Chapter 2

Semiconductors

2.1 INTRODUCTION

The physics of the materials used in the construction of various electronic devices is very important to understand the basic operation of the device. The materials used in the electronic devices are semiconductors. These are the materials which neither fall in the category of conductors nor insulators. Unlike the dielectric materials, they have some conductivity, which can be controlled by external agents. This advantage of controlling the conductivity of the material at designers’ requirements makes the material most suitable for electronic devices. In other words, the electron, the basic entity, is to be made to do work as per the requirement. The tiny energy associated with the tiny particle is contributed in the direction that is useful for a given application. The electrons, which are free and are not bound to any particular atom or molecule, are the charges that contribute to the conduction. In a semiconductor, unlike a conductor, there are positively charged holes that contribute conduction. So, in a semiconductor, there are two types of charge carriers, that contribute to conduction. They are negatively charged electrons and positively charged holes. Along with conduction current in a semiconductor, there is also a possibility of one more type of current known as diffusion current. This diffusion current may be once again due to electrons or holes. This type of current is not possible in a conductor. So, in a conductor, the current is only due to conduction of electrons. Whereas, in a semiconductor, the current consists of four components, namely, conduction current due to electrons, conduction current due to holes, diffusion current due to electrons and diffusion current due to holes. The behaviour of all these components of the current is to be known in detail to understand the electrical characteristics of the material that is used in the electronic devices. The most popularly used materials used in fabrication of the active devices are silicon and germanium.

The charge of negative electricity and the mass of the electron are known to be 1.6 × 10⁻¹⁹ C and 9.11 × 10⁻³¹ kg, respectively. Since 1 A of current is flow of 1 C/s, 1 × 10⁻¹² A (1 pA) represents the motion of approximately 6000000 electrons per second. But, 1 pA of current is in general, very minute. The radius of the electron is estimated to be 10⁻¹⁵ m. This being a very small value, the electron is assumed to be point in all the discussion to follow.

A semiconductor material has two types of charge carriers. They are electrons and holes. Holes are formed when an electron is missing from the covalent bond that is formed between neighbouring atoms of the semiconductor material. Since an electron is missing from the stable covalent bond, the empty space in the bond structure is referred to as hole. The electron is negatively charged and so, the hole is positively charged. The magnitude of the charge of a hole is same as that of the electron. These holes also contribute to conduction and the details of all of the current components will be taken up in detail later in the chapter.

The unit of energy, Joule, is in general used in mks system. Depending on the magnitude of the energy, the joule can be expressed in kilo, mega, micro, or pico in the discussion of basic electronic theory; since the energies associated with the electron are very small, expressing them in joules is difficult. Thus, a new unit of energy is used for such tiny magnitudes of energies.

One electron volt (eV) can be defined as

 

If an electron falls through a potential of 1 V, its kinetic energy will increase with decrease in potential energy. Thus,

 

This unit is unit of energy and can be used for any type of energy.

2.2 POTENTIAL BARRIER

The concept of potential barrier finds a very interesting and most useful application in semiconductor junctions. To understand this concept, consider two parallel electrodes A and B separated by distance d, with B at a negative potential V_d with respect to A as shown in Fig. 2.1.

The force f on charge q in presence of electric field is defined as charge times the electric field intensity ε. The same force may be expressed in terms of Newton’s second law of motion as its mass m times acceleration if the charge moves with a velocity v.

 

The work done per unit charge in moving a point charge from point A to point B is known as the potential V of the point B.

 

And, the potential energy U is charge times the potential,

 

According to the law of conservation of energy the total energy in a system remains constant. Thus, the two components of energy available in this system are potential energy U [Eq.(2.5)] and the kinetic energy associated with moving charge (½ mv²).

Referring to Fig. 2.1, consider an electron leaving the surface of A with an initial velocity v₀ in the direction towards B. Since the potential of A is zero, the potential energy at this electrode is zero and the total energy is entirely due to kinetic energy and is equal to

As the electrons move forward towards B, the potential, and thus the potential energy, increases. Due to conservation property of energy, increase in potential energy leads to decrease in kinetic energy and thus decrease the velocity with which the electron travels. Thus,

Thus, the velocity of the electron reduces as it travels toward B, as is true since the electron is moving in a repelling field. The final speed of the electron, as it reaches the electrode B, is independent of the variations of the field distribution between the two plates and depends only upon the magnitude of the potential difference V_d. If the electron is to reach the electrode B, its initial speed must be large enough so that

If not, Eq. (2.8) leads to the impossible result that velocity v is imaginary. Fig. 2.1 also shows the variation of the potential between the two electrodes A and B. It also gives a plot of energy with the distance. As can be seen from the figure, when an electron is released with an initial velocity such that the electron, as it travels forward, reduces its velocity and at a distance where the total energy is equal to the potential energy, the speed of the electron becomes zero and the electron will have to halt. Which means at this distance, there is no kinetic energy and the electron is at rest. Further, the electron cannot move in the forward direction to reach the electrode B and will have to reverse its direction toward electrode A. The distance travelled by the electron between the electrodes, x₀, depends on the total energy of the system or, in other words, on the initial kinetic energy of the electron. The higher the initial velocity, larger is the value of x₀. If the electron were to reach the other electrode, the initial energy should be more than the potential energy at the electrode B. This is as shown in Fig. 2.1.

From the above discussion it can be concluded that the electron never reaches the shaded portion of the energy diagram as shown in the figure. The electron travels forward with an initial velocity toward B until the electron hits the potential hill. On hitting this hill, it reverses its direction and travels in the opposite direction but can never reach the hill. No electron can thus exist in the shaded portion of the plot since if it exists, it is supposed to have negative kinetic energy, that is, negative velocity, which is an impossible condition practically. This potential hill is a fictitious barrier, which obstructs the electron from entering the other side of the barrier. Physically, there is nothing like this potential hill but the electron behaves as if it is bouncing back after colliding with the potential hill. This is the potential barrier, which is important in understanding the behaviour of the semiconductor junction.

2.3 ENERGY-BAND THEORY

The general structure of metals and semiconductors is crystalline in nature, that is, the atoms are arranged in arrays in the form of a definite fundamental pattern. The fundamental pattern may either consist of atoms or molecules depending up on the material under consideration. The energy associated with electrons of a single atom is not applicable to these crystalline structures. The atoms are not free as in a gas, but are shared by the neighbouring atoms. So, it is not possible to consider a single atom separately and discuss about the discrete energy levels of the electrons in the atom. The energy associated with electrons in a crystal is not the same as that when the atom is singled out. This is so because the potential characterising the crystalline structure is now a periodic function in space whose value at any point is appreciably by the presence of the neighbouring atoms. The levels of the outer or valence shell are affected and the inner shell electrons are not much disturbed since the neighbouring atoms share the electrons from the outermost shell. Or, in other words, the valence shell electrons are no more an inherent property of the base atom but are shared by more than one atom. The new energy levels are determined by quantum mechanics and sharing the valence shell electrons results in an energy-band of closely spaced energy states instead of widely separated energy levels of the isolated atom. That is, the energy levels are no more discrete and separated by large amounts of levels as in the case of an isolated atom, but are in very closely packed energy states such that it seems to be a continuous energy-band where electrons occupy different levels in the band.

For a better understanding of the concept of energy-band theory, consider group IVA elements [carbon (C), silicon (Si), germanium (Ge) and tin (Sn)] with four valence-shell electrons. There are N atoms in a considered crystal and so, there will be 4N electrons in the outer subshell of the system. Of these, 2N electrons are s type and remaining 2N are p type. Assume a situation where it is possible to vary the spacing between the neighbouring atoms without changing the basic structure of the crystal. When the interatomic distance is very large, the atoms can be considered to be isolated and the energy levels discrete, as in the case of a gas. The electrons in the inner shells are not considered since they are not affected. In the crystalline system thus formed, there are 2N s electrons occupying all the possible 2N levels. Of the possible 6N p states in the valence shell, only 2N levels are occupied and the remaining 4N levels of p electrons are vacant.

Now, if the interatomic spacing in the crystal is decreased, an atom will exert an electric force on its neighbours. A coupling between atoms is formed and the wave functions overlap, and the crystal becomes an electronic system that must follow Pauli exclusion principle. Pauli exclusion principle states that in an electronic system, no two electrons can have same four quantum numbers, n, l, m₁ and m_s. Or, in other words, no two electrons can occupy the same quantum state in an electronic system. So, when an electronic system is formed by bringing the atoms close to each other in a crystal, the discrete energy levels of the electrons of single atom spread out into an energy-band, which is nothing but a large number of closely packed energy levels. As the interatomic spacing is decreased, since N is a very large number, and the separation between levels is small, the total spread between the minimum and maximum energy may be several electron volts. As the spacing is further decreased, the energy bands of both the s electrons and p electrons spread out as shown in Fig. 2.2. It may be noted that the s electron-band and the p electron-band are separated by an energy gap (forbidden energy gap), which decreases as the interatomic spacing is decreased. Thus, as can be seen from the figure, the 2N electrons occupy all the s states, while only 2N states of the possible 6N states of the p type energy-band are occupied.

