A “thin” slab (thickness d << 1/α, where α is the
absorption coefficient) of GaAs semiconductor at T = 300 K is
uniformly doped with NA = 1016 cm-3 and ND = 0. The GaAs is
illuminated with a light source at t = 0 s, resulting in a uniform
generation rate of electron hole pairs g’ = 1020 cm-3s-1
through-out the slab of GaAs. The electric field is zero.
a) Give the analytic expression for the excess-carrier
concentration δn(t) versus the time t > 0 s.
b) The steady state excess carrier concentration is found to be
δn(∞) = 1014 cm-3. What is the minority carrier lifetime τn0 ?
A “thin” slab (thickness d << 1/α, where α is the absorption coefficient) of GaAs semiconductor...
2. A GaAs semiconductor at T 300 K is uniformly doped with NA 1016 cm3 and No 0. The GaAs is illuminated with a light source at t = 0 s, resulting in a uniform generation rate of electron hole pairs g' 1020 cm . The electric field is zero. a) Give the analytic expression for the excess-carrier concentration δn(t) versus the time t>0s b) The steady state excess carrier concentration is found to be on( is the minority carrier...
PartYour AnswerCorrect AnswerToleranceMarks Comment 4.0 1% 1.00 Correct Total: 1.0 A p-type semiconductor, doped to a uniform concentration of 5.3 x 1015 cm-3, is exposed to a light source which generates electron-hole pairs everywhere in the semiconductor at a rate of 8.8 x 1020 cm-3s-1. The excess carrier lifetime is 0.1 us and the intrinsic carrier concentration is 6.9 x 109 cm-3. Which of the following most accurately gives the hole concentration at steady state? Assume band-to-band transition. Please choose...
Semiconductor Physics: A semiconductor has a intrinsic concentration of 1010 cm-3 and has donor impurity concentration of 1015 cm-3. Light is shined (uniformly) upon the sample generating 5*1019 electron-hole pairs.cm-3s-1 with a minority carrier lifetime of 10-6 s. What is the net electron-hole recombination rate? I think at steady state, the electron-hole recombination rate is the same as the generation rate. But the current circumstances of the question make me doubt that assumption or conclusion (the question is 4 marks).
This is for solid state electronics: Problem 4: Excess electrons and holes are generated at the end of a Si bar (at x=0). The Si is homogeneously doped with P atoms to a concentration of 1017 cm-3. The minority carrier lifetime = 1 μs. The electron diffusion coefficient = 25 cm2/s The hole diffusion coefficient = 10 cm2/s The excess electron and hole concentrations at x=0 are equal to each other and are 1015 cm-3. (a) Write the formula (with...
3. A silicon step junction has uniform impurity doping concentrations of N. 5 x 1015 cm-3 and Nd = 1 x 1015 cm-, and a cross-sectional area of A-|0-4 cm2. Let tao -0.4 s and tpo 0.1 us. Consider the geometry in Figure.Calculate (a) the ideal reverse saturation current due to holes, (b) the ideal reverse saturation current due to electrons, (c) the hole concentration at a, if V V and (d) the electron current at x = x" +...
Problem 1 Using what we have leamed in chapter 1, derive, for a semiconductor, the expressions of The total current density Conductivity - Problem 2 Consider Germanium sample with the following characteristics the electron and hole mobility for Ge is 0.39 and 0.19 m2N.s The electron and hole effectives masses are 0.56me and 0.4 me The energy gap is 0.67 eV at T-27°C 1) 2) Find the intrinsic carrier concentration for Ge What is the resistivity of the Ge sample...
Problem 3 (25 points) Consider a silicon pn junction at T - 300 K, NA- 1016 cm3, ND-5x1016 cm-3. The minority carrier lifetimes are τα , τ,-1 us. The junction is forward biased with Va-0.5V The minority carrier diffusion coefficients are D 25 cm/s, Da- 10 cm2/s n,1.5x1010 cm3 kT 0.0267 Depletion region p-type n-type a) (5 points) Calculate the excess electron concentration as a function of x in the p-side (see the figure above) b) (10 points) Calculate the...
1. a. Find the main error in each of the band diagrams shown below. For all of the band diagrams Ny 1019/cm3, Ne- 1019/cm3, ni = 3 x 108/cm". E,-1.25 eV, T = 300 K. Ef Ef EFi Main error: Main error: Main error: Main error: Consider a semiconductor sample with the following characteristics: EG 1.25 eV, T 300 K, Nd 5 x 101*/cm3, Na 1014/cm3, N.-1019/cm3, N.-1019/cm3, ni-3 × 108/cm3. Assume complete ionization b. Find the equilibrium electron and...