Common Base (CB), and Common Collector (CC).

Working of Transistor

There are four possible ways of biasing the two P-N junctions (emitter junction and collector
junction) of the transistor.

Active mode: Also known as linear mode operation.

Saturation mode: Maximum collector current flows and the transistor acts as a closed switch from collector to emitter terminals.

Cut-off mode: Denotes operation like an open switch where only leakage current flows.

Inverse mode: The emitter and collector are interchanged.

Different modes of operation of a transistor

Operating mode Emitter base bias Collector base bias

1. 1. Active Forward Reverse
   2. Saturation Forward Forward
   3. Cut off Reverse Reverse
   4. Inverse Reverse Forward

• A transistor is mostly used in the active region of operation i.e., the emitter-base junction is forward biased and the collector-base junction is reverse biased.
• From the operation of the junction transistor, it is found that when the current in the emitter circuit changes.
• There is a corresponding change in collector current.
• In each state of the transistor, there is an input port and an output port. In general, each electrical quantity (V or I) obtained at the output is controlled by the input.

Transistor Configurations

A transistor can be connected in a circuit in the following three different configurations. Common base (CB), Common emitter (CE), and Common collector (CC) configuration. (1 ) CB configurations: Base is common to both emitter and collector.

• Input current = Ie Input voltage = VEB Output voltage = VCB Output current = IC
• With a small increase in emitter-base voltage VEB, the emitter current Ie increases rapidly due to small input resistance.

Input characteristics: If VCB = constant, the curve between Ie and VEB is known as input
characteristics.

It is also known as emitter characteristics:

The input characteristics of NPN transistors are also similar. but I and VEB both are negative and VCB is positive. The dynamic input resistance of a transistor is given by

⇒ \(R_i=\left(\frac{\Delta V_{E B}}{\Delta I_e}\right)_{V_{\text {ces-constant }}}\left\{R_i \text { is of the order of } 100 \Omega\right\}\)

Output characteristics: Taking the emitter current ie constant, the curve drawn between I C and VCB are known as output characteristics of CB configuration.

Dynamics output resistance \(\mathrm{R}_0=\left(\frac{\Delta \mathrm{V}_{\mathrm{CB}}}{\Delta \mathrm{i}_{\mathrm{C}}}\right)_{\mathrm{i}_{\mathrm{a}}=\text { constant }}\)

Note: Transistor as CB amplifier

⇒ \(\text { ac current gain } \alpha_c=\frac{\text { Small change in collector current }\left(\Delta i_c\right)}{\text { Small changeincollectorcurrent }\left(\Delta i_e\right)}\)

⇒ \(\text { dc current gain } \alpha_{\mathrm{dc}}(\text { or } \alpha)=\frac{\text { Collector current }\left(i_{\mathrm{c}}\right)}{\text { Emitter current }\left(i_e\right)}\)

value of dc lies between 0.95 to 0.99

⇒ \(\text { Voltage gain } \mathrm{A}_{\mathrm{v}}=\frac{\text { Change in output voltage }\left(\Delta \mathrm{V}_0\right)}{\text { Change in input voltage }\left(\Delta \mathrm{V}_{\mathrm{i}}\right)}\)

A v =ac × Resistance gain

⇒ \(\text { Power gain }=\frac{\text { Change in output power }\left(\Delta P_0\right)}{\text { Change in input power }\left(\Delta P_C\right)}\)

⇒ \(\text { Power gain }=\alpha_{\mathrm{ac}}^2 \times \text { Resistance gain }\)

Common Emitter(CE): The transistor is most widely used in the CE configuration. When a transistor is used in CE configuration, the input is between the base and the emitter and the output is between the collector and the emitter.

The variation of the base current B with the base-emitter voltage VBE is called the input characteristic. The output characteristics are controlled by the input characteristics. This implies that the collector current changes with the base current.

CE configurations: Emitter is common to both base and collector.

The graphs between voltages and currents, when the emitter of a transistor is common to input and output circuits, are known as CE characteristics of a transistor.

Input characteristics: The input characteristics curve is drawn between base current Ib and emitter-base voltage VEB, at constant collector-emitter voltage VCE.