As the spacing between the atoms is further reduced, the energy gap between the two energy levels decreases and tends to combine at a particular spacing value. That is, at this value of spacing, the two bands overlap and the energy gap vanishes. Now, there are 8N possible states in the energy-band, of which 4 N states are occupied and 4N are vacant. It may be noted at this juncture that the identity of the electron is lost—the s and p electrons are no more different and these electrons belong to the crystal as such. Simply, there are 4N electrons in the possible 8N states. Since the atom contributes four electrons to the crystal from its outermost shell, it is referred to as tetravalent and the energy band as valence band.

Reducing the spacing further is a very complex situation and it may be represented as shown in Fig. 2.2. As can be seen from the figure, the two bands, which merged with each other, will separate once again with a forbidden energy gap. Now, there are two energy bands, the lower one is referred to as valence band and the upper one as conduction band, separated by an energy gap. The valence band consisting of 4N states is totally occupied by the 4N electrons and the conduction band having 4N states is totally vacant. This is where the interatomic spacing of a tetravalent semiconductor material is found in nature. The energy gap between the energy-bands depends on the type of material and is very large for an insulator, small for a semiconductor and in case of a metal there is no energy gap and the two bands overlap.

2.4 CLASSIFICATION OF MATERIALS

A material may be classified as metal, insulator or semiconductor based on the energy-band structure. The conductivity of the material depends on the electrons available in the conduction band. The more the electrons are in this band, more is the conductivity of the material. The electrons available in this energy-band are not bound to the crystal structure but are free to move about randomly throughout the material. In all the discussion, the electrons referred to are these electrons, which are available in the conduction band, and the electrons in the lower energy-bands are of no use as far as conduction is concerned. The energy-band structure for all three types of materials is as shown in Fig. 2.3.

In an insulator, as shown in Fig. 2.3, the forbidden energy gap between the conduction band and valence band is very large, of the order of 6 eV (diamond, for example). The valence band is totally filled with electrons and there are no electrons available in the conduction band. That is, all the electrons are bound to the crystalline structure and there are no electrons that can contribute to conduction. If one requires conduction in a material, the electrons from the valence band are to be shifted to conduction band. In other words, an electron that is bound in the covalent bond structure of the material is to be given a suitable energy such that the bond is broken and the excited electron reaches the conduction band level. It may be noted that the electron can only exist in the valence band or the conduction band, but if an electron in the valence band receives energy which will launch it into the forbidden energy gap, it will exist only in the valence band. Until the electron receives sufficient energy to overcome the forbidden energy gap, and thus to jump from the valence band to the conduction band, it remains in the valence band. In case of an insulator, this energy gap is so large that shifting electrons from the lower band to the conduction band is not possible and so there are no electrons in the conduction band. Thus, there is no conduction possible in these materials.

A material that has a relatively small energy gap, of the order of 1 eV, is referred to as a semiconductor. The energy gap E_G is 0.785 and 1.21 eV for germanium and silicon, respectively, at 0 K. At this absolute temperature (0 K), these materials are also insulators since the valence band is full and conduction band is totally empty. There is no conduction possible with such materials even at extremely small temperatures. But, as the temperature increases, the electrons in the valence band receive energy in the form of thermal energy and since E_G is small, small energy increments are sufficient to shift the electrons from the lower to upper bands. As the temperature increases, the system receives more and more energy from thermal excitations and more and more electrons are displaced from the valence band to the conduction band. Thus, the conductivity of the material improves. In other words, the covalent bond between the atoms in the crystal is broken and the electrons are released. These electrons are not bound to any structure now and are free to wander anywhere in the material. Their motion is very random, but the average number of electrons crossing a given cross-section would be the same as that of the electrons crossing from the opposite direction and so, the net movement of electrons at any given instant of time is zero. When an electron is released from the valence band into the conduction band, it leaves back an empty pocket in the valence band, which is positively charged and referred to as hole. A hole in a semiconductor refers to the empty energy levels in an otherwise filled valence band. This positively charged hole also contributes to conduction. As one electron shifting from the valence band to the conduction band leaves behind one hole in the valance band, the number of holes in the valence band is exactly the same as that of the electrons in the conduction band. This is true for a material that is pure. That is, in such a material, all the atoms of the crystal are of the same type and no other impurity atom is available. Then the material is said to be intrinsic. Introducing impurity atoms in the crystal may also enhance the conduction of the material. Such a material is referred to as extrinsic. The band gap energy of a crystal is a function of interatomic spacing and depends on temperature. It decreases with increase in temperature.

A metal is a material where there is no energy gap and the valence band and the conduction band are overlapping, as shown in Fig. 2.3. The electrons do not pertain to any definite band and so applying a small electric field to the material excites the electrons to a higher energy level and these electrons move under the influence of the applied field to constitute conduction current. There is no possibility of a hole in a metal or a conductor since there is no differentiation of valence and conduction bands and the energy gap is absent. Electrons can exist in any level of energy in the band.

2.5 INTRINSIC AND EXTRINSIC SEMICONDUCTORS

An intrinsic semiconductor is the one that is pure and has no impurity atoms in its crystalline structure. An extrinsic semiconductor is the one in which impurities are added purposefully to increase the conductivity of the material. For example, either trivalent or pentavalent atoms are added to tetravalent atoms. Adding the impurities, the conductivity of the material increases to a great extent.

The conductivity of the material depends on the number of atoms in the conduction band of the material. For a good conductor the electron concentration n (number of electrons per unit volume) is very large (≈ 10²³ electrons per m³); for an insulator it is very small (≈ 10⁷) and for a semiconductor its value lies between these two values. The valence electrons in a semiconductor are not free to wander about as they are in a metal, but rather are trapped in a bond between two adjacent ions.

All the Group IVA elements (C, Si, Ge and Sn) have four electrons in the valence shell and exhibit properties of semiconductors, of which silicon and germanium are widely used in electronic devices. The crystalline structure is a three dimensional tetrahedron with one atom affixed at each vertex. The basic unit repeats in all the three directions in the structure. The structure is as shown in Fig. 2.4. The silicon or germanium material has four electrons in the valence shell and these electrons of each atom are shared by the neighbouring four atoms to form covalent bonds. Thus, an electron pair is formed, which is the covalent bond. Sharing four electrons from the neighbouring atoms along with the four atoms of its valence shell constitutes eight electrons in the system and thus stability is achieved in the structure. Since four electrons are donated to the neighbouring atoms to form covalent bonds, the left-over part of the atom is an ion positively charged four times to that of the magnitude of the charge carried by each electron. There is a binding force between the adjacent atoms due to this sharing of electrons forming the covalent bonds. These valence electrons are tightly bound to the nucleus and so they are not free to wander about in the material. Even though there are four electrons available in the valence band of the atom, the crystal conductivity is very low since these electrons are bound.

The above discussion refers to the case when the material is ideal at 0° K. In this condition, the material behaves as an insulator since there are no free electrons. At room temperature, the conduction of the material increases since some of the electrons gain energy from the thermal agitation and acquire sufficient energy to get transferred from valence to conduction band. The thermal energy is sufficient to break the covalent bond and the electron that is bound previously gets released and does not belong to any particular atom. This electron is free and can go anywhere in the material. While a negatively charged electron is released to be free, it leaves behind an empty space, that is positively charged. This empty location in the valence band of the atom is referred to as hole. This positively charged hole also contributes to conduction like the free electrons in a semiconductor. Thus, breaking a covalent bond in the structure creates two carriers of current∔charge carriers∔one electron and one hole. The concept of electron and hole charge carriers is as shown in Fig. 2.5.

The positively charged hole contributes to conduction. The electrons in the neighbouring covalent atoms move to occupy the empty spaces in the valence band, that is, the hole. The electron moving from a bond to fill a hole leaves a hole in its initial position. Hence, the hole effectively moves in the direction opposite to that of the electron. An electron from another covalent bond may now fill this hole, in its new position, and the hole will correspondingly move one more step in the direction opposite to the motion of the electron. This conduction of the electrons from one hole to the other is equal to holes moving in the opposite direction. This mechanism of hole conduction does not involve any free electrons. So, the conduction of electricity is entirely due to holes moving in the opposite direction to that of the motion of bound electrons. This motion of positively charged holes in the forward direction is also conduction and since it does not involve free electrons, it is the conduction current or drift current due to holes in a semiconductor. Since existence of holes in a metal is not possible, such conduction current is not possible in a conductor.

