Demystifying Semiconductors: The Building Blocks of Modern Electronics

Demonstrative Video


Basic Electronics Contents:



Semiconductor Fundamentals


Semiconductor Materials

  • 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

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Intrinsic Semiconductor


Extrinsic Semiconductor

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


p-type semiconductors

Summary

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Mass Action Law


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


Problem

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

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

Problem-1

A p-type semiconductor with a length of 5 cm, cross-sectional area of 1 cm2, 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:

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