Calculate the drift current density in (mA/m2) for the silicon at roomtemperature. If the intrinsic carrier concentration of (2x1016electron/m³) withthe doping concentration of (3x1016atoms/m³) of the phosphorus and(2x1016atoms/m³) of Boron. Given the mobility for the electrons (0.15m2/V.s)and the mobility for the holes (0.05m2/V.s) and the electric field (100V/m).85.75O 62.77O 82.77O 72.7280.77

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Answered: Calculate the drift current density in…

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter1: Atomic Structure

Section: Chapter Questions

Problem 6RQ: How many valence electrons are generally contained in materials used for insulators?

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Problem 2 In a P-type semiconductor. the Fermi level is 0.3 cV above the valance band at a room temperature of 30O °K. Determine the new position of the Fermi level for termperatures of (a) 350 K and (b) 400 K. The Fermi level in a P-type material is given by
An impurity such as antimony (Sb) has five electrons in its outer shell. When there are Sb impurities in Silicon,..... Please select one: a. the Crystal will be negatively charged b. an N-type material is formed c. both of the above d. None of the above
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Silicon is doped with a boron concentration of 4×1018/cm3. Is boron a donor or acceptor impurity? Find the electron and hole concentrations at 300 K. Is this material n-type or p-type? Find the electron and hole mobilities. What is the resistivity of the material?
A silicon diode is in connected to a DC voltage source with Forward biased, the net currentflowing through the diode is (25mA) where the applied voltage across the terminals of thediode is (820mV). Determine diode temperature, if Is "dark saturation current", the diodeleakage current density in the absence of light is 3.4 × 10−10 A
Consider an p-type silicon for which the dopant concentration ND = 1018/cm3. Find the electron and hole concentrations at T = 350 K. electron = 1018/cm³ & holes= 2.25 × 102/cm³ electron = 2.25 x 102/cm³ & holes= 2.25 x 104/cm3 electron = 17.22 × 104 /cm³ & holes= 1018/cm³ 3 electron 1018/cm³ & holes= 3 %3D 1018/cm3
A 2.5-cm long silicon sample has electron and hole densities as 1015/cm³ and 104/cm³, respectively. Assume that the electron and hole mobilities in silicon are 1100 cm²/(V-s) and 400 cm²/(V-s), respectively. Its cross-sectional area is 3.5 cm² and it is subjected to a 5-V across the length. Determine - (a) The electron and hole velocities in the sample. (b) Electron and hole currents.
Subject - Semi Conductor's At 300K, fermi level is found to be 0.2eV below the edge of conduction band for an n-type silicon semiconductor. Find the electron and hole concentrations. After doping again, fermi level is now founded 0.25eV below the edge of conduction band. Find the type and concentration of impurity atoms added at the later stage At 300K, Nc= 2.8x 1019 cm3
Q7/ Silicon has a conductivity of 5 x 10- (N.m)- when pure. How many indium atoms/m3 are required so that the conductivity of 200 (.m)- could be achieved in silicon using indium as an impurity? m2 Given that the mobility of holes in silicon is 0.05 and of electrons is V. Sec m 0.13 V. Sec
The conductivity for n-side is 1500 s/m and for p-side is 400 s/m in pn junction, while the conductivity of the pure silicon is 4 × 10-4 s/m. Find the barrier potential at 300°K, if the µn = 2.4µp?
An n-type semiconductor sample has an electron density of 6.25x10^18/cm^3 at 300K.If the additive-free concentration of the carriers is 2.5x10^13/cm^3, what is the hole density at this temperature?
The linear electron and hole concentration profiles in a 4 um wide region of silicon material is shown in the figure below. The silicon material is subjected to electron injection from the left and hole injection from the right as shown in the figure. Assume that the cross-sectional area of the material ?=1 ??2, electron mobility ??=1312.75 ??2/? ?, and hole mobility ??=463.33 ??2/? ?. Find the total current (to one decimal place) flowing through the material. ?=300?, ?=1.6×10−19 ?, ?=8.62×10−5 ??/?, ?=1.38×10−23 ?/?, and 1 ??=1.6×10−19 ?. Explain how depletion and diffusion capacitances differ

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