The concept of hole current can be explained more elaborately with the help of Fig. 2.6. There are large numbers of filled covalent bonds in the crystalline structure of the material, as shown. They are assumed to be in straight line with a filled position at 6, as in Fig. 2.6(a). This represents a hole. Now, if the electron from position 7 leaves to occupy the vacant position 6, position 6 is occupied and an unfilled hole is available at 7. Effectively, even though the electron has moved from 7 to 6, one feels that the hole has shifted from 6 to 7. This continues and the hole can move in the forward direction when the bound electron is moving in the reverse direction. Since conventional current direction is reverse to that of the electron movement, the direction of movement of hole and the conventional current flow would be the same.

In a pure intrinsic semiconductor, the number of electrons in the conduction band is same as the number of holes in the valence band. Thus, n is the electron concentration, number of electrons per unit volume, and p is the hole concentration, number of holes per unit volume, then,

 

where n_i is called the intrinsic concentration of the semiconductor material. This value of intrinsic concentration is a function of temperature, since the number of holes and electrons depends on the temperature. The higher the temperature, the greater is the value of n_i. To increase the conductivity of the material, one requires large number of charge carriers (free electrons or holes) in the material. So, when an intrinsic material is considered, its conductivity can be improved by increasing the temperature. But, increasing the temperature beyond a limit is not possible practically. Thus, increasing the charge carrier concentration by increasing the temperature is not possible. An alternative is to increase the free electrons in the conduction band or to increase the holes in the conduction band by some artificial means. This increases the conductivity of the material even at room temperature. One way out is to add impurity atoms to the crystal structure and make the material extrinsic or impure by adding atoms other than the parent atoms in the structure purposefully. A small percentage of trivalent (B, Ga, In) or pentavalent (P, As, Sb) atoms is doped to the intrinsic silicon or germanium to make it extrinsic and improve conductivity.

2.6 p- AND n-TYPE SEMICONDUCTOR MATERIALS

Donor impurities: A pentavalent atom has five electrons in the valence shell. When these atoms are doped with either silicon or germanium, they displace some of the parent atoms. Four electrons of the five in the valence shell form covalent bonds with the neighbouring atoms as in the case of a parent atom and one electron is left over. This left over electron, when in the conduction band, can contribute to conduction. The amount of energy required to free the excess electron is very small, of the order of 0.01 eV for Ge or 0.05 eV for Si. Thus, there are new electrons that are free and with energy levels in the range of conduction band. These electrons are very useful for conduction. In other words, the impurity atom donates one electron to the material, which is a charge carrier. Thus, the pentavalent impurity is referred to as donor impurity. The suitable pentavalent atoms are antimony, phosphorus and arsenic. The structure of the crystal lattice of such a donor is as shown in Fig. 2.7.

When a donor impurity is added to the semiconductor, discrete energy levels are introduced in the energy gap at a very small distance below the conduction band. These levels are discrete as the doped impurity is away from the crystalline structure. The energy difference of this impurity atom’s electron and the conduction band lower edge is very small, of the order of 0.01 eV for Ge or 0.05 eV for Si. Thus, at room temperature, the fifth electron is at an energy level that is in the conduction band and so is a very good carrier of current.

Such a material where donor atoms are doped to increase free electrons is known as n-type material. Thus, in this n-type material, the number of electrons is very large compared to holes. The initial concentrations of the electrons and holes are equal to the intrinsic concentration. After doping, the electron concentration increases to the extent of doping concentration, but the hole concentration would be low compared to the intrinsic concentration since the excess electrons from the impurity atoms can recombine with the holes. As a whole, the electron concentration is very large compared to hole concentration in an n-type material. Thus,

Acceptor impurities: When a trivalent atom having three electrons in the outermost shell is doped with a semiconductor material, a p-type material is formed. The three electrons available in the valence shell are shared by the neighbouring Si or Ge atoms to form covalent bonds. The fourth bond is not formed and there is a vacancy in the bond, which leads to the creation of a hole, as shown in Fig. 2.8. These vacant spaces, holes, are positive charge carriers, which can conduct current. These holes accept electrons for conduction to take place and so, the trivalent atoms, boron, gallium and indium are known as acceptor impurities.

Adding acceptor impurity atoms to the semiconductor leads to insertion of discrete energy levels in the forbidden energy gap of the semiconductor. This level is nearer to the upper edge of the valence band. Since very minute energy is sufficient for the electron in the valence band to jump into this discrete energy level, large numbers of electrons contribute for the conduction through the holes created by the impurity doping. The conductivity of the material is multiplied by adding impurity atoms to the semiconductor material.

The p-type material thus has a large number of holes compared to that of electrons in the material. Holes are referred to as majority carriers in a p-type material whereas electrons in the p-type material are known as minority carriers. In a p-type material, hole concentration is very large compared to electron concentration and thus,

 

Mass action law: By adding acceptor atoms the hole concentration is increased and by adding donor atoms the electron concentration is increased in a semiconductor material. At thermal equilibrium, if n is electron concentration and p is the hole concentration in a given type of material, whatever may be the concentration of the impurities added, the Mass Action Law states that

 

where n_i is the intrinsic concentration and is the function of temperature.

2.7 MOBILITY AND CONDUCTIVITY

Considering a metal, there are a large number of electrons that are free and are not bound to an atom or molecule. Since there is no energy gap in metals, each atom contributes an electron or more for conduction. When an electron is removed from an atom, the remaining portion of the atom is a positively charged ion. So, one can imagine the situation as large number of electrons roaming around in the material where the positive heavy masses of ions are stationary. This is known as electron gas theory of a metal. So, a cloud of electrons wandering around in the material in the vicinity of fixed positive ions represents the metal.

These free electrons in the metal are not stationary but are moving randomly in the material. The electrons moving in a given direction collide with the stationary ions and change their direction and proceed further. The distance between two collisions is very random and the average value of such distances is called as mean free path. Since the direction is random, the number of electrons crossing a given cross-section at a given time would be same as that of the number of electrons moving in the opposite direction and hence the net current flowing would be zero.

When an electric field e is now applied to a metal, the electrons would be accelerated in the direction opposite to that of the applied field and velocity would increase infinitely with time when there are no collisions with the static positive ions. However, due to the presence of these static ions, the collisions take place and the electron loses energy. When steady state is achieved, the velocity of the electrons becomes finite. This velocity, referred to as drift velocity, is in opposite direction to that of the applied field. The speed at time t between collision is at, where a is the acceleration and can be given as q ε/m. Thus, the average speed is proportional to the applied field and so,

 

where μ is called the mobility of electrons. Thus, steady state drift speed has been superimposed upon the random thermal motion of electrons. Such a directed flow of electrons constitutes a current. To calculate the magnitude of this current, consider a metal of volume A × L, where A is the cross-section area and L is the length of the material. Assume that there are N electrons in the volume considered and T is the time taken for an electron to travel a distance of L. Then the total number of electrons passing through any crosssection in unit time is N/T, or, the total charge crossing the cross-section, by definition, current can be given as

 

since L/T is the average drift speed v of the electrons. Current density is current per unit area of cross-section and so,

 

Thus,

 

Since N is total number of electrons in a volume LA of material considered, electron concentration n is

 

Therefore,

 

where ρ = nq is the charge density and v is the drift velocity.

From Eq. (2.14),

 

where σ = nqμ is the conductivity of the metal. The above equation is nothing but Ohm’s Law.

If now a semiconductor is considered, there are two types of charge carriers. The first type are free electrons, which are similar to those present in a conductor and, in addition to this, positively charged holes also contribute to current as discussed in the previous sections. So, now in a semiconductor, the total current would be superposition of the two independent components of currents due to electrons and holes. If J_n is electron current density in a semiconductor, it can be given as

 

where μ_n is the mobility of the electrons in the semiconductor.

Similarly, the hole current density J_p can be expressed as

 

where μ_p is the mobility of holes in the semiconductor material and p is the hole concentration.

Thus, the total current density can be given as

 

where σ = (nμ_n + pμ_p)q is the conductivity of the semiconductor.

