part02b

2.3 Carrier Properties

We have now met two characters, the electron and the hole. Before any equations, let us interview them: what charge do they carry, how heavy are they, how many of them are there, and how can we change that number? This section is a “collage” of general facts, and every later section leans on it.

2.3.1 Charge

Both carriers are charged, and the magnitude of the charge is the same for each:

• electron charge: $-q$
• hole charge: $+q$
• $q=1.60\times 10^{-19}\text{C}$ (three-place accuracy, MKS units).

The sign is always written explicitly (we never hide it inside $q$). Remember this convention, because a forgotten sign is the most common source of wrong answers in the next chapters.

2.3.2 Effective Mass

Charge is a single number. Mass is not so simple, and this is where the physics gets interesting.

Thought experiment 1: the electron in a vacuum. Put an electron of rest mass $m_0$ between two parallel plates with an electric field $\mathcal{E}$ between them. Newton’s second law gives

$$F=-q\mathcal{E}=m_0\frac{dv}{dt}\text{\hspace{1em}(2.2)}$$

where $v$ is the velocity and $t$ is time. The electron simply accelerates, and the acceleration is set by $m_0$ alone.

Thought experiment 2: the electron in a crystal. Now put the same field across a semiconductor crystal (Fig. 2.9b). Does Eq. (2.2) still describe the motion? The answer is no, for two reasons:

1. The electron collides with atoms and is periodically decelerated.
2. Even between collisions, besides our applied field the electron feels the complicated internal electric fields of the crystal, which Eq. (2.2) knows nothing about.

Strictly, only quantum mechanics can describe the motion. Here is the happy surprise: if the crystal is large compared with atomic dimensions, the complicated quantum description collapses into an equation of the same form as Newton’s law, provided we swap $m_0$ for an effective mass:

$$F=-q\mathcal{E}=m_n^{\ast }\frac{dv}{dt}\text{\hspace{1em}(2.3)}$$

where $m_n^{\ast }$ is the electron effective mass; for holes we write the same with $m_p^{\ast }$. All the crystal’s internal forces and quantum effects are “lumped into” the number $m^{\ast }$.

Analogy. Imagine running through a swimming pool versus through open air. Your actual body mass is the same, but you respond to your own push as if you were much heavier. The effective mass is the “apparent mass” of the carrier after the surroundings have had their say. It lets us treat electrons and holes as ordinary, classical-looking particles.

Caveats the book mentions: the acceleration can depend on direction in the crystal (effective mass has several components); different observables (cyclotron resonance, conductivity, density of states) use different groupings of these components; and $m^{\ast }$ drifts slightly with temperature. In this chapter we use only the density of states effective masses (Section 2.4.1), listed in Table 2.1.

Table 2.1: Density of states effective masses at 300 K

Material	$m_n^{\ast }/m_0$	$m_p^{\ast }/m_0$
Si	1.18	0.81
Ge	0.55	0.36
GaAs	0.066	0.52

Notice the striking number: electrons in GaAs are very light ($0.066m_0$), which is one reason GaAs is used for fast devices.

2.3.3 Carrier Numbers in Intrinsic Material

An intrinsic semiconductor is, loosely, an extremely pure sample with a negligible amount of impurity; more precisely, one whose properties are native to the material and not caused by additives. Define

$$n={\text{number of electrons/cm}}^3,p={\text{number of holes/cm}}^3.$$

In an intrinsic semiconductor at equilibrium

$$n=p=n_i\text{\hspace{1em}(2.4)}$$

and, at room temperature,

$$n_i\approx \{\begin{array}{ll}2\times 10^6{\text{cm}}^{-3}& \text{GaAs}\\ 1\times 10^{10}{\text{cm}}^{-3}& \text{Si}\\ 2\times 10^{13}{\text{cm}}^{-3}& \text{Ge}\end{array}$$

Why $n=p$? In a very pure material carriers are created only in pairs. Break a bond and you simultaneously create a free electron and a missing bond (hole). Lift an electron from the valence to the conduction band and you automatically leave a valence-band hole behind.

How big is $n_i$, really? Put numbers on it:

$$\begin{array}{rlrl}{\text{bonds per cm}}^3& =(5\times 10^{22})(4)& & \text{four bonds per Si atom}\\ & =2\times 10^{23}{\text{cm}}^{-3}& & \text{multiply}\\ \frac{n_i}{\text{bonds}}& =\frac{10^{10}}{2\times 10^{23}}=5\times 10^{-14}& & \text{fraction of bonds broken}\end{array}$$

That is fewer than one bond in $10^{13}$ broken. The book’s vivid picture: cover every university chalkboard in the world with the bonding model and you would probably draw just one broken bond.

2.3.4 Manipulation of Carrier Numbers: Doping

Intrinsic carrier numbers are tiny and fixed by nature. Doping is the deliberate addition of controlled amounts of specific impurity atoms to raise either the electron or the hole concentration. Nearly every semiconductor device is made this way.

Table 2.2: Common silicon dopants (most widely used marked with $\star$)

Donors (electron-increasing), Column V	Acceptors (hole-increasing), Column III
P $\star$	B $\star$
As $\star$	Ga
Sb	In
	Al

Donor action (bonding model, Fig. 2.10a). Substitute a Column V atom (5 valence electrons) for a Si atom. Four of its five electrons fit snugly into the four bonds. The fifth does not fit and is only weakly bound; at room temperature it is easily freed and wanders through the lattice as a carrier. Important: this does not create a hole. The leftover donor ion (${\text{P}}^{+}$) is fixed in the lattice, and no Si-Si bond was broken.

Acceptor action (Fig. 2.10b). Substitute a Column III atom (3 valence electrons). It cannot complete all four bonds, so it readily accepts an electron from a neighboring Si-Si bond. That completes its own bonding (making it a fixed negative ion, ${\text{B}}^{-}$) and leaves behind a wandering hole. Again only one carrier type increases.

Memory trick. Column V donates a spare electron (5 > 4: surplus). Column III accepts one (3 < 4: shortage). The ion left behind is immobile; it is the carrier that moves.

Loose end 1: how weakly is “weakly bound”? Breaking a Si-Si bond takes about 1 eV, and very few break at room temperature. So “weakly” should mean a binding energy of about $0.1$ eV or less. Let us estimate it using the hydrogen atom (Fig. 2.11, the pseudo-hydrogen atom).

Treat the donor core (${\text{P}}^{+}$) plus its fifth electron as a hydrogen atom, but with two substitutions because the electron now moves through a sea of Si atoms rather than a vacuum:

• ${\epsilon }_0\to K_s{\epsilon }_0$ (with Si dielectric constant $K_s=11.8$),
• $m_0\to m_n^{\ast }$.

Applying these to Eq. (2.1) with $n=1$:

$$E_B\approx \frac{m_n^{\ast }}{m_0}\frac{1}{K_s^2}{E}_H{|}_{n=1}\approx -0.1\text{eV}\text{\hspace{1em}(2.5)}$$

Let us verify the number using Table 2.1 ($m_n^{\ast }/m_0=1.18$):

$$\begin{array}{rlrl}{K}_s^2& =11.8^2=139.24& & \text{square the dielectric constant}\\ E_B& \approx 1.18\times \frac{1}{139.24}\times (-13.6\text{eV})& & \text{insert }{m}_n^{\ast }/m_0,K_s,E_H\\ & \approx -0.115\text{eV}& & \text{multiply out}\end{array}$$

So $|E_B|\sim 0.1$ eV. Measured values (Table 2.3) agree, and they are about $1/20$ of the Si band gap.

Table 2.3: Dopant-site binding energies in Si

Donor	$|E_B|$	Acceptor	$|E_B|$
Sb	0.039 eV	B	0.045 eV
P	0.045 eV	Al	0.067 eV
As	0.054 eV	Ga	0.072 eV
		In	0.16 eV

Why the binding is so weak: the dielectric constant enters squared in the denominator. The crystal screens the donor’s pull by a large factor, and the small effective mass shrinks the binding further.

Loose end 2: dopants in the energy band picture. An electron freed from a donor with exactly the binding energy becomes a conduction band electron with the lowest possible energy, $E_c$. Hence the bound electron must sit an energy $|E_B|$ below $E_c$:

$$E_D=E_c-|E_B|.$$

We draw donor levels as short dashes (Fig. 2.12) rather than a line, because a bound electron is localized near its donor site and does not wander through the crystal. Likewise acceptors add allowed levels slightly above $E_v$:

$$E_A=E_v+|E_B|.$$

What happens as temperature rises (Fig. 2.13):

• Donors: at $T\to 0$ K all donor levels are filled (no thermal energy to kick electrons up). With increasing $T$ electrons hop into the conduction band. At room temperature donors are essentially all ionized.
• Acceptors: at $T\to 0$ K all acceptor levels are empty. With increasing $T$ valence-band electrons hop onto acceptor sites, leaving holes behind. At room temperature essentially all acceptors hold an extra electron.

Doping non-elemental semiconductors (GaAs). The same ideas hold but there are two lattice sites:

• Column VI atoms (S, Se, Te) replacing Column V As act as donors.
• Column II atoms (Be, Mg, Zn) replacing Column III Ga act as acceptors.
• Column IV atoms (Si, Ge) are special: Si usually replaces Ga (donor, a popular n-type dopant), but under some conditions replaces As (acceptor). An impurity that can act as either is called amphoteric.

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Exercise 2.1: Energy Quiz

(a) $1\text{eV}$ equals how many joules?

$$\begin{array}{rlrl}1\text{eV}& =(1.60\times 10^{-19}\text{C})(1\text{V})& & \text{energy gained by a charge }q\text{ across 1 V}\\ & =1.60\times 10^{-19}\text{J}& & \text{C}\cdot \text{V}=\text{J}\end{array}$$

(b) $kT$ in eV at $300$ K:

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5}\text{eV/K})(300\text{K})& & \text{Boltzmann constant times }T\\ & =2.585\times 10^{-2}\text{eV}\approx 0.0259\text{eV}& & \text{multiply}\end{array}$$

This number, about $26$ meV, is the natural “thermal energy unit” for the rest of the book. Keep it in your head.

(c) Ionization energy of dopants in Si: it equals $|E_B|$ of the dopant site, roughly $0.1$ eV (measured $0.04$ to $0.07$ eV for common dopants).

(d) $E_G(\text{Si})=1.12$ eV at 300 K (Section 2.2.4).

(e) $E_G({\text{SiO}}_2)\approx 8$ eV (Section 2.2.4).

(f) Ionizing hydrogen from $n=1$ costs, from Eq. (2.1),

$$|E_H(n=1)|=\frac{13.6}{1^2}=13.6\text{eV}.$$

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2.3.5 Carrier-Related Terminology

Terminology is a classic stumbling block, so here is the glossary to memorize:

Term	Meaning
Dopants	Specific impurity atoms added in controlled amounts to raise the electron or the hole concentration
Intrinsic semiconductor	Undoped, extremely pure; properties native to the material
Extrinsic semiconductor	Doped; properties controlled by the added impurities
Donor	Impurity that increases the electron concentration (n-type dopant)
Acceptor	Impurity that increases the hole concentration (p-type dopant)
n-type material	Donor-doped; more electrons than holes
p-type material	Acceptor-doped; more holes than electrons
Majority carrier	Most abundant carrier: electrons in n-type, holes in p-type
Minority carrier	Least abundant carrier: holes in n-type, electrons in p-type

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2.4 State and Carrier Distributions

So far our knowledge has been qualitative. Real engineers need numbers: how many electrons are in a doped sample, and at what energies do they live? We build the answer from two ingredients, like counting people in a stadium:

1. How many seats exist at each energy? (density of states)
2. What fraction of those seats is occupied? (Fermi function)

Multiply the two and you have the carrier distribution.

2.4.1 Density of States

Section 2.2 said each band holds $4N$ states, but not how those states are spread in energy. That spread is the density of states. Quantum mechanics gives, for energies not too far from the band edges,

$$g_c(E)=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }(E-E_c)}}{{\pi }^2{\hslash }^3},E\ge E_c\text{\hspace{1em}(2.6a)}$$ $$g_v(E)=\frac{m_p^{\ast }\sqrt{2m_p^{\ast }(E_v-E)}}{{\pi }^2{\hslash }^3},E\le E_v\text{\hspace{1em}(2.6b)}$$

where $g_c$, $g_v$ are the densities of states in the conduction and valence bands.

