Physics for Information Science. Module 3, intrinsic semiconductors to the p-n junction.

Page 1 of 18

Module 3 at a glance

Module 1 classified materials by their band gap: conductors (bands overlap), insulators (large gap) and, in between, semiconductors. This module studies this middle class in detail, and also the device on which all modern electronics is built, the p-n junction.

The story has two halves.

1. Counting carriers. How many free electrons and holes does a semiconductor have at a given temperature? We derive the electron density in the conduction band and the hole density in the valence band, combine them to get the intrinsic carrier concentration, and see what doping and temperature do to the number of carriers and to the Fermi level.
2. Joining p and n. When a p-type region and an n-type region are joined, a depletion region and a potential barrier are created. Biasing changes this barrier, and the result is the diode equation and the V-I characteristic of the diode.

Everything in the second half depends on the first half. The diode equation is really a statement about how the number of electrons on each side of a barrier depends on the height of that barrier, and the carrier-density formulas tell us exactly that.

KTU syllabus. Intrinsic semiconductor, derivation of density of electrons in conduction band and density of holes in valence band, intrinsic carrier concentration, variation of intrinsic carrier concentration with temperature, extrinsic semiconductor (qualitative). Formation of p-n junction, Fermi level in semiconductors (intrinsic and extrinsic), energy band diagram of p-n junction, qualitative description of charge flow across a p-n junction, forward and reverse biased p-n junctions, diode equation (derivation), I-V characteristics of p-n junction.

What you will be able to do

• Explain how holes are formed and why they behave as positive carriers.
• Explain what effective mass and density of states mean, and why both are needed to count carriers.
• Derive $n=N_c\,e^{-(E_c-E_F)/kT}$ and $p=N_v\,e^{-(E_F-E_v)/kT}$, and give the physical reason for each step.
• Derive the intrinsic carrier concentration, state the law of mass action, and explain how $n_i$ depends on temperature and band gap.
• Describe n-type and p-type semiconductors with band diagrams, and describe how their carrier density changes over a wide range of temperature.
• Find the position of the Fermi level in intrinsic, n-type and p-type material, and explain why it moves with doping and temperature.
• Explain the formation of a p-n junction in terms of diffusion and drift, draw its band diagram, and explain what forward and reverse bias do.
• Derive the diode equation $I=I_s\left(e^{eV/kT}-1\right)$ and draw and explain the V-I characteristic.
• Solve the standard numerical problems on $n_i$, band gap and diode current, and a few more on doped carrier densities and conductivity.

How to use these notes

• Each topic is a page: idea → derivation or explanation → worked example → common mistakes.
• Green boxes show key ideas. Amber boxes marked Watch out show mistakes that make you lose marks. Boxes marked Note on the source show where the lecture notes were corrected. Boxes marked Beyond the syllabus give short extra points that make the syllabus topics easier to understand. They are not usually asked in exams on their own.
• The derivations of $n$, $p$, $n_i$, the intrinsic Fermi level and the diode equation are favourite long-answer questions. Learn the steps and the reason for each step, not just the results.
• The last three pages have the standard numerical problems, extra worked examples, and practice questions with a formula sheet.

Where each topic sits

Before you start. From Module 1: energy bands, the band gap $E_g=E_c-E_v$, the Fermi-Dirac function $f(E)=\dfrac{1}{1+e^{(E-E_F)/kT}}$, and conductivity $\sigma=ne\mu$. From maths: integration by substitution, and the laws of exponents and logarithms. Remember that $kT\approx0.0259\ \text{eV}$ at $300\ \text{K}$, because almost every estimate in this module compares an energy with $kT$.

Page 2 of 18

Intrinsic semiconductors and holes

The idea

An intrinsic semiconductor is a pure semiconductor, with no important impurities. Silicon (Si) and germanium (Ge) are the standard examples. Both belong to group IV of the periodic table. Each atom has four valence electrons, and in the crystal each atom shares one electron with each of its four nearest neighbours. These shared pairs are covalent bonds, and every valence electron is held in one of them.

At $0\ \text{K}$ every valence electron is held in its bond. In terms of bands, the valence band is completely full and the conduction band is completely empty. A full band cannot carry current (for every electron moving in one direction there is another moving in the opposite direction, and there is no empty state to move into), and an empty band has nothing to carry. So at absolute zero a pure semiconductor is an insulator.

As the temperature rises, the lattice vibrates and some electrons get enough thermal energy to break their covalent bond. The energy needed is the band gap: about $1.1\ \text{eV}$ for silicon and $0.66\ \text{eV}$ for germanium. A freed electron moves almost freely through the spaces between the atoms. It has become a conduction electron. In terms of bands, the electron has jumped from the valence band to the conduction band.

Thermal energy is shared randomly, so at room temperature ($kT\approx0.026\ \text{eV}$) only a tiny fraction of bonds are broken at any moment. That is why the result depends so strongly on temperature, and why the number of carriers depends exponentially on $E_g/kT$.

Holes

When an electron leaves the valence band, it leaves behind an empty state in the bond it came from, called a hole.

