Answer to Question #207531 in Mechanical Engineering for UUID

Question #207531

An Al-silicon dioxide-silicon MOS capacitor has an oxide thickness of 450 Å and a doping of NA 10¹5 cm³. The fixed oxide charge density is 3 x 10¹¹1 cm². Calculate (a) the flat-band voltage and (b) the threshold voltage.


1
Expert's answer
2021-06-24T11:53:02-0400

Part a

ϕFP=(0.0259)ln(1015151010)\phi_{FP}=-(0.0259) \ln(\frac{10^{15}}{15*10^{10}})

ϕms=ϕm(x+Es2e+ϕFP)=3.20(3.25+0.56+0.288)=0.898V\phi_{ms}=\phi_m'-(x'+\frac{E_s}{2e}+|\phi_{FP}|)=3.20-(3.25+0.56+0.288)=-0.898 V

Cox=εoxtox=3.98.851014450108=9.67108F/cm2C_{ox}=\frac{\varepsilon_{ox}}{t_{ox}}=\frac{3.9*8.8510^{-14}}{450*10^{-8}}=9.67*10^{-8}F/cm^2

vFB=ϕmsϕssCox=0.898310111.610199.67108=1.52Vv_{FB}=\phi_{ms}-\frac{\phi_{ss}}{C_{ox}}=-0.898-\frac{3*10^{11}*1.6*10^{-19}}{9.67*10^{-8}}=-1.52 V

Part b

VT=QSD(max)Cox+2ϕFP+VFBV_T=\frac{|Q'_{SD}(max)|}{C_{ox}}+2|\phi_{FP}|+V_{FB}

VdT=[411.78.8510140.2881.610191015]0.5=0.863μmV_{dT}=[\frac{4*11.7*8.85*10^{-14}*0.288}{1.6*10^{-19}*10^{15}}]^{0.5}=0.863 \mu m

QSD(max)=1.6101910150.863104=1.38108c/cm2|Q'_{SD}(max)|=1.6*10^{-19}*10^{15}*0.863*10^{-4}=1.38*10^{-8}c/cm^2

VT=1.381087.67108+20.2881.52=0.764VV_T=\frac{1.38*10^{-8}}{7.67*10^{-8}}+2*0.288-1.52 = -0.764 V


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Charan
09.07.21, 05:23

Thank you this is correct

Leave a comment

LATEST TUTORIALS
New on Blog