Football stadium analogy. The field is the band gap. Seats lie at various distances from the field; the number of seats at a given distance is the number of states at a given energy distance from the band edge. There are no seats on the field itself (no states in the gap) and the stands widen as you climb away from it.

Key points to remember:

• $g_c(E)=0$ at $E=E_c$ and grows like $\sqrt{E-E_c}$ upward; $g_v(E)=0$ at $E=E_v$ and grows like $\sqrt{E_v-E}$ downward (Fig. 2.14).
• The two curves differ only because $m_n^{\ast }\ne m_p^{\ast }$. If the masses were equal the two “stands” would be mirror images.
• $g_c(E)dE$ is the number of conduction band states per cm${}^3$ between $E$ and $E+dE$; likewise $g_v(E)dE$ for the valence band. So $g$ has units of states per cm${}^3$ per eV.

2.4.2 The Fermi Function

The density of states says how many seats exist. The Fermi function $f(E)$ says what fraction is filled, that is, the probability that an available state at energy $E$ is occupied by an electron at equilibrium:

$$f(E)=\frac{1}{1+e^{(E-E_F)/kT}}\text{\hspace{1em}(2.7)}$$

with

• $E_F$: the Fermi energy or Fermi level,
• $k=8.617\times 10^{-5}$ eV/K: Boltzmann’s constant,
• $T$: absolute temperature in K.

Case 1: $T\to 0$ K. Look at the exponent $(E-E_F)/kT$:

$$\begin{array}{rl}E<E_F:& \frac{E-E_F}{kT}\to -\infty \Rightarrow f(E)\to \frac{1}{1+e^{-\infty }}=1\\ E>E_F:& \frac{E-E_F}{kT}\to +\infty \Rightarrow f(E)\to \frac{1}{1+e^{+\infty }}=0\end{array}$$

So at absolute zero every state below $E_F$ is full and every state above is empty: a sharp cutoff (Fig. 2.15a). It is a bucket of water at rest: everything below the water line is wet.

Case 2: $T>0$ K. Three observations:

1. At $E=E_F$: $f(E_F)=\frac{1}{1+e^0}=\frac{1}{2}$. The Fermi level is the energy at which a state is exactly a coin-flip.
2. For $E\ge E_F+3kT$: $e^{(E-E_F)/kT}\gg 1$, so

$$f(E)\approx e^{-(E-E_F)/kT}.$$

The filled probability decays exponentially to zero and most states are empty. 3. For $E\le E_F-3kT$: $e^{(E-E_F)/kT}\ll 1$, so

$$f(E)\approx 1-e^{(E-E_F)/kT},\text{i.e. }1-f(E)\approx e^{(E-E_F)/kT}.$$

The empty-state probability $1-f$ decays exponentially to zero as you go further below, and most states are filled.

Raising $T$ merely “rounds the corner” of the step, over a window of a few $kT$ ($\approx 3kT=78$ meV at 300 K), tiny against the $1.12$ eV Si gap (Fig. 2.15b).

Universality. The Fermi function holds only at equilibrium, but it is universal: it applies equally to insulators, semiconductors, and metals. It is just the statistics of electrons.

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Exercise 2.2

Problem. The probability that a state is filled at the conduction band edge $E_c$ is exactly equal to the probability that a state is empty at the valence band edge $E_v$. Where is the Fermi level?

Solution.

The empty-state probability is $1-f(E)$, so the statement reads

$$f(E_c)=1-f(E_v)\text{given condition}$$

Write each side using Eq. (2.7):

$$f(E_c)=\frac{1}{1+e^{(E_c-E_F)/kT}}\text{definition of }f$$ $$1-f(E_v)=1-\frac{1}{1+e^{(E_v-E_F)/kT}}=\frac{e^{(E_v-E_F)/kT}}{1+e^{(E_v-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}\text{common denominator, then divide top and bottom by }{e}^{(E_v-E_F)/kT}$$

Equating the two:

$$\frac{1}{1+e^{(E_c-E_F)/kT}}=\frac{1}{1+e^{(E_F-E_v)/kT}}\text{set equal}$$ $$\frac{E_c-E_F}{kT}=\frac{E_F-E_v}{kT}\text{equal fractions with equal numerators have equal exponents}$$ $$E_c-E_F=E_F-E_v\text{multiply by }kT$$ $$E_F=\frac{E_c+E_v}{2}\text{solve for }{E}_F$$

Result: the Fermi level sits at midgap.

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Exercise 2.3: Fermi function versus temperature

Problem. For $T=100,200,300,400$ K compute and plot $f(E)$ versus $\Delta E=E-E_F$ for $-0.2\le \Delta E\le 0.2$ eV, all curves on one set of axes.

Solution. The book uses a MATLAB script. The steps in words and equations:

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5})T& & \text{thermal energy for each }T\\ \Delta E_j& =-5kT+0.1kT(j-1),j=1,\dots ,101& & \text{energy grid from }-5kT\text{ to }+5kT\\ f_j& =\frac{1}{1+e^{\Delta E_j/kT}}& & \text{Fermi function at each grid point}\end{array}$$

% Fermi Function Calculation, f(dE,T)
k = 8.617e-5;                 % Boltzmann constant (eV/K)
for ii = 1:4
    T  = 100*ii;
    kT = k*T;
    dE(ii,1) = -5*kT;
    for jj = 1:101
        f(ii,jj)    = 1/(1+exp(dE(ii,jj)/kT));
        dE(ii,jj+1) = dE(ii,jj) + 0.1*kT;
    end
end
dE = dE(:,1:jj);              % strip the extra dE value
plot(dE', f); grid;           % transpose to form data columns
xlabel('E-E_F (eV)'); ylabel('f(E)');

What the plot shows. Every curve passes through $f=1/2$ at $\Delta E=0$. The $T=100$ K curve is nearly a step; the $T=400$ K curve is the most gradual. Hotter means a fatter transition region, exactly as the $3kT$ rule predicts (note the grid for each curve spans $\pm 5kT$, so the cold curves reach the plot’s flat parts sooner).

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2.4.3 Equilibrium Distribution of Carriers

Now multiply seats by occupancy:

$$\begin{array}{rl}\text{electrons in conduction band:}& g_c(E)f(E)\\ \text{holes in valence band:}& g_v(E)[1-f(E)]\end{array}$$

(A hole is an unfilled state, hence the factor $1-f$.) Fig. 2.16 shows these for three Fermi-level positions. General features:

• All distributions are zero at the band edges (because $g=0$ there), peak very close to $E_c$ or $E_v$, and then decay rapidly. Most carriers live near the band edges.
• $E_F$ in the upper half of the gap (or higher): electrons far outnumber holes, because $f(E)$ is much larger than $1-f(E)$ at corresponding energies.
• $E_F$ near midgap: about equal numbers of both (assuming $g_c$ and $g_v$ are of comparable size, true for Si and Ge).
• $E_F$ in the lower half of the gap: holes dominate.
• The exponential-tail statements hold provided $E_c-3kT\ge E_F\ge E_v+3kT$.

Shorthand pictures: Fig. 2.17 draws many dots near the band edges and fewer farther in. Fig. 2.18, the “at a glance” picture, draws a dashed line near midgap labeled $E_i$ (the intrinsic Fermi level). A solid $E_F$ line above $E_i$ means n-type; below $E_i$ means p-type. In extrinsic diagrams the dashed $E_i$ is kept as a reference that splits the gap into upper and lower halves.

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2.5 Equilibrium Carrier Concentrations

This is the summit of the chapter. We now turn the qualitative picture into working formulas. The book compares the alternative forms of each formula to a toolbox: open-end, box and ratchet wrenches all turn the same bolt, but in a tight spot only one fits. Do not be alarmed by the algebra; every step is shown.

2.5.1 Formulas for $n$ and $p$

Since $g_c(E)dE$ is the number of conduction band states per cm${}^3$ in $[E,E+dE]$ and $f(E)$ is the probability each is filled, $g_c(E)f(E)dE$ is the number of electrons there. Adding up (integrating) over the whole band gives the electron concentration. The same logic with $g_v(1-f)$ gives holes:

$$n={\int }_{E_c}^{E_{top}}{g}_c(E)f(E)dE\text{\hspace{1em}(2.8a)}$$ $$p={\int }_{E_{bottom}}^{E_v}{g}_v(E)[1-f(E)]dE\text{\hspace{1em}(2.8b)}$$

Derivation of the $n$ formula. Insert Eq. (2.6a) and Eq. (2.7):

$$n=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{E_c}^{E_{top}}\frac{\sqrt{E-E_c}}{1+e^{(E-E_F)/kT}}dE\text{\hspace{1em}(2.9)}$$

Make the substitutions

$$\eta =\frac{E-E_c}{kT},{\eta }_c=\frac{E_F-E_c}{kT},E_{top}\to \infty \text{(i.e. upper limit }\eta \to \infty )\text{\hspace{1em}(2.10)}$$

Why may we push the upper limit to infinity? The integrand dies off exponentially within a few $kT$ of $E_c$, so extending the limit adds a negligible amount. Also, $E-E_F=(E-E_c)-(E_F-E_c)=kT(\eta -{\eta }_c)$ and $\sqrt{E-E_c}dE=(kT{)}^{3/2}\sqrt{\eta }d\eta$. Therefore

$$n=\frac{\sqrt{2}(m_n^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\frac{\sqrt{\eta }d\eta }{1+e^{\eta -{\eta }_c}}\text{\hspace{1em}(2.11)}$$

This integral has no closed form with simple functions; it is a tabulated function, the Fermi-Dirac integral of order 1/2:

$$F_{1/2}({\eta }_c)={\int }_0^{\infty }\frac{\sqrt{\eta }d\eta }{1+e^{\eta -{\eta }_c}}\text{\hspace{1em}(2.12)}$$

Define the effective densities of states

$$N_C=2{\left(\frac{2\pi m_n^{\ast }kT}{h^2}\right)}^{3/2},N_V=2{\left(\frac{2\pi m_p^{\ast }kT}{h^2}\right)}^{3/2}\text{\hspace{1em}(2.13)}$$

(equivalently $2(m^{\ast }kT/2\pi {\hslash }^2{)}^{3/2}$). Checking that the constants combine: $\frac{2}{\sqrt{\pi }}{N}_C=\frac{2}{\sqrt{\pi }}\cdot \frac{2(m_n^{\ast }kT{)}^{3/2}}{(2\pi {)}^{3/2}{\hslash }^3}=\frac{\sqrt{2}(m_n^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}$, exactly the prefactor in Eq. (2.11). Hence

$$n=N_C\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_c),{\eta }_c=\frac{E_F-E_c}{kT}\text{\hspace{1em}(2.14a)}$$ $$p=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v),{\eta }_v=\frac{E_v-E_F}{kT}\text{\hspace{1em}(2.14b)}$$

These hold for any Fermi level position. At 300 K the constants are $N_{C,V}=(2.510\times 10^{19}{\text{cm}}^{-3})(m_{n,p}^{\ast }/m_0{)}^{3/2}$.

Simplification. If $E_F\le E_c-3kT$ (so ${\eta }_c\le -3$), the “+1” in the denominator is negligible and

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_c)\approx e^{{\eta }_c}=e^{(E_F-E_c)/kT}\text{\hspace{1em}(2.15a)}$$

and likewise for $E_F\ge E_v+3kT$:

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\approx e^{(E_v-E_F)/kT}\text{\hspace{1em}(2.15b)}$$

So, if $E_v+3kT\le E_F\le E_c-3kT$:

\boxed{\;n=N_C\,e^{(E_F-E_c)/kT}\;}\tag{2.16a} \qquad \boxed{\;p=N_V\,e^{(E_v-E_F)/kT}\;}\tag{2.16b}

This is the same as replacing $f$ and $1-f$ by simple exponentials, valid when $E_F$ is in the gap and at least $3kT$ from either edge.

Degenerate vs nondegenerate (Fig. 2.19).

• Nondegenerate: $E_v+3kT\le E_F\le E_c-3kT$. Eqs. (2.16) work.
• Degenerate: $E_F$ closer than $3kT$ to a band edge, or inside a band. Then one must use the Fermi-Dirac integral, Eqs. (2.14).

