lec7v2a - 6.012 Electronic Devices and Circuits Lecture 7...

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6.012 - Electronic Devices and Circuits Lecture 7 - p-n Junctions: I-V Relationship - Outline Announcements Handout - Lecture Outline and Summary First Hour Exam - In 2 wks, Oct. 8, 7:30-9:30 pm; thru this week, PS #4 Review Depletion approximation for an abrupt p-n junction C q Depletion charge storage and depletion capacitance (recitation topic) DP (v AB ) = – AqN Ap x p = – A[2 e q( f b -v 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 b -V AB )] 1/2 I-V relationship for an abrupt p-n junction Assume: 1. Low level injection 2. All applied voltage appears across junction: 3. Majority carriers in quasi-equilibrium 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
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Abrupt p-n 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 < 0 p r ( x ) = Ì -x p Ô qN Dn for 0 < x < x n x n x -qN Ap Ô Ó 0 for x n < x Integrating the charge once gives the electric field Ï 0 for x < - x Ô p Ô - qN Ap ( x + x p ) for - x < x < 0 p Ô E ( x ) = Ì e Si Ô qN Dn ( x - x n ) for 0 < x < x n Ô e Si Ô Ó 0 for x n < x Clif Fonstad, 9/03 Lecture 7 - Slide 2
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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 < 0 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
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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 built-in 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
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The Depletion Approximation, cont.: An important case: asymmetrically doped junctions A p+-n junction (N Ap >> N Dn ): 2 e Si f b 2 q f b N Dn x >> x p , w ª x ª , ª n n qN E pk Dn e Si An n+-p junction (N Dn >> N Ap ): 2 q f b N Ap 2 e Si f b x p >> x n , w ª x ª qN Ap , ª p e E pk Si Note that in both cases the depletion region exists
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