• Why a hole acts as a positive charge. Before the bond broke, the region around it was electrically neutral: the negative charge of the electrons exactly balanced the positive nuclei. When one electron is removed, a net charge of $+e$ is left at that place. So the hole is a "missing negative charge", which behaves in every electrical way like a positive charge $+e$.
• Why a hole can move. A valence electron in a neighbouring bond needs very little energy to hop into the vacancy, because it moves from one valence-band state to another of almost the same energy. When it does this, the vacancy appears where that electron came from. If this is repeated, the vacancy moves through the crystal. In an electric field, the valence electrons hop mostly against the field, so the hole moves with the field, opposite to the electrons, exactly as a positive charge would.
• Why holes count as carriers. The current produced by all those valence electrons hopping in one direction is the same as the current produced by a single positive particle moving in the other direction. It is much simpler to follow one hole than many electrons, so we treat holes as real charge carriers in the valence band.

A useful picture is a row of cars in a car park with one empty space. As each car moves forward into the space in front, the empty space moves backwards. Watching the space is easier than watching every car.

Key idea. Breaking one covalent bond creates an electron-hole pair. In an intrinsic semiconductor the number of free electrons is always equal to the number of holes, $n=p$.

Generation, recombination and equilibrium

Two processes go on all the time:

• Generation: thermal energy breaks bonds and creates electron-hole pairs.
• Recombination: a conduction electron meets a hole and falls back into the bond, and both carriers disappear. The energy is released as heat (in silicon and germanium) or light (in materials such as GaAs, used in LEDs).

At a fixed temperature, the two rates become equal and the carrier numbers become steady. This is thermal equilibrium. If the temperature is raised, the generation rate increases, so the equilibrium carrier numbers also increase.

Properties of intrinsic semiconductors

Limitations

• The conductivity at room temperature is low.
• The current depends much more on temperature than on the applied voltage, so the behaviour of the device changes as the temperature changes.
• The conductivity cannot be chosen. It is fixed by the material and the temperature.
• So pure semiconductors are of little use in devices until they are doped (page 8).

Watch out. A hole is not a particle that "exists" on its own. It is a convenient way to describe the movement of many valence electrons into an empty state. Draw it as an open circle in the valence band, never in the conduction band. Also remember that the hole moves opposite to the electrons that fill it.

Page 3 of 18

Effective mass

To count carriers and to describe how they move, we need to know how an electron in a crystal responds to a force. The answer is the effective mass.

Why the ordinary mass is not enough

For a free body, Newton's second law gives the acceleration produced by a force $F$:

This is correct only if $F$ is the only force acting. An electron in a crystal is not free. It moves through the periodic potential of the positive ion cores, which push and pull it differently at every point of the lattice. When we apply an external force (for example $F=-eE$ from an electric field), the actual acceleration of the electron is the combined result of the external force and all the internal crystal forces.

We have two choices:

1. keep the true mass $m$ and add up all the internal forces (very complicated), or
2. keep only the external force and include the effect of the crystal in a modified mass.

The second choice gives

where $m^*$ is the effective mass. It is the mass the electron appears to have when it responds to an external force inside that particular band of that particular crystal.

What determines the effective mass

The effective mass depends on the shape of the band near the energy where the carriers are. In quantum mechanics it is decided by how the energy $E$ of a state changes with its wave vector $k$:

A sharply curved band gives a small effective mass (the carrier is easy to accelerate); a flat band gives a large one. You do not need this formula for the exam, but it explains the next two facts.

• In the conduction band, near its bottom $E_c$, the band curves upwards like $E=E_c+\dfrac{\hbar^2k^2}{2m_e^*}$. The curvature is positive, so the effective mass $m_e^*$ of the electron is positive. Depending on the material, it can be smaller or larger than the free-electron mass $m$. For example, it is about $0.26\,m$ in silicon (for conduction) and $0.067\,m$ in gallium arsenide.
• Near the top of the valence band the band curves downwards. The curvature is negative, so an electron there has a negative effective mass: an external force would make it accelerate opposite to the force. Physically, the crystal forces are stronger than the applied force for these electrons.

Why we use holes in the valence band

Negative-mass electrons are difficult to think about. But the valence band near its top is almost full, with only a few empty states. Describing the motion of the few holes is equivalent to describing the motion of all the valence electrons, and a hole near the top of the valence band behaves like a particle with

• positive charge $+e$, and
• positive effective mass $m_h^*$.

Two negatives (negative charge, negative mass) make a positive description that follows Newton's law in the normal way. So we write $m_e^*$ for the effective mass of conduction electrons and $m_h^*$ for the effective mass of holes, and both are positive numbers.

Key idea. Effective mass is not a change in the electron itself. It is a bookkeeping trick that lets us use $a=F/m^*$ for a carrier inside a crystal, with all the crystal forces hidden inside $m^*$.

Note on the source. The lecture notes said that the conduction-band effective mass "can be equal to or greater than" the free electron mass. In fact it is often smaller (as the examples show). What matters is that it is a property of the band, not of the electron itself.

Watch out. Do not write "the hole has negative mass". It is the valence-band electron near the band top that has a negative effective mass. The hole has a positive effective mass $m_h^*$.

Page 4 of 18

Density of states

The idea

To find how many electrons are in the conduction band, we need two things for every energy $E$:

1. how many states (places for an electron) are there at that energy, and
2. the probability that each state is occupied.

