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This probability interpretation and the a t t e n d a n t uncertainty tie in immediately with Heisenberg's Uncertainty Principle. T h e "plausibility" of Eqs.

Two such sets of quantities are m o m e n t u m and position. In fact. If the limit of Ax is zero.

It is found that when. In accordance with the Uncertainty Principle. T h e r e a s o n we have to look to the wave theory for an explanation of t h e p h e n o m e n a occurring at atomic distances is partly a result of the H e i s e n b e r g Uncertainty Principle.

Solution a At constant mass m. Wave Mechanics is. We summarize the preceding discussion as: W is t h e potential energy of the particle. A s was shown in the example. This is t h e equation of a wave and w h e n applied to electron waves. The wave function is allowed to b e a complex quantity and will. The electron is not to be r e g a r d e d as a wave but as a particle. Because t h e position. H e incorporated the quantization theory proposed by Planck and t h e wave-like n a t u r e of m a t t e r as p r o p o s e d by de Broglie.

This equation is as basic to wave mechanics as Newton's laws are to classical mechanics and as Maxwell's equations are to electromagnetic theory. This quantity is a hypothetical one. Calculations in q u a n t u m mechanics use the probability function to d e t e r m i n e the position.

It can b e said to be analogous to a classical particle that can be located in a given point in space. U n d e r such circumstances. These are: Postulate 1. T h e wave packet consists of the summation of constant energy wave functions. Postulate 2. A sketch of a wave packet is shown in Fig. This probability can be d e t e r m i n e d by a solution of Schrodinger's equation.

The wave packet is a q u a n t u m mechanical m e a s u r e that can b e assumed to b e localized at a given point in space.

A s we shall see later. A n y a t t e m p t to determine the first position will disturb the electron and cause it to trace out a different path. Over a small region where the electron is k n o w n to exist. W e then apply t h e solution to the study of t h e energy levels in t h e hydrogen atom.

Application t o the Potential W e l l We shall assume. This is equivalent to saying that it is certain that t h e particle is somewhere in space. T h e infinitely d e e p one-dimensional well is. In t h e next section. The solution t o E q. The total energy of t h e electron is kinetic.

Energy is therefore quantized i. If a particle is located at a certain point. In other words. W h e n this wavefunction is substituted in Schrodinger's equation. J -rtl For a particle in a three-dimensional box. Having established the significance and plausibility of Schrodinger's equation.

We will t h e n extend t h e results to t h e manyelectron sample. This example can b e modified to include a third region. This concept is applied in t h e study of the E s a k i diode. What was de Broglie's contribution to Wave Mechanics? Interpret the solution of Schrodinger's equation to the deep potential well and how it relates to the science of Wave Mechanics. Herein lies the analogy between a beam of electrons and a light beam.

Clearly identify the magnitudes of the kinetic energy at two separate radii. It describes the large number of energies that an electron can have in a region where the energy is confined.

Avogadro's number is 6. Express the energy in eV and J. For the electron in Problem 1. An electron is located in a one-dimensional potential energy well having width of 3 A. Determine a the kinetic energy of the electron in the ground state.

Determine the uncertainty in: For a particle that has a mass of 2 grams and energy 1. An electron has a de Broglie wavelength of A. In accordance with physics statistics. Calculate the energy in eV and Joules. We will use Schrodinger's equation to formulate the conditions existing in the hydrogen atom.

A t each of these two t o p bands. We then decided that Wave Mechanics' concepts are n e e d e d to explain t h e behavior of electrons in solids.

We then project the results. We establish the existence of discrete energy levels from which. In this chapter. This is the forbidden band. We will therefore consider a quantum-mechanical solution since it is not constrained by these limitations. The potential distribution is. Section 2. The electron. T h e potential energy negative is highest at infinite r and lowest at t h e nucleus. By substituting this solution in Schrodinger's radial equation. By this method.

