Pure \$Si\$ at \$500\ K\$ has equal number of electron \$\left({n}_{e}\right)\$ and hole \$\left({n}_{h}\right)\$ concentration of \$1.5 \times {10}^{16} \ {m}^{-3}\$. Doping by indium increases \${n}_{h}\$ to \$4.5 \times {10}^{22}\ {m}^{-3}\$. The doped semiconductor is of

\${n}_{i}^{2} = {n}_{e}{n}_{h}\$\${\left(1.5 \times {10}^{16}\right)}^{2} = {n}_{e}\left(4.5 \times {10}^{22}\right)\$\${n}_{e} = 0.5 \times {10}^{10}\$\${n}_{e} = 5 \times {10}^{9}\$\${n}_{h} = 4.5 \times {10}^{22}\$\${n}_{h} \gg {n}_{e}\$Semiconductor is \$p\$-type and \${n}_{e} = 5 \times {10}^{9} {m}^{-3}\$.

If the ratio of the concentration of electrons to that of holes in a semiconductors is \$\dfrac{7}{5}\$ and the ratio of currents is\$\dfrac{7}{4}\$, then the ratio of drift velocities is

The drift velocity of charge carriers in the material of constant cross-sectional area A is given as:\$v=\dfrac{I}{nAq}\$(here I is the current flowing through the material, n is the charge-carrier density, and q is the charge on the charge-carrier.)We have the ratio of I as \$\dfrac{7}{4}\$ and ratio of n as \$\dfrac{7}{5}\$.Thus we get the ratio of v as \$\dfrac{5}{4}\$

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>> $Electron Hole Concentration...$

Electron Hole Concentration Calculation

Q: Suppose a pure Si crystal has \$5\times 10^{28}\$ atoms \$m^{-3}\$. it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that \$n_{1}=1.5\times 10^{16}m^{-3}\$.

1 atom of Si doped out of \$10^{6}\$ atoms (1 ppm of As)\$ \Rightarrow \$ In \$ 5 \times 10^{28}\$ atoms net doped =\$\frac{5\times10^{28}}{10^{6}} = 5\times 10^{22} \$ atoms1 As atom creates \$1 e^{\circleddash }\$ excenso, no. of excen electron \$ = 5 \times 10^{22} = n_{e}\$also, \$ n_{e}n_{n}=n_{1}^{2}\$\$ \Rightarrow n_{n}=\frac{n_{1}^{2}}{n_{e}} = \frac{(1.5\times10^{16})^{2}}{5\times 10^{22}} = 4.5\times 10^{9}\$\$ \therefore \$ No. of holes formed \$ = 4.5 \times 10^{9}.\$as \$ n_{e} > n_{n} \rightarrow \$ It is a N - type semiconductor formed.

Q: Write down 5 difference between N-type and P-type semiconductors.

Difference between the P and N type semiconductors Sl No P-type Semi conductor N-type semi conductor 1P-type semiconductor is formed due to the dopping of III group elements i.e. Boron, Aluminium, Thallium.N-type semi conductor is formed due to dopping of Nitrogen, Phosporus, Arsenic, Antimony, Bismoth. 2These are also known as Trivalent semi conductors.These are also known pentavalent semiconductor. 3P-type semiconductors is positive type semiconductor it means it deficiency of 1 electron is required.N-type semiconductor is negative type semi-conductor it means excess of 1 electron is required. 4In P-type semiconductor majority charge carries are holes and minority charge carries are electrons.In N-type semiconductor majority charge carries are electrons and minority charge carries are hole. 5A hole indicates a missing electron. In this no. of holes is more than the no. of electrons.In N-type semiconductor the no. of holes is less than the no. of free electron.

Q: The number of silicon atoms per \$m^3\$ is \$5\, \times\, 10^{28}\$. This is doped simultaneously with \$5\, \times\, 10^{22}\$ atoms per \$m^3\$ of arsenic and \$5 \times\, 10^{20}\$ per \$m^3\$ atoms of indium. Calculate the number of electrons and holes. Given that \$n_i\, =\, 1.5\, \times\, 10^{16}\ /m^{3}\$. Is the material n-type or p-type?

Number of silicon atoms, \$N=5\times10^{28}\ atom/m^3\$Number of arsenic atoms, \$n_{AS}=5\times10^{22}\ atom/m^3\$Number of indium atoms, \$n_{IN}=5\times10^{22}\ atom/m^3\$\$n_i=1.5\times10^{16}\ electron/m^3\$\$n_e=5\times10^{22}-1.5\times10^6=4.99\times10^{22}\$Let number of holes be \$n_h\$In thermal equilibrium, \$n_en_h=n_i^2\$so, \$n_h=4.51\times10^9\$\$n_e>n_H\$, material is an n-type semiconductor.

Q: A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per \$5\times 10^7\$ silicon atoms. If the number density of atoms in the silicon specimen is \$5\times 10^{28}\$ atoms\$/m^3\$. Then the number of acceptor atoms in silicon will be.

Number density of atoms in Si\$=\displaystyle 5\times 10^{28}\frac{atoms}{m^3}=5\times 10^{22}\frac{atoms}{cm^3}\$Since \$1\$ atom of indium is doped in \$5\times 10^7\$ silicon atoms,\$\therefore\$ Total number of doped indium atoms\$\displaystyle =\frac{5\times 10^{22}}{5\times 10^7}=1\times 10^{15}\frac{atoms}{cm^3}\$\$\therefore\$ Number of acceptor atoms in silicon\$=\displaystyle 1\times 10^{15}\frac{atoms}{cm^3}\$.