The second is the Fermi-Dirac function from Module 1. The first is the density of states.

In a band, the allowed energy levels are so close together (Module 1: about $10^{-23}\ \text{eV}$ apart) that we treat them as continuous. Instead of counting individual levels, we ask: how many states per unit volume are there in a small energy range $dE$ around $E$? This number is written $Z(E)\,dE$, and $Z(E)$ is the density of states.

Density of states. The number of energy states per unit volume with energy between $E$ and $E+dE$ is

$Z(E)\,dE=\frac{4\pi}{h^3}\,(2m^*)^{3/2}\,E^{1/2}\,dE$

where $m^*$ is the effective mass of the carriers, $h$ is Planck's constant, and $E$ is measured from the edge of the band.

Where the formula comes from (outline)

You are not asked to derive this, but the outline shows why each factor is there.

• A carrier in a crystal of volume $V$ behaves like a particle in a box. Its allowed momenta form a fine grid in "momentum space", with one state for each volume $h^3/V$ of that space.
• States with momentum between $p$ and $p+dp$ fill a thin spherical shell of volume $4\pi p^2\,dp$. Dividing by $h^3/V$ gives the number of states. Dividing by $V$ gives the number per unit volume.
• Each state can hold two electrons of opposite spin, which gives a factor $2$.
• Finally, change from momentum to energy using $E=\dfrac{p^2}{2m^*}$, so $p=\sqrt{2m^*E}$ and $dp=\dfrac{\sqrt{2m^*}}{2\sqrt E}\,dE$. Putting everything together gives the $\dfrac{4\pi}{h^3}(2m^*)^{3/2}E^{1/2}$ above.

What it tells us

• $Z(E)\propto\sqrt E$: there are no states exactly at the band edge, and there are more and more states as you go deeper into the band. The shell in momentum space becomes larger as you go to larger momentum.
• A larger effective mass gives more states at a given energy (the factor $(m^*)^{3/2}$). Flat bands are "dense" in states.
• In the conduction band, energy is measured up from $E_c$, so $E$ is replaced by $E-E_c$:

• In the valence band, hole energy is measured down from $E_v$, so $E$ is replaced by $E_v-E$:

Inside the band gap there are no states at all in a pure crystal, so $Z(E)=0$ there.

How it is used

The number of electrons in a small energy range is

and the total is the integral of this over the band. The next two pages carry out that integral for electrons and for holes.

Watch out. The formula uses the effective mass: $m_e^*$ in the conduction band and $m_h^*$ in the valence band. Writing the free electron mass $m$ is a common mistake. Also, $E$ in the formula is the energy measured from the band edge, not the absolute energy.

Page 5 of 18

Density of electrons in the conduction band

Plan

The number of electrons per unit volume in a small energy range $dE$ is

Then add up (integrate) over the whole conduction band, from its bottom $E_c$ to $\infty$. The upper limit $\infty$ is not literally the top of the band. It is used because, as we will see, the occupation probability falls so fast with energy that states more than a few $kT$ above $E_c$ contribute nothing, and extending the integral to infinity makes it easy to calculate without changing the answer.

Step 1: the two ingredients

Let $dn$ be the number of electrons per unit volume with energy between $E$ and $E+dE$:

Number of states. Measuring energy from the band edge $E_c$ (the lowest energy a conduction electron can have):

Probability of occupation. The Fermi-Dirac function is

Why we may simplify it. In a semiconductor, the Fermi level lies in the gap, typically several tenths of an eV below $E_c$. For every energy in the conduction band, $E-E_F$ is therefore at least that large, which is many times $kT$ ($0.026\ \text{eV}$ at room temperature). For example, if $E_c-E_F=0.56\ \text{eV}$ (mid-gap silicon), then even at the band edge $\dfrac{E-E_F}{kT}\approx21.7$ and $e^{21.7}\approx2.6\times10^9$. Compared with that, the $1$ in the denominator is completely negligible:

This is the Boltzmann approximation: the electrons in the conduction band are so few, and so thinly spread over the available states, that the Pauli exclusion principle (which the $1$ in the Fermi-Dirac function represents) hardly matters, and they behave like a classical gas.

Note on the source. The lecture notes said "$(E-E_F)\gg kT$ for all $T$". The approximation is really a condition. It is true when the Fermi level is several $kT$ below the conduction band (a non-degenerate semiconductor), which is the normal case. It fails for very heavily doped material, where the Fermi level can reach or enter the band.

Step 2: set up the integral

Multiply the two ingredients and integrate over the band:

Take the constants outside. Split $e^{(E_F-E)/kT}=e^{E_F/kT}\,e^{-E/kT}$. The first factor does not depend on $E$:

Step 3: substitute

The integrand depends on the energy above the band edge, so we use that as the new variable. Let $x=E-E_c$, so $E=x+E_c$ and $dE=dx$. The limits $E=E_c$ and $E=\infty$ become $x=0$ and $x=\infty$. Then $e^{-E/kT}=e^{-E_c/kT}e^{-x/kT}$, and the constant factor $e^{-E_c/kT}$ joins $e^{E_F/kT}$ outside:

Step 4: the standard integral

With $a=\dfrac1{kT}$ this gives $\dfrac{\sqrt\pi}{2}(kT)^{3/2}$, so

Now simplify the constants: $\dfrac{2\pi^{3/2}(2m_e^*kT)^{3/2}}{h^3}=2\left(\dfrac{2\pi m_e^*kT}{h^2}\right)^{3/2}$, since $(\pi)^{3/2}(2m_e^*kT)^{3/2}=(2\pi m_e^*kT)^{3/2}$ and $h^3=(h^2)^{3/2}$.