In the solution of the three equations. Before proceeding any further. Chapter 2 Energy Bands and Current Carriers in Semiconductors profile of the hydrogen atom potential distribution. T h e solutions to this equation have spherical symmetry. We will p r o c e e d with an outline of the general steps in the solution of Schrodinger's e q u a t i o n for the hydrogen a t o m and then present the results of this solution.

Let us digress briefly to explain how a solution is obtained so as to identify the sources of these q u a n t u m n u m b e r s and their relation to the physical properties of the electron and to t h e electronic orbits. We will now consider a simplified Schrodinger equation and examine the results.

We begin by substituting the expression for the potential energy in Schrodinger's equation. T h e position of the p r o t o n is fixed at the origin. The simplest and most basic solution.

For this work. It was then postulated that t h e electron. H e concluded this from a study of atomic spectra at about the same time that the extra q u a n t u m n u m b e r.

A quantum-mechanical interpretation of the orbital radius of the hydrogen a t o m is that it is n o longer a fixed n u m b e r and t h e electron having a fixed orbit but that there is a high probability of locating the electron at a distance identified by a quantum number.

The word magnetic results from the fact that an electron in an orbit represents an accelerating charge and. The azimuthal number. T h e existence of these three q u a n t u m n u m b e r s was predicted earlier by Sommerfeld. A schematic representation of an isolated silicon a t o m is shown in Fig. Silicon has is 2s 2p 3s 3p for an atomic n u m b e r of In Table 2. Since the levels were originally assigned by spectral studies.

TABLE 2. To comply with t h e Pauli Exclusion Principle. The r e a d e r will question our reference to m a n y electrons. In fact.. The letter s refers to sharp. The forces referred to above are due to t h e electrons of the same atom. The smaller the value of I. Figure 2. They find themselves in a field of charge. All that must be d o n e is to assign to the electrons t h e energy levels d e t e r m i n e d by the one-electron model. In addition to this. E l e c t r o n Solid The transition from the one-electron p r o b l e m to the many-electron a t o m does not require new solutions.

The energy levels are actually depressed. Within a given shell. A m o r e complete m o d e l will b e included later. In Fig. The energy levels of t h e two atoms are slightly modified so as to a c c o m m o d a t e t h e Exclusion Principle. The electrons in the lower energy levels are initially shielded. We observe that for copper. The b a n d gap energy refers to the width of the forbidden b a n d separating the conduction and valence bands. This region is k n o w n as the forbidden band or band gap.

This effect is illustrated in Fig. The top b a n d is k n o w n as the conduction band. It is a p p a r e n t from the electron-volt width of t h e gap that it requires less energy.

The specific characteristics of t h e bands. N o such b a n d exists in copper. It is worthwhile noting that. This forbidden gap. Let us consider silicon in m o r e detail. A t higher temperatures. A great a m o u n t of energy is required to excite the electrons from the filled levels to t h e higher empty levels. Only two of t h e six states are filled with electrons.

T h e b a n d structure of silicon indicates the existence of a forbidden gap. R is the normal atomic spacing. Electrons at absolute zero t e m p e r a t u r e occupy the lowest levels of energy in accordance with the Pauli Exclusion Principle so that each is theoretically stacked. The 3p level has six states. A t very low t e m p e r a t u r e s and since electrons t e n d to occupy the lower levels first.

T h e energies of the states are very close to each other. A t still smaller spacing. A t large atomic spacing. A s t h e spacing is reduced.

Following this. F r o m studies of energy level splitting. Varying t h e spacing of t h e atoms. Of course. The process of energy b a n d formation in silicon is interesting and we will study the splitting in m o r e detail by considering the representation shown in Fig.

Si and Ge. Some experimental proof of the m o d e l is obtainable from studies of x-ray spectra. Q If each electron in a solid is permitted to occupy only one energy level. Society GaAs and GaP.

In order to obtain them. Q What particular aspects completely specify the electrical conducting properties of a solid?