Dynamic input resistance \(\mathrm{R}_{\mathrm{i}}=\left(\frac{\Delta \mathrm{V}_{\mathrm{BE}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE} \rightarrow \text { constant }}}\)

Output characteristics: Variation of collector current IC with VCE can be noticed for VCE between 0 to 1 V only. The value of VCE up to which the IC changes with VCE is called knee voltage. The transistors are operated in the region above knee voltage.

Dynamic output resistance \(\mathrm{R}_0=\left(\frac{\Delta \mathrm{V}_{\mathrm{CE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right)_{\mathrm{I}_{\mathrm{B}} \rightarrow \text { cons tant }}\)

Transistor as a device:

The transistor can be used as a device application depending on the configuration used (namely CB, CC, and CE), the biasing of the E-B and B-C junction, and the operation region namely cutoff, active region, and saturation.

When the transistor is used in the cutoff or saturation state it acts as a switch. On the other hand for using the transistor as an amplifier, it has to operate in the active region.

Transistor as a switch:

We shall try to understand the operation of the transistor as a switch by analyzing the behavior of the base-biased transistor in the CE configuration. Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we get \(V_{B B}=I_B R_B+V_{B E} \quad \text { and } \quad V_{C E}=V_{C C}-I_C R_C.\)

We shall treat V BB as the DC input voltage Vi and VCE as the DC output voltage Vo.So, we have
\(V_i=I_B R_B+V_{B E} \quad \text { and } \quad V_0=V_{C C}-I_C R_C .\)

Let us see how V 0 changes as Vi increases from zero onwards. In the case of the Si transistor, as long as input Vi is less than 0.6 V, the transistor will be in a cut-off state, and current I C will be zero. Hence V O = VCC.

When V i becomes greater than 0.6 V the transistor is in an active state with some current I C in the output path and the output V0 decreases as the term I CRC increases. With the increase of Vi, I C increases almost linearly and so V0 decreases linearly till its value becomes less than about 1.0 V.

Beyond this, the change becomes non-linear and the transistor goes into a saturation state. With further increase in V i the output voltage is found to decrease further towards zero though it may never become zero.

If we plot the V0 vs Vi curve, [also called the transfer characteristics of the base-biased transistor, we see that between cut off state and active state and also between active state and saturation state there are regions of non-linearity showing that the transition from cutoff state to active state and from active state to saturation state are not sharply defined.

As long as Vi is low and unable to forward bias the transistor, V0 is high (at VCC). If Vi is high enough to drive the transistor into saturation very near to zero. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.

This shows that if we define low and high states as below and above certain voltage levels corresponding to the cutoff and saturation of the transistor, then we can say that a low input switches the transistor off and a high input switches it on.

Transistor as an Amplifier (CE-Configuration): To operate the transistor as an amplifier it is necessary to fix its operating point somewhere in the middle of its active region. If we fix the value of V BB corresponding to a point in the middle of the linear part of the transfer curve then the DC base current I B would be constant and the corresponding collector current I C will also be constant.

The DC voltage V CE = VCC – I CRC would also remain constant. The operating values of VCE and I B determine the operating point, of the amplifier. If a small sinusoidal voltage with amplitude vs is superposed on the DC base bias by connecting the source of that signal in series with the VBB supply, then the base current will have sinusoidal variations superimposed on the value of IB.

As a consequence the collector current. has sinusoidal variations superimposed on the value of I C producing in turn corresponding change in the value of V0. We can measure the AC variations across the input and output terminals by blocking the DC voltages with larger capacitors.

In the description of the amplifier given above, we have not considered any AC signal. In general, amplifiers are used to amplify alternating signals. Now let us superimpose an ac input signal vi (to be amplified) on the bias VBB (dc). The output is taken between the collector and the ground.

The working of an amplifier can be easily understood if we first assume that vi = 0. Then applying Kirchhoff’s law to the output loop, we get \(V_{c C}=V_{C E}+I_C R_L\)

Likewise, the input loop gives V_cc = V_ce + I_cR_l

V_BB = V_BE + I_BR_B

when v_i is not zero, we get

V_BE + v_i= V_BE + I_BR_B + I_B (R_B + ri)

The change in V_BE can be related to the input resistance ri and the change in I B. Hence

⇒ \(\begin{aligned}
& v_i=\Delta l_B\left(R_B+r_i\right) \\
& =r \Delta I_B
\end{aligned}\)

The change in I B causes a change in I C. We define a parameter ac, which is similar to the dc defined in the equation as

