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Hello, this question is extremely important. Rating highestlevel, but you must answer all 5 parts of question to receive fullcredit. Please list all steps, graphs, etc. as they're needed andany intuition you used to complete problems. Thank you,8. An abrupt silicon p-n junction has NA = 2 x 10^14 cm^-3 on one side andND = 5 x 10^15 cm-3 on theother. At atemperature of 300Â°Ka) (4) Find the position of the Fermi levels in both the p and nregionsb) (4) Find the majority concentrations in each regionc) (4) Find the minority concentrations in each regiond) (4) Draw and label the equilibrium band diagrame) (4) Determine the size of the potential "hill"
Nc (cm-3) - Effective Density ofStates for electrons
NV (cm-3) - EffectiveDensity of States for holes
ni (cm-3) - IntrinsicCarrier Concentration
nn, pp (cm-3) -Majority Carrier Concentrations (electrons in n-type or holes inp-type)
np, pn (cm-3) -Minority Carrier Concentrations (electrons in p-type or holes inn-type)
ND (cm-3) - Donor DopingConcentration (density of donor atoms)
NA (cm-3) - Acceptor DopingConcentration (density of acceptor atoms)
EF (eV) - Fermi level in extrinsicmaterial
EFi (eV) - Fermi level in intrinsicmaterial
2.90 x 1019 cm-3
1.02 x 1019 cm-3
1.05 x 1019 cm-3
5.65 x 1018 cm-3
1.0 x 1010 cm-3
1.8 x 1013 cm-3
At 300Â° K
To understand the meaning of capacitance and ways of calculating capacitance.
When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the conductor to ground, and the conductor will have a nonzero potential V relative to ground. If more charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this conductor: C=q/V.
Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are often called electrodes), and the relevant voltage is the voltage between these two electrodes.
This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallel-plate capacitor, which consists of two plates each of area A separated by a small distanced with air or vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in mind that the voltage difference is the line integral of the electric field between the plates.
With the questionAn oceanographer is studying how the ion concentration in seawaterdepends on depth. She makes a measurement by lowering into thewater a pair of concentric metallic cylinders at the end of a cableand taking data to determine the resistance between theseelectrodes as a function of depth. The water between the twocylinders forms a cylindrical shell of inner radius ra, outerradius rb, and length L much larger than rb. The scientist appliesa potential difference ?V between the inner and outer surfaces,producing an outward radial current I. Let ? represent theresistivity of the water.We realize that we integrate resistivity(Length/area) but then fromthe solution shown we integrate dR= resistivity(dr/(2pirL)) Weunderstand everything but why the L goes to the bottom of thisintegration.Thanks.