ECE474S09_Lecture28 - ECE 474: Principles of Electronic...

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Unformatted text preview: ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University [email protected] Lecture 28: Chp. 04: Current State-of-Art Transistors   Photo-generated carriers     Find αr for “low level injection” Excess carriers   Diffusion and Drift Current     Einstein’s relation Find built-in E(x) and n(x) Lecture 28: Chp. 04: Current State-of-Art Transistors   Photo-generated carriers     Find αr for “low level injection” Excess carriers   Diffusion and Drift Current     Einstein’s relation Find built-in E(x) and n(x) Example in Lecture 25: Try to evaluate the constant of proportionality for recombination The constant of proportionality for recombination is αr. (p. 125) Doped Si @ 300K: Doped Si @ 300K + laser light: Given: τn = τp = 1 µsec. Majority carrier concentration: n0 = 1017 cm-3 Minority carrier concentration: p0 = 2.25 x 103 cm-3 n p δn = 101 x 1017 cm-3 => increase factor: 102 ≈ 100 x 1017 cm-3 => increase factor: 1016 = gop τn = (1025 cm-3 s-1)(1 x 10-6 s) = 100 x 1017 cm-3 = gop τn = 100 x 1017 cm-3 δp How well did my example problem meet this criteria? Badly. Now we will consider the problem from a couple of different angles: Effect of recombination. Effect of low level of injection. New Example: Find αr but change: gop = 1015 cm-3 s-1 Doped Si @ 300K: Doped Si @ 300K + laser light: Given: τn = τp = 1 µsec. n p δn = 1.0 x 1017 cm-3 + 1.0 x 109 cm-3 ≈ 1.0 x 1017 cm-3 = 2.25 x 103 cm-3 + 1.0 x 109 cm-3 ≈ 1.0 x 109 cm-3 = gop τn = (1015 cm-3 s-1)(1 x 10-6 s) = 1.0 x 109 cm-3 = gop τn = 1.0 x 109 cm-3 Majority carrier concentration: n0 = 1017 cm-3 Minority carrier concentration: p0 = 2.25 x 103 cm-3 δp “low level injection”: Putting extra carriers into a region OTHER than by intrinsic (thermal) or extrinsic (doping) is called “injection”. Why NOT thermal and doping: because they leave a material NEUTRAL If the extra carrier concentrations are low compared with the thermal and doped concentrations, this is called “low level injection”. Check the levels of injection in the new example on the next page. New Example: Find αr but change: gop = 1015 cm-3 s-1 Doped Si @ 300K: Doped Si @ 300K + laser light: Given: τn = τp = 1 µsec. n p δn = 1.0 x 1017 cm-3 + 1.0 x 109 cm-3 ≈ 1.0 x 1017 cm-3 = 2.25 x 103 cm-3 + 1.0 x 109 cm-3 ≈ 1.0 x 109 cm-3 = gop τn = (1015 cm-3 s-1)(1 x 10-6 s) = 1.0 x 109 cm-3 = gop τn = 1.0 x 109 cm-3 Low level injection of electrons Majority carrier concentration: n0 = 1017 cm-3 Minority carrier concentration: p0 = 2.25 x 103 cm-3 δp YES YES, but our material is n-type Suppose p0 was not small by comparison. “A more general expression..” p. 126 Eq’n (4-8) This expression still requires δn2 = small = “low level injection of electrons” YES, but our material is n-type n0 αrn0 - t / τn Different Example: Start by doing this: Minority carrier concentration: n0 = (2 x 106 cm-3)2/1015 cm-3 = 4 x 10-3 cm-3 Majority carrier concentration: p0 = 1.0 x 1015 cm-3 Start by doing this: δn(t=0) = Δn = gop τn = (gop cm-3 s-1)(1 x 10-8 s) = 1014 cm-3 Minority carrier concentration: n0 = (2 x 106 cm-3)2/1015 cm-3 = 4 x 10-3 cm-3 Majority carrier concentration: p0 = 1.0 x 1015 cm-3 δp(t=0) = Δp = gop τn = 1014 cm-3 n p = 4 x 10-3 cm-3 + 1014 cm-3 ≈ 1014 cm-3 = 1.0 x 1015 cm-3 + 1014 cm-3 = 1.1 x 1015 cm-3 This plot assumes low level injection for electrons, δn<< p0 ? δn(t=0) = Δn = gop τn = (gop cm-3 s-1)(1 x 10-8 s) = 1014 cm-3 Minority carrier concentration: n0 = (2 x 106 cm-3)2/1015 cm-3 = 4 x 10-3 cm-3 Majority carrier concentration: p0 = 1.0 x 1015 cm-3 δp(t=0) = Δp = gop τn = 1014 cm-3 n p = 4 x 10-3 cm-3 + 1014 cm-3 ≈ 1014 cm-3 = 1.0 x 1015 cm-3 + 1014 cm-3 = 1.1 x 1015 cm-3 ? I don’t quite agree. Lecture 28: Chp. 04: Current State-of-Art Transistors   Photo-generated carriers     Find αr for “low level injection” Excess carriers   Diffusion and Drift Current     Einstein’s relation Find built-in E(x) and n(x) Example from Lectures 26 and 27: (Si @300K, Doping yes, laser light no) 2 different n-type doping concentrations together in the same block of Si. n0 = 1017 cm-3 n0 = 1013 cm-3 z y x ...
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This note was uploaded on 04/14/2009 for the course ECE 474 taught by Professor Ayres during the Spring '09 term at Michigan State University.

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