Result

Meaning of $N_c$

The constant

is called the effective density of states in the conduction band. The result reads as if all the conduction band states were squeezed into a single level at $E_c$ holding $N_c$ states per unit volume, each occupied with the Boltzmann probability $e^{-(E_c-E_F)/kT}$. For silicon at $300\ \text{K}$ it is about $3\times10^{25}\ \text{m}^{-3}$ ($3\times10^{19}\ \text{cm}^{-3}$). Note that $N_c$ itself depends on temperature, as $T^{3/2}$, because at higher temperature the electrons spread over a wider slice of the band.

Key idea. The electron density depends exponentially on how far the Fermi level lies below the conduction band. If $E_F$ moves up by $kT$, $n$ increases by a factor $e\approx2.7$. If it moves up by $0.06\ \text{eV}$ (about $2.3\,kT$), $n$ increases tenfold.

Note on the source. One place in the lecture notes wrote the standard integral as $\dfrac{\sqrt\pi}{2a}$. The correct form is $\dfrac{\sqrt\pi}{2a^{3/2}}$. The two are the same only when $a=1$.

Watch out. Be careful with the sign in the exponent: $E_F<E_c$, so $E_F-E_c$ is negative and $n$ is a small fraction of $N_c$. The final factor $2$ and the power $\tfrac32$ both come from the algebra in step 4. Show that working in the exam.

Page 6 of 18

Density of holes in the valence band

The method is like the electron derivation, with two changes: the states are in the valence band, and we need the probability that a state is empty (holds a hole) instead of full.

Step 1: the two ingredients

Let $dp$ be the number of holes per unit volume with energy between $E$ and $E+dE$:

Number of states. Hole energies lie below the top of the valence band $E_v$, so measure from $E_v$ downwards. $E_v$ plays the same role for holes that $E_c$ plays for electrons: it is the edge of the band where the carriers collect.

Probability that a state holds a hole. A hole is the absence of an electron. A state is either occupied (probability $F$) or empty (probability $F_h$), and these must add up to one, so

Why we may simplify it. In the valence band $E<E_F$ by several $kT$, so $E-E_F$ is large and negative, and $e^{(E-E_F)/kT}\ll1$. The term in the denominator is negligible compared with $1$:

This is the same Boltzmann approximation, now for holes: very few valence states are empty, so holes also behave like a thin classical gas.

Step 2: set up the integral

Holes occupy energies from deep in the band up to $E_v$. As for electrons, we take the lower limit as $-\infty$, because states more than a few $kT$ below $E_v$ are almost certainly full and give no holes:

Step 3: substitute

Let $x=\dfrac{E_v-E}{kT}$, the depth below the band top in units of $kT$. Then $\dfrac{E}{kT}=\dfrac{E_v}{kT}-x$, $E_v-E=kTx$ and $dE=-kT\,dx$. The limits $E=-\infty$ and $E=E_v$ become $x=\infty$ and $x=0$:

The minus sign reverses the limits, and $(kT)^{1/2}\cdot kT=(kT)^{3/2}$ joins the other constants:

Step 4: the standard integral

With $a=1$, $\displaystyle\int_0^\infty x^{1/2}e^{-x}dx=\frac{\sqrt\pi}{2}$, so

Result

where

is the effective density of states in the valence band. It has the same meaning as $N_c$: the result is as if all the valence band states were concentrated at $E_v$. Because holes usually have a different effective mass from electrons, $N_v$ differs from $N_c$ (for silicon $N_v\approx1.0\times10^{19}\ \text{cm}^{-3}$ against $N_c\approx2.8\times10^{19}\ \text{cm}^{-3}$).

Key idea. The two results are mirror images. Electrons: exponent $-(E_c-E_F)/kT$, the distance from $E_F$ up to the conduction band. Holes: exponent $-(E_F-E_v)/kT$, the distance from $E_F$ down to the valence band. If the Fermi level moves up, $n$ increases and $p$ decreases by the same factor.

Watch out. Use $F_h=1-F$, not $F$, for holes. Remember that the effective mass changes to $m_h^*$, the band edge to $E_v$, and the order inside the exponent to $E_F-E_v$.

Page 7 of 18

Intrinsic carrier concentration

The idea

In a pure semiconductor, every broken bond makes one electron and one hole, so

where $n_i$ is the intrinsic carrier concentration (or intrinsic carrier density). It is the number of electrons (or holes) per unit volume in the pure material at a given temperature.

Derivation

Since $n=p=n_i$, we can write $n_i^2=np$. Using the product is the trick: it removes the Fermi level, which we do not yet know. Multiply the two results of the last two pages:

Combine the prefactors: $2\times2=4$, and $\left(\dfrac{2\pi kT}{h^2}\right)^{3/2}\times\left(\dfrac{2\pi kT}{h^2}\right)^{3/2}=\left(\dfrac{2\pi kT}{h^2}\right)^{3}$, leaving $(m_e^*)^{3/2}(m_h^*)^{3/2}=(m_e^*m_h^*)^{3/2}$.

