EECE 363 Quiz 2B 2008 Solutions

025 v and s 10410 12 fcm answer the following

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Unformatted text preview: 1010 cm-3, NA=1017 cm-3, ND=2×1016 cm-3, Dp=10 cm2/s, Dn=18 cm2/s, Lp=5 μm, Ln=10 μm, T=300 K, VT=0.025 V, and εs = 1.04×10-12 F/cm. Answer the following questions. (Suggested time to spend: 30 min) a) Calculate the diffusion currents due to holes and electrons Ip and In in the depletion region. (4 points) I p = Aq ( Dp Lp N D ( ) n i eV / VT − 1 2 )( = 2.5 × 10 −5 1.6 × 10 −19 I n = Aq ( ( ) (5 × 10 102 × 10 ) (1.5 × 10 ) (e )( 10 2 −4 0.65 / 0.025 16 ) = 176 μ A ) Dn 2 n i eV / VT − 1 Ln N A )( = 2.5 × 10 −5 1.6 × 10 −19 ) (10 × 10 18)(1× 10 ) (1.5 × 10 ) (e 10 2 −4 0.65 / 0.025 17 ) = 31.7 μ A b) What is the ratio of the junction capacitance with the 0.65-V forward bias to the same capacitance at equilibrium (i.e., zero bias)? Why does the capacitance change with the application of a bias? (4 points) ⎛N N V0 = VT ln⎜ A 2 D ⎜n ⎝i C j (V ) = C j0 1 ⎞2 ⎛ V ⎜1 − ⎟ ⎜ V⎟ 0⎠ ⎝ ( )( 16 ⎡ 17 ⎞ ⎟ = 0.025 × ln⎢ 10 × 2 × 10 ⎟ ⎢ 1.5 × 1010 2 ⎠ ⎣ , then ( C j (0.6...
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This document was uploaded on 02/16/2014 for the course EECE 363 at University of British Columbia.

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