MIT6_012F09_lec04

# MIT6_012F09_lec04 - 6.012 Electronic Devices and Circuits...

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Unformatted text preview: 6.012 - Electronic Devices and Circuits Lecture 4 - p-n Junctions: Electrostatics - Outline • Review Poisson's equation for φ o (x) given N d (x) and N a (x): ε d 2 φ o (x)/dx 2 = q [n i {e q φ o (x)/kT – e –q φ o (x)/kT } – N d (x) + N a (x)] Knowing φ o (x), we have n o (x) = n i e q φ o (x)/kT and p o (x) = n i e –q φ o (x)/kT Slowly varying profiles: quasi-neutrality holds In n-type, for example, n o (x) ≈ N d (x) – N a (x), p o (x) = n i 2 /n o (x) Given n o and/or p o , φ o (x) = (kT/q)ln[n o (x)/n i ] = – (kT/q)ln[p o (x)/n i ] • Abrupt p-n junction in TE (electrostatics) Abrupt profile: Take as an example an abrupt p-n junction with N a (x) – N d (x) ≡ N Ap for x < 0 and N d (x) – N a (x) ≡ N Dn for x > 0 Observe: 1. n o (x) and p o (x) depend exponentially on φ o (x) 2. φ o (x) is insensitive to the details of the charge profile, ρ (x) Depletion approximation: 0 for x < –x p and x > x n Approximate net charge, ρ (x) ≈ – q N Ap for –x p < x < 0 q N Dn for 0 < x < x n Integrate once to get E(x), and again to get φ (x) Find x p and x n by fitting φ o (x) to known ∆ φ crossing junction • Applying bias to a p-n junction (what happens?) Clif Fonstad, 9/22/09 Lecture 4 - Slide 1 Non-uniform doping in thermal equilibrium Reviewing from Lecture 3: In a non-uniformly doped sample in TE we have: g L (x,t) = 0, J e (x) = 0, J h (x) = 0, and ∂ / dt = 0. Also: n(x) = n o (x) and p(x) = p o (x) . Applying these conditions to the two current density equations gave: and = q μ e n o ( x ) E ( x ) + qD e dn o ( x ) dx " d # dx = D e μ e 1 n o ( x ) dn o ( x ) dx = q μ h p o ( x ) E ( x ) " qD h dp o ( x ) dx # d \$ dx = " D h μ h 1 p o ( x ) dp o ( x ) dx And Poisson ’s equation became: " d 2 # ( x ) dx 2 = dE ( x ) dx = \$ ( x ) % = q % p o ( x ) " n o ( x ) + N d ( x ) " N a ( x ) [ ] In the end, we had three equations in our three remaining unknowns, n o (x), p o (x), and φ (x). Clif Fonstad, 9/22/09 Lecture 4 - Slide 2 Non-uniform doping in thermal equilibrium, cont. The first two equations can be solved by integrating to get: and Ref : " ( x ) = 0 at all x where p o ( x ) = n o ( x ) = n i n o ( x ) = n i e μ e D e " ( x ) p o ( x ) = n i e # μ h D h " ( x ) Next use the Einstein relation: μ h D h = μ e D e = q kT Note: @ R.T. q kT " 40 V # 1 and kT q " 25 mV Using the Einstein relation we have: Finally, putting these in Poisson’s equation, a single equation for φ (x) in a doped semiconductor in TE materializes: n o ( x ) = n i e q " ( x ) kT and p o ( x ) = n i e # q " ( x ) kT d 2 " ( x ) dx 2 = # q \$ n i e # q " ( x )/ kT # e q " ( x )/ kT ( ) + N d ( x ) # N a ( x ) [ ] Clif Fonstad, 9/22/09 Lecture 4 - Slide 3 Doing the numbers: I. D to µ conversions, and visa versa To convert between D and µ it is convenient to say 25 mV, kT/q ≈ in which case q/kT ≈ 40 V-1 : 17 ˚C/62˚F Example 1: µ e = 1600 cm 2 /V-s, µ h = 600 cm 2 /V-s D e = μ e q kT ( ) = 1600/40 = 40 cm 2 / s D h =...
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MIT6_012F09_lec04 - 6.012 Electronic Devices and Circuits...

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