Combine the exponents: $(E_F-E_c)+(E_v-E_F)=E_v-E_c=-E_g$. The $E_F$ terms cancel:

Taking the square root:

Why $n_i$ depends only on $E_g$ and $T$

The Fermi level has cancelled out completely. This makes physical sense: in a pure crystal, the number of broken bonds is decided by how much energy it costs to break one ($E_g$) and how much thermal energy is available ($kT$), together with the number of states available to receive the carriers ($N_c$ and $N_v$, which depend on the effective masses and on $T$). Nothing else about the crystal comes in.

Law of mass action

The product $np$ does not depend on $E_F$:

So at a given temperature, for a given semiconductor, the product of the electron and hole densities is a constant. This is the law of mass action, named by analogy with chemistry, where the product of the concentrations of the reacting substances is fixed at equilibrium. Here the "reaction" is

Generation runs at a rate decided by the temperature, and recombination runs at a rate proportional to $np$ (an electron has to meet a hole). At equilibrium the two rates are equal, so $np$ is fixed.

The law holds for intrinsic and extrinsic semiconductors, because the derivation never assumed $n=p$ until the very end. If doping increases $n$, more electrons are available to recombine with holes, so $p$ must fall until $np$ is back to $n_i^2$. Page 9 uses this to find the minority carrier density in doped material.

How $n_i$ varies

• $n_i$ is independent of the Fermi level.
• $n_i$ decreases as the band gap $E_g$ increases, exponentially. A material with a wider gap has far fewer intrinsic carriers.
• $n_i$ increases steeply with temperature: $n_i\propto T^{3/2}e^{-E_g/2kT}$. Both factors increase with $T$, but the exponential completely dominates. The $T^{3/2}$ factor changes by only a few per cent over the range where $n_i$ changes by orders of magnitude.
• The factor $2$ in $e^{-E_g/2kT}$ shows that breaking one bond produces two carriers: the energy cost $E_g$ is shared between the electron and the hole.

At room temperature, $n_i\approx1.5\times10^{16}\ \text{m}^{-3}$ for silicon and $2.5\times10^{19}\ \text{m}^{-3}$ for germanium. The smaller gap of germanium gives it over a thousand times more intrinsic carriers. Compare these with the roughly $5\times10^{28}$ atoms per cubic metre in the crystal: in silicon at room temperature, with two bonds per atom, only about one bond in $10^{13}$ is broken.

A straight-line test

Taking logarithms of $n_i\propto T^{3/2}e^{-E_g/2kT}$:

So a graph of $\ln(n_i/T^{3/2})$ against $1/T$ is a straight line of slope $-E_g/2k$. Measuring the conductivity of a pure sample at several temperatures and plotting it in this way is a standard laboratory method to find the band gap.

Watch out. The exponent in $n_i$ is $-E_g/2kT$, with a 2. In $n$ or $p$ alone it is the distance to the band edge over $kT$, with no 2. Also, since the curve rises exponentially, heating by a few degrees changes $n_i$ a lot (see problem 1 on the numerical problems page: a $10\ \text{K}$ rise roughly doubles $n_i$ in silicon).

Page 8 of 18

Extrinsic semiconductors

Doping

Adding a controlled amount of impurity to an intrinsic semiconductor is called doping. The impurity is the dopant, and the result is an extrinsic semiconductor.

The dopants come from group V (pentavalent, 5 valence electrons) or group III (trivalent, 3 valence electrons). Their atoms are about the same size as silicon and germanium atoms, so they take the place of host atoms in the crystal (substitutional impurities) without disturbing it much. Very small amounts are enough. A typical doping of $10^{16}$ dopant atoms per $\text{cm}^3$ is only one dopant for every five million silicon atoms, yet it increases the carrier density about a million times above the intrinsic value.

Depending on the dopant, two types of extrinsic semiconductor are possible: n-type and p-type. Compared with intrinsic material, extrinsic semiconductors

• conduct much better,
• have a conductivity that can be fixed by choosing the doping concentration, and
• have a conductivity that depends much less on temperature over the normal working range (see the temperature regions below).

n-type semiconductors

Dopant: a pentavalent impurity such as phosphorus, arsenic or antimony.

• Four of the five valence electrons form covalent bonds with four neighbouring silicon atoms. The fifth electron has no bond to join. It is held only weakly by the electrostatic attraction of its parent ion, and this attraction is further weakened by the surrounding crystal (its dielectric constant is about 12 in silicon). Its ionisation energy is only about $0.01$ to $0.05\ \text{eV}$, compared with $1.1\ \text{eV}$ to break a bond.
• At ordinary temperatures, thermal energy easily frees this electron and it moves into the conduction band. It leaves behind a fixed positive ion (for example $\text{P}^+$), bound in the lattice and unable to move. Because the impurity atom donates an electron for conduction, it is called a donor.
• The donor atoms are far apart and do not interact, so their electrons occupy a single discrete donor level $E_D$, just below the conduction band. The small gap $E_c-E_D$ is the ionisation energy.
• Donating an electron does not create a hole in the valence band. The positive charge left behind is the fixed ion, not a mobile hole.