⇒ \(\beta_{\mathrm{ac}}=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{\mathrm{i}_{\mathrm{c}}}{i_{\mathrm{b}}}\)

which is also known as the AC gain Ai. Usually, β_ac is close to β_dc in the linear region of the output characteristics. The change in I C due to a change in I B causes a change in V_cE and the voltage drops across the resistor R L because VCC is fixed. These changes can be given by \(\Delta V_{c c}=\Delta V_{c E}+R_L \Delta I_c=0 \text { or } \Delta V_{c E}=-R_L \Delta I_c\)

The change in VCE is the output voltage v0. From the equation, we get \(v_0=\Delta V_{c E}=-\beta_{a c} R_{\mathrm{L}} \Delta l_{\mathrm{B}}\)

The voltage gain of the amplifier is \(A_v=\frac{v_0}{v_i}=\frac{\Delta V_{C E}}{r \Delta I_B}=-\frac{\beta_{a c} R_L}{r}\)

The negative sign represents that the output voltage is opposite with phase of the input voltage. From the discussion of the transistor characteristics, you have seen that there is a current gain AC in the CE configuration. Here we have also seen the voltage gain Av.

Therefore the power gain Ap can be expressed as the product of the current gain and voltage gain. Mathematically \(A_p=\beta_x \times A_v\) Since ac and Av are greater than 1, we get ac power gain. However, it should be realized that a transistor is not a power-generating device. The energy for the higher AC power at the output is supplied by the battery.

Note: Transistor as CE amplifier

• AC Currnet gain \(\beta_{\mathrm{ac}}=\left(\frac{\Delta \mathrm{i}_{\mathrm{c}}}{\Delta \mathrm{i}_{\mathrm{b}}}\right) \mathrm{V}_{\mathrm{cE}}=\text { constant }\)
• DC current gain \(\beta_{d c}=\frac{i_c}{i_b}\)
• Voltage gain: \(\mathrm{A}=\frac{\Delta \mathrm{V}_0}{\Delta \mathrm{V}_{\mathrm{i}}}=\beta_{x c} \times \text { Resistance gain }\)
• Power gain \(=\frac{\Delta P_0}{\Delta P_i}=\beta_{a c}^2 \times \text { Resistance }\)
• Trans conductance (g m): The ratio of the change in collector current to the change in emitter-base voltage is called trans conductance \(g_m=\frac{\Delta i_c}{\Delta V_{E B}} \text {. Also } g_m=\frac{A_v}{R_L} ; R_L=\text { Load resistance. }\)
• Relation between and b: \(\beta=\frac{\alpha}{1-\alpha} \text { or } \alpha=\frac{\beta}{1+\beta}\)

Solved Examples

Example 30. Let E, i_c, and i_B represent the emitter current, the collector current, and the base current respectively in a transistor. Then

1. i_c is slightly smaller than i_E.
2. i_c is slightly greater than i_E.
3. i_B is much smaller than i_E.
4. i_B is much greater than i_E.

Answer: (1,3)

Example 31. In a common base transistor amplifier, the input and the output resistance are 500Ω and 40kΩ and the emitter current is 1.0mA. Find the input and the output voltages. Given Ω = 0.95.
Solution: The input voltage is the emitter current multiplied by input resistance, that is, V in = iE × Rin = (1.0 × 10–3 A) × 500Ω = 0.5 V Similarly, the output voltage is V out = iC × Rout = Ω iE × Rout = 0.95 (1.0 × 10–3 A) × (40 × 103Ω) = 38 V.

Example 32. A P–N–P transistor is used in common–emitter mode in an amplifier circuit. A change of 40ΩA in the base current brings a change of 2mA in collector current and 0.04 V in base-emitter voltage. Find the: 1 input resistance (Rinp.), and 2 the base current amplification factor. If a load of 6kΩ is used, then also find the voltage gain of the amplifier.
Solution: \(\begin{aligned}
& \text { Given } \Delta \mathrm{I}_{\mathrm{B}}=40 \mu \mathrm{A}=40 \times 10^{-5} \mathrm{~A} \\
& \Delta \mathrm{I}_{\mathrm{c}}=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A} \\
& \Delta \mathrm{V}_{\mathrm{BE}}=0.04 \text { volt, } \mathrm{R}_{\mathrm{L}}=6 \mathrm{k} \Omega=6 \times 10^3 \Omega
\end{aligned}\)

