143f2010-11-soln

143f2010-11-soln - N.CHEUNG EE143 Fall 2010 Homework...

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Unformatted text preview: N.CHEUNG EE143, Fall 2010 Homework Assignment # 11 Solutions Note : | F | is sometimes labeled as | p | or | n | in equations (representing the value of | F | for p-type and n- type substrates respectively ) Problem 1 (a) NMOS | F | = 0.026 ln 2 10 16 1.45 10 10 = 0.37 V V FB = MS = 4.15 – (4.15+0.56+0.37) = -0.93V V Si at onset of strong inversion = 2 | F | = +0.74V x dmax = ( 2 s 2 F qN a ) 1/2 = 0.22 m Q dmax = N a x dmax = 4.4 10 11 q /cm 2 x ox = 9 10-6 cm implies C ox = 3.85 10-8 F/cm 2 Vox = Q dmax C ox = +1.82V V TN = V FB +V si +V ox = -0.93 + 0.74 + 1.82 = +1.63V (b) PMOS V FB = 4.15 – (4.15+0.56 -0.37) = -0.19V V Si at onset of strong inversion = -2 | F | = -0.74V ( minus because of opposite charge of depletion region) Vox = - Q dmax C ox = -1.82V ( minus because of opposite charge of depletion region) VTP = V FB +V si +V ox = -0.19 - 0.74 – 1.82 = -2.75V Problem 2 a.) p = 0.026 ln 5 10 15 1.45 10 10 = 0.332 volts MS = - 0.56 - p = - 0.892 volts V T = MS- Q f C ox + 2 p + 4 s qN a p C ox 1.0 = - 0.892+0.664+ 1 C ox [ - 1.6 10-19 3 10 10 + (4 1.04 10-12 1.6 10-19 5 10 15 0.332) 1/2 ] C ox = 2.33 10-8 F/cm 2 x ox =...
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143f2010-11-soln - N.CHEUNG EE143 Fall 2010 Homework...

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