Lect_02 - ECE 442 ECE 442 2. PN Junctions and Diodes J Jose...

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CE 442 ECE 442 2. PN Junctions and Diodes Jose E. Schutt-Aine lectrical & Computer Engineering Electrical & Computer Engineering University of Illinois [email protected] ECE 442–Jose Schutt Aine 1 1
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Definitions B : material dependent parameter = 5.4 10 31 for Si E G : Bandgap energy = 1.12 eV k : Boltzmann constant=8.62 10 -5 ev/K intrinsic carrier concentration n i : intrinsic carrier concentration At T = 300 K, n i = 1.5 10 10 carriers/cm 3 J p : current density A/m 2 q : electron charge D p : Diffusion constant (diffusivity) of holes p : mobility for holes = 480 cm 2 /V sec : mobility for electrons = 1350 cm 2 /V sec n N D : concentration of donor atoms n no : concentration of free electrons at thermal equilibrium N A : concentration of acceptor atoms p po : concentration of holes at thermal equilibrium Einstein Relation : : p n D Dk T V thermal voltage  ECE 442–Jose Schutt Aine 2 T np g q
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PN Junction When a p material is connected to an n-type material, a junction is formed oles from p pe diffuse to n pe region – Holes from p-type diffuse to n-type region – Electrons from n-type diffuse to p-type region – Through these diffusion processes, recombination takes place ome holes disappear from p- pe So e o es d sappea o p type – Some electrons disappear from n-type A depletion region consisting of bound charges is thus formed harges on both sides cause electric field otential ECE 442–Jose Schutt Aine 3 Charges on both sides cause electric field potential = V o 3
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PN Junction Potential acts as barrier that must be overcome for holes to diffuse into the n-region and electrons to diffuse into the p-region pen circuit: No external current Open circuit: No external current Junction built-in voltage From principle of detailed balance and equilibrium we get: 2 ln AD oT i NN VV n    :silicon permittivity s 8 11.7 1.04 10 F/m so   For Si, V o is typically 0.6V to 0.8V Charge equality in depletion region gives: AN AN n A pD x N xN pA nD qx AN qx AN A: cross-section of junction x p : width in p side 2 11 s ep n p o Wx x V  ECE 442–Jose Schutt Aine 4 x n : width in n side 4 dep qN N
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Example Find the barrier voltage across the depletion region of a silicon diode at T = 300 K with N D =10 15 /cm 3 and N A =10 18 /cm 3 . N 2 ln A D oT i NN VV n    Use 0 3 @ 300K, 10 1. 51 0/ cm i n 0.026 V T V  18 15 13 2 20 10 10 10 0.026ln 0.026ln 2.25 1.5 10 oo V   0.026 29.12 0.7571 volts V  0.7571 volts V ECE 442–Jose Schutt Aine 5 5
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PN Junction under Reverse Bias When a reverse bias is applied – Transient occurs during which depletion capacitance is charged to new bias voltage – Increase of space charge region – Diffusion current decreases – Drift current remains constant arrier potential is increased – Barrier potential is increased – A steady state is reached – After transient: steady-state reverse current = I S -I D ( I D is very small) reverse current ~ I ~10 -15 A ECE 442–Jose Schutt Aine 6 S Under reverse bias the current in the diode is negligible 6
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This note was uploaded on 05/05/2010 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

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Lect_02 - ECE 442 ECE 442 2. PN Junctions and Diodes J Jose...

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