Input Resistance,

⇒ \(R_{\text {inp. }}=\frac{\Delta V_{B E}}{\Delta I_B}=\frac{0.04}{40 \times 10^{-6}}=10^3 \Omega=1 \mathrm{k} \Omega\)

Current amplification factor,

⇒ \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{2 \times 10^{-3}}{40 \times 10^{-6}}=50\)

Voltage gain in common–emitter configuration, \(A_v=\beta \frac{R_L}{R_{\text {inp. }}}=50 \times \frac{6 \times 10^3}{1 \times 10^3}=300 .\)

Example 33. In an N–P–N transistor 10¹⁰ electrons enter the emitter in 10^-6 s. 2% of the electrons are lost in the base. Calculate the current transfer ratio and current amplification factor.
Solution: We know that current = charge/time

The emitter current (IE) is given by \(I_E=\frac{\mathrm{Ne}}{\mathrm{t}}=\frac{10^{10} \times\left(1.6 \times 10^{-19}\right)}{10^{-6}}=1.6 \mathrm{~mA}\)

The base current (IB) is given by \(I_B=\frac{2}{100} \times 1.6=0.032 \mathrm{~mA}\)

In a transistor, IE = IB + IC IC = IE – IB = 1.6 – 0.032 = 1.568 mA

Current transfer ratio \(=\frac{I_C}{I_E}=\frac{1.568}{1.6}=0.98\)

Current amplification factor= \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=\frac{1.568}{0.032}=49 .\)

Example 34. When the voltage between emitter and the base VEB of a transistor is changed by 5mV while keeping the collector voltage VCE fixed when then its emitter current changes by 0.15 mA. Calculate the input resistance of the transistor.
Solution: 33.33 ohm

Example 35. A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base–emitter voltage, the base current changes by 20A, and the collector current changes by 2 mA. The load resistance is 5k. Calculate (a) the factor, (B) the input resistance R BE’ (C) the transconductance, and (D) the voltage gain.
Solution: \(\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{2 \mathrm{~mA}}{20 \mu \mathrm{A}}=100\)

The input resistance \(R_{B E}=\frac{\Delta V_{B E}}{\Delta I_B}=\frac{20 \mathrm{mV}}{20 \mu \mathrm{A}}=1 \mathrm{k} \Omega\)

Transconductance \(=\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{V}_{\mathrm{BE}}}=\frac{2 \mathrm{~mA}}{20 \mathrm{mV}}=0.1 \mathrm{mho} .\)

The change in output voltage is RLC = (5 kW) (2mA) = 10V.

The applied signal voltage = 20 mV.

Thus, the voltage gain is, \(=\frac{10 \mathrm{~V}}{20 \mathrm{mV}}=500\)

Example 36. The a-c current gain of a transistor is = 19. In its common-emitter configuration, what will be the change in the collector current for a change of 0.4 mA in the base current? What will be the change in the emitter current?
Solution: By definition, the a-c current gain b is given by

⇒ \(\beta(\mathrm{a}-\mathrm{c})=\frac{\Delta_{\mathrm{iC}}}{\Delta_{\mathrm{iB}}} \quad \Delta_{\mathrm{ic}}=\beta \times \Delta_{\mathrm{BB}}=19 \times 0.4 \mathrm{~mA}=7.6 \mathrm{~mA}.\)

The emitter – -current is the sum of the base- -current and the collector current (iE = iB + iC)

⇒ \(\Delta_{\mathrm{iE}}=\Delta_{\mathrm{iB}}+\Delta_{\mathrm{iC}}=0.4 \mathrm{~mA}+7.6 \mathrm{~mA}=80 \mathrm{~mA} .\)

Example 37. A transistor is connected in a common-emitter (C-E) configuration. The collector supply is 8 V and the voltage drop across a resistor of 800 in the collector circuit is 0.5 V. If the current gain factor is 0.96, find the base current.
Solution: The alternating-current gain is \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{1-0.96}=24\)

The collector – current is

⇒ \(\mathrm{i}_{\mathrm{c}}=\frac{\text { voltage }- \text { drop across collector resistor }}{\text { resistance }}=\frac{0.5 \mathrm{~V}}{800 \Omega} \times 10^{-3} \mathrm{~A} \text {. }\)

But \(\beta=\frac{i_C}{i_B}\) where i B is base – current.