2.8 PROPERTIES OF SEMICONDUCTOR MATERIALS

Charge densities in a semiconductor: Consider a semiconductor that is doped with donor atoms of concentration N_D and acceptor atoms of concentration N_A. n is the electron concentration after doping and p is hole concentration. Now, since N_D number of donor atoms are added per unit cubic metre of the material, it releases N_D number of free electrons per unit volume. This means that it is leaving behind N_D positive ions in the system. Now, the total positive charge available in the system would be the number of holes p plus the number of positive ions added by doping of donor atoms, N_D, per unit volume. Similarly, if N_A acceptor atoms are added per unit volume, N_A holes and N_A negative ions per unit volume are released into the material. Thus, there are n plus N_A number of negative charges in the material per unit volume. To maintain the neutrality in the system, the number of positive charges should be the same as that of negative charges and thus,

 

To make an n-type material, one adds only donor atoms and not acceptor atoms into the semiconductor. In that case,

 

In an n-type material, it is well understood that the number of electrons is very large compared to the number of holes and so, p is very small compared to n and can be neglected. Thus,

 

That is, the electron concentration in an n-type material is approximately equal to the added impurity donor concentration.

On similar argument, in a p-type material,

 

That is, the hole concentration in a p-type material is approximately equal to the added impurity acceptor concentration.

Applying mass action law in an n-type material,

 

where n_n is the electron concentration and p_n is hole concentration in an n-type material. But since in an n-type material, N_D ≈ n_n

 

Or,                 p_n = n_i²/N_D                     …(2.28)

And similarly, in a p-type material,

 

Eqs. (2.28) and (2.29) give the minority carrier concentration in n-type and p-type materials, respectively.

It may be also observed that if in a material both donor and acceptor atoms are added, then the material would be intrinsic if the donor and acceptor atom concentration is same. The material can be recognised as n-type if in the material N_D >> N_A and as p-type if N_A >> N_d. If one wants to convert a p-type material to n-type, one must add donor atoms in large concentration such that N_D >> N_A is satisfied.

Intrinsic concentration:    As already discussed, the intrinsic concentration of the semiconductor is a function of temperature and increases with temperature. Thus, increasing the temperature increases the new electron hole pairs in the material and thus the conductivity of the material improves. The definite relation of the intrinsic concentration with temperature is

 

where E_G0 is the energy gap at 0 K in eV, k is the Boltzmann’s constant in eV/K and A₀ is a constant independent of temperature.

The constants E_G0, μ_n, μ_p and so on for Si and Ge materials are as listed in Table 2.1. From the table, it can be observed that only 1 atom in about 10⁹ atoms contributes free electrons and holes to the crystal because of broken bonds in germanium. For silicon, the ratio is further less, 1 in 10¹².

 

Table 2.1 Electrical Properties of Germanium and Silicon

Property	Ge	Si
Atomic number	32	14
Atomic weight	72.6	28.1
Density g/cm3	5.32	2.33
Dielectric constant (relative)	16	12
Atoms/cm3	4.4 × 1022	5.0 × 1022
EG0, eV at 0 K	0.785	1.21
EG, eV at 300 K	0.72	1.1
ni· at 300 K, cm−3	2.5 × 1013	1.5 × 1010
Pi at 300 K, Ωcm−3	45	230000
μn cm2/v-s	3,800	1300
μp cm2/v-s	1800	500
Dn, cm2/s = μnVt	99	34
Dp, cm2/s = μpVt	47	13
σi at 300°K (Ω-cm)−1	0.0224	4.35 × 10−6
Percentage increase in conductivity per degree rise in temperature	6	8
Dn/Dp (approximately)	2	3
Cut in Voltage Vr, V [Diodes]	0.2	0.6
Variation of I0 with temperature per °C	11	8
Permissible Temperature of Operation (Transistors)	100°C	200°C

The energy gap: The forbidden energy gap between the valence and conduction bands is not fixed for a given semiconductor and is a function of temperature. The dependence of energy gap on temperature is as follows:

 

for silicon and that for germanium is

 

The mobility: The mobility of electrons and holes in a semiconductor depends not only on temperature but also on the applied electric field as follows:

 

in the temperature range of 100 − 400 K. The value of m is as follows:

 

m	=	2.5 for electrons in Si
	=	2.7 for holes in Si
	=	1.66 for electrons in Ge
	=	2.33 for holes in Ge            …(2.34)

The mobility is independent of electric field ε when ε < 10³ V/cm in n-type silicon. When 10³ < ε < 10⁴ V/cm, μ_n varies approximately as ε^−1/2. For higher fields, μ_n is inversely proportional to ε.

2.9 HALL EFFECT AND ITS APPLICATIONS

The type of extrinsic semiconductor and the mobility of the material can be estimated with the help of Hall effect. The Hall effect can be explained as follows.

Consider a material, either a semiconductor or a metal, as shown in Fig. 2.9. Assume that the specimen is carrying a current I and is subjected to a transverse magnetic field B. Then, an electric field ε is induced in the direction perpendicular to both B and I. This phenomenon is known as Hall effect. This is very useful in semiconductors to determine the type of material, carrier concentration and mobility of the material.

As shown in Fig. 2.9, if current I is assumed to be flowing in positive x direction, and the applied magnetic field B is transverse to x axis, that is, z direction, then the induced electric field will have a force in the negative y direction by the Hall effect. This induced electric field in the specimen exerts force downwards on the charge carriers flowing in the material. If the material is a metal, the charge carriers are electrons flowing in the negative y-direction since the current is assumed to be flowing in the positive x direction. So, by the induced field these charge carriers will experience a force downwards in the negative y direction and one expect accumulation of electrons on the lower edge of the sample. Since the electrons assemble on the bottom of the specimen, equal amount of positive charge has to be deposited on top of the specimen. Thus, the electric field induced by the virtue of Hall effect will have positive field on the top whereas it will be negative at the bottom. This induces a field in the negative y direction.

When a semiconductor sample is subjected to Hall effect, the induced force due to the field will be on the charge carriers as in the case of a metal. If the semiconductor is n-type, the induced force will be exerted on both positive holes and negative electrons in the material. Since the electrons dominate the holes in an n-type semiconductor, the induced field would be same as in the case of a metal and will be in the negative y-direction. But if the specimen is a p-type semiconductor, since there are large number of positive holes, the induced field will force these holes towards the lower edge of the sample and so positive charge is accumulated on the bottom with an equal amount of negative charge on the top. This leads to the fact that the induced electric field is having a direction in the positive y-direction.

If an unknown semiconductor is subjected to the Hall effect, as explained in the earlier paragraphs, the type of material could be determined. If the induced field is in the positive direction of y axis, then the material is p type and if the induced field is in negative y-direction then the sample is n-type.

In an equilibrium state, the induced electric field ε due to Hall effect must exert a force on the carrier, which must balance the magnetic force. Thus,

 

where q is the magnitude of the charge on the carrier and v is the drift velocity with which the carrier is moving.

The electric field

 

where V_H is Hall voltage and d is the distance between the faces 1 and 2 of the specimen. The current density J can be given as

 

Then,

 

From the above equation, if V_H, B and I are measured, ρ can be determined. The Hall coefficient R_H can be defined as

 

Thus,

The conductivity σ can be given as

 

or,

 

From Eq. (2.44), if R_H is determined from Hall effect and conductivity measured for a given semiconductor, the mobility can be easily determined.

2.10 THERMISTOR AND SENSISTOR

The conductivity of the semiconductor depends on the number of free electrons and holes available for conduction in a material. The number of the charge carriers depends on temperature in an intrinsic material and so the conductivity of the semiconductor depends on temperature. As the temperature increases, the electron-hole pairs increase, thus increasing the conductivity of the semiconductor. The conductivity of the material can also be improved by doping the intrinsic material with either trivalent of pentavalent atoms to form p-type or n-type semiconductor. The variation of conductivity of a material with respect to a given parameter like temperature or light intensity is known as conductivity modulation.

From Eq. (2.30), it is seen that the conductivity increases approximately by 6% per degree increase in temperature for germanium and the rate is 8% for silicon. In fact, such a large variation of the conductivity with temperature is a drawback in some circuits. But, this phenomenon is very useful in some applications. A semiconductor that is designed for such an application is called a thermistor. The thermistor is a resistor made up of semiconductor instead of conventional carbon or other materials. An ordinary resistor exhibits resistance constant value irrespective of temperature. But a thermistor, being a temperature sensitive resistor, exhibits a resistance that decreases with increase in temperature. This thermistor finds applications in thermometry, microwave power measurement, thermal relay and other thermally controlled devices. Since silicon and germanium are too sensitive to impurities, they are not used as thermistor materials, but sintered mixtures of oxides such as NiO, Mn₂O₃ and Co₂O₃ are used.

