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Use Eq. (2) below to calculate the intrinsic number density...
Use the equations for n and p on page 3 to obtain an expression for the...
Use the equations for n and p on page 3 to
obtain an expression for the intrinsic carrier concentration
ni . HINT: Use the mass action
law: np=n2i.
For a semiconductor material, the number of free electrons per unit volume in the conduction band is given by the expression: and the number of holes per unit volume in the valence band is given by: exp kT where m and mp are the effective masses of the electrons and holes.
4. Estimate the number per cm of intrinsic electrons, n, and holes. p, and calculate the...
4. Estimate the number per cm of intrinsic electrons, n, and holes. p, and calculate the electrical conductivity of Si crystal at 200 C. Take the effective mass of electrons as 0.33me and holes as 0.55m, and μ.-0.14 mm 0.038 m'N.s, Eg 1.12 eV. Please pay attention to the units used
You have an intrinsic semiconductor. (a) When temperature T = 0[K], obtain the density of electrons...
You have an intrinsic semiconductor. (a) When temperature T = 0[K], obtain the density of electrons n in the conduction band and that of holes p in the valence band; (b) When T = 300[K], obtain the mathematical relationship between n and p (e.g., n=p, n>p, or n<p); (c) When T = 300[K], obtain the mathematical relationship between the np product and the intrinsic carrier concentration ni.
Consider a semiconductor material X, with the following parameters at a room temperature of 300K: Energy...
Consider a semiconductor material X, with the following parameters at a room temperature of 300K: Energy bandgap of Eg = 1.15 ev, density of states at the Conduction band edge of Nc = 4.8e+23, effective density of states at the Valance band edge of Nv = 1e+25, drift mobilities of the electrons and holes, ue and uh, such that ue =0.4 and uh = 0.02. (1) What is the intrinsic concentration and conductivity of 'material x' at room temperature 300K?...
2. Review of density of states Calculate the total number of available states in the conduction b...
2. Review of density of states Calculate the total number of available states in the conduction band of silicon between energies Ee and Ec+4kT. The density of states at an energy E in conduction band and valence band are given by: If the total number of available states calculated bertween E. and E.+4kT is equal to the total number of available states calculated between Ev and E-ykT. Find the value of 'y' (bandgap of Silicon is 1.12 eV).
2. Review...
Question 8 Pure silicon at room temperature has an electron number density of about 5 ×...
Question 8 Pure silicon at room temperature has an electron number density of about 5 × 1015 m3 and an equal density of holes In the valence band. Suppose that one of every 10° silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, n or p? (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and...
In class Monday we established that the number density of free electrons in silicon was 1.09E+16 electrons per cubic meter. Now calculate the number of free electrons per silicon atom. The density of...
In class Monday we established that the number density of free electrons in silicon was 1.09E+16 electrons per cubic meter. Now calculate the number of free electrons per silicon atom. The density of silicon is 2.33 Mg/m3 ; the atomic mass of silicon is 28.085 g/mole. Consider silicon which has a band gap of 1.11 eV and a measured conductivity of 0.00034 /ohmm at 300K. Its electron mobility is 0.145 m^2/(V x sec) and its hole mobility is 0.050 m^2/(V...
Band structure Consider a one-dimensional semiconductor crystal consisting of 11 atoms with nearest- neighbor atoms separated...
Band structure Consider a one-dimensional semiconductor crystal consisting of 11 atoms with nearest- neighbor atoms separated by a 5 . The band structure for electrons in the conduction band is given by Ec(k) = 101(k-0.2n)2-A(k-02n)"] + 2.25 [eV] and the band structure for holes in the valence band is given by where the wavevector k s in units ofA-1. The allowed wavevectors are--< k 즈 al (a) Is this a direct or indirect gap semiconductor? What is the energy gap...
b. At TR, calculate the number of free electrons (n) per mº for Ag assuming the...
b. At TR, calculate the number of free electrons (n) per mº for Ag assuming the electron mobility of Ag is; He. Ag = 0.012 m?NV-S c. For the same number of free electrons determined in b., if you were able to purify the Ag material and increase the electron mobility to 0.15 m²N - S, what is the new conductivity of Ag, Ag? Would this conductivity be greater or lesser than for the Ag in b.? LAANI UJ points...
Please explain part b in details thx! Question 2 At 300 K, the bandgap of GaP is 2.26 eV and the effective density of states at the conduction and valence band edge are 1.8 x 1019 cm23 and 1.9 x 1019...
Please explain part b in details thx!
Question 2 At 300 K, the bandgap of GaP is 2.26 eV and the effective density of states at the conduction and valence band edge are 1.8 x 1019 cm23 and 1.9 x 1019 cm3, respectively. (a) Calculate the intrinsic concentration of GaP at 300K (7 marks) Calculate the GaP effective mass of holes at 300K. (b) (8 marks) (c The GaP sample is now doped with donor concentration of 1021 cm3 with...
Use the equations for n and p on page 3 to
obtain an expression for the intrinsic carrier concentration
ni . HINT: Use the mass action
law: np=n2i.
For a semiconductor material, the number of free electrons per unit volume in the conduction band is given by the expression: and the number of holes per unit volume in the valence band is given by: exp kT where m and mp are the effective masses of the electrons and holes.
4. Estimate the number per cm of intrinsic electrons, n, and holes. p, and calculate the electrical conductivity of Si crystal at 200 C. Take the effective mass of electrons as 0.33me and holes as 0.55m, and μ.-0.14 mm 0.038 m'N.s, Eg 1.12 eV. Please pay attention to the units used
2. Review of density of states Calculate the total number of available states in the conduction band of silicon between energies Ee and Ec+4kT. The density of states at an energy E in conduction band and valence band are given by: If the total number of available states calculated bertween E. and E.+4kT is equal to the total number of available states calculated between Ev and E-ykT. Find the value of 'y' (bandgap of Silicon is 1.12 eV).
2. Review...
Question 8 Pure silicon at room temperature has an electron number density of about 5 × 1015 m3 and an equal density of holes In the valence band. Suppose that one of every 10° silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, n or p? (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and...
Band structure Consider a one-dimensional semiconductor crystal consisting of 11 atoms with nearest- neighbor atoms separated by a 5 . The band structure for electrons in the conduction band is given by Ec(k) = 101(k-0.2n)2-A(k-02n)"] + 2.25 [eV] and the band structure for holes in the valence band is given by where the wavevector k s in units ofA-1. The allowed wavevectors are--< k 즈 al (a) Is this a direct or indirect gap semiconductor? What is the energy gap...
b. At TR, calculate the number of free electrons (n) per mº for Ag assuming the electron mobility of Ag is; He. Ag = 0.012 m?NV-S c. For the same number of free electrons determined in b., if you were able to purify the Ag material and increase the electron mobility to 0.15 m²N - S, what is the new conductivity of Ag, Ag? Would this conductivity be greater or lesser than for the Ag in b.? LAANI UJ points...
Please explain part b in details thx!
Question 2 At 300 K, the bandgap of GaP is 2.26 eV and the effective density of states at the conduction and valence band edge are 1.8 x 1019 cm23 and 1.9 x 1019 cm3, respectively. (a) Calculate the intrinsic concentration of GaP at 300K (7 marks) Calculate the GaP effective mass of holes at 300K. (b) (8 marks) (c The GaP sample is now doped with donor concentration of 1021 cm3 with...
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