Q What phenomena cause energy levels to split? Q When atoms are brought together to form a solid. Properties such as conductivity. The top two bands of semiconductors in which electrons can exist are the conduction band and the valence band.

These levels are so close to each other that bands are effectively formed. E Determine the electron configuration. Each level. This resulted in a fourth quantum number m for the spin of two electrons. The number of electrons available in the conduction band depends on temperature. This band is the forbidden band. These two bands are separated by a band in which states are available but no electrons can exist. E How many of the electrons in an atom of silicon. The precise w c Figure 2.

In this section. Assuming a one-dimensional hydrogen crystal. The electronic potential energy of t h e hydrogen atom is given by E q. It consists of a regular array of square-well potentials. E l e c t r o n interaction with t h e core is purely coulombic in nature.

It is necessary to r e p e a t h e r e that our information is not obtained from the expressions for the wavefunctions but in the expressions that result from applying the b o u n d a r y conditions 0 0 Q 0 w w. A n electron having a certain mass. Kronig and P e n n e y used a onedimensional m o d e l of the potential energy distribution in a solid. A t o m s are fixed in position. Subject to the b o u n d a r y conditions. Non-ideal effects. E l e c t r o n to electron interaction is precluded.

In spite of its crude similarity to Fig. The dashed curve that is superimposed is the free particle solution. Those ranges of energies result in t h e forbidden bands. The implication is that the greater t h e electron energy. Chapter 2 Energy Bands and Current Carriers in Semiconductors and from the equations resulting from substitution of the expressions for the wave functions in Schrodinger's two equations. A t the higher values of energy. The values k Figure 2.

O n e can conclude that at the values of energies corresponding to these discontinuities. For the sake of comparison. These values of k m a r k the boundaries of the energy zones that m a k e u p the forbidden bands. The two i m p o r t a n t solutions consist of an expression on t h e left-hand side of each of the equations.

The o t h e r energy ranges are those that are either occupied by electrons or that could be occupied by electrons. Certain values of energy. A plot of these two equations is exhibited in Fig. T h e following observations can b e m a d e from the plots: P e n n e y m o d e l plot displays discontinuities and perturbations w h e n c o m p a r e d to the parabolic shape of t h e free particle solution. These modifications are greatest at t h e lower values of energy.

In silicon and germanium. In gallium arsenide. We will refer to this p r o p e r t y in relation to the effective mass later in this chapter. P e n n e y m o d e l results have exhibited two properties that are extremely important. In a real three-dimensional crystal. P e n n e y model bears a very general resemblance to the actual conditions existing in a crystal.

O n e distinguishing feature of semiconductors is the location of the conduction b a n d energy minimum with respect to the valence b a n d m a x i m u m on t h e E-k diagrams. In this manner. It is important to realize that the one-dimensional K r o n i g. Direct a n d Indirect S e m i c o n d u c t o r The actual b a n d structures of semiconductors are m u c h m o r e complex than those shown in Fig.

We have shown in Fig. T h e new curves are shown sketched in Fig. In indirect semiconductors. G a A s is known as a direct band gap semiconductor. The application to which the direct semiconductors b e c o m e i m p o r t a n t is the optical device.

These four electrons occupy the 3s and 3p levels in the energy b a n d representation. The bonding is a result of the fact that each a t o m shares four outermost-orbit electrons with four adjacent atoms.

A p h o t o n by itself cannot excite an electron from the t o p of the valence b a n d of an indirect semiconductor to the b o t t o m of t h e conduction b a n d because t h e p h o t o n has sufficient energy to cause the transition but does not possess the necessary m o m e n t u m for this transition.

A n electron moving b e t w e e n the valence b a n d and the conduction b a n d of an indirect semiconductor can occur through a defect in the semiconductor or by the action of phonons. In a direct b a n d gap semiconductor. In this case. T h e conductivity of metals decreases with increasing t e m p e r a t u r e.

