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Unformatted text preview: 6.012  Electronic Devices and Circuits Lecture 7  pn Junctions: IV Relationship  Outline • Announcements Handout  Lecture Outline and Summary First Hour Exam  In 2 wks, Oct. 8, 7:309:30 pm; thru this week, PS #4 • Review Depletion approximation for an abrupt pn junction C q Depletion charge storage and depletion capacitance (recitation topic) DP (v AB ) = – AqN Ap x p = – A[2 e q( f bv AB ){N Ap N Dn /(N Ap +N Dn )}] 1/2 dp (V AB ) ≡ ∂ q DP / ∂ v AB  V AB = A[ e q{N Ap N Dn /(N Ap +N Dn )}/2( f bV AB )] 1/2 • IV relationship for an abrupt pn junction Assume: 1. Low level injection 2. All applied voltage appears across junction: 3. Majority carriers in quasiequilibrium with barrier 4. Negligible SCL generation and recombination Relate minority populations at QNR edges, x p and x n , to v AB Use excesses at x p , x n to find hole and electron currents in QNRs Connect currents across SCL to get total junction current, i D • Features and limitations of model Engineering the minority carrier injection across a junction Deviations at low and high current levels, and large reverse bias Clif Fonstad, 9/03 Lecture 7  Slide 1 Abrupt pn junctions: The Depletion Approximation: an informed first estimate of r (x) Assume full depletion for x p < x < x n , where x p and x are two n unknowns yet to be determined. This leads to: r (x) Ï 0 for x <  x Ô p qN Dn Ô qN Ap for x < x < p r ( x ) = Ì x p Ô qN Dn for 0 < x < x n x n x qN Ap Ô Ó for x n < x Integrating the charge once gives the electric field Ï for x <  x Ô p Ô qN Ap ( x + x p ) for  x < x < p Ô E ( x ) = Ì e Si Ô qN Dn ( x  x n ) for 0 < x < x n Ô e Si Ô Ó for x n < x Clif Fonstad, 9/03 Lecture 7  Slide 2 The Depletion Approximation, cont.: Insisting E(x) is continuous at x = 0 yields our first equation relating x n and x p , our unknowns: N Ap x p = N Dn x n Integration again gives the electrostatic potential: Ï f for x <  x p Ô p Ô f + qN Ap ( x + x p ) 2 for  x < x < p p Ô f ( x ) = Ì 2 e Si 2 Ô f qN Dn ( x  x n ) for 0 < x < x n n Ô 2 e Si Ô Ó f n for x < x n Requiring that the potential be continuous at x = 0 gives us our second relationship between x n and x p : 2 qN Dn 2 f + qN Ap x = f x p p n 2 e Si 2 e Si n Clif Fonstad, 9/03 Lecture 7  Slide 3 The Depletion Approximation, cont.: Combining our two equations and solving for x p and x n gives: x p = 2 e Si f b q N Dn N Ap N Ap + N Dn ( ) , x n = 2 e Si f b q N Ap N Dn N Ap + N Dn ( ) where we have introduced the builtin potential, f b : ˆ kT Ê kT N Ap ˜ = kT N Dn N Ap f b ≡ f f p = ln N Dn   ln ln n q n i Ë Á q n i ¯ q n i 2 We also care about the total width of the depletion region, w : 2 e Si f b ( N Ap + N Dn ) w = x + x = p n q N Ap N Dn And we want to know the peak electric field,  E pk  : qN Ap x p qN Dn x n 2 q f b N Ap N Dn = = = E pk e Si e Si e Si ( N Ap + N Dn ) Clif Fonstad, 9/03 Lecture 7  Slide 4 The Depletion Approximation, cont.: An important case:...
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This note was uploaded on 07/20/2009 for the course CSAIL 6.012 taught by Professor Prof.cliftonfonstadjr. during the Fall '03 term at MIT.
 Fall '03
 Prof.CliftonFonstadJr.

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