2.5.2 Alternative Expressions for $n$ and $p$

An intrinsic semiconductor is surely nondegenerate, so Eqs. (2.16) apply. Setting $n=p=n_i$ and $E_F=E_i$:

$$n_i=N_C{e}^{(E_i-E_c)/kT}\text{\hspace{1em}(2.17a)}$$ $$n_i=N_V{e}^{(E_v-E_i)/kT}\text{\hspace{1em}(2.17b)}$$

Solve for the constants:

$$N_C=n_i{e}^{(E_c-E_i)/kT}\text{\hspace{1em}(2.18a)}$$ $$N_V=n_i{e}^{(E_i-E_v)/kT}\text{\hspace{1em}(2.18b)}$$

Substitute into Eqs. (2.16):

$$\begin{array}{rlrl}n& =n_i{e}^{(E_c-E_i)/kT}{e}^{(E_F-E_c)/kT}=n_i{e}^{(E_F-E_i)/kT}& & \text{(2.19a)}\\ p& =n_i{e}^{(E_i-E_v)/kT}{e}^{(E_v-E_F)/kT}=n_i{e}^{(E_i-E_F)/kT}& & \text{(2.19b)}\end{array}$$

Only one constant ($n_i$) and two energies appear, and the symmetry is easy to remember: swap $E_F\leftrightarrow E_i$ to go from $n$ to $p$. Intuition: the farther $E_F$ sits above $E_i$, the more electrons; the farther below, the more holes.

2.5.3 $n_i$ and the $np$ Product

Multiply Eqs. (2.17a) and (2.17b):

$$n_i^2=N_C{N}_V{e}^{-(E_c-E_v)/kT}=N_C{N}_V{e}^{-E_G/kT}\text{\hspace{1em}(2.20)}$$ $$n_i=\sqrt{N_C{N}_V}{e}^{-E_G/2kT}\text{\hspace{1em}(2.21)}$$

This tells you $n_i$ is controlled by the exponential of the band gap over thermal energy: wide gap means tiny $n_i$, and increasing $T$ rapidly increases $n_i$ (Fig. 2.20 shows $n_i$ versus $T$ for Ge, Si, GaAs).

Now multiply Eqs. (2.19a) and (2.19b):

$$np=n_i^2{e}^{(E_F-E_i)/kT}{e}^{(E_i-E_F)/kT}=n_i^2\text{\hspace{1em}(2.22)}$$

The $np$ product holds for any equilibrium, nondegenerate sample, doped or not. If you know one carrier concentration, you instantly get the other. Think of a seesaw: push electrons up and holes must go down, with the product pinned at $n_i^2$.

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Exercise 2.4: $n_i$ versus temperature in Si

Combine Eq. (2.21) with Eq. (2.13), using $N_{C,V}=2.510\times 10^{19}(m^{\ast }/m_0{)}^{3/2}(T/300{)}^{3/2}$:

$$n_i=2.510\times 10^{19}{\left(\frac{m_n^{\ast }}{m_0}\frac{m_p^{\ast }}{m_0}\right)}^{3/4}{\left(\frac{T}{300}\right)}^{3/2}{e}^{-E_G/2kT}$$

Problem (a). Confirm the Si curve of Fig. 2.20 using

$$E_G(T)=1.17-\frac{4.730\times 10^{-4}{T}^2}{T+636},\frac{m_n^{\ast }}{m_0}=1.028+6.11\times 10^{-4}T-3.09\times 10^{-7}{T}^2,\frac{m_p^{\ast }}{m_0}=0.610+7.83\times 10^{-4}T-4.46\times 10^{-7}{T}^2,$$

with $E_G$ in the exponent replaced by $E_G-E_{ex}$, $E_{ex}=0.0074$ eV.

Solution. The MATLAB script just evaluates the formula at each temperature. Doing it by hand at $T=300$ K shows what the script does:

$$\begin{array}{rlrl}{E}_G(300)& =1.17-\frac{(4.730\times 10^{-4})(300{)}^2}{300+636}=1.17-\frac{42.57}{936}=1.1245\text{eV}& & \text{band gap fit}\\ E_G-E_{ex}& =1.1245-0.0074=1.1171\text{eV}& & \text{exciton correction}\\ \frac{m_n^{\ast }}{m_0}& =1.028+0.1833-0.0278=1.1835& & \text{electron mass ratio}\\ \frac{m_p^{\ast }}{m_0}& =0.610+0.2349-0.0401=0.8048& & \text{hole mass ratio}\\ (\frac{m_n^{\ast }}{m_0}\frac{m_p^{\ast }}{m_0}{)}^{3/4}& =(0.9524{)}^{0.75}=0.9641& & \text{mass factor}\\ kT& =(8.617\times 10^{-5})(300)=0.025851\text{eV}& & \text{thermal energy}\\ \frac{E_G-E_{ex}}{2kT}& =\frac{1.1171}{0.051702}=21.607& & \text{exponent}\\ e^{-21.607}& =4.13\times 10^{-10}& & \text{exponential}\\ n_i& =(2.510\times 10^{19})(0.9641)(1)(4.13\times 10^{-10})\approx 1.0\times 10^{10}{\text{cm}}^{-3}& & \text{multiply}\end{array}$$

This matches the accepted value; that is exactly why $E_{ex}$ was tuned to $0.0074$ eV. The script repeats this for $T=200,225,\dots ,700$ K.

Problem (b). Compare with the Sproul-Green experimental fit (valid $275\le T\le 375$ K):

$$n_i=9.15\times 10^{19}{\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}$$

Solution. At $T=300$ K:

$$\begin{array}{rlrl}\frac{0.5928}{kT}& =\frac{0.5928}{0.025851}=22.932& & \text{exponent}\\ e^{-22.932}& =1.10\times 10^{-10}& & \text{exponential}\\ n_i& =(9.15\times 10^{19})(1)(1.10\times 10^{-10})\approx 1.006\times 10^{10}{\text{cm}}^{-3}& & \text{multiply}\end{array}$$

Script output (both agree within about 2%):

$T$ (K)	(a) $n_i$ (cm${}^{-3}$)	(b) $n_i$ (cm${}^{-3}$)
200	$5.246\times 10^4$
275	$1.059\times 10^9$	$1.051\times 10^9$
300	$1.000\times 10^{10}$	$1.006\times 10^{10}$
325	$6.798\times 10^{10}$	$6.887\times 10^{10}$
350	$3.565\times 10^{11}$	$3.623\times 10^{11}$
375	$1.518\times 10^{12}$	$1.542\times 10^{12}$
400	$5.449\times 10^{12}$
500	$2.716\times 10^{14}$
600	$3.988\times 10^{15}$
700	$2.865\times 10^{16}$

Look how violent the growth is: from 300 K to 700 K, $n_i$ increases about $2.9\times 10^6$ times.

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2.5.4 Charge Neutrality Relationship

So far no formula knows how much dopant we added. Charge neutrality supplies the missing link.

Take a uniformly doped sample at equilibrium and look at any small piece far from the surface. It must be electrically neutral; otherwise there would be an electric field, hence carrier motion, hence current, contradicting equilibrium. The charged objects are electrons ($-q$), holes ($+q$), ionized donors $N_D^{+}$ ($+q$) and ionized acceptors $N_A^{-}$ ($-q$). Net charge per cm${}^3$:

$$qp-qn+q{N}_D^{+}-q{N}_A^{-}=0\text{\hspace{1em}(2.23)}$$ $$p-n+N_D^{+}-N_A^{-}=0\text{\hspace{1em}(2.24)}$$

At room temperature essentially all shallow dopants are ionized, so $N_D^{+}=N_D$ and $N_A^{-}=N_A$ (total donors and acceptors per cm${}^3$), giving the standard form

$$\boxed{p-n+N_D-N_A=0}\text{\hspace{1em}(2.25)}$$

2.5.5 Carrier Concentration Calculations

Assume nondegenerate and totally ionized. Then we have two equations and two unknowns ($n$, $p$), with $n_i$, $N_A$, $N_D$ known.

$$\begin{array}{rlrl}p& =\frac{n_i^2}{n}& & \text{from }np=n_i^2(2.26)\\ \frac{n_i^2}{n}-n+N_D-N_A& =0& & \text{insert in (2.25)}(2.27)\\ n^2-n(N_D-N_A)-n_i^2& =0& & \text{multiply by }n\text{ and rearrange}(2.28)\end{array}$$

This is a quadratic in $n$. By the quadratic formula (keeping the positive root since concentrations cannot be negative):

$$n=\frac{N_D-N_A}{2}+{\left[{\left(\frac{N_D-N_A}{2}\right)}^2+n_i^2\right]}^{1/2}\text{\hspace{1em}(2.29a)}$$

and, by the same steps with $n\leftrightarrow p$, $N_D\leftrightarrow N_A$,

$$p=\frac{N_A-N_D}{2}+{\left[{\left(\frac{N_A-N_D}{2}\right)}^2+n_i^2\right]}^{1/2}\text{\hspace{1em}(2.29b)}$$

Four special cases:

1. Intrinsic ($N_A=N_D=0$): $n=p=n_i$.
2. Doped, one species dominant ($N_D-N_A\approx N_D\gg n_i$, or $N_A-N_D\approx N_A\gg n_i$). This is the common practical case, since Si dopings are rarely below $10^{14}$ cm${}^{-3}$ while $n_i\approx 10^{10}$ cm${}^{-3}$. The square root becomes $\sqrt{(N_D/2{)}^2}=N_D/2$, so

$$n\approx N_D,p\approx \frac{n_i^2}{N_D}(N_D\gg N_A,N_D\gg n_i)\text{\hspace{1em}(2.30)}$$ $$p\approx N_A,n\approx \frac{n_i^2}{N_A}(N_A\gg N_D,N_A\gg n_i)\text{\hspace{1em}(2.31)}$$

Example. Si at 300 K with $N_D=10^{15}$ cm${}^{-3}$: $n\approx 10^{15}$ cm${}^{-3}$ and $p=n_i^2/n=10^{20}/10^{15}=10^5$ cm${}^{-3}$. The minority carriers are ten orders of magnitude fewer than the majority ones. 3. High temperature ($n_i\gg |N_D-N_A|$): the roots reduce to $n=p\approx n_i$. Every semiconductor becomes intrinsic when hot enough. 4. Compensated ($N_A\approx N_D$, both nonzero): donors and acceptors cancel each other, even producing intrinsic-like material when $N_D-N_A=0$. Both $N_A$ and $N_D$ must be kept in the formulas (common in as-grown GaAs).

Use the full Eqs. (2.29) only when $|N_D-N_A|\sim n_i$; otherwise Eqs. (2.30), (2.31) suffice.

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Exercise 2.5

Problem. A Si sample is doped with $10^{14}$ boron atoms/cm${}^3$. Find the carrier concentrations at (a) 300 K and (b) 470 K.

Solution (a), 300 K.

$$\begin{array}{rlrl}& \text{Boron is an acceptor, so }{N}_A=10^{14}{\text{cm}}^{-3}& & \text{Table 2.2}\\ & N_D\text{not mentioned}\Rightarrow N_D\ll N_A& & \text{inferred}\\ & n_i=1.00\times 10^{10}{\text{cm}}^{-3}\ll N_A& & \text{at 300 K, so (2.31) applies}\\ p& \approx N_A=10^{14}{\text{cm}}^{-3}& & \text{(2.31a)}\\ n& =\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{14}}=10^6{\text{cm}}^{-3}& & \text{(2.31b)}\end{array}$$

Solution (b), 470 K. From Fig. 2.20, $n_i\approx 10^{14}$ cm${}^{-3}$ at 470 K. Now $n_i\sim N_A$, so the full Eq. (2.29b) is needed:

$$\begin{array}{rlrl}p& =\frac{N_A}{2}+{\left[{\left(\frac{N_A}{2}\right)}^2+n_i^2\right]}^{1/2}& & N_D=0\\ & =0.5\times 10^{14}+{\left[(0.5\times 10^{14}{)}^2+(10^{14}{)}^2\right]}^{1/2}& & \text{insert numbers}\\ & =0.5\times 10^{14}+{\left[1.25\times 10^{28}\right]}^{1/2}& & \text{add under root}\\ & =0.5\times 10^{14}+1.118\times 10^{14}=1.62\times 10^{14}{\text{cm}}^{-3}& & \text{take root, add}\\ n& =\frac{n_i^2}{p}=\frac{10^{28}}{1.62\times 10^{14}}=6.18\times 10^{13}{\text{cm}}^{-3}& & \text{np product}\end{array}$$

Notice how heating has erased most of the doping advantage: electrons are now more than a third as numerous as holes.