When a metal is considered, its conductivity is also not a constant value with respect to temperature. It varies with temperature. The conductivity decreases with increase in temperature in a metal. Increase of temperature in a metal results in greater thermal agitation of the ions and this results in decrease of mean free path of the free electrons. Since mean free path depends on the mobility of the material, the mobility decreases. The conductivity of the metal is directly proportional to the mobility and thus the conductivity decreases. Thus, the resistance of the metal increases with increase in temperature. This rate of increase is of the order of 0.4% per degree rise in temperature. Thus, a metal has a positive coefficient of resistance while a semiconductor has a negative coefficient of resistance.

A very heavily doped semiconductor behaves in a manner similar to that of a metal. Thus, this semiconductor also exhibits positive coefficient of resistance like that of a metal due to decrease in carrier mobility with increase in temperature. Such a device is known as sensistor. A sensistor is a resistor made up of a heavily doped semiconductor having positive coefficient of resistance. This can have a temperature coefficient of resistance of the order of 0.7% per degree rise in temperature over the range of − 60 to + 150 °C.

The conductivity of a semiconductor can be improved by creating new electron-hole pairs in the system. Various possible schemes have been discussed in the above sections. Another scheme is to achieve conductivity modulation by light intensity variations. Such a material whose conductivity can be modulated by light can be referred to as a photoconductor. When the radiated light is exposed to the covalent bonds of the semiconductor, the photons strike the bonds to break them. Breaking one covalent bond releases two charge carriers, one electron and one hole. The number of covalent bonds broken would depend on the number of photons incident on the material. Thus, larger the intensity of the light, larger is the conductivity of the material. Such a material exhibiting photoconductivity modulation can be either referred to as a photoconductor or a photoresistor. Such a device finds large applications in the measurement of illumination (light meter), recording modulation intensity (sound track recording), on-off relay (digital control systems) and so on. The photoconducting device with the widest application is the cadmium sulfide cell.

2.11 TIME VARIATION OF EXCITED MINORITY CARRIERS

In equilibrium, at room temperature, in a semiconductor, the number of free electrons and holes remains constant. This does not mean that the given electron is always present and will contribute to conduction. The free electrons and holes are not stationary but wander throughout the material randomly due to thermal energy available at room temperature. In this process, recombination and generation of electrons take place continuously, but at any given time, the number of free electrons or holes is unaltered. To create charge carriers, one has to break a covalent bond. To break a covalent bond, energy from an external agency is expended and so, one loses energy in creating an electron-hole pair. On the other hand, when an electron recombines with a hole, one loses two charge carriers (one electron and one hole) and energy is gained by the external agency. In a system, generation and recombination take place continuously and the lifetime of a charge carrier is the time between generation and recombination. The term mean lifetime of a charge carrier is the average of the lifetimes of these carriers. τ_n and τ_p are the mean lifetimes of an electron and a hole, respectively.

Consider an n-type semiconductor bar. Due to thermal agitation, electron-hole pairs are generated—let the rate of generation be g. On the other hand, equal number of electron-hole pairs disappear due to recombination after thermal equilibrium is achieved. That is, the rate of generation is same as the rate of recombination at a given temperature at equilibrium. The recombination is due to electron falling into empty valence band, resulting in loss of a pair of charge carriers. Let p₀ and n₀ be the concentrations of holes and electrons at room temperature at equilibrium. At a time t = t′ (as shown in Fig. 2.10) the bar is subjected to illumination, which can break the covalent bonds. Thus, new electron-hole pairs are generated and in this condition the rate of generation is more than recombination and thus the electron and hole concentration increases until new equilibrium is achieved. The additional electrons and holes generated would be same in number. Let and be the new concentrations at equilibrium after excitation. Thus,

 

 

Although the increase in hole concentration is same as that in electron concentration, since n-type bar is considered, there are large numbers of electrons and adding a few more electrons leads to very small rate of increase in electrons. But, when holes, the minority carriers, are considered—since they are very small in number—adding a few more leads to a large rate of change in concentration. So, the increase in the electrons can be neglected and the minority carrier behaviour is studied.

After a steady state is reached, the excitation is turned off, say, at t = 0. When the excitation is switched off, the rate of generation decreases and the rate of recombination would be more than that of generation, thus leading to loss of charge carriers. To understand how the rate of recombination affects the charge concentration, the following analysis is done.

Assuming that the mean lifetime τ_p is independent of hole concentration,

 

p/τp	=	decrease in hole concentration per second due to recombination,	…(2.46)
g	=	increase in hole concentration per second due to thermal agitation.	…(2.47)

At equilibrium, since no charge can be created or destroyed, the above two rates are equal. Or, the rate of change of hole concentration would be zero,

 

But after the excitation is switched off at t = 0, the rate of change of hole concentration would be the difference between the generation rate and recombination rate. Thus,

 

When no radiation is falling on the sample, the generation rate would be p₀/τ_p and therefore,

 

If p′ denotes the excess or injected carrier density, defined as increase in minority carrier concentration above equilibrium level, then, p′ would be a function of time and

 

Since p₀ is constant, substituting the above equation in Eq. (2.50),

 

The minus sign in the above equation indicates that there is a decrease in the hole concentration. Solving the above differential equation for t ≥ 0,

Thus, the excess minority carrier concentration after the external excitation is switched off decays towards the equilibrium value exponentially with the time constant as the mean lifetime of the carriers.

It is important to note here that all the electrons encountering a hole will not recombine. In fact, recombination results in release of energy and loss of two charge carriers. In order to recombine, when an electron collides with a hole both should have the same momentum and they must travel in the opposite directions. This is a very stringent condition and very rare. So, recombination by such direct encounters is very small.

In practice, recombination is not by direct collision as explained above, but the mechanism involves recombination centres or traps. These traps contribute electronic states in the energy gap of the semiconductor. Such a location acts effectively as a third body that can satisfy the conservation of momentum requirement. These new states are due to imperfections in the crystal. Metallic impurities in the semiconductor are capable of introducing energy states in the forbidden gap. Not only the volume imperfections, but also surface imperfection leads to the traps in the crystal. So, purposefully, gold is introduced in a controlled fashion into silicon to achieve the desired traps in the crystal.

2.12 CONCEPT OF DIFFUSION CURRENT

In a semiconductor, in addition to the conduction current due to electrons and holes, there are two more components of current due to diffusion of holes and electrons. This diffusion is not observed in metals. The concept of diffusion is as explained below.

Consider a p-type semiconductor bar where the doping is not uniform throughout the material, but is very large towards one end of the bar and the concentration decreases as one moves towards the opposite end. That is, the doping of acceptor atoms is controlled and is non-uniform. Assume that the bar is oriented in x direction with maximum doping end at x = 0. As seen from the figure, the concentration of the impurity in the semiconductor is a function of x and decreases with increase in x. So, there is a concentration gradient dp/dx and thus the density of carriers is also not constant. Now, consider an imaginary cross-section of the bar at some x where x > 0 as shown in the figure. To the left of this imaginary cross-section the concentration of the charge carriers, holes, is large compared to that to the right. This is because the concentration is maximum at x = 0 and decreases with increase in x. These holes are not stationary as seen previously and are in random motion due to thermal agitation. Because of this random motion, the holes cross the imaginary cross-section in both directions. In a given time interval, there is a probability that the number of holes crossing from left to right would be more than that of holes crossing from right to left. This is a statistical phenomenon since the concentration gradient occurs in the semiconductor. Since the number of charge carriers crossing the cross-section is not the same in both the directions, there is a net transport of charge carriers in the direction of positive x direction. This leads to a current flow in the positive x direction as shown in Fig. 2.11. It should be noted that this current is not due to the repulsions of the charge carriers, but is purely a statistical phenomenon. This current is referred to as diffusion current.

As one expects, this diffusion current is proportional to the concentration gradient in the semiconductor. Thus, diffusion current density is proportional to the concentration gradient. The proportionality constant is split into components, as seen below, for convenience,

 

where q is the charge associated with each charge carrier and D_p is the diffusion constant for holes measured in square metres per second. Since p decrease with increase in x, the concentration gradient dp/dx is negative. Since the hole current is in positive direction, the negative sign in the equation is inserted.

Similarly, if pentavalent atoms non-uniformly dope a semiconductor, a concentration gradient of n-type material can be achieved. In such a semiconductor, diffusion current due to electrons can be achieved. The constituent diffusion current density of electrons can be given as

 

where D_n is the diffusion constant of electrons. Minus sign is not required in this equation since the conventional current and the concentration gradient coincide.

Thus, in a semiconductor various components of current are possible. The electron current density due to conduction and diffusion can be given as

 

The hole current density due to conduction and diffusion is

 

Einstein relationship: Both the mobility and the diffusion are statistical phenomena. So, D and μ are not independent. The relationship between them is given by Einstein equation:

 

where V_T is the volt equivalent of temperature and is defined as

 

where is the Boltzmann’s constant in joules per kelvin. At room temperature, T = 300 K and V_T = 26 mV.

