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Lecture Notes – 03ECE 531 Semiconductor Devices

Dr. Andre Zeumault

Outline

1 Density of States – Electronic 11.1 Derivation in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Density of States Effective Mass – Derivation . . . . . . . . . . . . . . . . . . . . . . 31.3 Conductivity Effective Mass – Derivation . . . . . . . . . . . . . . . . . . . . . . . . 6

2 The Fermi Function 72.1 Temperature Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Boltzmann Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Probability of Unoccupied States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Carrier Concentrations 93.1 Free Electron Concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.1 Boltzmann Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.1.2 Full Solution Using the Fermi-Dirac Integral . . . . . . . . . . . . . . . . . . 10

3.2 Free Hole Concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Ionized Donor Concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.4 Ionized Acceptor Concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.5 Degenerate vs Non-Degenerate Semiconductors . . . . . . . . . . . . . . . . . . . . . 123.6 Charge Neutrality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.7 Equilibrium Carrier Concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.8 Determining the Position of the Fermi Energy . . . . . . . . . . . . . . . . . . . . . . 14

3.8.1 Intrinsic Semiconductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.8.2 Extrinsic Semiconductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.9 Temperature Dependence of the Carrier Concentrations . . . . . . . . . . . . . . . . 15

1 Density of States – Electronic

The electronic density of states is a measure of the energy levels available to a carrier. It isdefined as the number of states per unit volume per unit energy or more simply, the concentrationof states per unit energy. Therefore, the following quantity gives the concentration of states in adifferential energy segment from E to E + dE:

g(E)dE

1.1 Derivation in 3D

Whenever a dispersion relation is known, the density of states can be derived from it. We knowfrom quantum mechanics that energy levels are quantized. We can denote such a state located atsome energy level En by:

δ(E − En)

Here, δ is the Dirac delta function, having units of inverse energy. So far, we just have a singleenergy level at En. To decorate energy space and account for all the various energy levels that mayexist required additional delta functions:

δ(E − El), δ(E − Em), δ(E − En), ...

Each state is non-overlapping in energy and can hold at most 2 electrons, each with opposite spin.Because of this, we can represent the totality of states using a summation.∑

n

δ(E − En)

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Dr. Andre Zeumault

To convert this into a definition of the density of states, we need to multiply by 2 to account forthe degeneracy of each state and divide by the sample volume, Ωr.

g(E) =2

Ωr

∑n

δ(E − En) (density of states)

So far, n is just an index that is used to identify a particular energy level. Since these alsocorresponds to states that the electron can occupy, they also index a particular Quantum state.Inthe language of Quantum Mechanics, we would appropriately call n a quantum number. So far, theonly such index we’ve introduced is the wave vector, k, since we know that the dispersion relationis E = h2k2

2m∗ . Thus, we can replace n with k without loss of generality.

g(E) =2

Ωr

∑k

δ(E − Ek)

Now, it should be understood that, in order to evaluate the density of states for any particularsituation, one must know the dispersion relation – the function relating E and k. In the parabolicapproximation,

E =h2k2

2m∗DOS

We can substitute this expression into the expression for the density of states,

g(E) =2

Ωr

∑k

δ

(E − h2k2

2m∗DOS

)Equivalently,

g(E) =2

Ωr

∑k

δ

(h2

2m∗DOS

[2m∗DOSE

h2 − k2

])Using the property of delta functions,

δ(ax) =1

|a|δ(x)

We obtain,

g(E) =2m∗DOSh2

2

Ωr

∑k

δ

(2m∗DOSE

h2 − k2

)That’s some progress, but how do we proceed? It turns out that this summation over discretevalues of k is quite difficult, so we make an approximation. We assume that the spacing between kpoints is small enough to treat the summation as an integration. But what do we mean by spacingbetween k points?

The spacing between k-points is determined by the size of the crystal along the length inquestion, Lx, Ly, Lz as well as the boundary conditions we impose, the total volume of the crystalbeing Ωr = LxLyLz. For periodic boundary conditions, the starting point and ending pointof the sample are treated as the same point in space. In the other alternative, we impose a hard

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Dr. Andre Zeumault

boundary condition as though the electron encounters the vacuum of space at its edge. Thisforces the wave function to vanish at the physical boundaries of the crystal. In either case, it canbe shown that the following distances between k points are satisfied,

• periodic boundary conditions: ∆k = 2πL

• hard boundary conditions: ∆k = πL

It is important to be aware of these distinctions. Periodic boundary conditions are typically as-sumed, therefore we will adopt the same convention in this class.