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2.5.6 Determination of $E_F$

Eqs. (2.14), (2.16), (2.19) give a one-to-one link between $E_F$, $n$ and $p$: know any one and you can compute the other two.

(1) Exact position of $E_i$. In an intrinsic sample $n=p$. Using Eq. (2.16) with $E_F=E_i$:

$$\begin{array}{rlrl}{N}_C{e}^{(E_i-E_c)/kT}& =N_V{e}^{(E_v-E_i)/kT}& & n=p(2.33)\\ \frac{E_i-E_c}{kT}+\ln N_C& =\frac{E_v-E_i}{kT}+\ln N_V& & \text{take }\ln \\ 2E_i& =E_c+E_v+kT\ln \frac{N_V}{N_C}& & \text{collect }{E}_i\\ E_i& =\frac{E_c+E_v}{2}+\frac{kT}{2}\ln \frac{N_V}{N_C}& & (2.34)\end{array}$$

Since $N_V/N_C=(m_p^{\ast }/m_n^{\ast }{)}^{3/2}$ by Eq. (2.13):

$$\boxed{E_i=\frac{E_c+E_v}{2}+\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}}\text{\hspace{1em}(2.36)}$$

So $E_i$ is exactly at midgap only if $m_p^{\ast }=m_n^{\ast }$ or $T\to 0$. In Si at 300 K, $m_p^{\ast }<m_n^{\ast }$ so $E_i$ lies slightly below midgap (computed below, about $0.0073$ eV). That is usually small enough to be ignored when sketching.

(2) Doped semiconductors (nondegenerate, fully ionized). Solve Eqs. (2.19) for $E_F-E_i$:

$$\boxed{E_F-E_i=kT\ln \frac{n}{n_i}=-kT\ln \frac{p}{n_i}}\text{\hspace{1em}(2.37)}$$

Insert the appropriate concentrations. With $n\approx N_D$ (n-type) and $p\approx N_A$ (p-type):

$$E_F-E_i=kT\ln \frac{N_D}{n_i}(N_D\gg N_A,N_D\gg n_i)\text{\hspace{1em}(2.38a)}$$ $$E_i-E_F=kT\ln \frac{N_A}{n_i}(N_A\gg N_D,N_A\gg n_i)\text{\hspace{1em}(2.38b)}$$

More donors push $E_F$ up away from $E_i$; more acceptors push it down (Fig. 2.21). There are maximum nondegenerate dopings in Si at room temperature: about $N_D\approx 1.6\times 10^{18}$ cm${}^{-3}$ and $N_A\approx 9.1\times 10^{17}$ cm${}^{-3}$. Beyond these the material is degenerate, which is why “highly doped” ($n^{+}$, $p^{+}$) and “degenerate” are often used as synonyms.

Practical rule. If unsure, assume nondegenerate and compute $E_F$. If the answer falls inside the degenerate zone ($E_F$ within $3kT$ of an edge), redo the calculation with the full Fermi-Dirac machinery.

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Exercise 2.6

Problem. For each condition of Exercise 2.5, locate $E_i$, compute $E_F-E_i$, and draw a dimensioned band diagram. Use $E_G(\text{Si})=1.08$ eV and $m_p^{\ast }/m_n^{\ast }=0.71$ at 470 K.

Solution (a), 300 K ($N_A=10^{14}$ cm${}^{-3}$).

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}=\frac{3}{4}(0.0259)\ln \frac{0.81}{1.18}& & \text{(2.36), Table 2.1}\\ & =0.01943\times (-0.3761)=-0.0073\text{eV}& & \text{evaluate}\end{array}$$

So $E_i$ is $0.0073$ eV below midgap. Next, since $N_A\gg n_i$, use (2.38b):

$$\begin{array}{rlrl}{E}_i-E_F& =kT\ln \frac{N_A}{n_i}=0.0259\ln \frac{10^{14}}{10^{10}}& & \text{(2.38b)}\\ & =0.0259\times 9.210=0.239\text{eV}& & \ln 10^4=9.21\end{array}$$

($E_F$ is $0.239$ eV below $E_i$.) Dimensions for the diagram with $E_G=1.12$ eV: $E_c-E_i=0.56+0.0073=0.567$ eV; $E_i-E_v=0.553$ eV; $E_F-E_v=0.553-0.239=0.314$ eV.

Solution (b), 470 K. Here $kT=0.0405$ eV, $p=1.62\times 10^{14}$, $n_i=10^{14}$ cm${}^{-3}$.

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}(0.0405)\ln (0.71)=0.0304\times (-0.3425)=-0.0104\text{eV}& & \text{(2.36)}\end{array}$$

$E_i$ is $0.0104$ eV below midgap. Because $N_A\sim n_i$, use Eq. (2.37) with the exact $p$:

$$\begin{array}{rlrl}{E}_i-E_F& =kT\ln \frac{p}{n_i}=0.0405\ln \frac{1.62\times 10^{14}}{10^{14}}& & \text{(2.37)}\\ & =0.0405\times 0.4824=0.0195\text{eV}& & \ln 1.62=0.4824\end{array}$$

$E_F$ is only $0.0195$ eV below $E_i$: the hot sample is nearly intrinsic. Dimensions with $E_G=1.08$ eV: $E_c-E_i=0.54+0.0104=0.550$ eV; $E_i-E_v=0.530$ eV; $E_F-E_v=0.530-0.0195=0.511$ eV.

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2.5.7 Carrier Concentration Temperature Dependence

We have scattered facts about temperature already; now let us stitch them into one story. Consider a phosphorus-doped Si sample with $N_D=10^{15}$ cm${}^{-3}$ and follow its electron concentration $n$ as we heat it from nearly 0 K (Fig. 2.22).

Three regimes:

Regime	Roughly	What happens	$n$
Freeze-out	below about 100 K	Too little thermal energy to free the weakly bound fifth electrons; only a fraction of donors are ionized. Band-to-band excitation is utterly negligible.	$n=N_D^{+}$, rising from $0$ at $T=0$ K
Extrinsic (the normal operating range of most devices)	about 150 K to 450 K	Essentially all donors are ionized; thermally generated band-to-band electrons are still far fewer than $N_D$.	$n\approx N_D$, flat
Intrinsic	high $T$	Band-to-band excitation grows exponentially until it swamps the fixed donor supply.	$n\approx n_i$, rising steeply

Analogy. Imagine a reservoir ($N_D$ donors) with a small dam. At low temperature the dam barely leaks (freeze-out). At moderate temperature it spills fully and the water level sits at a plateau (extrinsic). At very high temperature a flood from the whole river system (band-to-band excitation) overwhelms the reservoir (intrinsic).

Two mechanisms feed the electron count: electrons donated by dopants, and electrons excited from the valence band (broken Si-Si bonds). At $T=0$ K both are zero. Warming first “defrosts” donor electrons; later the band-to-band mechanism takes over.

Why materials with wide gaps run hotter. The larger $E_G$, the larger the energy needed to excite across the gap, so the intrinsic regime begins at a higher temperature. Since the start of the intrinsic region is the top of the normal operating range, GaAs devices can operate hotter than Si devices, which can operate hotter than Ge devices. Taking the onset as $n_i=N_D=10^{15}$ cm${}^{-3}$ and reading Fig. 2.20, the maximum operating temperatures are approximately

$$T_{max}\approx 385\text{K (Ge)},540\text{K (Si)},>700\text{K (GaAs)}.$$

This is why GaAs and SiC ($E_G>2$ eV) are pursued for high-temperature electronics.

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2.6 Summary and Concluding Comments

We have described the carriers in a semiconductor at “rest” (equilibrium). What you now own:

• Two visualization models: the bonding model (great for seeing broken bonds, holes, dopants) and the energy band model (a kind of nonverbal sign language that encodes energies).
• A picture of the carriers: classical-like particles with charge $-q$ (electron) and $+q$ (hole) and effective masses $m_n^{\ast }$, $m_p^{\ast }$. In pure material they are equal in number and few; doping selectively raises one type.
• A chain of quantitative tools: densities of states, the Fermi function, the symmetric nondegenerate forms for $n$ and $p$, the $np$ product, charge neutrality, and the simplified room-temperature results.
• Qualitative feel: carriers cluster near band edges, $n_i$ rises steeply with $T$, and majority carriers sit at $N_D$ (or $N_A$) over the useful temperature range.
• A warning: do not “plug and chug.” Every formula has assumptions (equilibrium, nondegenerate, fully ionized dopants) and you should know their limits before using them.
• Reference numbers worth memorizing: $E_G(\text{Si})=1.12$ eV, $n_i(\text{Si})=10^{10}$ cm${}^{-3}$ at 300 K, $kT=0.0259$ eV at 300 K.

Table 2.4: Carrier Modeling Equation Summary

Density of states and Fermi function

$$g_c(E)=\frac{m_n^{\ast }\sqrt{2m_n^{\ast }(E-E_c)}}{{\pi }^2{\hslash }^3}(E\ge E_c),g_v(E)=\frac{m_p^{\ast }\sqrt{2m_p^{\ast }(E_v-E)}}{{\pi }^2{\hslash }^3}(E\le E_v),f(E)=\frac{1}{1+e^{(E-E_F)/kT}}$$

Carrier concentration relationships

$$\begin{array}{rlrlrl}n& =N_C\frac{2}{\sqrt{\pi }}{F}_{1/2}\left(\frac{E_F-E_c}{kT}\right),& p& =N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}\left(\frac{E_v-E_F}{kT}\right)& & \text{(any }{E}_F)\\ N_C& =2{\left(\frac{2\pi m_n^{\ast }kT}{h^2}\right)}^{3/2},& N_V& =2{\left(\frac{2\pi m_p^{\ast }kT}{h^2}\right)}^{3/2}& & \\ n& =N_C{e}^{(E_F-E_c)/kT},& p& =N_V{e}^{(E_v-E_F)/kT}& & \text{(nondegenerate)}\\ n& =n_i{e}^{(E_F-E_i)/kT},& p& =n_i{e}^{(E_i-E_F)/kT}& & \text{(nondegenerate)}\end{array}$$

$n_i$, $np$ product, and charge neutrality

$$n_i=\sqrt{N_C{N}_V}{e}^{-E_G/2kT},np=n_i^2,p-n+N_D-N_A=0$$

$n$, $p$, and Fermi level computational relationships

$$\begin{array}{rlrl}n& =\frac{N_D-N_A}{2}+{\left[{\left(\frac{N_D-N_A}{2}\right)}^2+n_i^2\right]}^{1/2},& p& =\frac{N_A-N_D}{2}+{\left[{\left(\frac{N_A-N_D}{2}\right)}^2+n_i^2\right]}^{1/2}\\ E_i& =\frac{E_c+E_v}{2}+\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }},& E_F-E_i& =kT\ln \frac{n}{n_i}=-kT\ln \frac{p}{n_i}\\ n& \approx N_D,p\approx \frac{n_i^2}{N_D}(N_D\gg N_A,N_D\gg n_i),& E_F-E_i& =kT\ln \frac{N_D}{n_i}\\ p& \approx N_A,n\approx \frac{n_i^2}{N_A}(N_A\gg N_D,N_A\gg n_i),& E_i-E_F& =kT\ln \frac{N_A}{n_i}\end{array}$$

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Problems: Chapter 2

Here are all 23 end-of-chapter problems, each with a full solution. Constants used throughout: $kT=0.025851$ eV at 300 K ($0.0259$ in rounded form), $n_i(\text{Si},300\text{ K})=10^{10}$ cm${}^{-3}$, $q=1.602\times 10^{-19}$ C.

