ECE474S09_Lecture24

# ECE474S09_Lecture24 - ECE 474 Principles of Electronic...

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ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

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Lecture 24: Chp. 04: Current State-of-Art Transistors Photo-generated carriers Introduction of Diffusion Current
Lecture 24: Chp. 04: Current State-of-Art Transistors Photo-generated carriers Chp. 03: Thermal: undoped, “intrinsic”, fig 3-17 Chp. 03: From dopants Na and Nd, “extrinsic”, Eq’n 3-43 + fig. 3-23 Chp. 04: shine laser light on sample Introduction of Diffusion Current

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Undoped Si @ 300K
Undoped Si @ 300K: Temperature A b s o r p ti o n

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Undoped Si @ 300K: A b s o r p ti o n Result: a steady state concentration of e- and holes, n i =p i = 1.5 x 10 10 cm -3
Undoped Si @ 300K: Review question: kT = 0.0259 eV, not 1.1 eV. How is this possible? A b s o r p ti o n

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Undoped Si @ 300K: There is a probability f(E) for an electron to have energy E = E C . => f(E C ) = [1 + exp(E C -E F )/kT] -1 . For undoped, E F E i = E gap /2 => f(E C ) = [1 + exp(E C -E i )/kT] -1 . A b s o r p ti o n
Undoped Si @ 300K: NEW: define: gi = the generation rate of electron- hole pairs (EHPs) per sec due to Temperature. P. 81 A b s o r p ti o n Result: a steady state concentration of e- and holes, n i =p i = 1.5 x 10 10 cm -3

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Undoped Si @ 300K: A b s o r p ti o n R e c o m b i n a ti o n Result: a steady state concentration of e- and holes, n i =p i = 1.5 x 10 10 cm -3
Undoped Si @ 300K: R e c o m b i n a ti o n Result: a steady state concentration of e- and holes, n i =p i = 1.5 x 10 10 cm -3 NEW: define: ri = the recombination rate of electron-hole pairs (EHPs) per sec P. 81

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Undoped Si @ 300K: A b s o r p ti o n R e c o m b i n a ti o n Result: a steady state concentration of e- and holes, n i =p i = 1.5 x 10 10 cm -3
Doped Si @ 300K

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Doped Si @ 300K Minority Majority = e- OR Majority = holes
Doped Si @ 300K Example: N d = 10 17 cm -3 P atoms Fully ionised: N d = N d + Analyse this.

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Doped Si @ 300K Example: N d = 10 17 cm -3 P atoms Fully ionised: N d = N d + Majority carrier concentration n 0 N d + = 10 17 cm -3 E F -E i = kT ln[n 0 /n i ] = 0.0259eV ln 10 17 cm -3 1.5 x 10 10 cm -3 = 0.407 eV E F E i E V E C E gap = 1.1 eV 0.407 eV
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