In order to further evaluate the summation over k, we need a useful tool to convert a summationover discrete values of k to an integral over a 3D k-space. Since we know the actual spacingbetween k-points to be 2π

Lx, 2πLy

, 2πLz

in the kx, ky and kz directions respectively, the k-space volume

is Ωk = 2πLx

2πLy

2πLz

= 8π3

Ωr. Thus, each infinitesimal increment d3~k must be normalized by the actual

k-space volume, since this sets the total number of points which can fit into the finite k-spacevolume. Thus, the conversion between the summation and the integration is as follows:∑

k

=1

Ωk

∫d3~k =

Ωr

8π3

∫d3~k

Making appropriate substitutions, we have,

g(E) =Ωr

8π3

2m∗DOSh2

2

Ωr

∫kδ

(2m∗DOSE

h2 − k2

)d3~k

If we assume that the k-space is isotropic,

d3~k = 4πk2dk

After substitution,

g(E) = 4πΩr

8π3

2m∗DOSh2

2

Ωr

∫ ∞0

δ

(2m∗DOSE

h2 − k2

)k2dk

To evaluate this integral, it is helpful to use an additional subtitution,

2k2dk = kd(k2)

Making the appropriate substitution, we have:

g(E) =1

24π

Ωr

8π3

2m∗DOSh2

2

Ωr

∫ ∞0

δ

(2m∗DOSE

h2 − k2

)kd(k2)

This integral can be easily evaluated as:

g(E) =1

24π

Ωr

8π3

2m∗DOSh2

2

Ωr

√2m∗DOSE

h2

After simplification, we have:

g(E) =

√2 (m∗DOS)3/2

π2h3

√E

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Dr. Andre Zeumault

Figure 1: (a) Fermi surface of the conduction band electrons in Silicon, showing the 6-fold valleydegeneracy. (b) Close up of the elliptical shape of one of the six valleys. The ellipse has twoequivalent directions (the transverse direction) and one longitudinal direction. Neamen 4th edition.

This is an important result. It gives the density of states of a charge carrier (at this point wehaven’t distinguished between electrons and holes) having a parabolic, isotropic dispersion relation.

1.2 Density of States Effective Mass – Derivation

Having introduced the concept of density of states, we can derive the density of states effectivemass equation for silicon, given in the previous lecture. In order to derive the density of stateseffective mass for silicon, we must first visualize the constant energy surfaces of silicon (i.e. FermiSurfaces, Figure 1).

Silicon has a 6-fold valley degeneracy, meaning there are 6 equivalent points in k space wherean electron has the same energy. In addition, the shape of the Fermi surface for each of the valleysis ellipsoidal. Thus, the effective mass of an electron depends on the direction of travel. Since theellipse has two equivalent directions (the transverse direction) there will be two equivalent massesand one distinct mass (longitudinal direction). We denote them as follows:

• longitudinal effective mass: m∗l

• transverse effective mass: m∗t

The energy of an electron in any of the 6 ellipses is given by:

E =h2k2

x

2m∗t+h2k2

y

2m∗t+h2k2

z

2m∗l

Normalizing both sides by the energy, E:

1 =h2k2

x

2m∗tE+

h2k2y

2m∗tE+

h2k2z

2m∗lE

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Dr. Andre Zeumault

Figure 2: Spherical Fermi surface. Neamen 4th edition.

We can compare the form of this equation to that of an ellipse in 3D momentum space:

1 =k2x

a2+k2y

b2+k2z

c2

If we compare the expression for the electron energy to the equation for an ellipse, we see that inthe case of the Fermi Surface of a conduction band electron in silicon:

a = b =

√2m∗tE

h

c =

√2m∗lE

h

Since we know that the volume of an ellipse is:

VEllipse =4

3πabc

It can be shown that the volume of a single Fermi surface is:

VFS =4

3π

√8(m∗t )

2m∗lE3

h3

Since there are a total of 6 surfaces, the total volume of the Fermi Surface is:

VFS,total = 6 ∗ 4

3π

√8(m∗t )

2m∗lE3

h3

The density of states effective mass essentially maps the volume of the true Fermi Surface (Figure 1),which is anisotropic in nature (i.e. directionally dependent), to an isotropic one – a sphere (Figure 2.Thus, the volume of the two are equated and what results is a defining relation for the density ofstates effective mass.