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Problem 2.1: $E_G$ versus $T$

Band gap model:

$$E_G(T)=E_G(0)-\frac{\alpha T^2}{T+\beta }$$

with, for Si, $E_G(0)=1.170$ eV, $\alpha =4.730\times 10^{-4}$ eV/K, $\beta =636$ K.

(a) Plot $E_G$ versus $T$ for $0\le T\le 600$ K and note $E_G(300)$.

$$\begin{array}{rlrl}{E}_G(0)& =1.170& & \alpha T^2=0\\ E_G(100)& =1.170-\frac{(4.73\times 10^{-4})(10^4)}{736}=1.170-0.0064=1.1636\text{eV}& & \\ E_G(200)& =1.170-\frac{(4.73\times 10^{-4})(4\times 10^4)}{836}=1.170-0.0226=1.1474\text{eV}& & \\ E_G(300)& =1.170-\frac{(4.73\times 10^{-4})(9\times 10^4)}{936}=1.170-0.0455=1.1245\text{eV}& & \\ E_G(400)& =1.170-\frac{(4.73\times 10^{-4})(1.6\times 10^5)}{1036}=1.170-0.0731=1.0969\text{eV}& & \\ E_G(500)& =1.170-\frac{(4.73\times 10^{-4})(2.5\times 10^5)}{1136}=1.170-0.1041=1.0659\text{eV}& & \\ E_G(600)& =1.170-\frac{(4.73\times 10^{-4})(3.6\times 10^5)}{1236}=1.170-0.1378=1.0322\text{eV}& & \end{array}$$

The plot is a curve that is flat near $T=0$ (slope $\propto T^2$), then bends downward and becomes nearly linear above 300 K. At 300 K the gap is $1.12$ eV, the number everyone remembers.

(b) Compare with $E_G(T)=1.205-2.8\times 10^{-4}T$ for $T>300$ K.

$$\begin{array}{rl}T=300:& 1.205-0.0840=1.121\text{eV}(\Delta =-0.0035\text{ vs. exact})\\ T=400:& 1.205-0.1120=1.093\text{eV}(\Delta =-0.0039)\\ T=500:& 1.205-0.1400=1.065\text{eV}(\Delta =-0.0009)\\ T=600:& 1.205-0.1680=1.037\text{eV}(\Delta =+0.0048)\end{array}$$

Over $300\le T\le 600$ K the linear form is within about $5$ meV ($<0.5\%$) of the precise fit: an excellent simplification in that region (but it badly overshoots below 300 K where the true curve flattens).

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Problem 2.2: Bonding model visualizations

• (a) Missing atom: erase one circle (a Si core) together with its eight line-ends (its four shared pairs); the four neighbors are left with a dangling line-end each (Fig. 2.4a).
• (b) Electron: a broken bond: one line segment removed, with a small dot (the electron, charge $-q$) wandering through the lattice.
• (c) Hole: the same broken bond viewed as a gap in the line: an empty bond site (a “positive” charge $+q$) that moves when a neighboring bond electron hops into it.
• (d) Donor: a Column V atom (e.g., P) substituted for a Si circle; four of its five valence electrons complete the four bonds, and the fifth is drawn as an extra dot loosely attached to a $+$ core (${\text{P}}^{+}$), later freed to wander (Fig. 2.10a).
• (e) Acceptor: a Column III atom (e.g., B) substituted for a Si circle; only three bonds are complete, one bond is short an electron. A neighboring bond electron hops in, giving the core a $-$ charge (${\text{B}}^{-}$) and creating a wandering hole (Fig. 2.10b).

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Problem 2.3: Energy band model visualizations

Reference diagram: horizontal lines $E_c$ (top) and $E_v$ (bottom), energy vertical, position horizontal.

• (a) Electron: a filled dot just above $E_c$ in the conduction band.
• (b) Hole: an open circle just below $E_v$ in the valence band.
• (c) Donor sites: a row of short dashes at $E_D=E_c-|E_B|$, about $0.045$ eV (P in Si) below $E_c$.
• (d) Acceptor sites: a row of short dashes at $E_A=E_v+|E_B|$, about $0.045$ eV (B in Si) above $E_v$.
• (e) Donor freeze-out at $T\to 0$: all donor dashes carry dots (electrons), and the conduction band is empty. The electrons are frozen on their donors.
• (f) Acceptor freeze-out at $T\to 0$: all acceptor dashes are empty, the valence band is completely full, and no holes exist (the holes are “frozen” on acceptors).
• (g) Carrier energy distributions: many dots clustered near $E_c$ (electrons) and many open circles near $E_v$ (holes), thinning out rapidly with distance from the edges (Fig. 2.17).
• (h) Intrinsic: equal numbers of dots near $E_c$ and open circles near $E_v$; a dashed $E_i$ line near midgap (Fig. 2.18, left).
• (i) n-type: a solid $E_F$ line above $E_i$; many electrons, few holes.
• (j) p-type: a solid $E_F$ line below $E_i$; many holes, few electrons.
• (k) Nondegenerate: $E_F$ in the gap at least $3kT$ from both $E_c$ and $E_v$.
• (l) Degenerate: $E_F$ within $3kT$ of a band edge, or inside the conduction (n${}^{+}$) or valence (p${}^{+}$) band.

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Problem 2.4: Si-doped GaAs

(a) Remove a Ga atom and an As atom from the GaAs bonding diagram (Fig. P2.4). Each atom takes its bonding electrons with it: removing Ga takes its 3 valence electrons (3 of the four lines to its As neighbors), and removing As takes its 5 (one of its four neighbors’ lines each side). Draw the two circles erased with their line-ends; each of the neighbors then shows a dangling line-end.

(b) Insert a Si atom (4 valence electrons, 4 line-ends) in each vacancy, reconnecting to the four neighbors.

(c) Si on a Ga site.

$$\begin{array}{rlrl}\text{Ga valence}& =3,\text{Si valence}=4& & \text{Column III vs. IV}\\ \text{Excess electrons per substituted Si}& =4-3=+1& & \text{one spare electron}\end{array}$$

One electron does not fit in the bonds and is released: Si on Ga is a donor, so the GaAs is n-type.

(d) Si on an As site.

$$\begin{array}{rlrl}\text{As valence}& =5,\text{Si valence}=4& & \text{Column V vs. IV}\\ \text{Electrons per substituted Si}& =4-5=-1& & \text{one bond is an electron short}\end{array}$$

A bond is left incomplete, so a neighboring bond electron is captured and a hole results: Si on As is an acceptor, so the GaAs is p-type. Si in GaAs is therefore amphoteric.

(e) Band diagrams.

• (i) Si on Ga: draw a dashed donor level just below $E_c$ (filled with an electron at low $T$; ionized at room temperature).
• (ii) Si on As: draw a dashed acceptor level just above $E_v$ (empty at low $T$; filled by a valence-band electron at room temperature, leaving a hole).

In both cases $E_G=1.42$ eV.

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Problem 2.5: States per cm${}^3$ in $\gamma kT$

Problem. Find the number of conduction band states per cm${}^3$ between $E_c$ and $E_c+\gamma kT$, $\gamma$ an arbitrary constant.

Solution. Integrate the density of states over that interval:

$$\begin{array}{rlrl}N& ={\int }_{E_c}^{E_c+\gamma kT}{g}_c(E)dE& & \text{states in the range}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{E_c}^{E_c+\gamma kT}(E-E_c{)}^{1/2}dE& & \text{insert (2.6a)}\\ & =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}\cdot \frac{2}{3}(\gamma kT{)}^{3/2}& & \int u^{1/2}du=\frac{2}{3}{u}^{3/2}\\ & =\boxed{\frac{2\sqrt{2}(m_n^{\ast }\gamma kT{)}^{3/2}}{3{\pi }^2{\hslash }^3}}{\text{cm}}^{-3}& & \text{since }m\sqrt{m}=m^{3/2}\end{array}$$

(Use consistent units, then convert to cm${}^{-3}$.)

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Problem 2.6: Fermi function questions

(a) At $E=E_F$:

$$f(E_F)=\frac{1}{1+e^0}=\frac{1}{2}.$$

(b) With $E_F=E_c$, find the occupation probability at $E_c+kT$:

$$\begin{array}{rlrl}f(E_c+kT)& =\frac{1}{1+e^{(E_c+kT-E_c)/kT}}& & E_F=E_c\\ & =\frac{1}{1+e^1}=\frac{1}{3.718}=0.269& & e=2.718\end{array}$$

(c) Given: $P(\text{filled at }{E}_c-kT)=P(\text{empty at }{E}_v+kT)$. Where is $E_F$?

$$\begin{array}{rlrl}f(E_c-kT)& =1-f(E_v+kT)& & \text{given}\\ \frac{1}{1+e^{(E_c-kT-E_F)/kT}}& =\frac{1}{1+e^{(E_F-E_v-kT)/kT}}& & \text{use }1-f(E)=\frac{1}{1+e^{-(E-E_F)/kT}}\\ E_c-kT-E_F& =E_F-E_v-kT& & \text{equal exponents}\\ E_F& =\frac{E_c+E_v}{2}& & \text{the }kT\text{ cancels}\end{array}$$

$E_F$ is at midgap.

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Problem 2.7: Peak of the carrier distributions

Problem. For a nondegenerate semiconductor, show the electron distribution peaks at $E_c+kT/2$ and the hole distribution at $E_v-kT/2$.

Solution. Conduction band: let $x=E-E_c$. Nondegenerate means $f(E)\approx e^{-(E-E_F)/kT}$, so

$$\begin{array}{rlrl}{g}_c(E)f(E)& =C\sqrt{x}{e}^{-(E_c-E_F)/kT}{e}^{-x/kT}\propto \sqrt{x}{e}^{-x/kT}& & C\text{ constant}\\ \frac{d}{dx}\left(\sqrt{x}{e}^{-x/kT}\right)& =e^{-x/kT}\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{kT}\right)=0& & \text{peak: derivative zero}\\ \frac{1}{2\sqrt{x}}& =\frac{\sqrt{x}}{kT}& & \text{set bracket to zero}\\ x& =\frac{kT}{2}& & \text{solve}\\ \Rightarrow E_{peak}& =E_c+\frac{kT}{2}& & \end{array}$$

Valence band: let $y=E_v-E$. Then $1-f(E)\approx e^{(E-E_F)/kT}=e^{-(E_F-E_v)/kT}{e}^{-y/kT}$ and $g_v\propto \sqrt{y}$, so $g_v(1-f)\propto \sqrt{y}{e}^{-y/kT}$, whose peak is at $y=kT/2$ by the identical calculation: $E_{peak}=E_v-kT/2$. $■$

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Problem 2.8: Population at $E_c+5kT$

Problem. Relative to the peak, what is the electron population per unit energy at $E=E_c+5kT$?

Solution. The distribution is $h(x)\propto \sqrt{x}{e}^{-x/kT}$ with the peak at $x=kT/2$ (Problem 2.7).

$$\begin{array}{rlrl}\frac{h(5kT)}{h(kT/2)}& =\frac{\sqrt{5kT}{e}^{-5}}{\sqrt{kT/2}{e}^{-1/2}}& & \text{evaluate both}\\ & =\sqrt{10}{e}^{-4.5}& & \sqrt{5/0.5}=\sqrt{10}\\ & =(3.162)(0.01111)=0.0351& & e^{-4.5}=0.01111\end{array}$$

So the population at $E_c+5kT$ is only about 3.5% of the peak value. The carriers really do huddle close to the band edge.

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Problem 2.9: Distributions in Si at 300 K with $E_F=E_c-E_G/4$

Setup. (The problem statement reads $E_F$ a quarter of the gap below $E_c$; with $E_G=1.12$ eV.) Then

$$E_c-E_F=\frac{1.12}{4}=0.28\text{eV}=\frac{0.28}{0.025851}kT=10.8kT>3kT\Rightarrow \text{nondegenerate}.$$

Likewise $E_F-E_v=0.84$ eV $=32.5kT$.

