Consider a n-type Si crystal at room temperature (300K) doped with 6 x 1016 cm3 arsenic impurity atoms and with certain number of shallow holes. Find out the equilibrium electron concentration, hole concentration and Fermi level EF with respect to E and the conduction band edge Ec. For Si at 300K, the value of n; is 1.45 x 1010 cm3 and k =1.38 x 10-23 J/K, lev =1.60 x 10-19J. The band gap energy, E, of Si is 1.2eV. Solution: n= Nd = 6 x 1016 cm³. In equilibrium condition, hole concentration = 3.5 x 103 cm3. EF - E1 = 0.396eV Ec - EF = 0.164eV.
Consider a n-type Si crystal at room temperature (300K) doped with 6 x 1016 cm3 arsenic impurity atoms and with certain number of shallow holes. Find out the equilibrium electron concentration, hole concentration and Fermi level EF with respect to E and the conduction band edge Ec. For Si at 300K, the value of n; is 1.45 x 1010 cm3 and k =1.38 x 10-23 J/K, lev =1.60 x 10-19J. The band gap energy, E, of Si is 1.2eV. Solution: n= Nd = 6 x 1016 cm³. In equilibrium condition, hole concentration = 3.5 x 103 cm3. EF - E1 = 0.396eV Ec - EF = 0.164eV.
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Consider a n-type Si crystal at room temperature (300K) doped with 6 x
1016 cm-3 arsenic impurity atoms and with certain number of shallow
holes. Find out the equilibrium electron concentration, hole concentration
and Fermi level EF with respect to Ei, and the
For Si at 300K, the value of ni is 1.45 x 1010 cm-3 and k = 1.38 x 10-23 J/K,
1eV = 1.60 x 10-19J. The band gap energy, Eg, of Si is 1.2eV.
Solution:
n @ Nd = 6 x 1016 cm-3.
In equilibrium condition, hole concentration = 3.5 x 103 cm-3.
EF – EI = 0.396eV
EC – EF = 0.164eV.
Transcribed Image Text:Consider a n-type Si crystal at room temperature (300K) doped with 6 x
1016 cm3 arsenic impurity atoms and with certain number of shallow
holes. Find out the equilibrium electron concentration, hole concentration
and Fermi level EF with respect to E and the conduction band edge Ec.
For Si at 300K, the value of n; is 1.45 x 1010 cm3 and k =1.38 x 10-23 J/K,
lev =1.60 x 10-19J. The band gap energy, E, of Si is 1.2eV.
Solution:
n= Nd = 6 x 1016 cm³.
In equilibrium condition, hole concentration = 3.5 x 103 cm3.
EF - E1 = 0.396eV
Ec - EF = 0.164eV.
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