How an n-type semiconductor behaves with temperature:

• At $0\ \text{K}$ the donors are not ionised. Their electrons sit in $E_D$, the conduction band is empty and the valence band is full. The material is an insulator.
• As the temperature rises, the donors ionise and their electrons enter the conduction band. Because $E_c-E_D$ is so small, by about $100\ \text{K}$ essentially all donors are ionised. More heating cannot produce any more donor electrons.
• At high enough temperatures, intrinsic behaviour takes over. Many electrons are excited from the valence band across the full gap, creating electron-hole pairs, and they finally outnumber the donor electrons.

Electrons far outnumber holes, so electrons are the majority carriers and holes are the minority carriers (the few holes come from thermal generation). Hence the name n-type (negative carriers).

p-type semiconductors

Dopant: a trivalent impurity such as boron, gallium or indium.

• With only three valence electrons, the impurity is one short of completing four covalent bonds with its neighbours. One bond is left incomplete.
• An electron from a neighbouring bond can jump into this vacancy with only a little energy (about $0.01$ to $0.05\ \text{eV}$), and complete the fourth bond. This leaves a hole in the valence band where the electron came from. The impurity atom, now having an extra electron, becomes a fixed negative ion (for example $\text{B}^-$). Because it accepts an electron from the valence band, it is called an acceptor.
• The acceptor atoms do not interact, so their states form a discrete acceptor level $E_A$, just above the valence band.
• Accepting an electron creates a hole without creating a conduction electron. The hole can move freely through the valence band, but the negative ion is fixed in its place by its bonds.

How a p-type semiconductor behaves with temperature:

• At $0\ \text{K}$ the acceptor levels are empty and the valence band is full. The conduction band is also empty. The material is an insulator.
• With a little heating (energy of the order of $0.01\ \text{eV}$), valence electrons jump into the acceptor levels, leaving holes in the valence band. At normal temperatures the acceptor levels are saturated (all filled), and a few electrons are also excited into the conduction band.
• At high temperatures, thermally generated electron-hole pairs dominate and far outnumber the holes produced by the acceptors.

Holes far outnumber electrons, so holes are the majority carriers and electrons are the minority carriers. Hence p-type (positive carriers).

Three temperature regimes

The temperature behaviour described above has three clear stages, shown here for n-type silicon with $10^{15}$ donors per $\text{cm}^3$.

1. Freeze-out (low temperature). There is too little thermal energy even to ionise the donors. Most donor electrons are "frozen" in $E_D$, and $n$ increases rapidly as the temperature increases.
2. Extrinsic or saturation region (roughly $100$ to $500\ \text{K}$ for this doping). All donors are ionised, so $n\approx N_D$, and $n_i$ is still far below $N_D$. The carrier density is constant, independent of temperature. Devices are designed to work in this flat region, and this is why doping gives stable conductivity.
3. Intrinsic region (high temperature). $n_i$ has grown until it is comparable to and then greater than $N_D$. Electron-hole pairs across the gap now dominate, and the material behaves as intrinsic: $n\approx p\approx n_i$, rising steeply with temperature. Doped devices stop working properly here, which is why silicon devices have a maximum operating temperature.

Heavier doping or a wider band gap pushes the intrinsic region to a higher temperature. This is one reason why silicon (gap $1.1\ \text{eV}$) replaced germanium (gap $0.66\ \text{eV}$) in most electronics: germanium becomes intrinsic at a much lower temperature.

Summary

Watch out. An n-type semiconductor is not negatively charged, and a p-type is not positively charged. Each extra electron is balanced by a positive donor ion, and each hole by a negative acceptor ion, so the crystal as a whole is neutral. "n" and "p" tell the sign of the majority mobile carriers, not the net charge.

Page 9 of 18

Carrier densities in doped semiconductors

Beyond the syllabus. The syllabus treats extrinsic semiconductors qualitatively. This page shows, with two simple equations that you already have, how many majority and minority carriers a doped sample has. It makes the law of mass action, the Fermi level and the p-n junction much easier to understand.

Two equations

Take an n-type sample with $N_D$ donors per unit volume, all ionised (the extrinsic region of the previous page).

1. Charge neutrality. The sample as a whole is neutral. The positive charges are the holes and the donor ions. The negative charges are the electrons:

2. Law of mass action. At any temperature,

Majority and minority carriers

In the normal case $N_D\gg n_i$. Then $p$ is tiny, the $p$ in the neutrality equation is negligible, and

For a p-type sample with $N_A$ acceptors per unit volume, in the same way,

• The majority carrier density is decided by the doping and hardly depends on temperature.
• The minority carrier density is decided by $n_i^2$ and so rises steeply with temperature. This is why the reverse current of a diode, which is carried by minority carriers, is so sensitive to temperature (page 15).
• Doping increases the majority carriers and decreases the minority carriers below the intrinsic value. Adding donors does not only add electrons. It also removes holes, because the extra electrons recombine with them.

Example. Silicon with $n_i=1.5\times10^{10}\ \text{cm}^{-3}$ is doped with $N_D=10^{16}\ \text{cm}^{-3}$ donors. Then

The electrons outnumber the holes by a factor of about $4\times10^{11}$.