Step 1: carrier concentrations.

$$\begin{array}{rlrl}{N}_C& =(2.510\times 10^{19})(1.18{)}^{3/2}=3.22\times 10^{19}{\text{cm}}^{-3}& & \text{Table 2.1}\\ N_V& =(2.510\times 10^{19})(0.81{)}^{3/2}=1.83\times 10^{19}{\text{cm}}^{-3}& & \\ n& =N_C{e}^{-(E_c-E_F)/kT}=3.22\times 10^{19}{e}^{-10.81}=3.22\times 10^{19}(2.02\times 10^{-5})=6.5\times 10^{14}{\text{cm}}^{-3}& & \\ p& =N_V{e}^{-(E_F-E_v)/kT}=1.83\times 10^{19}{e}^{-32.5}=1.5\times 10^5{\text{cm}}^{-3}(\text{check: }{n}_i^2/n=1.5\times 10^5)& & \end{array}$$

Step 2: normalized distribution (from Problem 2.10a, for $x=E-E_c$):

$$n(E)=n\cdot \frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{x}{e}^{-x/kT},\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}=\frac{2}{(1.7725)(0.004156)}=271.5{\text{eV}}^{-3/2}$$

so

$$n(E)=(6.5\times 10^{14})(271.5)\sqrt{x}{e}^{-x/kT}=1.76\times 10^{17}\sqrt{x}{e}^{-x/kT}{\text{cm}}^{-3}{\text{eV}}^{-1}(x\text{ in eV})$$

and for holes ($y=E_v-E$):

$$p(E)=(1.5\times 10^5)(271.5)\sqrt{y}{e}^{-y/kT}=4.1\times 10^7\sqrt{y}{e}^{-y/kT}{\text{cm}}^{-3}{\text{eV}}^{-1}.$$

Step 3: values for the plot (peak at $kT/2=0.0129$ eV from the band edge):

distance from edge (eV)	$n(E)$ (cm${}^{-3}$eV${}^{-1}$)	$p(E)$ (cm${}^{-3}$eV${}^{-1}$)
0	0	0
0.0129 (peak)	$1.2\times 10^{16}$	$2.8\times 10^6$
0.05	$5.7\times 10^{15}$	$1.3\times 10^6$
0.10	$1.2\times 10^{15}$	$2.7\times 10^5$
0.20	$3.4\times 10^{13}$	$8\times 10^3$

(Each entry is the coefficient times $\sqrt{x}{e}^{-x/kT}$; for instance at $0.05$ eV, $\sqrt{0.05}{e}^{-1.934}=0.2236\times 0.1445=0.0323$.) The electron curve is a skewed hump about $4\times 10^9$ times taller than the hole hump, because $E_F$ lies in the upper half of the gap.

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Problem 2.10: Temperature variation of the electron distribution

(a) Show $\frac{g_c(E)f(E)}{n}=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{E-E_c}{e}^{-(E-E_c)/kT}$.

$$\begin{array}{rlrl}{g}_{c}f& =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3}\sqrt{E-E_c}{e}^{-(E-E_F)/kT}& & \text{nondegenerate }f\\ n& =N_C{e}^{-(E_c-E_F)/kT}& & \text{(2.16a)}\\ \frac{g_{c}f}{n}& =\frac{m_n^{\ast }\sqrt{2m_n^{\ast }}}{{\pi }^2{\hslash }^3{N}_C}\sqrt{E-E_c}{e}^{-(E-E_c)/kT}& & \text{the }{E}_F\text{ factors cancel}\end{array}$$

Now use $N_C=\frac{2(m_n^{\ast }kT{)}^{3/2}}{(2\pi {)}^{3/2}{\hslash }^3}$:

$$\frac{m_n^{\ast 3/2}\sqrt{2}}{{\pi }^2{\hslash }^3}\cdot \frac{(2\pi {)}^{3/2}{\hslash }^3}{2(m_n^{\ast }kT{)}^{3/2}}=\frac{\sqrt{2}(2\pi {)}^{3/2}}{2{\pi }^2(kT{)}^{3/2}}=\frac{4{\pi }^{3/2}}{2{\pi }^2(kT{)}^{3/2}}=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}$$

which gives the stated result. $■$

(b) Plot for $T=300,600,1200$ K. Define $\varphi (x)=\frac{2}{\sqrt{\pi }(kT{)}^{3/2}}\sqrt{x}{e}^{-x/kT}$ in eV${}^{-1}$. The peak (at $x=kT/2$) is ${\varphi }_{max}=\frac{2}{\sqrt{2\pi }}\frac{e^{-1/2}}{kT}=\frac{0.484}{kT}$:

$T$ (K)	$kT$ (eV)	peak position (eV)	${\varphi }_{max}$ (eV${}^{-1}$)
300	0.02585	0.0129	18.7
600	0.05170	0.0259	9.36
1200	0.10340	0.0517	4.68

Selected values $\varphi (x)$ (eV${}^{-1}$):

$x=E-E_c$ (eV)	300 K	600 K	1200 K
0.05	8.77	8.16	4.68
0.10	1.79	4.39	4.08
0.20	0.053	0.90	2.19
0.40	$\approx 0$	0.027	0.45

Discussion. All three curves are normalized to the same area (1 eV${}^{-1}$ per electron). Heating makes the distribution lower and broader: the peak moves outward ($kT/2$) and shrinks ($\propto 1/kT$), and a long high-energy tail develops. At 300 K nearly every electron is within about $0.1$ eV of the edge; at 1200 K a visible fraction is $0.3$ to $0.4$ eV up. (Same total number $n$, spread over more energies.)

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Problem 2.11: Hole formulas (2.14b) and (2.16b)

Solution. Start from Eq. (2.8b) and mirror the electron derivation.

$$\begin{array}{rlrl}p& ={\int }_{-\infty }^{E_v}{g}_v(E)[1-f(E)]dE& & \text{(2.8b), bottom}\to -\infty \\ 1-f(E)& =\frac{1}{1+e^{(E_F-E)/kT}}& & \text{algebra: }1-\frac{1}{1+e^a}=\frac{1}{1+e^{-a}}\\ p& =\frac{m_p^{\ast }\sqrt{2m_p^{\ast }}}{{\pi }^2{\hslash }^3}{\int }_{-\infty }^{E_v}\frac{\sqrt{E_v-E}}{1+e^{(E_F-E)/kT}}dE& & \text{insert (2.6b)}\end{array}$$

Let ${\eta }^{\prime }=(E_v-E)/kT$ and ${\eta }_v=(E_v-E_F)/kT$. Then $E_F-E=(E_F-E_v)+(E_v-E)=kT({\eta }^{\prime }-{\eta }_v)$, $\sqrt{E_v-E}dE\to (kT{)}^{3/2}\sqrt{{\eta }^{\prime }}d{\eta }^{\prime }$ (sign absorbed by reversing the limits) and the lower limit $-\infty \to {\eta }^{\prime }=\infty$:

$$p=\frac{\sqrt{2}(m_p^{\ast }kT{)}^{3/2}}{{\pi }^2{\hslash }^3}{\int }_0^{\infty }\frac{\sqrt{{\eta }^{\prime }}d{\eta }^{\prime }}{1+e^{{\eta }^{\prime }-{\eta }_v}}=N_V\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\text{(2.14b)}$$

where $N_V=2(2\pi m_p^{\ast }kT/h^2{)}^{3/2}$ (the constants combine exactly as for $N_C$). If $E_F\ge E_v+3kT$, then ${\eta }_v\le -3$, the $+1$ is negligible in the denominator $1+e^{{\eta }^{\prime }-{\eta }_v}$ and

$$\frac{2}{\sqrt{\pi }}{F}_{1/2}({\eta }_v)\approx \frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}{\int }_0^{\infty }\sqrt{{\eta }^{\prime }}{e}^{-{\eta }^{\prime }}d{\eta }^{\prime }=\frac{2}{\sqrt{\pi }}{e}^{{\eta }_v}\frac{\sqrt{\pi }}{2}=e^{(E_v-E_F)/kT}$$

hence

$$p=N_V{e}^{(E_v-E_F)/kT}\text{(2.16b)}.■$$

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Problem 2.12: Hypothetical constant density of states

Given $g_c(E)=N_C/kT=\text{const}$ for $E\ge E_c$.

(a) With $E_F\le E_c-3kT$, $f\approx e^{-(E-E_F)/kT}$, so

$$g_c(E)f(E)=\frac{N_C}{kT}{e}^{-(E-E_F)/kT}=\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}{e}^{-(E-E_c)/kT}.$$

The sketch is a pure exponential decay starting at its maximum at $E=E_c$ (no square-root rise and no peak away from the edge), falling by $1/e$ every $kT$.

(b) Electron concentration:

$$\begin{array}{rlrl}n& ={\int }_{E_c}^{\infty }\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}{e}^{-(E-E_c)/kT}dE& & \text{integrate the distribution}\\ & =\frac{N_C}{kT}{e}^{(E_F-E_c)/kT}[-kT{e}^{-(E-E_c)/kT}{]}_{E_c}^{\infty }& & \text{integral of exponential}\\ & =N_C{e}^{(E_F-E_c)/kT}& & \text{bracket}=kT\end{array}$$

This reproduces Eq. (2.16a) exactly (the constant was chosen to do so). Without the approximation on $f$ the integral is elementary too:

$$n=\frac{N_C}{kT}{\int }_{E_c}^{\infty }\frac{dE}{1+e^{(E-E_F)/kT}}=N_C\ln (1+e^{(E_F-E_c)/kT})$$

(which tends to $N_C{e}^{(E_F-E_c)/kT}$ when $E_F\ll E_c$, and to $N_C(E_F-E_c)/kT$ for $E_F\gg E_c$).

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Problem 2.13: $N_C$, $N_V$

(a) Verify $N_{C,V}=(2.510\times 10^{19}{\text{cm}}^{-3})(m^{\ast }/m_0{)}^{3/2}$ at 300 K.

$$\begin{array}{rlrl}kT& =(8.617\times 10^{-5})(1.602\times 10^{-19})(300)=4.141\times 10^{-21}\text{J}& & \text{in joules}\\ 2\pi m_{0}kT& =(6.2832)(9.109\times 10^{-31})(4.141\times 10^{-21})=2.370\times 10^{-50}\text{kg J}& & \text{numerator}\\ h^2& =(6.625\times 10^{-34}{)}^2=4.389\times 10^{-67}{\text{J}}^2{\text{s}}^2& & \text{denominator}\\ \frac{2\pi m_{0}kT}{h^2}& =5.40\times 10^{16}{\text{m}}^{-2}& & \text{divide}\\ {\left(5.40\times 10^{16}\right)}^{3/2}& =1.255\times 10^{25}{\text{m}}^{-3}& & \text{power }3/2\\ N& =2\times 1.255\times 10^{25}=2.51\times 10^{25}{\text{m}}^{-3}=2.51\times 10^{19}\mathbf{c}{\mathbf{m}}^{-3}& & \text{factor 2, and }10^6{\text{ m}}^{-3}{\text{ per cm}}^{-3}\end{array}$$

For general $m^{\ast }$, the quantity $(m^{\ast }{)}^{3/2}$ scales out as $(m^{\ast }/m_0{)}^{3/2}$. $■$

(b) Table at 300 K ($N_C$ uses $m_n^{\ast }$, $N_V$ uses $m_p^{\ast }$):

Material	$m_n^{\ast }/m_0$	$(m_n^{\ast }/m_0{)}^{3/2}$	$N_C$ (cm${}^{-3}$)	$m_p^{\ast }/m_0$	$(m_p^{\ast }/m_0{)}^{3/2}$	$N_V$ (cm${}^{-3}$)
Si	1.18	1.282	$3.22\times 10^{19}$	0.81	0.729	$1.83\times 10^{19}$
Ge	0.55	0.408	$1.02\times 10^{19}$	0.36	0.216	$5.42\times 10^{18}$
GaAs	0.066	0.01696	$4.26\times 10^{17}$	0.52	0.375	$9.41\times 10^{18}$

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Problem 2.14: $n_i$ comparisons

(a) At what temperature does $n_i$ of (i) Si and (ii) GaAs equal $n_i(\text{Ge},300\text{ K})\approx 2\times 10^{13}$ cm${}^{-3}$?

