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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>>Class 12
>>Physics
>>Semiconductor Electronics: Materials, Devices and Simple Circuits
>>Intrinsic and Extrinsic Semiconductors
>> $p-type and n-type semiconductor$

p-type and n-type semiconductor

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 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: 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: 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: Let \$n_p\$ and \$n_e\$ the number of holes and conduction electrons in an intrinsic semiconductor. Then:

In an intrinsic semicondutor, number of holes carriers are equal to number of electrons as no doping is there in intrinsic material.In extrinsic materials, due to doping, numbers of holes are different to those of electrons.So, \$n_p=n_e\$

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