F2010_ECE132_Midterm

F2010_ECE132_Midterm - ECE 132 Midterm Name...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 132 Midterm Name __________________________ November 4, 2010 1) a) A semiconductor is doped with an acceptor dopant which is very close (0.1kT) to the valence band. It has dopant density NA. Calculate the Fermi level relative to EV and then draw it on the diagram below. (EG = 1.1eV; NA = 1017 cm‐3; NC = 2.8∙1019; NV = 1.8∙1019; ni = 1.5∙1010) Ec EG 0.1kT EA, NA Ev Page 1 of 8 1) (Continued) b)The semiconductor from part (a) has had two donor dopants of equal concentration ND added where ND1=ND2=ND. So there are now three doping levels, two donors and one acceptor. ED1 and ED2 are 0.1kT and 5kT from the conduction band edge respectively. (EG = 1.1eV; ND=1016 cm‐3; NA = 1017 cm‐3; NC = 2.8∙1019; NV = 1.8∙1019; ni = 1.5∙1010) (i) (ii) Calculate the Fermi level and draw it on the figure below. 0.1kT 5kT Calculate the equilibrium concentration of valence band holes, p0. Energies not drawn to any precise scale Ec ED1, ND1 ED2, ND2 EG 0.1kT EA, NA Ev (iii) Calculate the value of n0 and explain why it can be ignored in the preceding calculations. Page 2 of 8 2) In a semiconductor the diffusion of excess carriers is described by the diffusion equation, given below. ∂δp ∂ 2δp δp − = Dp ∂t ∂x 2 τ p a) For a region which has no generation or recombination of carriers, and given boundary conditions δp(x=0) and δp(x1), find an expression for the steady state excess hole concentration δp(x) between 0 and x1. δp(x) δp(x1) δp(x=0) x1 x Page 3 of 8 2) (Continued from previous page) b) c) Based on your answer to part (a), Find an expression for the hole diffusion current JP(x) between 0 and x1 which results from this carrier distribution. What is the direction of the hole diffusion current JP in the region 0 to x1? Page 4 of 8 3) a) A silicon PN junction is doped with NA =1017 acceptors/cm3 on the P side and ND =1016 donors/cm3 on the N side. Assuming complete ionization and ignoring minority carriers (n>>p or p>>n) at 300K : ; ; i) Calculate the built in voltage, Vbi. ; P ; ; N Page 5 of 8 3) (Continued from previous page) ii) Calculate the depletion region width, W. iii) Find the maximum magnitude of the electric field, Eo, and sketch the electric field as a function of position (E vs. x). Page 6 of 8 3) (Continued from previous page) b) A similar setup in silicon has an intrinsic (undoped) region placed between the P and N regions, known as a P‐i‐N junction. For the same doping configuration as in part (a), NA =1017 acceptors/cm3 and ND =1016 donors/cm3, and with the same assumptions at 300K : i) Calculate the built in voltage, Vbi. P i N Page 7 of 8 3) (Continued from previous page) ii) Calculate the depletion region width, W. iii) Find the maximum magnitude of the electric field, Eo, and sketch the electric field as a function of position (E vs. x). Page 8 of 8 ...
View Full Document

Ask a homework question - tutors are online