(i) Si. Use the Sproul-Green fit and solve numerically:

$$2\times 10^{13}=9.15\times 10^{19}{\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}\Rightarrow {\left(\frac{T}{300}\right)}^2{e}^{-0.5928/kT}=2.19\times 10^{-7}$$

$T$ (K)	$(T/300{)}^2{e}^{-0.5928/kT}$
400	$6.0\times 10^{-8}$
428	$2.13\times 10^{-7}$
430	$2.31\times 10^{-7}$

So $T\approx 429$ K ($\approx 156{}^{\circ }$C). Cross-check against the table of Exercise 2.4: $n_i=5.4\times 10^{12}$ at 400 K and $4.7\times 10^{13}$ at 450 K, bracketing $2\times 10^{13}$.

(ii) GaAs. Scale from $n_i(300)=2.25\times 10^6$ using $n_i\propto T^{3/2}{e}^{-E_G/2kT}$. The required ratio is $2\times 10^{13}/2.25\times 10^6=8.9\times 10^6$, i.e. $\ln =16.0$. Including the temperature dependence of the GaAs gap, $E_G(T)=1.519-\frac{5.405\times 10^{-4}{T}^2}{T+204}$ eV:

$$\ln \frac{n_i(T)}{n_i(300)}=\frac{3}{2}\ln \frac{T}{300}+\frac{E_G(300)}{2k\cdot 300}-\frac{E_G(T)}{2kT}$$

with $E_G(300)=1.4225$ eV so the second term is $27.51$. At $T=550$ K: $0.909+27.51-13.74=14.69$. At $T=600$ K: $1.040+27.51-12.35=16.20$. Interpolating gives $T\approx 595$ K ($\approx 320{}^{\circ }$C). (If the gap were held fixed at $1.42$ eV the answer would be about $650$ K; the shrinking gap makes GaAs go intrinsic a bit sooner.)

(b) $E_{GA}=1$ eV, $E_{GB}=2$ eV, same masses (so same $N_C{N}_V$), at 300 K:

$$\begin{array}{rlrl}\frac{n_{iA}}{n_{iB}}& =\frac{\sqrt{N_C{N}_V}{e}^{-E_{GA}/2kT}}{\sqrt{N_C{N}_V}{e}^{-E_{GB}/2kT}}=e^{(E_{GB}-E_{GA})/2kT}& & \text{prefactors cancel}\\ & =e^{1/(2\times 0.025851)}=e^{19.34}\approx 2.5\times 10^8& & \text{insert numbers}\end{array}$$

One extra eV of gap costs a factor of a quarter of a billion in carriers.

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Problem 2.15: $n_i(T)$ for Ge

Problem. Confirm Fig. 2.20 for Ge using $n_i=(1.76\times 10^{16})T^{3/2}{e}^{-0.392/kT}$.

Solution. Evaluate at several temperatures. Example at 300 K:

$$\begin{array}{rlrl}{T}^{3/2}& =300^{1.5}=5196& & \text{power}\\ \frac{0.392}{kT}& =\frac{0.392}{0.025851}=15.16& & \text{exponent}\\ e^{-15.16}& =2.60\times 10^{-7}& & \text{exponential}\\ n_i& =(1.76\times 10^{16})(5196)(2.60\times 10^{-7})=2.4\times 10^{13}{\text{cm}}^{-3}& & \text{product}\end{array}$$

in agreement with the Ge value $\approx 2\times 10^{13}$. Results by the same steps:

$T$ (K)	200	300	400	500	600	700
$n_i$ (cm${}^{-3}$)	$6.6\times 10^9$	$2.4\times 10^{13}$	$1.6\times 10^{15}$	$2.2\times 10^{16}$	$1.3\times 10^{17}$	$4.9\times 10^{17}$

Plotted on a log scale versus $T$ (or $1/T$) these points trace the Ge curve of Fig. 2.20. (The exponent $0.392$ eV is $E_G/2$ with $E_G\approx 0.78$ eV extrapolated to 0 K, and the prefactor absorbs the $T^{3/2}$ of $\sqrt{N_C{N}_V}$.)

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Problem 2.16: Concentration questions with a twist

(a) $p$-type Si, with the given $N_A$, $T\approx 0$ K. At absolute zero there is no thermal energy: the acceptors are un-ionized (every acceptor site empty of the extra electron) and no electron crosses the gap. This is the freeze-out regime:

$$p=0,n=0.$$

(Total ionization, $p=N_A$, applies only near room temperature.)

(b) Given $N>n_i$, full ionization, $n=N$ and $p=n_i^2/N$. Electrons are the majority carrier, with $n$ equal to the impurity concentration, so the impurity is a donor (an acceptor would give $p=N$ instead).

(c) $n=10^5$ cm${}^{-3}$ at 300 K in equilibrium:

$$p=\frac{n_i^2}{n}=\frac{(10^{10}{)}^2}{10^5}=10^{15}{\text{cm}}^{-3}.$$

(The sample is strongly p-type.)

(d) $E_F-E_i=+0.259$ eV in Si at 300 K:

$$\begin{array}{rlrl}\frac{E_F-E_i}{kT}& =\frac{0.259}{0.0259}=10& & \text{exactly 10 }kT\\ n& =n_i{e}^{10}=10^{10}(2.203\times 10^4)=2.2\times 10^{14}{\text{cm}}^{-3}& & \text{(2.19a)}\\ p& =n_i{e}^{-10}=10^{10}(4.54\times 10^{-5})=4.5\times 10^5{\text{cm}}^{-3}& & \text{(2.19b)}\end{array}$$

(e) Ge near room temperature, nondegenerate, $n_i=10^{13}$ cm${}^{-3}$ (as given), $n=2p$, $N_A=0$:

$$\begin{array}{rlrl}np& =n_i^2& & \text{np product}\\ 2p^2& =10^{26}\Rightarrow p=\sqrt{5\times 10^{25}}=7.07\times 10^{12}{\text{cm}}^{-3}& & \text{insert }n=2p\\ n& =2p=1.41\times 10^{13}{\text{cm}}^{-3}& & \\ p-n+N_D-N_A& =0\Rightarrow N_D=n-p+N_A& & \text{charge neutrality}\\ N_D& =1.41\times 10^{13}-7.07\times 10^{12}+0=7.07\times 10^{12}{\text{cm}}^{-3}& & \end{array}$$

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Problem 2.17: Carrier concentrations in Si

For all parts: nondegenerate, totally ionized dopants.

(a) $T=300$ K, $N_A=0$, $N_D=10^{15}$ cm${}^{-3}$. Since $N_D\gg n_i$:

$$n\approx N_D=10^{15}{\text{cm}}^{-3},p=\frac{n_i^2}{n}=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}.$$

(b) $T=300$ K, $N_A=10^{16}$, $N_D=0$:

$$p\approx N_A=10^{16}{\text{cm}}^{-3},n=\frac{10^{20}}{10^{16}}=10^4{\text{cm}}^{-3}.$$

(c) $T=300$ K, $N_A=9\times 10^{15}$, $N_D=10^{16}$ (compensated). Net doping:

$$N_D-N_A=10^{16}-9\times 10^{15}=10^{15}{\text{cm}}^{-3}\gg n_i$$ $$n\approx N_D-N_A=10^{15}{\text{cm}}^{-3},p=\frac{10^{20}}{10^{15}}=10^5{\text{cm}}^{-3}.$$

(d) $T=450$ K, $N_A=0$, $N_D=10^{14}$. First get $n_i(450\text{ K})$ from the Exercise 2.4 formula:

$$\begin{array}{rl}{E}_G& =1.17-\frac{4.73\times 10^{-4}(450{)}^2}{450+636}=1.0818\text{eV}\\ \frac{m_n^{\ast }}{m_0}& =1.028+0.2750-0.0626=1.2404,\frac{m_p^{\ast }}{m_0}=0.610+0.3524-0.0903=0.8720\\ (1.2404\times 0.8720{)}^{3/4}& =1.0606,(450/300{)}^{3/2}=1.837\\ kT& =0.038777\text{eV},\frac{E_G-0.0074}{2kT}=\frac{1.0744}{0.077553}=13.854,e^{-13.854}=9.63\times 10^{-7}\\ n_i& =(2.510\times 10^{19})(1.0606)(1.837)(9.63\times 10^{-7})\approx 4.7\times 10^{13}{\text{cm}}^{-3}\end{array}$$

Now $n_i$ is comparable to $N_D$, so use Eq. (2.29a):

$$\begin{array}{rl}n& =\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}=5\times 10^{13}+\sqrt{2.5\times 10^{27}+2.22\times 10^{27}}\\ & =5\times 10^{13}+6.87\times 10^{13}=1.19\times 10^{14}{\text{cm}}^{-3}\\ p& =\frac{n_i^2}{n}=\frac{2.22\times 10^{27}}{1.19\times 10^{14}}=1.9\times 10^{13}{\text{cm}}^{-3}\end{array}$$

(Check: $n-p=1.0\times 10^{14}=N_D$.)

(e) $T=650$ K, $N_A=0$, $N_D=10^{14}$. Again compute $n_i$:

$$\begin{array}{rl}{E}_G& =1.17-\frac{4.73\times 10^{-4}(650{)}^2}{1286}=1.0146\text{eV}\\ \frac{m_n^{\ast }}{m_0}& =1.2946,\frac{m_p^{\ast }}{m_0}=0.9305,(1.2946\times 0.9305{)}^{3/4}=1.150,(650/300{)}^{3/2}=3.189\\ kT& =0.056011\text{eV},\frac{1.0146-0.0074}{2kT}=8.991,e^{-8.991}=1.245\times 10^{-4}\\ n_i& =(2.510\times 10^{19})(1.150)(3.189)(1.245\times 10^{-4})\approx 1.15\times 10^{16}{\text{cm}}^{-3}\end{array}$$

Now $n_i\gg N_D$, so the sample is essentially intrinsic:

$$n=\frac{N_D}{2}+\sqrt{{\left(\frac{N_D}{2}\right)}^2+n_i^2}\approx 5\times 10^{13}+1.146\times 10^{16}=1.15\times 10^{16}{\text{cm}}^{-3},p=n-N_D=1.14\times 10^{16}{\text{cm}}^{-3}.$$

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Problem 2.18: Position of $E_i$ and $E_F$ for Problem 2.17

Use $E_i-\frac{E_c+E_v}{2}=\frac{3}{4}kT\ln (m_p^{\ast }/m_n^{\ast })$ and Eq. (2.37)/(2.38).

(a) 300 K, $N_D=10^{15}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.0259)\ln \frac{0.81}{1.18}=-0.0073\text{eV}\\ E_F-E_i& =kT\ln \frac{N_D}{n_i}=0.0259\ln 10^5=0.0259(11.513)=0.298\text{eV}\end{array}$$

Diagram ($E_G=1.12$): $E_c-E_i=0.567$, $E_c-E_F=0.269$, $E_F-E_v=0.851$ eV, $E_i$ at $0.0073$ eV below midgap.

(b) 300 K, $N_A=10^{16}$: $E_i$ again $0.0073$ eV below midgap;

$$E_i-E_F=kT\ln \frac{N_A}{n_i}=0.0259\ln 10^6=0.0259(13.816)=0.358\text{eV}$$

Diagram: $E_i-E_v=0.553$, so $E_F-E_v=0.553-0.358=0.195$ eV above $E_v$.

(c) 300 K, net $n=10^{15}$: same as (a): $E_F-E_i=0.298$ eV (the carrier numbers, not the individual dopings, set $E_F$). $E_i$ is $0.0073$ eV below midgap.

(d) 450 K, $E_G=1.08$ eV, $kT=0.038777$ eV, $n=1.19\times 10^{14}$, $n_i=4.7\times 10^{13}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.038777)\ln \frac{0.8720}{1.2404}=0.02908(-0.3523)=-0.0102\text{eV}\\ E_F-E_i& =kT\ln \frac{n}{n_i}=0.038777\ln \frac{1.19\times 10^{14}}{4.7\times 10^{13}}=0.038777(0.929)=0.036\text{eV}\end{array}$$

Diagram: $E_c-E_i=0.540+0.0102=0.550$, $E_c-E_F=0.514$, $E_F-E_v=0.566$ eV.