2.13 CONTINUITY EQUATION IN SEMICONDUCTORS

As discussed in Sec. 2.11, if the semiconductor material is disturbed from equilibrium, the concentrations of holes and electrons vary with time. However, the carrier concentration in the body of a semiconductor is a function of distance, too. The continuity equation is a differential equation that governs the relationship of the carrier concentration both with time and distance. This equation is derived on the basis of conservation of charges property.

Consider an infinitesimal element of p-type semiconductor having a volume L × dx. Let the average hole concentration in the volume be p. The problem is assumed to be one dimensional and the current I_p in the element is flowing in x direction. If I_p is the current entering into the element at x and I_p + dI_p at x + dx at the same time, dI_p coulombs of charge per second are leaving in addition to the input. Hence, the charge carriers in the element decrease within the volume. Since magnitude of the charges is q, dI_p/q is the decrease in the number of holes per second within the element. If the volume is Adx, then the decrease in hole concentration per second can be given as

If g = p₀/τ_p is the increase of holes per unit volume per second due to thermal agitation, and p/τ_p is the decrease of holes per unit volume per second due to recombination, as we have seen in Sec. 2.11, the rate of hole concentration change due to thermal agitation and recombination can be given as (p₀ − p)/τ_p. Thus, taking into account the conservation of charges from the above two concepts, the rate of change of hole concentration can be stated as

The above equation is called the continuity equation. This can also be referred to as law of conservation of charge. This law can also be applied to electrons and n is substituted for p in the above equation.

2.14 INJECTED MINORITY CARRIERS

The minority carrier behaviour is seen to be exponentially decaying with time when an external excitation is removed and the time constant is the lifetime of the carrier. Now, the behaviour of the minority carrier with distance is to be considered when introduced at one end of a semiconductor. This also decays exponentially with distance, the time constant being diffusion length this time.

Consider a uniformly doped n-type bar of semiconductor as shown in Fig. 2.13. Let the concentration of the donor atoms be N_D and thus n ≈ N_D is uniform throughout the bar under consideration. Assume that at x = 0, radiation falls as shown in the figure. The photons of the radiation collide with the covalent bonds of the semiconductor bar on this face and break them to release new electron-hole pairs. Since the bar is n-type, the holes are minority carriers and their behaviour is to be studied now.

Consider a low-level injection, which means that the excess minority carriers created in the bar are very small compared to the electron concentration, that is, p′ << n. The excess minority carriers are not stationary at this face but are diffused towards the other end of the bar and as they diffuse, recombination takes place with the free electrons and the minority carrier concentration decreases. Since the drift current is proportional to the charge carrier concentration, the drift current due to holes can be neglected as the amount of holes in the bar is very small compared to electron concentration, that is, p₀ + p′ << n. So, the hole current is entirely due to diffusion. The other components of electron current are present in the bar. Then,

From the continuity equation (2.61), at steady state, rate of change of hole concentration is zero, ∂p/∂t = 0, which leads to

Diffusion length for holes is defined as

 

Thus, the above differential equation can be rewritten as

The general solution to such a second order differential equation can be written as

 

where K₁ and K₂ are constants. When a very long bar is considered, as x increases, the second term increases if K₂ were to be a given value other than zero. That means as x tends to infinity, the hole concentration increases with distance, which is an impossible condition. Thus, K₂ = 0. If at x =0, the excess concentration of holes is assumed to be p′(0), then K₁ = p′(0). Then,

 

Thus, the minority carrier concentration decreases exponentially with L_p as constant. The diffusion length can be redefined as the length at which the concentration falls to 1/e times that of the initial value at x = 0. It can also be seen as the average distance that an injected hole travels before recombining with an electron.

Diffusion current: From Eq. (2.54) for current density of diffusion current due to holes and remembering J_p = I_p/A,

Thus, the diffusion current falls exponentially, as is the case with hole concentration. Now, the majority carrier electron diffusion current is

 

The excess minority and majority carriers would be the same since breaking a bond creates a electron-hole pair. Thus, p′ = n′ or,

 

Since p₀ and n₀ are independent of x,

 

Thus, the electron-hole current is

 

Drift current: Since the semiconductor bar is open circuited, the net current flowing in the bar, that is, the sum of drift current due to electrons and diffusion current due to holes and electrons, would be zero. Thus, if I_nd is the drift current due to electrons,

or

Thus, the electron drift current also decreases exponentially with distance.

As seen, if a drift current exists in the bar, this current would be due to a potential drop, that is, there should be some electric field existing in the bar. This can thus be calculated as

When an electric field is present in the bar, it should also cause drift current in the bar due to holes. Hence, the drift current due to holes, I_pd, can be given as

Since p << n in a n-type bar, I_pd << I_p. Thus, the hole drift current is very small compared to the hole diffusion current, justifying the assumption made in the beginning of this discussion under low-level injection.

2.15 GRADED SEMICONDUCTOR JUNCTION

Consider a semiconductor in which the doping of the impurity atoms is non-uniform. Such a semiconductor can be referred to as graded semiconductor. Assume that the considered bar is doped with trivalent atoms to form the p-type semiconductor. Assume that there is no external excitation given to the bar and the bar is at steady state. Even though the net current in the bar is zero, there is movement of the charge carriers, holes, randomly in the bar due to thermal agitation. Since p is not uniform in the bar, one can expect diffusion current due to holes to exist in the bar. In order to make the total current due to holes zero at steady state, there should be equal and opposite drift current due to holes in the bar (same is the case with electrons—total electron current should be zero). If a drift current is to be flowing in a material, there should be potential drop across the bar establishing an electric field. So, by virtue of non-uniform doping of impurities in the semiconductor, a potential drop is induced into the bar. This potential is investigated now.

Setting the total current density due to holes, J_p equal to zero, from Eq. (2.57),

 

Since V_T = D_p/μ_p,

The relation between the electric field and potential is ε = − dV/dx. Thus,

 

To find out the potential drop between two points 1 and 2, the above equation is to be integrated within the limits x₁ and x₂. If p₁ is the concentration at x₁ and p₂ is the concentration of holes at x₂,

 

V₂₁ is the potential drop between 2 and 1 as shown in Fig. 2.14. Thus, the potential drop between the two points depends only on the concentrations of holes at the two points, but does not depend on the position of the two points under consideration. The potential is independent of the distance between the two points. Equation (2.80) can be recast as follows to estimate the concentration at a given point to establish the defined potential:

This is the Boltzmann’s relationship of kinetic gas theory.

Carrying out the discussion for electrons as is done in the above for holes, where the electron current density is zero, leads to derivation of the following equation:

 

From Eqs. (2.81) and (2.82),

 

This equation states that the product np is a constant independent of x, and hence of the amount of doping, under thermal equilibrium. For an intrinsic semiconductor, n = p = n_i and hence,

 

which is the mass action law.

Step-graded junction: Consider a junction formed between two semiconductors as shown in Fig. 2.14. The junction is formed between p-type and n-type semiconductors and the doping in the material is assumed to be uniform. Such a junction is referred to as step-graded junction since the concentration of impurities is suddenly changed from p to n type in a step. Let the concentration of impurities towards the left of the junction be N_A and to the right the concentration of n-type material be N_D. The above discussion shows that the potential developed due to concentration gradient depends on the concentration but is independent of the distance. The potential drop between any point on the left of the junction and any point to the right of the junction can be given as

 

where V₀ is referred to as contact difference potential, p_p0 is the hole concentration in a p-type semiconductor at steady state and p_n0 is the hole concentration in n-type semiconductor at steady state. From Eq. (2.29),

 

Thus,

The above equation gives the value of the contact difference potential developed when a step-graded junction is formed between two semiconductors. This contact potential is very important in the understanding of the behavior of the p-n semiconductor junction, which is the topic for the next chapter.

2.1    Prove that the concentration n of free electrons per cubic meter of a metal is given by

n	=	, where
d	=	density, kg/m3
v	=	valence free electrons per atom
A	=	atomic weight
m	=	weight of atom of unit atomic weight, kg
A0	=	Avogadros’ number, molecules/mole.

Solution    The weight of an atom is considered to be equal to the atomic mass times the mass of an atom of unit atomic weight.

If n = no. of atoms per molecule, then AM = Molecular weight of one mole in grams, and

2.2    If a donor type impurity is added to the extent of 1 atom per 10⁸ Ge atoms, calculate the value of resistivity.

Solution    If there is 1 donor atom per 10⁸ Ge atoms, then,

Hence, n = N_D = 4.4 × 10¹⁴

Therefore, p << n, neglecting p.

