6.012 Recitation 7: From N+P Diode to MOS Structure

Recitation 7 From n+ p diode to MOS structure Recitation 7: From n+p diode to MOS structure Consider a p-n diode with very asymmetric doping:  xno =  xpo = 2s φB Na q(Na + Nd )Nd 2s φB Nd q(Na + Nd )Na xno  xpo The SCR is entirely on the lightly doped side.  xdo  xpo = 2s φB qNa Q is the charge per unit area (charge/cm2 ) and ρ(x) is the space charge density (charge/cm3 ). 1 Recitation 7 From n+ p diode to MOS structure Now if we insert an insulator layer (oxide) in between n+ and p, we obtain our MOS structure. 2 Recitation 7 From n+ p diode to MOS structure Same doping level: Nd = 1020 cm−3 , Na = 1016 cm−3 How does the depletion region xdo compare with xpo above? We can take some intuitive guess first. Then let us look at the problem quantitatively. QG = qNa xdo Coulomb/cm2 Electric field is the integration of charge density/s or ox : At the boundary, ox E(0− ) = s E(0+ ) E(0− ) → Eox , E(0+ ) → Es s Eox = Es  3Es ox 3 Recitation 7 From n+ p diode to MOS structure VB,O is potential drop across semiconductor body, while Vox,O is potential drop across oxide. (Note that unit wise: potential, voltage and bias are the same). But φB is fixed (same as n+ p junction). φB = φn+ − φp = Vox,o + VB,O 1 qNa xdo qNa 2 = Eox · tox + Es · xdo = · tox + x 2s do 2 ox ox . xdo can be solved from above equation: tox   s 2Cox φB xdo = 1+ −1 s qNa Cox Define oxide capacitance Cox = Si surface potential: φ(0) = φs = φp + 1 qNa 2 x 2 s do This is very important to calculate n(0), p(0) carrier concentration at the surface!! Now we see that xdo should be less than xpo in the n+ p junction case, because much less potential drop (VB,O compared to φB ) a lot of potential dropped across the oxide. 4 Recitation 7 From n+ p diode to MOS structure Now let us do some calculation: n+ -polysilicon gate, 1020 cm−3 and p-type substrate NA = 1017 cm−3 . tox = 100 A = 10 nm = 1 × 10−6 cm xdo = ? surface carrier concentration n(0), p(0) 3.9 × 8.85 × 10−14 F/cm ox Cox = = = 3.45 × 10−7 F/cm2 capacitance per unit area tox 1 × 10−6 cm φB = φn+ − φp = 550 mV − (−420 mV) = 970 mV = 0.97 V   2 φ 2Cox s B xdo = 1+ −1 Cox s qNa   11.9 × 8.85 × 10−14 Fcm−1 2 × (3.45 × 10−7 )2 × (0.97 V) F2 /cm4 = 1+ −1 3.45 × 10−7 Fcm−2 11.7 × 8.85 × 10−14 × 1.6 × 10−19 × 1017 cm−3   2 × 0.97 1+ −1 = 2.98 × 10−6 cm · 8.85 × 1.6 × 10−2 ˚ = 2.98 × 10−6 cm × (3.83 − 1) = 8.45 × 10−6 cm = 84.5 nm = 845 A On the contrary, if there is no oxide:   2 × 11.7 × 8.85 × 10−14 F · cm−1 × 0.97 2s φB xpo = = = 11.2 × 10−6 cm = 112 nm qNa 1.6 × 10−19 × 1 × 1017 1 1.6 × 10−19 C × 1 × 1017 cm−3 1 qNa 2 xdo = −420 mV + times(8.45 × 10−6 )2 2 s 2 11.7 × 8.85 × 10−14 , F · cm−1 = −0.42 + 0.55 = 0.13 V φ(0) = φs = φp + n(0) = ni e(qφ(0)/kT) = 1010 · 10130 mV/60 mV = 1.46 × 1012 cm−3 n2i p(0) = = 6.85 × 107 cm−3 n(0) 5 Recitation 7 From n+ p diode to MOS structure Yesterday, we also talked about what happens when we apply a bias (voltage) VGB means body ground, gate in reference to body (Compare to VD in diode case). It does not matter whether n or p for this definition. Since there is oxide, no matter VGB > 0 or VGB < 0, there is no current. =⇒ still consider like “thermal equilibrium” n(0) = ni e(qφ(0)/kT) Tomorrow, we will see more detailed discussion on what will happen to the surface carrier density n(0) or p(0) when we apply VGB . The surface carrier density will determine whether the MOSFET channel can conduct or not. 6