Week7HW Solutions

# Solution r np ni2 n n1 p p p1 n n1

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Unformatted text preview: conduction band is MUCH more efficient than the probably of capturing a hole from the valence band (which is necessary for recombination). 2) When computing SRH recombination rates, we usually focus on defects with energy levels near the middle of the bandgap. Beginning with the SRH expression, show that states near the middle of the bandgap have the largest effect on the SRH recombination rate. Solution: R= np − ni2 ( n + n1 )τ p + ( p + p1 )τ n n1 = ni e( ET − Ei ) k BT ′ p1 = ni e( Ei − ET ) k BT ′ Need to evaluate: ∂ R ∂ ET = 0 ′ −2 ⎛ ∂n ⎞ ∂p ∂R = np − ni2 −1 ⎡ n + n1 τ p + p + p1 τ n ⎤ × ⎜ 1 τ p + 1 τ n ⎟ = 0 ⎣ ⎦ ∂ET ∂ET ⎠ ′ ′ ′ ⎝ ∂ET ( )( ) ( ) ( ) ECE- 606 2 Spring 2013 Mark Lundstrom 2/24/2013 ⎛ ∂n1 ∂p1 ⎞ ⎜ ∂E ′ τ p + ∂E ′ τ n ⎟ = 0 ⎝T ⎠ T ∂ n1 n ∂ p1 p = 1 = − 1 ∂ ET k BT ∂ ET k BT ′ ′ n τ 2 E′ − E k T n1τ p = p1τ n 1 = e ( T i ) B = n p1 τp ET = Ei + ′ k BT ⎛ τ n ⎞ ln ⎜ ⎟ 2 ⎝τp⎠ If τ n is not too different from τ p , then the most effective recombination centers will be located near the intrinsic level. 3) Defect states come in two flavors. A “donor- like” state is positive when empty and neutral when filled. An acceptor- like state is neutral when empty and negative when filled. These states have different cross sections for electrons and holes. For example, when empty, a donor- like state has a large cross section for electron capture because there is a Coulombic attraction (a typical number might be 3 × 10−13 cm2). When a donor like state is filled, it...
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## This note was uploaded on 01/15/2014 for the course ECE 606 taught by Professor Staff during the Fall '08 term at Purdue.

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