 

σ	=	nqμ21 = 4.4 × 1014 × 1.6 × 10−19 × 3800
	=	0.1268 (Ω-cm)−1

∴ Resistivity

2.3

1. Find the resistivity of intrinsic silicon at 300 K
2. If a donor type impurity is added to the extent of 2 atoms per 10⁸ Si atoms, find the resistivity

Solution

1. Conductivity s_i = n_iq (μ_n + μ_p) for an intrinsic semiconductor.

    

   ∴   σi	=	1.6 × 10−19 × 1.5 × 1010 (1300 + 500)
   	=	4.32 × 10−6 (Ω-cm)−1

   Hence, resistivity

2. 

   Hence, n = N_D = 2.5 × 10¹⁴ atoms/cm³

   

   Hence, p << n, so p can be neglected.

    

   ∴  σ	=	nqμn = 1.6 × 10−19 × 2.5 × 1014 × 1300
   	=	0.052 (Ω-cm)−1

   Hence, resistivity

2.4

1. Find the concentration of holes and electrons in p-type Ge at 300 K if the conductivity is 60 (Ω-cm)⁻¹.
2. Repeat part (a) for n-type silicon if the conductivity is 0.13 (Ω-cm)⁻¹.

Solution

1. We know that s = nqm_n + pqm_p

   for p-type semiconductor p >> n

   ∴ σ_p ≈ pqμ_p

   

   Hence, from mass action law.

   
2. For an n-type semiconductor n >> p

       ∴ σ_n = nqμ_n + pqμ_p ≈ nqμ_n

   

2.5    Show that for an intrinsic Ge the conductivity increases by 5.5% per °C.

Solution    For an intrinsic Ge σ = n_iq (μ_n + μ_p).

Assume that μ is independent of temperature.

and,     

⇒     

Taking natural logarithm

Differentiating this equation

For Ge,

∴ Change in conductivity is 5.5% per °C.

2.6

1. Show that the conductivity of a semiconductor is minimum when it is lightly doped with p-type impurity such that
2. Show that the minimum conductivity is
3. Use the above result to find the minimum conductivity of silicon.
4. Find the corresponding doping level.

Solution

1. Conductivity σ = nq μ_n + pq μ_p
   

   To find minima, find and equate it to zero.

   
2. Substituting (2) in (1)
   
3. For silicon,
   
4. We have charge balance equation

    

   N_D + p = N_A + n

   Since this is a p-type semiconductor,

   

   On substituting the values

    

   N_A = 1.48 × 10¹⁰ atoms/cm³

2.7     A sample of Ge has n-type impurity concentration of 3 × 10¹⁴ donors/cm³ and p-type impurity concentration of 4 × 10¹⁴ acceptors/cm³. Find n and p at room temperature.

Solution

 

	NA	=	4 × 1014, ND = 3 × 1014
∴	ND + p	=	NA + n	…(1)
⇒	p − n	=	NA − ND = 4 × 1014 − 3 × 1014
		=	1 × 1014	…(2)

From mass action law np = n_i²

 

From (2) and (3), solving for p and n,

        p = 0.933 × 10¹⁴ holes/cm³.

        n = 6.698 × 10¹² Electrons/cm³.

2.8    Find the factor by which conductivity of pure Ge will change when the temperature is rised from 300 K to 600 K. μ_p(300) = 1800, μ_n(600) = 300, μ_n(300) = 3900, μ_n(600) = 1300.

Solution    We know that

 

σi	=	niq(mn + mp)
	=	q(mn + mp)(A0T3e−EG0/KT)½

2.9    Show that the p-type semiconductor has same conductivity as that of intrinsic semiconductor if

Solution    For a p-type semiconductor

 

σp	=	nqμn + pqμp
	=	q(nμn + pμp)

Using in the above equation,

2.10

1. How much donor impurity should be added to pure Ge, so that its resistivity drop to 10% of its original value.
2. Find p and n of p-type Ge if its resistivity is 0.01 Ω-cm.
3. Find n and p of a n-type Si if its resistivity is 10 Ω-cm.

Solution

1. 10% of 44.6 is 4.46, Since for Ge, p_i = 44.6.

   Hence, conductivity

   ∴ 0.224 = N_D qμ_n

   
2.      σ_p = N_Aq μ_n
   

   Since N_A >> n_i, so p = N_A = 3.47 × 10¹⁷

   
3.         
   
   

2.11    Find the conductivity of germanium with:

1. donor impurity of 1 part in 10⁶.
2. acceptor impurity of 1 part in 10⁷.
3. when both the above impurities are present simultaneously, given that n_i for Ge at 300 K is 2.5 × 10¹³/cm³, μ_n and μ_p for Ge are 3800 and 1800 cm²/V-s and the number of Ge atoms/cm³ = 4.4 × 10²².

Solution

1.     

   ∴          n ≈ N_D = 4.4 × 10¹⁶/cm³

   

   ∴ n >> p Hence, we can neglect p in calculating the conductivity.

           σ_n = nqμ_n = 4.4 × 10¹⁶ × 1.6 × 10⁻¹⁹ × 3800

               = 26.752 (Ω-cm)⁻¹

2.     
   p ≈ N_A ≈ 4.4 × 10¹⁵

   Hence,

   

   ∴ p >> n, hence we can neglect n while calculating the conductivity

    

   ∴	σp	=	pq μp = 4.4 × 1015 × 1.6 × 10−19 × 1800
   		=	1.267 (Ω-cm)−1

    

3. With both donor and acceptor impurities present,

   N_D = 4.4 × 10¹⁶/cm³, N_A = 4.4 × 10¹⁵/cm³

    

   Since    ND + p	=	NA + n
   n − p	=	ND − NA = 4.4 × 1016 − 0.44 × 1016
   	=	3.96 × 1016         …(1)

   From mass action law np = n_i² = (2.5 × 10¹³)² = 6.25 × 10²⁶                 …(2)

   Solving (1) and (2),

   we will get n and p. But in this particular problem, N_D and N_A are comparable, N_A holes combine with equal number of free electrons leaving the material n-type with net concentration N_D − N_A = 3.96 × 10¹⁶/cm³ = N′_D

   Hence, n = 3.96 × 10¹⁶/cm³

    

   σ	=	N′D q μn = 3.96 × 1016 × 1.6 × 10−19 × 3800
   	=	24.08 (Ω-cm)−1

    

2.12    An n-type Si bar is 2 cm long and has a cross section of 2 mm × 2 mm. When a 1 V battery is connected across it, a current of 8 mA flows. Find (a) doping level and (b) drift velocity.

Solution

1.  

   

   Since this is an n-type Si bar, we can use approximate formula. So, n = N_D.

   
2. Drift velocity

    

   v	=	με
   	=	1300 ×
   	=	650 cm/sec.

2.13    A sample of n-type semiconductor has conductivity of 100 mho/cm at 27°C. Find the conductivity at 127°C. n_i at 27°C is 2 × 10¹⁰ atoms/cm³ and the mobility of free electrons varies with absolute temperature. According to the equation μ_n = 2.1 × 10⁹ × T⁻²⁵ and E_G0 = 1.1 eV.

Solution

 μ_n (27) = 2.1 × 10⁹ × (300)−²⁵ = 1347 cm²/V-s

μ_n(127) = 2.1 × 10⁹ × (400)^−2.5 = 656 cm²/V-s

∴ ni (127)	=	317.64 ni (27)
	=	317.64 × 2 × 1010
	=	6.34 × 1012
at 27°C, σ	=	NDq μn
100	=	ND × 1.6 × 10−l19 × 1347
ND	=	4.64 × 1017 atoms/cm3

At 127°C, n_i increases to 6.34 × 10¹² but still N_D >> n_i;, so n = N_D

 

σ	=	NDq μn = 4.64 × 1017 × 1.6 × 10−19 × 656
	=	48.70 (Ω-cm)−1

2.14

1. Find the magnitude of the Hall voltage V_H in an n-type Ge bar, having majority carrier concentration N_D = 10¹⁷/cm³, assume B_z = 0.3 Wb/m², d = 3 mm, and ε_x = 6 V/cm.
2. What happens to V_H if an identical p-type germanium bar having N_A = 10¹⁷/cm³ is used in part (a).

Solution

1. Hall voltage

    

   VH	=	Bvd = Βμεxd
   	=	0.3 × 3800 × 6 × 102 × 3 × 10-3 × 10−4
   	=	205.2 mV.

2. If accepted atoms are added, then the polarity of V_H going to be changed, but the value remains the same to 205.2 mV.

