Principles of Electronic Devices 1 - Assignment 10 | ECEN 5355, Assignments of Electrical and Electronics Engineering

Download Principles of Electronic Devices 1 - Assignment 10 | ECEN 5355 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! 1. Prove that the small-signal capacitance of an MOS capacitor C, biased into depletion is given by equation: s d ox x C C ε + = 1 1 That is, show that C is equal to the capacitance of a series connection of two capacitors: (1) a capacitor made with one plate in the bulk of the silicon and the other plate at the oxide-silcion interface, and (2) a capacitor that has its plates separated by the oxide. (Hint. Use Gauss’s law to express the charge oxoxQ EΔ=Δ ε . Then, show that the voltage across the capacitor is oxoxoxoxoxox ttV εε /EE Δ+Δ=Δ and evaluate VQC ΔΔ= / .) The capacitance is defined as: dQ dVdV dQC GG 1 == Where the gate voltage is given by: oxsFBG VVV ++= φ And the oxide voltage, Vox, equals: ox ox d oxoxox t Q tV ε == E The depletion layer charge, Qd, and the surface potential, φs, equal dad xqNQ = and s da s xqN ε φ 2 2 = , So that the capacitance becomes: s d ox ox d s d ox xt dQ d dQ dV C εε φ + = + = 11 2. Take VFB = 0.5 V and use s d ox x C C ε + = 1 1 To show the behavior of the overall capacitance C for an MOS system in the depletion region. Sketch a plot of C/Cox versus VG. Consider that the silicon oxide is 100 nm thick and the silicon is p-type with 1 Ω-cm resistivity. Locate the flatband capacitance using asox ox s D ox FB Nq kTtL C C εεε 2 1 1 1 + = + = The doping density corresponding to a resistivity of 1 Ω-cm is Na = 1.5 x 1016 cm-3. The bulk potential is then: i a tF n N V ln=φ = 0.358 V The threshold voltage is given by: ox Fas FFBT C qN VV φε φ 22 2 ++= = 1.96 V In depletion, i.e. for 0 < φs < 2 φF and VFB < VG < VT, the capacitance is obtained from: s d ox x C C ε + = 1 1 where a ss d qN x φε2 = and ox sas sFBG C qN VV φε φ 2 ++= where ox ox ox t C ε = = 34.5 nF/cm2 The flatband capacitance is obtained from: sFBox FB CC C , 11 1 + = = 31.1 nF/cm2 = 0.9 Cox where kT Nq C assFB ε2 , = =313 nF/cm 2 The high-frequency capacitance in inversion equals: