Chapter 5 Solutions[1]

# Chapter 5 Solutions[1] - Problem 5.1 Solution Given...

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Problem 5.1 Solution: Given Thickness, t= 7kA o =7E3*1E-10 m=7E-7m = 7E-5 cm Resistivity, ρ=2.8µΩ -cm Sheet resistance= R s = ρ/t = (2.8 x 10 -6 ) / (7 x 10 -5 ) = 0.04Ω/□ cm So, Sheet resistance is 0.04 Ω/□ cm Problem 5.2 Solution: Given Base resistance = 2 KΩ R s = 160 Ω/□ Width= 8μm and Contact size= 5 μm * 5μm a) Given width Bias of 0.4 μm R = R s L d / W d Ld/Wd = R/Rs=2k/160=12.5 sq Ld=12.5*8μm=100μm R = R s L d / (W d + W b ) = 160*100μm/(8μm+0.4μm)=1.905k ΔR = 2K – 1.905K = 95 Ω b) Change in resistance due to non-uniform current flow is given by: Here K = W e / (W e W c )=8.4μ/(8.4μ-5μ)=2.47 W e = W d + W b = 8.4 μm K = 2.47 ΔR = (160 / π) *(1 / 2.47) ln{ (2.47+1)/(2.47 -1)} + ln {(2.47 2 -1)/2.47 2 }] = 8.658 Ω The non-uniform current flo w resistance change is 8.658 Ω

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c) Contact Resistance, Assuming Al-Cu-Si metallization c = 750 μm 2 ) R c = ((√R s ρ c ) / W c )coth(L c (√R s / ρ c )) = (√ (160 x 750) / 5µ ) x coth (5 µ √(160 / 750)) = 69.282 x 1.0199 = 70.66 Ω Problem 5.4 Solution: 1. Standard bipolar base = 6 μ m consider 2-3 times the minimum width width= 6x(2-3)=12μm
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## This document was uploaded on 09/28/2011.

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Chapter 5 Solutions[1] - Problem 5.1 Solution Given...

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