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Dr. Andre Zeumault

Figure 3: A depiction of the transport process. An electron, shown moving opposite to an appliedelectric field experiencing random scattering and drift.

The volume of a spherical Fermi Surface of radius k =

√2m∗

DOSE

h is:

VFS,Sphere =4

3π

(√2m∗DOSE

h

)3

=4

3π

√8(m∗DOS)3E3

h3

Equating the total volumes of the 6 Fermi surfaces, VFS,total with that of the spherical Fermi surface,VFS,sphere, leads to the following defining relation for the density of states effective mass of electronsin silicon.

m∗DOS,e = 62/3((m∗t )

2m∗l)1/3

For Silicon, m∗t = 0.19me and m∗l = 0.98me, thus giving a density of states effective mass ofelectrons of m∗DOS,e = 1.08me.

1.3 Conductivity Effective Mass – Derivation

In this section, we derive the conductivity effective mass for conduction band electrons in silicon.Before we begin, we need a simple picture depicting the process of carrier transport in a semicon-ductor. Carrier motion can be thought of as a superposition of random collisions plus drift alongthe direction of the applied electric field.

The conductivity effective mass originates from associating the average thermal energy of acharge carrier in 3D over each of its momentum variables. In essence, the average carrier momentumin each direction is equivalent. The usual expression for the parabolic band dispersion in 3D isgiven as the following, where a single mass has been associated with each axis of travel:

E =h2k2

x

2m∗cond,e+

h2k2y

2m∗cond,e+

h2k2z

2m∗cond,e

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Dr. Andre Zeumault

When we compare this equation to the actual band dispersion for silicon, within the parabolicapproximation:

E =h2k2

x

2m∗t,e+h2k2

y

2m∗t,e+h2k2

z

2m∗l,e

We see that they must be equal:

E =h2k2

x

2m∗cond,e+

h2k2y

2m∗cond,e+

h2k2z

2m∗cond,e=h2k2

x

2m∗t,e+h2k2

y

2m∗t,e+h2k2

z

2m∗l,e

If we take thermal averages of both sides, we arrive at the following:

〈E〉 =h2⟨k2x

⟩2m∗cond,e

+h2⟨k2y

⟩2m∗cond,e

+h2⟨k2z

⟩2m∗cond,e

=h2⟨k2x

⟩2m∗t,e

+h2⟨k2y

⟩2m∗t,e

+h2⟨k2z

⟩2m∗l,e

Next, we assume that the thermal averaged momenta are equivalent, that is:⟨k2x

⟩=⟨k2y

⟩=⟨k2z

⟩This last step is justified on the basis of the Equipartition Theorem used in statistical physics,which distributes average thermal energy equally among each individual degree of freedom. Thisallows us to simplify terms and arrive at the following:

1

2m∗cond,e+

1

2m∗cond,e+

1

2m∗cond,e=

1

2m∗t,e+

1

2m∗t,e+

1

2m∗l,e

After simplification, this leads to the following expression for the conductivity effective mass ofelectrons in silicon.

m∗cond,e =3

2mt,e∗

+ 1ml,e∗

2 The Fermi Function

The Fermi function gives the probability of finding an electron at a given energy (it can be derivedfrom statistical mechanics and the fact that electrons are Fermions). It is defined as follows:

f(E) =1

1 + eE−EFkbT

(Fermi Function)

The quantity EF is referred to as the Fermi energy. From the definition of the Fermi function,it is evident that the Fermi energy is the energy at which the probability of finding an electronicstate occupied by an electron is equal to 1/2.

2.1 Temperature Dependence

A plot of the Fermi function at various temperatures is given in Figure 4.At absolute zero (T = 0K), the Fermi function reduces to a step function. At such a condition,

all states below the Fermi level are occupied, and all states above the Fermi level are empty. As

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Dr. Andre Zeumault

Figure 4: Plot of the Fermi function as a function of energy for different temperatures. Neamen4th edition.

temperature increases, the Fermi function broadens, stretching over a larger energy. The probabilityof finding an electron at an energy below the Fermi energy decreases and the probability of findingan electron at an energy above the Fermi energy increases. Why? This should make intuitive sensesince at higher temperatures, carriers have more thermal energy and are therefore more likely tooccupy higher energy states. A corollary to this is that the broadening of the Fermi function withtemperature can wash out any fine structure that may exist in the density of states. This is oneof the reasons why it is usually necessary to cool semiconductors in order to study their exoticproperties related to their electronic structure.

