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ELCT363_Lecture15_Direct&Indirect_Bandgap_Excess_Carriers

ELCT363_Lecture15_Direct&Indirect_Bandgap_Excess_Carriers

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ELCT 363: Intro. to Microelectronics Lecture 15; Slide Mandal/Fall 2011 Ø There will be a Quiz (Quiz #2) on Tuesday, October 11 § Quiz #2 will cover Lecture #8 to Lecture #14 Ø Homework #2 Solution has been posted in the Blackboard Announcement 1
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ELCT 363: Intro. to Microelectronics Lecture 15; Slide Mandal/Fall 2011 -- Photon Electron Hole For the direct photon into e-h pair transformation, Eph = Eg; where Eph = h  is the photon energy The mass of the photon is negligibly small, therefore: pph  0; pe  ph; Therefore after the absorption the e and the h must have equal momentum pph = | pe - ph|, p is the momentum, p = m × v Interaction between incoming photon and generated electron and hole must satisfy the energy and momentum conservation laws Concept of k -Space and E vs. k Diagram 2
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ELCT 363: Intro. to Microelectronics Lecture 15; Slide Mandal/Fall 2011 The momentum of the quantum-mechanical particle of the mass m and velocity v is given by p = m  v = where is the wave vector of the particle and is the dimensionless Planck’s constant, = h/2 k λ π 2 = k The kinetic energy of a free particle using a classical mechanics is: 2 2 v m E = Since mv = k m k E 2 2 2 = i.e. the E(k) dependence of free electron is parabolic. Simple parabola equation: y= ax2 Momentum Energy 2 2 2 2 1 2 p m m k E = = 3
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ELCT 363: Intro. to Microelectronics Lecture 15; Slide Mandal/Fall 2011 Direct bandgap semiconductor Ø In a direct bandgap semiconductor , the valence band maxima and the conduction band minima occur at the same momentum space.
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