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Intrinsic temperature(Ti) is defined as the temperature at which the intrinsic concentration ni equals the doping concentration N. Calculate the value of Ti for Si with doping concentration ND = 10^15 cm−3 and assuming Nc and Nv to be independent of temperature. [Given EG,Si = 1.12 eV, Nc=2.8 × 10^19 cm−3, and Nv=10^19 cm−3. At 300 K kBT=0.026 eV.]. Enter answer without units.
1016 phosphorus atoms/cm3 are introduced in Si. Assuming complete ionization, EG,Si = 1.12 eV, Nc=3.2 x 1019(T/300)1.5 cm−3 and Nv=1.8 x 1019 (T/300)1.5 cm−3, kBT=0.026 eV (at 300 K) and ni = 1010 /cm3 (at 300 K), find out electron concentration in the conduction band at 600 K. If your answer in the format a * 1016 , enter only a without units upto four significant digits.
Intrinsic temperature (Ti) is defined as the temperature at which the intrinsic concentration ni equals the doping concentration N. Calculate the value of Ti for GaAs with doping concentration ND = 10^14 cm−3 and assuming Nc and Nv to be independent of temperature. [Given EG,GaAs = 1.43 eV, Nc=4.7 × 10^17 cm−3 and Nv=7 × 10^18 cm−3. At T=300 K kBT=0.026 eV.] Enter answer without units.
The position of the Fermi level for a silicon sample is at E = Ei + 0.45 eV at 300 K. The sample is uniformly illuminated, thereby generating an additional 10 per second. The minority carrier lifetime in this sample is 1 μs. Assume n for Si is 10 and kT = 26 meV at 300 K. For this illuminated sample,
at 300 K the position of electron quasi Fermi level (in eV) is at E = Ei +
at 300 K the position of hole quasi Fermi level (in eV) is at E = Ei +
A phosphorus doped (1016 atoms/cm3 ) Si sample has resistivity of Ω-cm. This sample is now doped with additional boron atoms to increase the resistivity of the doped sample to . Assume that due to this additional boron doping electron and hole mobilities are not changed and electron mobility is three times of the hole mobility throughout. The concentration of the additional boron atoms required is within the range from
Consider a p-type silicon sample with Ef = E = Ei + 0.35 eV. The built-in (contact) potential across the junction of these two materials (in Volts) is ………
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