ee606_s04_hw_solution1

# ee606_s04_hw_solution1 - Discussion on Selected Homework...

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Discussion on Selected Homework Problems for ECE 606 Homework 6 (6.4) (a) The semiconductor is concluded to be in equilibrium because the Fermi level is invariant as a function of position. (b) J N = 0 and J P = 0 at both x = L/ 2 and x = L/ 2. The net elec- tron and hole currents ar always zero everywhere under equilibrium conditions. (d) There is an electron diffusion current at both x = L/ 2 and x = L/ 2. With dn dx < 0, J N | diff at x = L/ 2 flows in the x direction. Conversely, dn dx > 0 at x = L/ 2 and J N | diff at this point flows in the + x direction. (e) There is an electron drift current at both x = L/ 2 and x = L/ 2. Since both n and ε are non-zero, it follows that J N | droft = n negationslash = 0. The drift component of the current has the same direction as the electric field. Thus J N | drift is in the + x direction at x = L/ 2 and in the x direction at x = L/ 2. Note that the diffusion and drift currents are in the opposite directions at the cited points, as must be the case since the net carrier currents are zero. Homework 5 Question 2 (Paper Question) (Sketch of solutions for the derivations) R = integraldisplay E o c dE c integraldisplay E o c E o w dE w g c ( E c ) f c ( E c ) g w ( E w )[1 f w ( E w )] s ( E c , E w ) 1

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= c n integraldisplay E o c g c ( E c ) f c ( E c ) dE c { integraldisplay E o c E o w g w ( E w ) dE w integraldisplay E o c E o w g w ( E w ) f w ( E w ) } dE w = c n [ n 3 D { integraldisplay E o c E o w g w ( E w ) dE w n 2 D } ] = c n [ n 3 D { N T 2 D n 2 D } ] = c n n 3 D ( P 2 D ) G = integraldisplay E c dE c integraldisplay E o c E o w dE w g c ( E c )[1 f c ( E c )] g w ( E w ) f w ( E w ) s ( E c , E w ) = e n integraldisplay E c dE c g c ( E w )(1 f c ( E c )) integraldisplay E o c E o w g w ( E w
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