2.2 Boltzmann Approximation

It can be seen that, depending on the energy level(s) of interest, certain simplifying assumptionsmay be made regarding the Fermi function. At very high temperatures, or when the energy ranges ofinterest are sufficiently higher than the thermal energy, kbT , we can replace the Fermi function withthe Maxwell-Boltzmann distribution. This distribution is contrasted with the Fermi function inFigure 5. Interestingly, the Maxwell-Boltzmann distribution is a statistical distribution of classical(distinguishable) particles. As shown, the form of the Boltzmann distribution is exponential withenergy. Mathematically, it is easy to show this as follows:

f(E) =1

1 + eE−EFkbT

≈ 1

eE−EFkbT

= e−E−EF

kbT

The condition for applicability of the Boltzmann approximation is as follows:

E − EF > 3kBT (Boltzmann Criterion)

This ensures that the exponential term dominates the denominator of the Fermi function. TheBoltzmann approximation is useful in that it simplifies calculations of carrier concentrations. Thiswill become clear in the next section. You should ponder a bit on the implications that thisapproximation has. How is it that electrons in a semiconductor can be well approximated usingclassical statistics?

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Dr. Andre Zeumault

Figure 5: Boltzmann approximation. Plot of the Maxwell-Boltzmann distribution and the Fermifunction. Neamen 4th edition.

2.3 Probability of Unoccupied States

So far, we’ve discussed the Fermi function, which gives the occupancy probability of a given energylevel. What if we are interested in the probability that a state is unoccupied? Since a state mustbe either occupied or unoccupied, that turns out to be simply

1− f(E)

This result will be used to calculate hole concentrations, since holes in the valence band are equiv-alent to unoccupied electronic states in the valence band.

3 Carrier Concentrations

Now that we have developed the concepts of density of states and the Fermi function, we can beginto define and calculate the total concentration of the various charges and charge carriers that existin a semiconductor. Typically, there are 4 types of charges that exist in a semiconductor.

• free electrons: n [cm−3]

• free holes: p [cm−3]

• ionized acceptors: N−A [cm−3]

• ionized donors: N+D [cm−3]

3.1 Free Electron Concentration

The concentration of free electrons, n is computed by summing all of the occupied electronic states.

n =

∫ ∞−∞

g(E)f(E)dE

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Dr. Andre Zeumault

Since the density of states is zero below the conduction band energy, we can redefine the limits ofthe integration as follows:

n =

∫ ∞EC

g(E)f(E)dE

When we first derived the density of states, it was for a free electron dispersion. The minimumenergy was therefore zero. In a solid, the minimum energy is the conduction band energy, and it isconvention to reference energies relative to the conduction band edge. Thus, the density of statesbecomes:

g(E) =

√2(m∗DOS,e

)3/2

π2h3

√E − EC

Using the definitions of the density of states and the Fermi function, the expression for the freeelectron concentration becomes:

n =

√2(m∗DOS,e

)3/2

π2h3

∫ ∞EC

√E − EC

1 + eE−EFkbT

dE

3.1.1 Boltzmann Approximation

Rather than evalutate this fully, for now we will make use of the Boltzmann approximation.

n ≈

√2(m∗DOS,e

)3/2

π2h3

∫ ∞EC

√E − ECe

−E−EFkbT dE

Rearranging this integral using u = E−ECkbT

substitution, we have:

n ≈ (kbT )3/2 e−EC−EF

kbT

√2(m∗DOS,e

)3/2

π2h3

∫ ∞0

√ue−udu

We recognize this integral to have the following simple solution:∫ ∞0

√ue−udu =

√π

2

Therefore, the free electron concentration within the Boltzmann approximation is given by:

n ≈ (kbT )3/2

√2(m∗DOS,e

)3/2

π2h3

√π

2e−EC−EF

kbT

Rather than writing out all of the various constants, we can introduce a new definition, known asthe effective conduction band density of states as NC as follows:

NC =

√2π

2

(m∗DOS,ekbT

)3/2

π2h3

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Dr. Andre Zeumault

This leads to the following simple result for the electron concentration:

n = NCe−EC−EF

kbT

3.1.2 Full Solution Using the Fermi-Dirac Integral

Prior to the Boltzmann approximation, we had the following integral to solve:

n =

√2(m∗DOS,e

)3/2

π2h3

∫ ∞EC

√E − EC

1 + eE−EFkbT

dE

To solve this integral, we need a useful result known as the Fermi-Dirac Integral of order 1/2,given by the following:

F1/2(ηc) =

∫ ∞0

η1/2dη

1 + eη−ηc

Rearranging our integral to match the form of the Fermi-Dirac integral yields the following:

n =

√2(m∗DOS,ekbT

)3/2

π2h3

∫ ∞0

η1/2dη

1 + eη−ηc

=

√2(m∗DOS,ekbT

)3/2

π2h3 F1/2(ηc)

Where ηc = EF−ECkbT

We recognize this result as the following, in terms of the effective density ofstates of the conduction band:

n =2√πNCF1/2(ηc)

The half order integral is solved numerically for various values of its argument. It can be thoughtof as a look-up table.

3.2 Free Hole Concentration

The derivation for the free hole concentration is very similar to that of the electrons. This can bedone as an exercise in your free time. The main results are presented here.

NV =

√2π

2

(m∗DOS,hkbT

)3/2

π2h3 (VB Eff. DOS)

p = NV eEV −EF

kbT (Boltz. Approx.)

p =2√πNV F1/2(ηv) (Full Solution)

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Dr. Andre Zeumault

3.3 Ionized Donor Concentrations

When dopants are added into a semiconductor, it is not a given that all of them will be ionized.Those dopants which are ionized are referred to as electrically active, these can be used tomodify the electronic conductivity of a semiconductor. Dopants which are not ionized are practicallyuseless. To estimate the concentration of ionized dopants, one proceeds using Fermi-Dirac statistics,although in most cases, the density of states for a dopant is a delta function in energy, which greatlysimplifies things.

Donors become ionized when unoccupied and are electrically neutral when occupied.The concentration of ionized donors is denoted as N+

D to distinguish it from the total dopantconcentration ND. The plus sign is used to denote ionization. The donor level energy is denotedED, and the donor ionization energy is given by EI,D = EC − ED.

N+D = ND(1− f(ED)) =

ND

1 + eEF−ED

kbT

The neutral dopant concentration can be denoted as N0D and is defined mathematically as:

N0D = NDf(ED) =

ND

1 + eED−EF

kbT

From these two definitions, it is clear that the total dopant concentration is due to the sum ofionized and neutral dopants.

ND = N0D +N+

D

We acknowledge that as the Fermi energy moves lower, a larger fraction of donors become ionized.

3.4 Ionized Acceptor Concentrations

Acceptors become ionized when occupied and are electrically neutral when empty.The concentration of ionized acceptors is denoted as N−A to distinguish it from the total dopant

concentration NA. The minus sign is used to denote ionization. The acceptor level energy is denotedEA, and the acceptor ionization energy is given by EI,A = EA − EV .

N0A = NAf(EA) =

NA

1 + eEA−EF

kbT

The neutral dopant concentration can be denoted as N0A and is defined mathematically as:

N−A = NA(1− f(EA)) =NA

1 + eEF−EA

kbT

From these two definitions, it is clear that the total dopant concentration is due to the sum ofionized and neutral dopants.

NA = N0A +N−A

Here we see that as the Fermi energy moves higher, a larger fraction of acceptors become ionized.

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Dr. Andre Zeumault

3.5 Degenerate vs Non-Degenerate Semiconductors

A degenerate semiconductor is one in which the Boltzmann approximation does not apply. Thisoccurs when the Fermi energy is sufficiently close to or above the band edge – above the conductionband for electrons, and below the valence band for holes. Whenever a semiconductor is degenerate,the full Fermi-Dirac integral must be used to calculate carrier concentrations. The term non-degenerate is used for semiconductors in which the Fermi energy is far from the band edges. Insuch a case, we can invoke the Boltzmann approximation when computing carrier concentrations.

3.6 Charge Neutrality

An important equilibrium constraint imposed on a semiconductor is charge neutrality. Chargeneutrality dictates that the net charge density in a semiconductor, averaged over its volume is zero.

ρ = e(p+N+

D − n−N−A

)= 0 (charge neutrality)

When we discuss devices, the charge neutrality condition will be modified to cover the entire devicegeometry.

3.7 Equilibrium Carrier Concentrations

For non-degenerate semiconductors, simple relationships can be established relating the carrierconcentrations in terms of the doping concentrations.