Covalent Bond Model Section 2. If the imparted energy is exactly equal to t h e b a n d gap energy. It is of interest to realize that it is the covalent bonding that imparts hardness to G r o u p I V semiconductors. A t the n o r m a l atomic spacing. With this in mind. This translates into a higher b a n d gap energy and a higher melting point w h e n c o m p a r e d to G r o u p I V semiconductors having the same atomic n u m b e r and lattice constant.

In III-V semiconductors. In the energy b a n d model. They b e c o m e free electrons. While there are approximately 1 0 free electrons per c m in silicon at r o o m t e m p e r a t u r e.

In metals. A n o t h e r analogy is represented by the cylinders shown in Fig. The car that was moved upstairs is available for motion and it is analogous to our electron in t h e conduction band. The vacancy that was created on the ground floor is analogous to the hole and it moves in a direction opposite to the motion of the cars. For every electron that leaves the valence band. The b o t t o m cylinder in Fig. By having the vacancy occupied by a n o t h e r electron.

W h e n an electric field is applied to the silicon. O n e crude analogy is that of a two-level parking garage where. This vacancy is the hole. A metal is thus said to consist of an array of positive ions s u r r o u n d e d by closed-shell electronic orbits immersed in a gas of free electrons. These cylinders represent the valence and conduction bands. Until a car is m o v e d to the first floor from the ground floor.

By tipping b o t h cylinders to the right side in Fig. T h e small volume of liquid and t h e bubble move in opposite directions. Current Carriers—Electrons and Holes Section 2. Electrons are thermally excited from the valence b a n d to the conduction band.

W h e n an electric field is applied. The force of gravity is analogous to an applied electric field. In this state. A n increase in the energy of a hole is caused by the raising of the energy level of the electron. In the energy b a n d diagram. This applies equally well to conduction b a n d and valence b a n d electrons. A s electrons move u p within the valence band.

W h e n t h e energy of t h e electron is increased. In the b a n d picture. Electron a n d Hole Energies The generation of an electron-hole pair in the bands is shown in Fig. A n electron located at the level of the b o t t o m of the conduction b a n d at rest has potential energy E. In using t h e mass of an electron in v a c u u m or free space.

The forces in t h e crystal are k n o w n as lattice forces. Such a mass. In o r d e r to account for the lattice forces in the equation of motion. The motion of an electron in t h e conduction b a n d of a semiconductor is analogous to that of a free particle. The motion of an electron opposite to the direction of the electric field constitutes a current in the direction of the field since current is defined as being opposite to t h e direction of t h e motion of an electron.

These particles are the holes. It is obvious from Fig. Based on E q. The expression for t h e energy of an electron in the lattice is obtained by using E q. In conclusion. T h e curves n e a r the m i n i m u m of the conduction b a n d and n e a r t h e m a x i m u m of the valence b a n d are nearly parabolic so one can safely assume that the mass of the electron is constant in those two regions.

A n electron negative charge n e a r the t o p of the valence band. In being accelerated. We show in Fig. They are labeled conductivity effective masses. A n electric field must b e applied to accelerate the electrons in the partially filled bands. T h e r e must b e energy b a n d s that are partially filled with electrons.

For conduction of current to take place. We refer to the E-k diagrams of Fig. In conductors. We n o t e from the table that the effective masses of electrons and holes in gallium arsenide are smaller than those in silicon. For energy bands to b e c o m e partially filled with electrons. B o t h t h e electrons in t h e conduction b a n d and t h e holes in the valence can now b e accelerated.

A t low temperatures. Applying an electric field to the solid accelerates the electrons in the valence b a n d and increases their kinetic energy. This increase in energy is n o w h e r e sufficient to m o v e the electron across the gap into t h e conduction band.

Give an example of each. The gap is 1. The negative-mass. The effective mass of an electron is inversely proportional to the second derivative of the expression for energy as a function of the wave vector.

