Electrical Sciences · Lecture

Semiconductors

Electronics

Prof. Mithun Mondal BITS Pilani, Hyderabad Campus
SECTION 01

Overview

Demonstrative Video

Basic Electronics Contents:

  • Semiconductors

  • PN Junction Diode

  • Bipolar Junction Transistors (BJT)

Semiconductor Fundamentals

Semiconductor Materials

  • Semiconductor: are special class of elements having a conductivity between that of a good conductor and that of an insulator

  • Semiconductor materials fall into one of two classes:

    • Single crystal - germanium (Ge) and silicon (Si) having repetitive crystal structure

    • Compound - gallium arsenide (GaAs), cadmium sulphide (CdS), gallium nitride (GaN), and gallium arsenide phosphide (GaAsP) are constructed of two or more semiconductor materials of different atomic structure.

  • The three semiconductors used most frequently in the construction of electronic device are Ge, Si, and GaAs.

  • Best conductors (silver, copper, and gold) have one valence electron , whereas the best insulators have eight valence electrons

  • Best semiconductors have four valence electrons .

  • Many years ago, Ge was only material suitable for making semiconductor devices.

  • But Ge devices had a fatal flaw (their excessive reverse current, discussed later) that engineers could not overcome.

  • Next to oxygen, silicon is the most abundant element on the earth.

  • The advantages of Si immediately made it the semiconductor of choice.

  • Without it, modern electronics, communications, and computers would be impossible.

  • An isolated silicon atom has 14 protons and 14 electrons

  • core net charge 14 because it contains 14 protons in the nucleus and 10 electrons in the first two orbits.

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Energy Band

  • The range of energies possessed by electrons of the same orbit in a solid is known as energy band .

  • In a single isolated atom, electrons revolving in any orbit possess a definite energy.

  • However in a solid, an atom is greatly influenced by the closely packed neighbouring atoms.

  • Because of this the electrons in the same orbit have a range of energies rather than a single energy.

  • Although there are number of energy bands in solids, but we are more concerned with the following:

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  • Valence band: Outermost orbit electrons in an atom, highest energy level.

  • Conduction band: Loose electrons responsible for current conduction.

  • Forbidden Energy Gap: Gap between valence and conduction bands.

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  • Silicon Crystal Structure:

    • Silicon atoms arrange themselves in an orderly pattern called a crystal.

    • Each silicon atom shares its electrons with four neighbouring atoms.

    • This sharing allows each atom to have a total of eight electrons in its valence orbit.

    • Covalent bonds form between atoms due to the equal and opposite forces of attraction.

    • Billions of covalent bonds hold the silicon crystal together, giving it solidity.

  • Eight Electrons and Chemical Stability:

    • Each atom in a silicon crystal has eight electrons in its valence orbit.

    • This configuration results in chemical stability and solid material.

    • Elements tend to combine and share electrons to achieve eight electrons in their outer orbit.

    • The reason why eight electrons provide stability is not fully understood, but it is observed in various materials.

  • Scientific Laws and Explanations:

    • There are advanced physics equations that partially explain the stability of eight electrons in different materials.

    • The significance of the number eight remains unexplained, similar to other observed laws in physics, such as the law of gravity and Coulomb’s law.

    • When the valence orbit has eight electrons, it is saturated because no more electrons can fit into this orbit

  • Ambient Temperature and Vibrating Atoms:

    • Ambient temperature is the temperature of the surrounding air.

    • At temperatures above absolute zero ( \(-273^{\circ}\) ), the heat energy causes the atoms in a silicon crystal to vibrate.

    • Higher ambient temperatures result in stronger mechanical vibrations.

    • The warmth felt when picking up a warm object is due to the vibrating atoms.

  • Generation of Free Electrons and Holes:

    • Vibrations in a silicon crystal can dislodge electrons from the valence orbit.

    • The released electron gains enough energy to move into a larger orbit, becoming a free electron.

    • The departure of the electron creates a vacancy in the valence orbit called a hole.

    • Holes behave like positive charges and can attract and capture nearby electrons.

    • The presence of holes is a critical difference between conductors and semiconductors.

  • Role of Doping:

    • At room temperature, thermal energy produces only a few holes and free electrons.

    • To increase the number of holes and free electrons, the crystal needs to be doped.

  • Recombination and Lifetime:

    • In a pure silicon crystal, thermal energy generates an equal number of free electrons and holes.

    • Free electrons move randomly throughout the crystal.

    • Occasionally, a free electron and a hole come close to each other, resulting in recombination.

    • Recombination is the merging of a free electron and a hole.

    • The time between the creation and disappearance of a free electron is called the lifetime.

    • The lifetime can vary from a few nanoseconds to several microseconds, depending on the crystal’s quality and other factors.