2.15    The Hall-effect is used to determine the mobility of holes in a p-type silicon bar. Assume the bar resistivity is 200000 Ω-cm, the magnitude of B_z is 0.4 Wb/m², and d = w = 3 mm. The measured values of the current and Hall voltage are 100 μA and 100 mV, respectively. Find μ_p·

Solution    We have

2.16    A sample of Ge is doped to the extent of 10¹⁴ donor atoms/cm³ and 7 × 10¹³ acceptor atoms/cm³. At the temperature of the sample the resistivity of pure Ge is 60 Ω-cm. If the applied electric field is 2 V/cm, find the total conduction current density.

Solution    We know that for a pure Ge

Given that	ND	=	1014/cm3 and NA = 7 × 1013 atoms/cm3
∴	ND − NA	=	10 × 1013 − 7 × 1013 = 3 × 1013 = p − n
	np	=	ni2 = (1.86 × 1013)2

∴	σ	=	nqμn + pqμp
		=	1.6 × 10−19 (3.88 × 1013 × 3800 + 0.89 × 1013 × 1800)
		=	0.02615 (Ω-cm)−1
∴ Current density	J	=	σε
		=	0.02615 × 2
		=	52.307 mA/cm2

2.17    Given a 20 Ω-cm n-type Ge bar with material lifetime of 100 μ sec, cross-section of 1 mm², and length of 1 cm. One side of the bar is illuminated with 10¹⁵ photons/sec. Assume that each incident photon generated one electron-hole pair and these are distributed uniformly throughout the bar. Find the bar resistance under continuous light excitation at room temperature.

Solution    When there is no radiation, ρ = 20 Ω-cm

Bar resistance

When the bar is illuminated, 10¹⁵ electron-hole pairs are created every second which last for 100 μ sec on average, that is, in 100 μ sec, 10¹¹ electron-hole pairs are created and destroyed. Thus, the excess number of holes and electrons will stabilise to 10¹¹ in the sample. But the sample volume is

1 cm × 1 mm × 1 mm = 0.01 cm³

p ′ = excess hole concentration

Similarly, n ′ = excess electron concentration = 10¹³ electrons/cm³

 

∴ New electron concentration	=	n = 0.8 × 1013 + 1013
	=	1.8 × 1013
New hole concentration	=	p = 7.84 × 1013 + 1013
	=	8.84 × 1013

∴     σ	=	(nμn + pμp)q
	=	(8.84 × 1013 × 3800 + 1.8 × 1013 × 1800) 1.6 × 10−19
	=	0.05894 (Ω-cm)−1

2.18

1. Radiation is incident on one end of n type molecule which causes minority carrier to be injected as shown in fig 2.13. Show that the equation of conservation of charge is
   
2. Verify that the concentration is given by the Equation

    

   p = + (p_o − )e^−t/τ

Solution

1. No. of holes per second generated thermally (increase).

   = No. of holes per second recombining (decrease).

   = No. of holes per second injected due to light (increase).

   Hence,            

   
2. The solution of the differential equation is

    

   	p	=	+ Ae−t/τ
   	at t	=	0, p = + A = po
   	⇒ A	=	po −
   ∴	P	=	+ (po − )e−t/τ

2.19    The hole concentration is a semiconductor specimen is shown.

1. Find an expression for and sketch the hole current density J_p (x) for the case in which there is no externally applied electric field.
2. Find an expression for and sketch the built-in electronic field that must exist if there is no net hole. Current associated with the distribution shown.
3. Find the value of the potential between the points x = 0 and x = w if

Solution

1.  

   
2. for x > w,        
   

   ∴ Integrating within the limits from x = 0 to x = w.

   
3. Since     

   ∴	V	=	V(w) − V(0)
   		=	26mV × ln 103 = 180 mV

   The sketches for J_p(x) and ε(x) will be as shown.

2.20

1. For an open-circuited graded semiconductor verify the Boltzmann equation for electrons.
2. For the step graded semiconductor derive the expression for contact potential V₀, starting with J_n = 0.

Solution

1.  

   
   

   We know that         D_n = μ_nV_T

   Using this in the above equation,

   

   Integrating within the limits from x₁ to x₂,

   
   
2. From the above result,

   Let         n₁ = n_po

   

   = thermal equilibrium concentration in p-type region of electrons.

   n₂= n_no and, = thermal equilibrium electron concentration in n-type region.

   = N_D.

   

2.21    Consider a step graded Ge semiconductor with N_D = 2000 N_A and with N_A corresponding to 1 acceptor atom per 10⁸ Ge atoms. Find the value of contact potential V₀ at room temperature.

Solution

 

2.1 If we were living at an ambient temperature of 1000 K in a world, would silicon and germanium still be conductors.

 

2.2 In a semiconductor at room temperature the intrinsic carrier concentration and resistivity are 1.5 × 10¹⁶/m² and 2 × 10³ σ-m, respectively. It is converted into an extrinsic semiconductor with a doping concentration of 10²⁰/m. Calculate

1. minority carrier concentration
2. resistivity

2.3 A specimen of metal has 7.87 × 10²⁸ free electrons/m³. The mobility of electrons is 34.8 cm²/v-s. Compute

1. conductivity of the metal
2. If an electric field of 30 V/cm is applied across the specimen, calculate the drift velocity of electron and the current density.

2.4 The resistivity of p-type Si specimen is 0.12 Ω-m. Find electron and hole concentration.

 

2.5 1 kΩ resistor is to be fabricated as a narrow strip of p-type Si 4mm thick. If the strip is 20 μm wide and 400 μm long, what concentration of acceptor atoms is required?

 

2.6 A p-type material has an acceptor ion concentration of 1 × 10¹⁶ per cm³. Its intrinsic carrier concentration is 1.48 × 10¹⁰/cm³. The hole and electron mobilities are 0.05 m²/V-s and 0.13 m²/V-s. Calculate the conductivity of the material.

 

2.7 The Hall effect is used to determine the mobility of holes in a p-type Si bar. Assume the bar resistivity is 300,000 Ω-cm. The magnetic field B_z = 0.1 Wb/m² and d = ω = 6 mm. The measured value of the current and voltage are 10 μA and 60 mV. Find μ_p.

 

2.8

1. Find the resistivity of Ge and Si at 600 K.
2. If donor impurity is added to an extent of impurity atom per 5 × 10⁷ Ge atoms, at room temp, find the resistivity of Ge.

2.9 A Si bar 0.1 cm long and 100 μm² in cross-sectional area is doped with 10¹⁷/cm³ antimony. Find the current at 300 K with applied voltage of 10 V.

 

2.10 Find the concentration of holes and electrons in p-type Ge at 300 K if the conductivity is 300 (Ω-cm)⁻¹.

1. What is the distinction between an intrinsic and an extrinsic semiconductor? Give two examples of elements used for p-type doping and n-type doping.
2. Explain the phenomena of drift and diffusion. Describe the recombination process in the semiconductor. How are these parameters related in the continuity equation?
3. How do you explain the fact that the semiconductors have negative temperature coefficient of resistance.
4. Sketch the energy band diagrams for intrinsic, p-type and n-type semiconductors.
5. What properties of semiconductor are determined from Hall effect?
6. Explain why a constant difference of potential must develop across a p-n junction.
7. What is conductivity modulation? How are thermistors and sensitors different from resistors? Explain.
8. What is a photo resistor? Explain how conductivity modulation achieved with such a resistor.
9. Bring out the effect of temperature on semiconductors and explain the concept of movement of a hole.
10. Describe the Hall effect phenomenon in semiconductors.
11. What is potential barrier? Explain its significance in a p-n junction.
12. Explain the effect of temperature on an intrinsic semiconductor.
13. Give the equation for conductivity in an extrinsic semiconductor. Discuss the effect of temperature on the energy gap and mobility of a semiconductor.
14. Derive an equation for minority carrier concentration behaviour after the external excitation is removed.
15. Discuss in detail about the low-level injection of minority carriers into a semiconductor.
16. Derive the equation for the contact potential of a step-graded semiconductor junction.
17. Define mobility and conductivity and give expressions for both in a semiconductor.
18. Define intrinsic concentrations of holes. What is the relationship between this density and the intrinsic concentration for electrons? What do these equal at 0 K?
19. State mass action law for extrinsic semiconductor. Give the expressions for minority carrier concentrations in extrinsic semiconductors of p and n type?
20. What is the relationship between mobility and diffusion constant? Give its equation and discuss.
21. Define mean lifetime and mean drift velocity and bring out their significance in a semiconductor.
22. What is Hall coefficient? Give its equation. How is it useful in determination of mobility of a semiconductor?
23. What are recombination centres? Why are they required in a semiconductor? How are they introduced into a semiconductor?
24. Explain qualitatively why a contact potential is developed in a graded semiconductor. Derive its equation for step-graded junction.
25. What is a photoconductor? Explain the process of photogeneration in detail.