⇒ \(i_B=\frac{i_C}{\beta}=\frac{0.625 \times 10^{-3} \mathrm{~A}}{24}=26 \times 10^{-6} \mathrm{~A}-26 \mu \mathrm{A} .\)

Feedback amplifier and transistor oscillator:

In an oscillator, we get AC output without any external input signal. A portion of the output power is returned (feedback) to the input in phase with the starting power (this process is termed positive feedback) The feedback can be achieved by inductive coupling (through mutual inductance) or LC or RC networks.

Suppose switch S1 is put on to apply proper bias for the first time. A surge of collector current flows in the transistor. This current flows through the coil T2 where terminals are numbered 3 and 4

This current does not reach full amplitude instantaneously but increases from X To Y, as The inductive coupling between coil T2 and coil T1 now causes a current to flow in the emitter circuit (note that this is the ‘feedback’ from input to output). As a result of this positive feedback, this current (in T1 emitter current) also increases from X’ to Y’.

The current in T2 (collector current), connected in the collector circuit acquires the value Y when the transistor becomes saturated. This means that the maximum collector current is flowing and can increase no further. Since there is no further change in collector current, the magnetic field around T2 ceases to grow.

As soon as the field becomes static, there will be no further feedback from T2 to T1. Without continued feedback, the emitter current begins to fall. Consequently, the collector current decreases causing the magnetic field to decay around the coil T2. Thus, T1 is now seeing a decaying field in T2 (opposite from what it saw when the field was growing at the initial start operation).

This causes a further decrease in means that both IE and IC cease to flow. Therefore, the transistor has reverted to its original state (when the power was first switched on). The whole process now repeats itself. The transistor is driven to saturation, then to cut-off, and then back to saturation.

The time for a change from saturation to cut-off and back is determined by the constant of the tank circuit or tuned circuit (inductance L of Coil T2 and C connected in parallel to it). The resonance frequency (v) of this tuned circuit determines the frequency at which the oscillator will oscillate. \(v=\frac{1}{2 \pi \sqrt{L C}}\)

Analogue Circuits and Digital Circuits and Signal:

There are two types of electronic circuits: analog circuits and digital circuits: In analog circuits, the voltage (or current) varies continuously with time. Such a voltage (or current) signal is called an ‘analog signal’. The figure shows a typical voltage analog signal varying sinusoidally between 0 and 5V.

On the other hand, in digital circuits, the voltage (or current) has only two levels, either zero or some constant value of voltage. A signal having only two levels of voltage (or current) is called a ‘digital signal’.

Figure shows a typical digital signal in which the voltage at any time is either 0 or 5V. In digital circuits, the binary number system is used, according to which the two levels of the (digital) signal are represented by the digits 0 and 1 only. The digital circuits are the basis of calculators, computers, etc.

Note: Voltage Signal:

Analog voltage signal: The signal which represents the continuous variation of voltage with time is known as analog voltage signal.

Digital voltage signal: The signal that has only two values. i.e., either a constant high value of voltage or zero value is called a digital voltage signal.

Decimal and Binary Number System:

Decimal number system: In a decimal number system, we have ten digits i.e. 0,1,2,3,4,5,6,7,8,9 A decimal number system has a bse of ten (10)

LSD = Least significant digit

MSD = Most significant digit

Binary number system: A number system that has only two digits i.e. 0 (Low) and 1 (High) is known as a binary system. The base of the binary number system is 2.

Each digit in a binary system is known as a bit and a group of bits is known as a byte.

The electrical circuits that operate only in these two states i.e. (On or High) and 0 (i.e. Off or Low) are known as digital circuits.

Decimal to binary conversion

Divide the given decimal number by 2 and the successive quotients by 2 till the quotient becomes
zero.

The sequence of remainders obtained during divisions gives the binary equivalent of a decimal
number.

The most significant digit (or bit) of the binary number so obtained is the last remainder and the
least significant digit (or bit) is the first remainder obtained during the division.

For Example: Binary equivalence of 61

Binary to decimal conversion: The least significant digit in the binary number is the coefficient of 2 with power zero. As we move towards the left side of LSD, the power of 2 goes on increasing.