A negative-mass. Suggest an application for indirect semiconductors in relation to their absorption and emission of photons. The b a n d gap energy of an insulator.

Trace on Fig. The conductivity at K is small. Bands b e c o m e partially filled if either the n u m b e r of electrons is not sufficient to fill the b a n d or. Why is the mass of an electron negative for an electron located at the top of the valence band?

Why are the effective masses of holes and electrons smaller in GaAs than in Si? Why is the ability to conduct electricity. Why is the conductivity of insulators. A n electric field applied to the m e t a l at a t e m p e r a t u r e of K easily moves electrons to higher empty levels and causes a current to flow. C and D. The direction of acceleration of each wavepacket.

An electric field is applied to the material in such a direction as to cause a force in the negative k-direction.

For the equation determined in a. Determine the expression for the total energy of the electron in the lowest energy state. Sketch as accurately as possible: The diagram for a free electron is shown dotted. Chapter 2 Problems 51 2. In semiconductors. The expression for t h e density of states in a metal.

So that we can d e t e r m i n e current-voltage relationships. We also found that these b a n d s of energy that electrons may occupy are separated by forbidden energy gaps w h e r e n o electron of t h e solid can exist. The density of states refers to the n u m b e r of available electron states for unit volume per unit energy at a certain energy level.

A n analogy should help clarify the significance of the variation of N E and E. This curve is not continuous but is m a d e u p of a set of discrete points with the adjacent states so close to each other that.

The densities of electrons and holes will be d e t e r m i n e d by using this function t o g e t h e r with a probability of occupancy of a state function. We will assume that the function given by E q. Classical mechanics states that at absolute zero all electrons have zero energy.

Its significance is clearer if we state the following: To obtain the total n u m b e r of energy states per unit volume in a given energy range. Wave Mechanics theories obviously do not agree with those of classical mechanics. A t the higher elevations. Section 3. A sketch of the distribution of states is shown in Fig.

The n u m b e r of seats per m e t e r of elevation at the higher elevations is larger than t h e n u m b e r of seats per m e t e r of elevation at the lower elevations. Let us consider an oval-shaped stadium.

A t absolute zero. The question then is: W h a t is the distribution of t h e electrons with respect to energy as t h e t e m p e r a t u r e is raised above zero. E is the Fermi energy. E is t h e dividing energy below which all states are occupied with p p p p p p. The Fermi statistics describe what occurs as t h e t e m p e r a t u r e is raised above zero. Since we observe from E q. We also n o t e d that t h e Pauli Exclusion Principle states that a q u a n t u m state can b e occupied by only two electrons that have opposite spin.

According to Wave Mechanics. T h e Fermi-Dirac distribution predicts that as t h e t e m p e r a t u r e increases. We will discuss shortly t h e physical significance of E. Because of the Pauli Exclusion Principle. In a later chapter. N o t e t h e rounding-off of the curve at the higher energy levels. Temperature is measured in K.

You may t h e n question our statements h e r e concerning occupancy of E. This is the same definition for E we arrived at in A p p e n d i x C.

Figure 3. In its almost p u r e form. In the following example. A n increase of the energy of a hole in the valence band. These states are distributed t h r o u g h o u t t h e bands. The states of i m p o r t a n c e in semiconductor devices. We will assume that the expression for the density of states given by E q.

To obtain the densities of states n e a r the b a n d edges. Using the above expressions. U p o n comparing f E from Eqs. Since the electrons occupy states in the conduction b a n d and holes represent unoccupied states in t h e valence band. The hole density. We will arbitrarily state that E q. This assumption. For energy levels below E. Let us briefly digress to investigate the condition for which t h e approximation of E q. We have replaced the t o p of t h e conduction b a n d by plus infinity and the b o t t o m of the valence band by minus infinity.