  • Main Points inside a Silicon Crystal:

    1. Creation of Free Electrons and Holes:

      • Thermal energy generates some free electrons and holes.

    2. Recombination of Free Electrons and Holes:

      • Other free electrons and holes recombine.

    3. Temporary Existence of Free Electrons and Holes:

      • Some free electrons and holes exist temporarily, waiting for recombination.

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Intrinsic Semiconductor

  • An extremely pure semiconductor is called an intrinsic semiconductor.

  • At absolute zero temperature, the valence band of an intrinsic semiconductor is completely filled, and the conduction band is completely empty.

  • When heat energy is supplied (e.g., at room temperature), some valence electrons are lifted to the conduction band, creating free-moving electrons and holes.

  • Intrinsic semiconductors have a negative temperature coefficient of resistance, meaning their resistivity decreases and conductivity increases with a rise in temperature.

Extrinsic Semiconductor

  • Extrinsic Semiconductor:

    • An intrinsic semiconductor, on its own, is not useful for electronic devices.

    • To make it conductive, a small amount of suitable impurity is added, resulting in an extrinsic (impure) semiconductor.

    • The process of adding impurities to a semiconductor is called doping, which needs to be closely controlled.

    • Extrinsic semiconductors are classified based on the type of impurity added:

      • n-type semiconductor: When a donor impurity is added, creating excess free electrons.

      • p-type semiconductor: When an acceptor impurity is added, creating excess holes.

Hole flow through a semiconductor

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Intrinsic semiconductor: equal number of free electrons and holes.
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n-type semiconductors

  • When a small amount of pentavalent impurity is added to a pure semiconductor, it creates an extrinsic semiconductor known as an n-type semiconductor.

  • Pentavalent impurities such as arsenic and antimony provide a large number of free electrons in the semiconductor crystal.

  • These impurities are called donor impurities because each atom donates one free electron to the semiconductor crystal.

  • The pentavalent impurity atoms fit into the crystal structure, with their four valence electrons forming covalent bonds and the fifth electron being free.

  • The small amount of pentavalent impurity results in a large number of free electrons available for conduction.

p-type semiconductors

  • When a small amount of trivalent impurity is added to a pure semiconductor, it creates an extrinsic semiconductor known as a p-type semiconductor.

  • Trivalent impurities such as boron, gallium and indium provide a large number of free holes in the semiconductor crystal.

  • These impurities are called acceptor impurities because each atom creates a hole that can accept an electron from the semiconductor crystal.

  • The trivalent impurity atoms fit into the crystal structure, with their three valence electrons forming covalent bonds but leaving one covalent bond incomplete, creating a hole.

  • The small amount of trivalent impurity results in a large number of holes available in the semiconductor.

Summary

  • In an n-type semiconductor , a small amount of pentavalent impurity is added, providing a large number of free electrons. These impurities are donor impurities.

  • In a p-type semiconductor , a small amount of trivalent impurity is added, providing a large number of free holes. These impurities are acceptor impurities.

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Mass Action Law

  • Under thermal equilibrium, the product of the number of holes and electrons is constant and independent of the amount of donor and acceptor impurity doping.

    \[\boxed{n \cdot p = n_i^2}\]
    • \[\begin{aligned} n & = \text{number of free electrons per unit volume}\\ p & = \text{number of holes per unit volume}\\ n_i & = \text{intrinsic concentration}\\ \end{aligned}\]

  • While considering the conductivity of the doped semiconductors, only the dominant majority charge carriers have to be considered

Charge densities in doped semiconductors

  • N-type semiconductor:

    \[\begin{aligned} N_D & = \text{conc. of donor atoms} \\ n_N & = \text{electron conc. in N-type} \\ p_N & = \text{hole conc. in N-type} \end{aligned}\]
    • \[\begin{aligned} n_N & = N_D + p_N \approx N_D \\ p_N & = \dfrac{n_i^2}{n_N}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_D} \end{aligned}\]

  • P-type semiconductor:

    \[\begin{aligned} p_P & = N_A + n_P \approx N_A \\ n_p & = \dfrac{n_i^2}{p_P}~(\text{from mass action law})\\ & \approx \dfrac{n_i^2}{N_A} \end{aligned}\]
    \[\begin{aligned} N_A & = \text{conc. of acceptor atoms} \\ p_p & = \text{hole conc. in P-type} \\ n_P & = \text{electron conc. in P-type} \end{aligned}\]

Extrinsic Conductivity

  • \[\sigma_N = qn_N\mu_n \approx qN_D\mu_n \quad \text{since}~n_N \approx N_D\]
    Conductivity of N-type semiconductor:
  • \[\sigma_P= qn_P\mu_p \approx qN_A\mu_p \quad \text{since}~p_P \approx N_A\]
    Conductivity of P-type semiconductor:

  • If conc. of donor atoms added to a P-type semiconductor exceeds the conc. of acceptor atoms i.e. \(N_D >> N_A\) then P-type is converted to N-type

  • Similarly, if \(N_A >> N_D\) , N-type converted to P-type

Problem

  • Find the conductivity of silicon

    1. in intrinsic condition at a room temp. of \(300^{\circ}\) K

    2. with donor impurity of 1 in \(10^8\)

    3. with acceptor impurity of 1 in \(5\times 10^{7}\)

    4. with both impurities present simultaneously

    Given that

    • \[\begin{aligned} & n_i ~\text{for silicon at}~ 300^{\circ}~\mathrm{K}~= 1.5 \times 10^{10}~\mathrm{cm}^{-3} \\ &\mu_n = 1300~\mathrm{cm^2/V-s}\\ &\mu_p =500~\mathrm{cm^2/V-s}\\ &\text{number of Si atoms per}~\mathrm{cm^3} = 5\times 10^{22} \end{aligned}\]

Solution

  1. \[\begin{aligned} \sigma_i &=q n_i\left(\mu_n+\mu_p\right) =\left(1.6 \times 10^{-19}\right)\left(1.5 \times 10^{10}\right)(1300+500) \\ &=4.32 \times 10^{-6} \mathrm{~S} / \mathrm{cm} \end{aligned}\]
    In intrinsic condition,
  2. \[\begin{aligned} p &=\frac{n_i^2}{n} \approx \frac{n_i^2}{N_D} =\frac{\left(1.5 \times 10^{10}\right)^2}{5 \times 10^{14}}=0.46 \times 10^6 \mathrm{~cm}^{-3} \end{aligned}\]
    \[\begin{aligned} \sigma &=n q \mu_n=N_D q \mu_n \\ &=\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(1300) \\ &=0.104 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]
    \(p\) \(p \ll n\) Therefore, Further, Hence, Number of silicon atoms
\[\begin{aligned} n &=\frac{n_i^2}{p} \approx \frac{n_i^2}{N_A} =\frac{\left(1.5 \times 10^{10}\right)^2}{10^{15}}=2.25 \times 10^5 \mathrm{~cm}^{-3} \end{aligned}\]
\[\begin{aligned} N_A{ }^{\prime} &=N_A-N_D=10^{15}-5 \times 10^{14}=5 \times 10^{14} \mathrm{~cm}^{-3} \\ \sigma &=N_A{ }^{\prime} q \mu_p =\left(5 \times 10^{14}\right)\left(1.6 \times 10^{-19}\right)(500) \\ &=0.04 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]
\[\begin{aligned} \sigma &=p q \mu_P=N_A q \mu_P =\left(10^{15} \times 1.6 \times 10^{-19} \times 500\right) \\ &=0.08 \mathrm{~S} / \mathrm{cm} . \end{aligned}\]
\(n\) \(p \gg n\) Hence, Further, (c)

Drift and Diffusion Currents

  • Flow of charge (current) through a semiconductor or PN junction diode has two components:

    1. Drift current

    2. Diffusion current

  • Drift current :

    • Drift current arises from the movement of carriers in response to an applied electric field.

    • Positive carriers (holes) move in the same direction as the electric field

    • Negative carriers (electrons) move in the opposite direction.

    • The net motion of charged particles generates a drift current that is in the same direction as the applied electric field.

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\[\text{Drift current density,}~J ~\mathrm{A/cm^2} = \begin{cases} J_n = qn\mu_n E ~\Rightarrow \text{due to free electrons} \\ J_p = qp\mu_p E ~\Rightarrow \text{due to holes} \\ \end{cases}\]
\[\begin{aligned} n &= \text{number of free electrons per cubic centimetre} \\ p & = \text{number of holes per cubic centimetre} \\ \mu_n & = \text{mobility of electrons in} ~\mathrm{cm}^2 / \mathrm{V}-s \\ \mu_p & = \text{mobility of holes in} ~\mathrm{cm}^2 / \mathrm{V}-\mathrm{s} \\ E & = \text{applied electric field intensity in}~ \mathrm{V} / \mathrm{cm} \\ q & = \text{charge of an electron} =1.6 \times 10^{-19} ~\text{coulomb} \\ \end{aligned}\]

  • Diffusion current :

    • Electric current can flow even in the absence of applied voltage provided a concentration gradient exists

    • When the number of either electrons or holes is greater in one region than that of other region

    • Charge carrier move from higher to lower concentration of same type of charge carrier