Where the Fermi level sits

Combining $n=N_D$ with $n=N_c\,e^{-(E_c-E_F)/kT}$ gives the position of the Fermi level directly:

and similarly $E_F-E_v=kT\ln\left(\dfrac{N_v}{N_A}\right)$ for p-type.

Example. For the silicon above at $300\ \text{K}$ with $N_c=2.8\times10^{19}\ \text{cm}^{-3}$:

The Fermi level is $0.205\ \text{eV}$ below the conduction band, well above mid-gap (which is about $0.56\ \text{eV}$ below $E_c$). Each tenfold increase in $N_D$ moves it up by $kT\ln10\approx0.06\ \text{eV}$.

Key idea. Doping does not change $n_i$ (a property of the pure material). It changes how the fixed product $n_i^2$ is shared between electrons and holes, and the Fermi level moves to whatever position gives that sharing.

Watch out. $p=n_i^2/N_D$ uses $n_i$ squared. If you forget the square, the answer will be wrong by a factor of $10^{10}$.

Page 10 of 18

Fermi level in semiconductors

The Fermi level is the energy at which the occupation probability is $\tfrac12$. In a semiconductor no states exist there (it lies in the gap), but its position is still very important. By the results of pages 5 and 6, $n$ depends on how far $E_F$ is below $E_c$, and $p$ on how far it is above $E_v$. So we can read the carrier balance from the Fermi level.

Intrinsic semiconductor

In an intrinsic semiconductor $n=p$. Put the two results:

Cancel the common factors $2$ and $\left(\dfrac{2\pi kT}{h^2}\right)^{3/2}$:

Collect the exponentials on one side. Dividing both sides by $(m_e^*)^{3/2}e^{(E_v-E_F)/kT}$ and adding the exponents, $(E_F-E_c)-(E_v-E_F)=2E_F-E_c-E_v$:

Take natural logarithms:

Solve for $E_F$:

• At $T=0\ \text{K}$ the second term vanishes: $E_F=\dfrac{E_c+E_v}{2}$. The Fermi level is exactly in the middle of the band gap.
• At higher temperatures it shifts slightly: up if $m_h^*>m_e^*$, down if $m_h^*<m_e^*$. The physical reason: the band with the larger effective mass has more states (larger $N_c$ or $N_v$), so to keep $n=p$ the Fermi level must move a little away from that band.
• The shift is tiny. For silicon, with density-of-states masses $m_e^*\approx1.08\,m$ and $m_h^*\approx0.81\,m$, at $300\ \text{K}$ it is $\tfrac34\times0.02585\times\ln(0.81/1.08)\approx-0.006\ \text{eV}$, against a gap of $1.12\ \text{eV}$. So the Fermi level of an intrinsic semiconductor is usually taken to be at the middle of the gap.

n-type semiconductor

The donor level $E_D$ lies just below the conduction band edge $E_c$.

• At $T=0\ \text{K}$ the donor level is full and the conduction band is empty, so the Fermi level (the boundary between filled and empty states) lies midway between $E_D$ and $E_c$:

• Low temperatures. As the temperature rises, donor electrons move into the conduction band and the donor level starts to empty. The Fermi level moves down towards $E_D$. When the donors are fully ionised (around the depletion temperature $T_d$, so called because the donor level has been depleted of its electrons), it has fallen to about the donor level, $E_{Fn}\approx E_D$.
• Extrinsic region. Now $n\approx N_D$ is fixed, but $N_c$ increases as $T^{3/2}$ and $kT$ also increases, so by $E_c-E_F=kT\ln(N_c/N_D)$ the Fermi level keeps falling, roughly linearly with $T$.
• High temperatures. Once intrinsic carriers dominate, the material behaves as intrinsic. Above the intrinsic temperature $T_i$ the Fermi level reaches the middle of the gap: $E_{Fn}=E_{Fi}$.

p-type semiconductor

The acceptor level $E_A$ lies just above the valence band edge $E_v$.

• At $T=0\ \text{K}$ the acceptor level is empty and the valence band is full, so the Fermi level lies midway between $E_v$ and $E_A$:

• Low temperatures. Holes in the valence band come only from the acceptors. As valence electrons fill the acceptor levels, the Fermi level moves up towards $E_A$. When the acceptors are saturated (around the saturation temperature $T_s$), it reaches about the acceptor level, $E_{Fp}\approx E_A$.
• Extrinsic region. With $p\approx N_A$ fixed, $E_F-E_v=kT\ln(N_v/N_A)$ increases with $T$, so the Fermi level keeps rising, roughly linearly.
• High temperatures. Above $T_i$ the material is intrinsic and $E_{Fp}=E_{Fi}$, at mid-gap.

Effect of doping level

At a fixed temperature in the extrinsic region:

• more donors push the Fermi level closer to the conduction band ($E_c-E_F=kT\ln(N_c/N_D)$ becomes smaller);
• more acceptors push it closer to the valence band.

If the doping is so heavy that the Fermi level reaches the band edge or enters the band, the semiconductor is called degenerate, and the Boltzmann approximation (and so all the formulas above) is no longer valid. Tunnel diodes (Module 4) use such degenerate doping.