Starting from the Boltzmann approximation:

n = NCeEF−EC

kbT

p = NV eEV −EF

kbT

The product of n and p is therefore,

np = NCNV e− EG

kbT = n2i

In the last step we’ve introduced a new term ni to denote the intrinsic carrier concentrationin a pure semiconductor.

ni =√NCNV e

− EG2kbT

The intrinsic carrier concentration depends primarily on the bandgap energy, temperature and theeffective masses of electrons and holes. Since, for an intrinsic semiconductor, EF = Ei, n = niand p = ni, using the above expressions for n and p, we can arrive at the following alternativeexpressions for n and p.

n = nieEF−Ei

kbT

p = nieEi−EF

kbT

The use of these expressions requires equilibrium conditions and a non-degenerate semiconductor –Boltzmann approximation. When we combine these results with that of charge neutrality, we canarrive at the following relationships for n and p of an extrinsic semiconductor, provided that the

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Dr. Andre Zeumault

doping concentration is not so high that it invalidates the Boltzmann approximation.

n =ND −NA

2+

[(ND −NA

2

)2

+ n2i

]1/2

p =NA −ND

2+

[(NA −ND

2

)2

+ n2i

]1/2

3.8 Determining the Position of the Fermi Energy

We are often concerned with the exact positioning of the Fermi energy. This moves with externalvoltages and directly determines the charge carrier concentrations.

3.8.1 Intrinsic Semiconductor

In an intrinsic semiconductor, we can find the intrinsic Fermi level position by equating the electronand hole concentrations through the identity:

n = p = ni

NCeEF−EC

kbT = NV eEV −EF

kbT

Solving for EF = Ei yields the following:

EF = Ei =EC + EV

2+kbT

2lnNV

NC

The intrinsic level is very close to the average between the conduction band and the valence band,referred to as midgap, albeit a slight modification that is dependent on the differences between theeffective density of states of the conduction band and valence band. Since these only depend onthe carrier masses, if they are the same mass, then the intrinsic level is exactly at midgap.

3.8.2 Extrinsic Semiconductor

For a doped (extrinsic) semiconductor, finding the Fermi energy involves functions of n and p.From the alternative expressions of n and p, we can define the following:

EF = Ei + kbT lnn

ni

= Ei − kbT lnp

ni

Since we’ve already solved for n and p for an extrinsic semiconductor in the Boltzmann approxi-mation:

n =ND −NA

2+

[(ND −NA

2

)2

+ n2i

]1/2

p =NA −ND

2+

[(NA −ND

2

)2

+ n2i

]1/2

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Dr. Andre Zeumault

Figure 6: A plot of the Fermi energy as a function of dopant concentration for the case of donorsand acceptors exclusively. Neamen 4th edition.

The Fermi level position is given by:

EF = Ei + kbT ln

ND−NA

2 +

[(ND−NA

2

)2+ n2

i

]1/2

ni

(ND > NA)

EF = Ei − kbT ln

NA−ND

2 +

[(NA−ND

2

)2+ n2

i

]1/2

ni

(NA > ND)

For the simplifying conditions:

EF = Ei + kbT lnND

ni(ND >> NA, ni)

EF = Ei − kbT lnNA

ni(NA >> ND, ni)

To summarize, a plot of the Fermi energy as a function of doping concentration is shown in Figure 6.

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Dr. Andre Zeumault

Figure 7: A plot of the electron concentration as a function of temperature. Neamen 4th edition.

3.9 Temperature Dependence of the Carrier Concentrations

We end this chapter with a discussion of the full temperature dependence of the free electronconcentration. At very high temperatures, a large concentration of carriers are thermally excitedabove the bandgap energy, EG. If the concentration of these carriers is much higher than thatof the dopants, the carrier concentration will essentially follow the intrinsic carrier concentration.This is referred to as the intrinsic region. This is indicated by the dashed line in Figure 7. Asthe temperature reduces, eventually the intrinsic carrier concentration will be below the extrinsicconcentration. If the dopants (donors in this case) are fully ionized, then the electron concentrationwill essentially be constant. This is referred to as the extrinsic region. As the temperature isfurther reduced, eventually the Fermi energy will cross over donor level, and dopant ionization willreduce. In this region, the introduction of carriers in the band will be thermally activated withan activation energy that corresponds to the dopant ionization energy. This is referred to as thefreeze-out region, because there is insufficient thermal energy available to ionize dopants.

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