(e) 650 K, $E_G=1.015$ eV, $kT=0.056011$ eV, $n=1.15\times 10^{16}$, $n_i=1.146\times 10^{16}$:

$$\begin{array}{rl}{E}_i-E_{mid}& =\frac{3}{4}(0.056011)\ln \frac{0.9305}{1.2946}=0.04201(-0.3303)=-0.0139\text{eV}\\ E_F-E_i& =kT\ln \frac{n}{n_i}=0.056011\ln (1.0044)\approx 0.0002\text{eV}\end{array}$$

So $E_F\approx E_i$ (intrinsic behaviour). Diagram: $E_c-E_i=0.5075+0.0139=0.521$, $E_i-E_v=0.494$ eV, and the $E_F$ and $E_i$ lines coincide.

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Problem 2.19: MATLAB program for $n$, $p$, $E_F-E_i$

(a) Program (nondegenerate Si, complete ionization). It reuses the $n_i(T)$ computation of Exercise 2.4(a).

% Carrier concentrations in Si (nondegenerate, fully ionized dopants)
k  = 8.617e-5;                       % eV/K
A  = 2.510e19;  Eex = 0.0074;
T  = input('Temperature T (K) = ');
ND = input('Donor concentration ND (cm^-3) = ');
NA = input('Acceptor concentration NA (cm^-3) = ');
 
EG  = 1.17 - 4.730e-4*T^2/(T+636);
mnr = 1.028 + 6.11e-4*T - 3.09e-7*T^2;
mpr = 0.610 + 7.83e-4*T - 4.46e-7*T^2;
ni  = A*(mnr*mpr)^0.75*(T/300)^1.5*exp(-(EG-Eex)/(2*k*T));
 
dN = (ND-NA)/2;
n  = dN + sqrt(dN^2 + ni^2);         % Eq. (2.29a)
p  = ni^2/n;                          % np product
EFEi = k*T*log(n/ni);                % Eq. (2.37)
 
fprintf('ni = %.3e  n = %.3e  p = %.3e  EF-Ei = %.4f eV\n', ni, n, p, EFEi);

(b) Running it reproduces the answers computed above: for example,

Case	$T$	$N_D$	$N_A$	$n$ (cm${}^{-3}$)	$p$ (cm${}^{-3}$)	$E_F-E_i$ (eV)
2.17(a)	300	$10^{15}$	0	$1.0\times 10^{15}$	$1.0\times 10^5$	0.298
2.17(b)	300	0	$10^{16}$	$1.0\times 10^4$	$1.0\times 10^{16}$	$-0.358$
2.17(c)	300	$10^{16}$	$9\times 10^{15}$	$1.0\times 10^{15}$	$1.0\times 10^5$	0.298
2.17(d)	450	$10^{14}$	0	$1.19\times 10^{14}$	$1.9\times 10^{13}$	0.036
2.17(e)	650	$10^{14}$	0	$1.15\times 10^{16}$	$1.14\times 10^{16}$	0.0002

Subtle point: the program’s $E_F-E_i$ is relative to $E_i$ itself, so to compare with the band diagrams you add the small shift of Eq. (2.36).

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Problem 2.20: Maximum nondegenerate doping in Si at 300 K

Problem. Verify $N_D\approx 1.6\times 10^{18}$ and $N_A\approx 9.1\times 10^{17}$ cm${}^{-3}$.

Solution. The nondegenerate limit is $E_F=E_c-3kT$ (donors) or $E_F=E_v+3kT$ (acceptors). At that point Eq. (2.16) with $e^{-3}=0.0498$ gives

$$\begin{array}{rlrl}{n}_{max}& =N_C{e}^{-3}=(3.22\times 10^{19})(0.0498)=1.60\times 10^{18}{\text{cm}}^{-3}& & \text{(2.16a)}\\ p_{max}& =N_V{e}^{-3}=(1.83\times 10^{19})(0.0498)=9.1\times 10^{17}{\text{cm}}^{-3}& & \text{(2.16b)}\end{array}$$

and with total ionization $N_D=n_{max}$, $N_A=p_{max}$ (the minority carriers are negligible). These agree with the text values. $■$

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Problem 2.21: Plot $E_F-E_i$ versus doping

Solution. Use Eq. (2.29) for the exact $n$ (or $p$) and Eq. (2.37), so the plot is correct even near $N\sim n_i$:

k = 8.617e-5; T = 300; kT = k*T; ni = 1e10;
N  = logspace(10,19,200);            % doping (cm^-3)
nD = N/2 + sqrt((N/2).^2 + ni^2);    % donor-doped, ND=N
EF_donor    = kT*log(nD/ni);         % EF-Ei  (>0)
EF_acceptor = -kT*log(nD/ni);        % Ei-EF = kT ln(p/ni), p = nD by symmetry
semilogx(N, EF_donor, N, EF_acceptor); grid
xlabel('N_A or N_D (cm^{-3})'); ylabel('E_F - E_i (eV)');

Selected values from Eq. (2.38) ($kT=0.0259$ eV):

$N_D$ or $N_A$ (cm${}^{-3}$)	$10^{12}$	$10^{14}$	$10^{15}$	$10^{16}$	$10^{17}$	$10^{18}$
$|E_F-E_i|$ (eV)	0.119	0.239	0.298	0.358	0.417	0.477

The curves rise logarithmically (a straight line per decade, each decade adding $kT\ln 10=0.0596$ eV) and flatten to $0$ near $N\sim n_i$. Accuracy check against Fig. 2.21: at the nondegenerate limit $N_D=1.6\times 10^{18}$, $E_F-E_i=0.0259\ln (1.6\times 10^8)=0.489$ eV, while $E_c-E_i=0.567$ eV, leaving $E_c-E_F=0.078\approx 3kT=0.0777$ eV, so the plotted solid line meets the $3kT$ boundary exactly where it should.

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Problem 2.22: GaAs considerations

(a) For GaAs $m_n^{\ast }=0.066m_0\ll m_p^{\ast }=0.52m_0$. Since $g\propto m^{\ast 3/2}$, at equal distances from the edges

$$\frac{g_v}{g_c}={\left(\frac{0.52}{0.066}\right)}^{3/2}=(7.88{)}^{3/2}\approx 22.$$

The sketch of Fig. 2.14 for GaAs therefore has the valence band curve (going downward) about 22 times steeper than the conduction band curve: the conduction band “stadium” has far fewer seats.

(b) $E_F$ for intrinsic GaAs must make $n=p$. With far fewer states in the conduction band, the Fermi level has to rise toward $E_c$ to fill the same number of electrons as the holes in the larger valence band. Hence $E_i$ lies above midgap (consistent with Eq. 2.36 since $m_p^{\ast }>m_n^{\ast }$ makes the log term positive).

(c) Position of $E_i$ at 300 K:

$$\begin{array}{rlrl}{E}_i-\frac{E_c+E_v}{2}& =\frac{3}{4}kT\ln \frac{m_p^{\ast }}{m_n^{\ast }}& & \text{(2.36)}\\ & =\frac{3}{4}(0.025851)\ln (7.88)=0.01939\times 2.064& & \text{insert}\\ & =0.040\text{eV}& & \end{array}$$

$E_i$ is about $0.040$ eV above midgap (roughly $1.5kT$).

(d) Maximum nondegenerate doping at 300 K:

$$\begin{array}{rl}{N}_C& =(2.510\times 10^{19})(0.066{)}^{3/2}=4.26\times 10^{17}{\text{cm}}^{-3},N_V=(2.510\times 10^{19})(0.52{)}^{3/2}=9.41\times 10^{18}{\text{cm}}^{-3}\\ N_{D,max}& =N_C{e}^{-3}=(4.26\times 10^{17})(0.0498)=2.1\times 10^{16}{\text{cm}}^{-3}\\ N_{A,max}& =N_V{e}^{-3}=(9.41\times 10^{18})(0.0498)=4.7\times 10^{17}{\text{cm}}^{-3}\end{array}$$

So n-type GaAs becomes degenerate at a much lower doping than n-type Si: the light electron mass means few conduction band states.

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Problem 2.23: $E_F$ versus temperature, $N_A=10^{14}$ cm${}^{-3}$ Si

(a) Qualitative argument as $T\to 0$. At very low $T$ the sample is in freeze-out: valence band full, acceptor levels $E_A=E_v+|E_B|$ empty, conduction band empty. The only carriers possible are holes created by moving a valence electron onto an acceptor (so that $p=N_A^{-}$), and those few holes obey $p=N_v{e}^{(E_v-E_F)/kT}$ while the acceptor ionization requires $E_F$ to lie near $E_A$. Charge neutrality $p=N_A^{-}$ with $N_A^{-}\ll N_A$ forces $E_F$ to lie between $E_v$ and $E_A$; in the limit $T\to 0$ it sits midway,

$$E_F\to \frac{E_v+E_A}{2}=E_v+\frac{|E_B|}{2}\approx E_v+0.02\text{eV}(\text{B in Si}).$$

(b) Program. In the range $200\le T\le 500$ K freeze-out is irrelevant (dopants fully ionized), so use $n_i(T)$ from Exercise 2.4 and Eq. (2.29b):

k=8.617e-5; A=2.510e19; Eex=0.0074; NA=1e14;
T  = 200:5:500;
EG = 1.17 - 4.730e-4*T.^2./(T+636);
mn = 1.028 + 6.11e-4*T - 3.09e-7*T.^2;
mp = 0.610 + 7.83e-4*T - 4.46e-7*T.^2;
ni = A*(mn.*mp).^0.75.*(T/300).^1.5.*exp(-(EG-Eex)./(2*k*T));
p  = NA/2 + sqrt((NA/2)^2 + ni.^2);       % Eq. (2.29b), ND = 0
EiEF = k*T.*log(p./ni);                   % Ei - EF, Eq. (2.37)
plot(T, EiEF); grid; xlabel('T (K)'); ylabel('E_i - E_F (eV)');

Typical output:

$T$ (K)	$n_i$ (cm${}^{-3}$)	$p$ (cm${}^{-3}$)	$E_i-E_F$ (eV)
200	$5.2\times 10^4$	$1.0\times 10^{14}$	0.368
300	$1.0\times 10^{10}$	$1.0\times 10^{14}$	0.239
400	$5.4\times 10^{12}$	$1.003\times 10^{14}$	0.100
500	$2.7\times 10^{14}$	$3.3\times 10^{14}$	0.008

(c) Conclusion. For fixed $N_A$, $E_F$ starts near the valence band edge at very low $T$, rises through the gap as $T$ increases, and approaches $E_i$ (about midgap) from below at high temperature, where the sample turns intrinsic. In the extrinsic range the $kT\ln (N_A/n_i)$ shrinks because $n_i$ increases faster than $kT$.

(d) Increasing $N_A$ in decades from $10^{14}$ to $10^{18}$ cm${}^{-3}$. At each temperature $E_i-E_F$ grows by $kT\ln 10$ per decade (0.0596 eV at 300 K, so $0.239\to 0.298\to 0.358\to 0.417\to 0.477$ eV at 300 K), pushing $E_F$ closer to $E_v$. The temperature at which the sample goes intrinsic ($n_i\sim N_A$) moves higher, widening the extrinsic range; at $N_A=10^{18}$ cm${}^{-3}$ the level reaches $E_F-E_v\approx 0.076$ eV $\approx 2.9kT$, so the sample sits at the edge of degeneracy (the limit $9.1\times 10^{17}$ cm${}^{-3}$ of Problem 2.20), and at low $T$ the freeze-out and dopant-band effects become important.

(e) Donors instead of acceptors. The picture is the mirror image: $E_F$ starts just below $E_c$ (midway between $E_D$ and $E_c$ as $T\to 0$), moves down toward $E_i$ from above as $T$ rises, sits $kT\ln (N_D/n_i)$ above $E_i$ in the extrinsic range, and moves up by $kT\ln 10$ per decade of $N_D$, with degeneracy near $N_D\approx 1.6\times 10^{18}$ cm${}^{-3}$. Only small asymmetries (the $\frac{3}{4}kT\ln (m_p^{\ast }/m_n^{\ast })$ offset of $E_i$) distinguish the two cases.

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Closing thought. You began this chapter with a block of silicon that looked like an impossible crowd of $5\times 10^{22}$ atoms per cm${}^3$ and ended with a handful of lines, $np=n_i^2$, charge neutrality, and $E_F-E_i=kT\ln (n/n_i)$, that tell you exactly who carries current and how to change it. Everything in the chapters ahead, from drift and diffusion to p-n junctions, starts with these numbers. Well done.