Key idea. Doping pushes the Fermi level towards the band with the majority carriers: up for n-type, down for p-type. Heating always pulls it back towards the middle of the gap, because at high temperature every semiconductor becomes intrinsic.

Watch out. In the intrinsic derivation, the masses appear as $\ln(m_h^*/m_e^*)$, with holes on top. If you write the ratio upside down, the direction of the shift is reversed. Also remember that $T_d$ belongs to n-type (donors depleted) and $T_s$ to p-type (acceptors saturated).

Page 11 of 18

Conductivity of semiconductors

Beyond the syllabus. Module 1 gave $\sigma=ne\mu$ for a metal. In a semiconductor, both electrons and holes carry current, and this short page shows how to combine them. It explains in numbers why doping is so important.

Two kinds of carrier

In an electric field $E$, electrons drift against the field and holes drift with it. Because their charges are opposite, both produce current in the same direction, so the two contributions add:

where $\mu_e$ and $\mu_h$ are the electron and hole mobilities. Comparing with $J=\sigma E$:

• Intrinsic: $n=p=n_i$, so $\sigma_i=n_ie(\mu_e+\mu_h)$.
• n-type: $n\approx N_D\gg p$, so $\sigma\approx N_De\mu_e$.
• p-type: $p\approx N_A\gg n$, so $\sigma\approx N_Ae\mu_h$.

Electrons usually have a higher mobility than holes (in silicon $\mu_e\approx0.135$ and $\mu_h\approx0.048\ \text{m}^2/\text{V s}$), because their effective mass for conduction is smaller.

Worked examples

Intrinsic silicon at room temperature, $n_i=1.5\times10^{16}\ \text{m}^{-3}$:

Intrinsic germanium, $n_i=2.5\times10^{19}\ \text{m}^{-3}$, $\mu_e=0.39$, $\mu_h=0.19\ \text{m}^2/\text{V s}$:

about five thousand times better than silicon, because of its smaller gap.

Doped silicon with $N_D=10^{22}\ \text{m}^{-3}$ ($10^{16}\ \text{cm}^{-3}$):

That is about half a million times the intrinsic value, from one dopant atom for five million silicon atoms. Its resistivity is $1/216\approx4.6\times10^{-3}\ \Omega\,\text{m}$.

Temperature dependence

• Intrinsic: $\sigma_i$ follows $n_i$, so it rises steeply (exponentially) with temperature. The mobilities fall slowly as lattice vibrations increase, but this is completely overcome by the rise in $n_i$. Hence the negative temperature coefficient of resistance of semiconductors.
• Extrinsic region: $n\approx N_D$ is constant, so $\sigma$ changes only through the mobility, which falls slightly with temperature. Over this range a doped semiconductor behaves rather like a metal, with a small positive temperature coefficient.

Watch out. In $\sigma=e(n\mu_e+p\mu_h)$ the two terms add, even though electrons and holes move in opposite directions. Opposite charge and opposite velocity give current in the same direction.

Page 12 of 18

Forming a p-n junction

Why junctions matter

On its own, a p-type or n-type crystal just conducts like a poor conductor, with the same resistance whichever way the current flows. It has no important use as such. Nearly all semiconductor devices combine the two types. The simplest is a single p-n junction, or p-n junction diode: one crystal that is p-type on one side and n-type on the other, made by special manufacturing processes (diffusing or growing dopants into a single crystal). Simply pressing two pieces together does not work, because the surfaces would not make a perfect crystal contact.

Diffusion and the depletion region

The p-side has many holes and few electrons. The n-side has many electrons and few holes. At the moment the junction is formed:

1. Diffusion. Any particles that are more concentrated in one place than another spread out, just as ink spreads in water. So electrons diffuse from n to p and holes diffuse from p to n. This movement caused by a difference in concentration is called diffusion, and the current it carries is the diffusion current.
2. Recombination. Near the junction, the diffusing electrons and holes meet and recombine: conduction electrons fall into holes and become bound valence electrons. Both carriers disappear.
3. A region without mobile carriers. A thin region on both sides of the junction is left without mobile carriers. This is the depletion region (depleted of carriers). It is about $1\ \mu\text{m}$ wide, narrower for heavier doping, and because it has no free carriers, it behaves like an insulator.
4. Uncovered ions. What is left in the depletion region are the fixed ions. On the p-side, each acceptor that has lost its hole is now an uncovered negative ion. On the n-side, each donor that has lost its electron is an uncovered positive ion. These ions cannot move, so the depletion region has a net charge on each side and is also called the space charge region. The total charge on the two sides is equal and opposite.

Barrier potential

The double layer of ions (positive on the n-side, negative on the p-side) produces an electric field inside the depletion region, pointing from the positive donor ions (n-side) to the negative acceptor ions (p-side).

This field acts on any carrier that wanders into the region:

• it pushes electrons back towards the n-side, and
• it pushes holes back towards the p-side.

So it opposes further diffusion of the majority carriers. As more carriers diffuse and recombine, the ion layers grow, the field becomes stronger, and diffusion is resisted more and more. The process is self-limiting.

The potential difference across the depletion region is the barrier potential (junction potential or built-in potential) $V_0$: about $0.7\ \text{V}$ for silicon and $0.3\ \text{V}$ for germanium. A majority electron on the n-side needs an energy $eV_0$ to climb over this barrier.