For Example: (11111100101)2 = 1 × 210 + 1 × 29 + 1 × 28 + 1 × 27 + 1 × 26 + 1 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 2021

Logic Gates:

A logic gate is a digital circuit that works according to some logical relationship between input and output voltages. It either allows a signal to pass through or stops it. The logic gates are the building blocks of digital circuits. There are three basic logic gates.

1. OR gate
2. AND gate
3. NOT gate

The OR Gate:

The OR gate is a device that has two input variables A and B and one output variable Y, and follows the Boolean expression, A + B = Y, read as’ A OR B equal Y’. Its logic symbol is shown in the figure.

The possible combinations of the inputs A and B and the output Y of the OR gate can be known with the help of an electrical circuit, shown in figure. In this circuit, two switches A and B (inputs) are connected in parallel with a battery and a bulb Y (output).

The AND Gate:

The AND gate is also a two-input and one-output logic gate. It combines the inputs A and B to give the output Y, according to the Boolean expression A.B=Y Read A AND B Equals Y’

The NOT Gate:

The NOT gate has only one input and one output. It combines the input A with the output Y, according to the Boolean expression A Y=, read as ‘NOT A equals Y’.

It means that Y is a negation (or inversion) of A. Since there are only two digits 0 and 1 in the binary system, we have, Y = 0, if A = 1 and Y = 1 if A = 0. The logic symbol of the NOT gate is shown in the figure.

The possible combinations of the input A and the output Y of the NOT gate can be known with the help of an electric circuit, shown in figure. In this circuit, a switch A (input) is connected in parallel to a battery and a bulb Y(output).

The working of the circuit is as follows: If switch A is open (A = 0), the bulb will glow (Y = 1). If switch A is closed (A = 1), the bulb will not glow (Y = 0). These two possible combinations of input A and output Y are tabulated in the figure, which is the truth table of the NOT gate.

Combinations of gates:

Various combinations of the three basic gates, namely, OR, AND, and NOT, produce complicated digital circuits, which are also called ‘gates’. The commonly used combinations of basic gates are the NAND gate, and NOR, gate. These are also called universal gates.

The NAND gate:

This gate is a combination of AND and NOT gates. If the output Y’ of the AND gate is connected to the input of the NOT gate, as shown in the figure, the gate so obtained is called the NAND gate. The logic symbol of the NAND gate is shown in the figure.

The Boolean expression for the NAND gate is \(\overline{A \cdot B}=Y\) read as ‘A AND B negated equals Y’

The truth table of the NAND gate can be obtained by logically combining the truth tables of AND and NOT gates. In the figure, the output Y’ of the truth table of the AND gate has been negated (NOT operation) to obtain the corresponding outputs Y for the NAND gate. The resulting table is the truth table of the NAND gate.

The NOR Gate:

The NOR gate is a combination of OR and NOT gates. If the output Y’ of the OR gate is connected to the input of the NOT gate, as shown in the figure, the gate so obtained is the NOR gate.

The Boolean expression for the NOR gate is

⇒ \(\overline{\mathrm{A}+\mathrm{B}}=\mathrm{Y}\)

read as ‘A OR B negated equals Y’ :

Example 1. The truth table of the NOR gate can be obtained by logically combining the truth tables of OR and NOT gates. In Figure (a), the outputs Y’ of the truth table of the OR gate have been negated to obtain the corresponding outputs Y for the NOR gate.

1. An OR gate and an AND gate respectively
2. An AND gate and a NOT gate respectively
3. An AND gate and an OR gate respectively
4. An OR gate and a NOT gate respectively

Question 2. The truth table for the following is

Answer: 2.

Example 3. In circuit in the following figure the value of Y is

1. zero
2. 1
3. Fluctuates between 0 and 1
4. Indeterminate as the circuit cannot be realized

Answer: 1. zero

Example 4. The output Y for the following logic gate circuit will be

1. AB
2. \(\bar{A} \cdot \bar{B}\)
3. \(\overline{A+B}\)
4. \(\overline{A \cdot B}\)

Answer: 4. \(\overline{A \cdot B}\)

Example 5. The following truth table belongs to which one of the four gates

1. OR
2. NAND
3. XOR
4. NOR

Answer: 4. NOR

Example 6. The given circuit is for the gate

1. NOR
2. NAND
3. NOT
4. XOR

Answer: 1. NOR

Example 7. The truth table of the logic circuit shown

Answer: 2.