Such semiconductors are said to b e degenerate. Before we p r o c e e d with evaluating the integral. We conclude that for a n o n. The second question concerns the effective masses We confirmed at t h e end of C h a p t e r 2 that the effective mass. The reason is to simplify the integration. E and E. In Eqs. In these small regions. To integrate E q. The justification is that since the exponential in t h e expressions decays very rapidly as we move away from the conduction and valence b a n d edges.

Relative masses of electrons and holes are shown in Table 2. The terms TV. The values for TV. By substituting for the constants in E q. Calculations of t h e intrinsic carrier density are illustrated by E x a m p l e 3. Mainly because of t h e larger b a n d gap. We define this density as n and nf. We will illustrate these effects in E x a m p l e 3. High t e m p e r a t u r e s and smaller b a n d gaps favor large values of intrinsic carrier density.

By substituting for t h e constant K in Eq. These same levels form t h e conduction and valence b a n d s in silicon. Using Eq. The reason for this uncertainty in the calculated value is a result of the very complex relations from which the values of the effective masses are determined. E is approximately halfway b e t w e e n E and E. It is important to indicate that there are discrepancies among various sources in the values of the intrinsic carrier density.

T h e second t e r m in t h e right-hand side of Eqs. For all practical purposes. In t h e following example. There is no sacrifice of accuracy in this approximation. We investigate t h e strength of t h e validity of this assumption for an intrinsic semiconductor by equating t h e expressions for t h e intrinsic electron and hole densities in Eqs. Q Briefly explain why the narrower the band gap.

Why is the Boltzmann approximation used? For what values of E. Is this true? E is Eq. In the equations derived so far. The value of k is given as 8. Typical dopants from C o l u m n V elements are phosphorous. This process cannot b e effective in good conductors. The atomic dimensions and the electronic structures of these impurities are similar to those of the semiconductor. E Calculate the intrinsic carrier density of Ge.

It is important to indicate here. These impurities are also k n o w n as dopants and the process of adding t h e m is k n o w n as doping the semiconductor. The result of the doping generates either of two types of extrinsic semiconductors. They m a y be N-type or P-type.

These have five valence electrons in their outermost orbit and are k n o w n as donor impurities because they have one electron in excess of what is n e e d e d for the covalent bonding.

The addition of as little as one impurity a t o m to a million semiconductor atoms has considerable effect on the conducting properties of the semiconductor. F Ans: A sketch of t h e resulting structure is shown in Fig. The question is then. Using the b a n d picture. E a c h impurity a t o m occupies the space of one silicon a t o m among. The hydrogen electron in its ground state. Because t h e hydrogen electron is in t h e field of t h e hydrogen nucleus.

To obtain a similar quantity of energy for our d o n o r electron. The i m p o r t a n t difference b e t w e e n this ionization mechanism to p r o d u c e d o n o r electrons and the intrinsic process to p r o d u c e electrons is that the ionized impurities are fixed charges in t h e lattice and n o holes are produced. The question 0. A n energy b a n d representation of the condition of d o n o r atoms in silicon is shown in Fig.

Using the above information. The phosphorus donor atoms have energy about 0. M o r e accurate values for ionization energies are listed in Table 3. Chapter 3 Intrinsic and Extrinsic Semiconductors lattice. A n order to magnitude value for E is obtained by assuming the masses to be equal so that E becomes. This condition is illustrated in Fig. The answer is n o and we shall investigate this in a later section. T h e acceptors' atoms are located at energy levels slightly above the valence band.

Is the total resulting density of electrons. A sketch of the resulting covalent b o n d a r r a n g e m e n t is shown in Fig. The c o m m o n dopants are: The absence of an electron from t h e top of the valence b a n d generates a hole.

This type of d o p a n t is k n o w n as an acceptor impurity. T h e atoms of these dopants have t h r e e electrons in their outermost orbit and w h e n a d o p a n t a t o m replaces a semiconductor atom. T h e condition for ionization of acceptor atoms is analogous to that of donor atoms.