    • Movement of charge carrier resulting in a current called diffusion current

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\[\text{Diffusion current density}~J~\mathrm{A/cm^2} = \begin{cases} J_p = -qD_p\dfrac{dp}{dx} \Rightarrow \text{hole}\\ \\ J_n = -qD_n\dfrac{dn}{dx} \Rightarrow \text{electron} \end{cases}\]

  • hole density \(p(x)\) decreases with increasing \(x\) , hence negative sign

  • \(J_p\) is positive in \(+x\) direction

  • \(dn/dx\) and \(dp/dx\) are concentration gradients

  • \(D_n\) and \(D_p\) are diffusion coefficients in cm 2 /s

  • \[= \begin{cases} J_p=q p \mu_p E-q D_p \dfrac{\mathrm{d} p}{\mathrm{~d} x} \\ \\ J_n=q n \mu_n E-q D_n \dfrac{\mathrm{d} n}{\mathrm{~d} x} \end{cases}\]
    Total current :

  • Einstein Relationship :

    • Relationship between mobility and diffusion coefficient of a particular type of charge carrier in the same semiconductor

    • Higher the charge carrier mobility, greater will be its tendency to diffuse

    • Used to determine \(D_{p,n}\) by experimentally measuring \(\mu_{p,n}\)

    • \[\frac{D_p}{\mu_p}=\dfrac{D_n}{\mu_n}=\dfrac{k T}{q}=V_T\]

Problem-1

A p-type semiconductor with a length of 5 cm, cross-sectional area of 1 cm 2 , and a doping concentration of \(1 \times 10^{16} \, \text{cm}^{-3}\) is subjected to an electric field of 100 V/m. Calculate the drift current in the semiconductor.

Solution:

\[\begin{aligned} \text{Length of the semiconductor}~L & = 5~\mathrm{cm} = 0.05~\mathrm{m} \\ \text{Cross-sectional area}~A &= 1~ \mathrm{cm^2} = 1 \times 10^{-4}~\mathrm{m^2}\\ \text{Doping concentration} ~n_p & = 1 \times 10^{16}~ \text{cm}^{-3}\\ \text{Electric field}~E &= 100~\mathrm{V/cm} \end{aligned}\]
Given Data:
\[\begin{aligned} J_{\text{drift}} & = q \mu_p n_p E \\ & = (1.6 \times 10^{-19} \, \text{C})(0.14 \, \text{m}^2/\text{Vs})(1 \times 10^{16} \, \text{cm}^{-3})(100 \, \text{V/m}) \\ & =2.24 \times 10^{-3} \, \text{A/m}^2 \\ I_{\text{drift}}& = J_{\text{drift}} \times A \\ & = (2.24 \times 10^{-3} \, \text{A/m}^2)(1 \times 10^{-4} \, \text{m}^2)\\ & =2.24 \times 10^{-7}~\mathrm{A} \end{aligned}\]

Problem-2

A silicon bar with a length of 2 mm and a cross-sectional area of \(0.2 \, \text{mm}^2\) has an excess minority carrier concentration of \(3 \times 10^{14} \, \text{cm}^{-3}\) . The diffusion coefficient for minority carriers in silicon is \(10^{-9} \, \text{m}^2/\text{s}\) . Calculate the diffusion current across the silicon bar.

Solution:

Given Data:

\[\begin{aligned} \text{Length of the silicon bar}~L& = 2~ \text{mm} = 2 \times 10^{-3}~ \text{m}\\ \text{Cross-sectional area}~A& = 0.2 ~\text{mm}^2 = 2 \times 10^{-7} \, \text{m}^2\\ \text{Excess minority carrier concentration}~n&= 3 \times 10^{14} \, \text{cm}^{-3}\\ \text{Diffusion coefficient for minority carriers}~D &= 10^{-9} \, \text{m}^2/\text{s} \end{aligned}\]
\[\begin{aligned} J_{\text{diffusion}} & = q D \frac{{dn}}{{dx}} \\ \frac{{dn}}{{dx}} & = \frac{{n}}{{L}} = \frac{{3 \times 10^{14} \, \text{cm}^{-3}}}{{2 \times 10^{-3} \, \text{m}}} = 1.5 \times 10^{17} \, \text{cm}^{-3}\text{m}^{-1} \\ J_{\text{diffusion}} &= (1.6 \times 10^{-19} \, \text{C})(10^{-9} \, \text{m}^2/\text{s})(1.5 \times 10^{17} \, \text{cm}^{-3}\text{m}^{-1}) \\ & = 2.4 \times 10^{-11} \, \text{A/m}^2 \\ I_{\text{diffusion}} &= J_{\text{diffusion}} \times A \\ & = (2.4 \times 10^{-11} \, \text{A/m}^2)(2 \times 10^{-7} \, \text{m}^2)\\ &= 4.8 \times 10^{-18} \, \text{A} \end{aligned}\]