EE3001:Solid State Devices
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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 +
Consider a p-type silicon sample with Ef = E = Ei + 0.25 eV. The built-in (contact) potential across the junction of these two materials (in Volts) is ………
An abrupt silicon p . The built-in potential is V = 8 V. For silicon, assume F/cm. The doping concentration in the n – layer is
An abrupt p+n junction has a built-in potential of 0.75 V and the maximum electric field at a reverse bias of 1 V is 1*10^4 V/cm. The maximum electric field will be 2*10^4 V/cm at a reverse bias voltage (in Volts) of
In a p-n junction, N = 10 /cm
The built-in potential at 300 K (in Volts) is
The built-in potential at 400 K (in Volts) is
An abrupt silicon p . The built-in potential is V = 4 V. For silicon, assume F/cm. The doping concentration in the n – layer is
An abrupt p n junction has a built-in potential of 0.75 V and the maximum electric field at a reverse bias of 2 V is 1 x
In a p-n junction, N = 10 /cm
The built-in potential at 300 K (in Volts) is
The built-in
potential at 400 K (in Volts) is
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 Ec −Ef at 600K. Enter answer without units upto four significant digits
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