The electron that occupies t h e available space in the acceptor bonding comes from the electrons that are in the valence b a n d of the semiconductor and. Langford Lane. The ionization energy is defined as the energy required to ionize a donor or an acceptor atom. Kidlington 0X5 1GB. The Boulevard. Chapter 3 E Intrinsic and Extrinsic Semiconductors c s 0. M o r e will be said later about recombinations.

Reprinted from S. Ge and Si at K. These impurities are randomly available and usually act as centers of recombination generation or traps. Listed in Table 3. Mobility and Impurity Levels in GaAs. Because they exist a r o u n d t h e middle of the energy gap. Sze and J.

In addition to t h e controlled addition of donor and acceptor impurities. In moving through the lattice. We stated earlier that the density of electrons. We label the extrinsic values of the electron and hole density at t h e r m a l equilibrium as n and p respectively so that we can write. In P-type semiconductors. T h i s carrier density results from the transfer of electrons to the conduction b a n d as they receive thermal energy.

In an earlier section. This product is therefore a constant equally valid for intrinsic as well as for d o p e d semiconductors.

Thermal equilibrium is therefore defined as the state in which a process is accompanied by an equal and opposite process. The answer is that as carriers are continuously generated. T h e question that comes to mind is: A s long as t h e thermal energy is available. We concluded that at r o o m t e m p e r a t u r e t h e intrinsic carrier density in silicon is approximately 1. In t h e next chapter. The r e a d e r should try solving for the minority carrier density to appreciate this caution.

T h e holes in the valence b a n d and the ionized donor atoms constitute the positive charges. The negative charges are m a d e u p of t h e electrons in the conduction b a n d and the acceptor atoms which are ionized. Assuming t h e r m a l equilibrium. For an N-type semiconductor. By replacing p in E q.

We notice that u p to about K. We note. This range extends from about to K. The material begins to b e c o m e intrinsic again and the electron density b e c o m e s m u c h larger than N because of t h e large n u m b e r of electrons that are thermally excited from the valence b a n d to the conduction band. The region K is k n o w n as the freeze-out region. In the intrinsic region of Fig. In the logarithmic sketch. This concentration of electrons remains constant at the value of N u p to r o o m t e m p e r a t u r e.

A b o v e K the intrinsic carrier density begins to increase rapidly and eventually reaches a value greater than N. Calculate the density of the majority and of the minority carriers at: We have d e n o t e d this intrinsic Fermi level by t h e symbol E.

Since n is useful in calculating carrier densities in extrinsic semiconductors. Calculations for the locations of the Fermi level in extrinsic silicon are carried out in t h e following example.

T h e location of E. E provides a useful reference for extrinsic semiconductors. We recall that E the Fermi level for an intrinsic semiconductor. I n each of these two equations. For a P-type semiconductor. It moves toward E when acceptors are added. Solution From Example 3. As the temperature is increased. Keeping in mind that the Fermi level for an intrinsic semiconductor is located approximately midway between the bottom of the conduction band and the top of the valence band. Why are N and N so much smaller than this number?

The Fermi level moves up towards the conduction band when donors are added and moves closer to the valence band when acceptors are added. A semiconductor that has been doped is known as an extrinsic semiconductor. When both donors and acceptor impurities are added.

As the temperature of a sample of semiconductor that is doped with donor impurities is increased. By referring to the periodic table. These impurities may be elements either from Column V of the periodic table. When donor impurities are added to an intrinsic semiconductor. The changes in the carrier densities are translated into changes in conductivity. Briefly define the phrase "thermal equilibrium"? What is a degenerate semiconductor?

The configuration of electrons in a certain element is 4s 4p. A donor atom that loses an electron and an acceptor atom that gains an electron are said to be ionized.

In intrinsic semiconductors. Phosphorus and arsenic are donors.

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