Q: A semiconductor has equal electron and hole concentration of \$ 6 \times 10^8 / m^3 \$. On doping with certain impurity , electron concentration increases to \$ 9 \times 10^{12} / m^3 \$. Calculate the new hole concentration.

\$ n_e n_h = n^{2}_{i} \$ \$ \Rightarrow \$ \$ n_h = \dfrac{n^{2}_{i}}{n_e} = \dfrac{(6 \times 10^8)^2}{9 \times 10^{12}} = 4 \times 10^4 \, per \, m^3 \$.

Q: Explain the term 'doping'

Doping is the process of adding some impurity atoms in a pure or (intrinsic) semiconductor so as to increase the conductivity of a semiconductor. Doping can be done in two ways:n-type dopant is added, or a pentavalent dopant is added to an intrinsic semiconductor, to form an n-type semiconductor.When p-type dopant or a trivalent dopant atom is added to an intrinsic semiconductor, a p-type semiconductor is formed.

Q: In an intrinsic semiconductor, the number of electrons in the conduction band is ________ the number of holes in the valence band.

In an intrinsic semiconductor, the number density of electrons is equal to the number density of holes i.e \$ne=nh\$.Since there is no doping, no extra hole or electron is produced.

Q: Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are \$0.36m^2V^{-1}s^{-1}\$ and \$0.17m^2V^{-1}s^{-1}\$. The electron and hole densities are each equal to \$2.5\times 10^{19}m^3\$. The electrical conductivity of germanium is

Given in question,Mobilities of electrons\$\mu_e=0.36\times m^2V^{-1}s^{-1}\$Mobilities o holes\$\mu_h=0.17\times m^2V^{-1}s^{-1}\$densities of electron\$=\$ densities of holes\$=2.5\times10^{19}m^{-3}\$As we know, conductivity,\$\sigma = \dfrac{1}{p}=e({\mu}_e n_e+{\mu}_n n_n)\$\$=1.6\times 10^{-19}[0.36\times2.5\times10^{19}+0.17\times2.5\times 10^{19})]\$\$=2.12 Sm^{-1}\$So the electrical conductivies of germanium is\$2.12Sm^{-1}\$

Physics

example

Applications of mass-action law

According to Mass-action Law:

n_{e} n_{h} = n_{i}^{2}

For intrinsic semiconductors,

n_{e} = n_{h}

For n-type semiconductors,

n_{e} \approx N_{D}

For p-type semiconductors,

n_{h} \approx N_{A}

For extrinsic semiconductors where approximation is not valid, electrical neutrality must be used along with mass action law.

n_{e} n_{h} = n_{i}^{2}

N_{D} + n_{h} = N_{A} + n_{e}

We have two equations and two variables and hence, the problems can be solved.

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Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times 1 0^{19} m^{3}$ . The electrical conductivity of germanium is

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Electron Hole Concentration Calculation

Suppose a pure Si crystal has 5×1028 atoms m−3. it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n1​=1.5×1016m−3.

Write down 5 difference between N-type and P-type semiconductors.

The number of silicon atoms per m3 is 5×1028. This is doped simultaneously with 5×1022 atoms per m3 of arsenic and 5×1020 per m3 atoms of indium. Calculate the number of electrons and holes. Given that ni​=1.5×1016 /m3. Is the material n-type or p-type?

Pure Si at 500 K has equal number of electron (ne​) and hole (nh​) concentration of 1.5×1016 m−3. Doping by indium increases nh​ to 4.5×1022 m−3. The doped semiconductor is of

A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per 5×107 silicon atoms. If the number density of atoms in the silicon specimen is 5×1028 atoms/m3. Then the number of acceptor atoms in silicon will be.

A semiconductor has equal electron and hole concentration of 6×108/m3. On doping with certain impurity , electron concentration increases to 9×1012/m3. Calculate the new hole concentration.

If the ratio of the concentration of electrons to that of holes in a semiconductors is 57​ and the ratio of currents is47​, then the ratio of drift velocities is

Explain the term 'doping'

In an intrinsic semiconductor, the number of electrons in the conduction band is ________ the number of holes in the valence band.

Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are 0.36m2V−1s−1 and 0.17m2V−1s−1. The electron and hole densities are each equal to 2.5×1019m3. The electrical conductivity of germanium is

Suppose a pure Si crystal has $5 \times 1 0^{28}$ atoms $m^{- 3}$ . it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_{1} = 1.5 \times 1 0^{16} m^{- 3}$ .

Pure $S i$ at $500 K$ has equal number of electron $(n_{e})$ and hole $(n_{h})$ concentration of $1.5 \times 10^{16} m^{- 3}$ . Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3}$ . The doped semiconductor is of

A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per $5 \times 1 0^{7}$ silicon atoms. If the number density of atoms in the silicon specimen is $5 \times 1 0^{28}$ atoms $/ m^{3}$ . Then the number of acceptor atoms in silicon will be.

A semiconductor has equal electron and hole concentration of $6 \times 1 0^{8} / m^{3}$ . On doping with certain impurity , electron concentration increases to $9 \times 1 0^{12} / m^{3}$ . Calculate the new hole concentration.

If the ratio of the concentration of electrons to that of holes in a semiconductors is $\frac{7}{5}$ and the ratio of currents is $\frac{7}{4}$ , then the ratio of drift velocities is

Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times 1 0^{19} m^{3}$ . The electrical conductivity of germanium is