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MOS Capacitor under Applied Bias ■ Oxide doesn’t permit any steady-state current between the n+ poly gate and the substrate. Therefore, if we wait long enough for transient currents to die out, the electron and hole currents are zero -Jn = 0 ■ and Jp = 0 Even though the structure isn’t in equilibrium, the absence of current implies that we can relate potential to carrier concentration in the silicon substrate (since that’s all we assumed in deriving the 60 mV rule.) ,, ,,,,, ,,,,, metal interconnect to gate + _ φmn+ gate oxide εox = 3.9 εo n+ polysilicon gate + _ VGB +V _ ox 0 + _VB ,,,,,,,,, ,,,,,,,,, p-type εs = 11.7 εo x + φpm _ metal interconnect to bulk ■ Flatband condition: cancel built-in drop by applying the flatband voltage V GB = – ( φ n+ – φ p ) = V FB = – 970 mV for Na = 1017 cm-3 EE 105 Spring 1997 Lecture 7 MOS Electrostatics in Flatband ■ When VGB = VFB, the gate is shifted from its thermal equilibrium potential (φn+) to a new value of VFB + φn+ = -(φn+ - φp) + φn+ = φp, which is the same potential as the p-type bulk. Therefore, there is no potential drop across the MOS structure in flatband ρ (x) charge density -tox 0 x E(x) electric field -tox 0 x φ (x) 250mV potential -tox 0 x -250mV VGB + φn+ -500mV -750mV -1.0V ■ If we continue to make the gate-bulk voltage more negative, the gate will take on a negative charge QG < 0. The substrate has a positive charge, which comes from holes that are attracted by the negative gate charge EE 105 Spring 1997 Lecture 7 MOS Capacitor in Accumulation ■ Charge density, electric field, and potential in accumulation: VGB < VFB, where VFB = - 0.97 V for this example. ρ(x) - QG (accumulated holes) charge density -tox 0 x QG E(x) -tox electric field 0 x + QG Eox = ε ox φ(x) -tox potential 0 x -250mV VGB + φn+ VGB - VFB -500mV -750mV -1.0V -1.25V EE 105 Spring 1997 Lecture 7 MOS Capacitor in Depletion ■ Now we make VGB > VFB. Note that thermal equilibrium falls into this range of applied bias. ρ (x) QG charge density Xd 0 -tox -qNa x E (x) electric field 0 -tox Xd x φ (x) 1V VGB + φn+ potential VGB 500 mV -tox 0 φs = 185 mV Xd x -500 mV ■ Surface potential at oxide/silicon interface is now positive --> n-type (slightly, ns = 1013 cm-3). EE 105 Spring 1997 Lecture 7 The Threshold Voltage VTn ■ Keep increasing VGB --> surface potential keeps increasing. At some point, the surface is n-type (i.e., we say that it is inverted) and the electron charge makes a significant contribution to the charge density. How do we model this phenomenon? We approximate that onset of inversion as the point where the electron concentration ns at the surface is the same as the hole concentration Na in the bulk. (In other words, “the surface is as n-type as the bulk is p-type.”) The gate-bulk potential at the onset of inversion is called the threshold voltage, VTn. To find the threshold voltage, we need to consider the electrostatics in depletion (no electrons at the surface at the onset of inversion) -- with the surface potential equal to the opposite of the bulk potential: φ s, max = – φ p φ (x) 1.5 V VTn + φn+ 1V φs,max = - φp = 420 mV Vox 500 mV VTn - VFB Xd,max -tox 0 -500 mV x VB,max - φp EE 105 Spring 1997 Lecture 7 Threshold Voltage Expression ■ We can solve for the threshold voltage: V T – V FB = V ox ′ + V B, max ■ The drop across the depletion region is V B, max = φ s, max – φ p = – φ p – φ p = – 2φ p ■ The drop across the oxide for VGB = VTn is – Q B, max – Q B, max V ox ′ = E ox ′t ox = ---------------------- t ox = ---------------------C ox ε ox ■ The bulk charge in inversion is found from the depletion width Xd,max – 2φ p Q B, max = – qN a X d, max = – qN a --------------------------------------- = – 2qε s N a ( – 2φ p ) ( ( 1 ⁄ 2 )qN a ) ⁄ ε s where the relationship between the depletion width Xd,max and the drop across the depletion region φs,max - (φp) = -φp - φp = -2φp can be found from Poisson’s Equation. EE 105 Spring 1997 Lecture 7 Threshold Voltage (p-type Substrate) ■ The threshold voltage is the sum of the flatband voltage (which cancels the builtin potential drop from gate to bulk), the drop across the oxide at the onset of inversion, and the maximum potential drop across the depletion region 1 V Tn = V FB – 2 φ p + --------- 2qε s N ( – 2 φ p ) a C ox EE 105 Spring 1997 Lecture 7 The Inverted MOS Capacitor (VGB > VTn) ■ We consider the surface potential as fixed (“pinned”) at φs,max = - 2 φp φ(x) 1.5 V 1.0 V Vox 500 mV - tox 0 VGB - VFB φs,max = 420 mV Xd,max x −2 φp - 500 mV ■ What is the inversion charge QN? see Section 3.7 for the derivation consider: bulk charge is constant for VGB > VTn --> all of the additional charge in the silicon is stored in the inversion layer, once inversion occurs. The inversion layer is separated from the gate by the gate oxide; we can relate the inversion charge (per cm2) to the applied voltage over VTn through Cox the capacitance (per cm2) of the oxide Q N = – C ox ( V GB – V Tn ) EE 105 Spring 1997 Lecture 7 Charge Storage in the MOS Structure ■ Three regions of operation: Accumulation: qG = Cox (vGB - vFB) ... parallel plate capacitor ■ Depletion: qG = - qB(vGB), with the bulk (depletion) charge in the silicon being a nonlinear function of vGB Inversion: qG = - qN - qB,max , where qB,max = qB(vGB = VT) is the depletion charge at the onset of inversion and Sketch of the gate charge as a function of gate-bulk voltage: QG on rsi e inv n letio dep −1 −2 ati ul VFB = − 0.97 V −QB,max −QB(VGB) 0 on −QN(VGB) 1 2 VGB (V) VTn = 0.6 V m cu ac EE 105 Spring 1997 Lecture 7 MOS Capacitance ■ The capacitance of the MOS structure is defined as dq G C = ------------dv GB V GB ■ From sketch, determine the slope and plot as the capacitance ion qG rs ve in n letio dep −2 −1 −QB,max −qB(vGB) 1 0 on ati ul 2 vGB (V) 2 VGB (V) VTn = 0.6 V VFB = − 0.97 V um −qN(vGB) cc a (a) C/Cox accumulation inversion 1.0 0.8 ple de 0.6 n tio 0.4 VFB = − 0.97 V −2 0.2 VTn = 0.6 V −1 1 0 (b) EE 105 Spring 1997 Lecture 7 Physical Interpretation of MOS Capacitance ■ Accumulation: parallel plate capacitor --> C = Cox ■ Depletion: increment in gate charge is mirrored at bottom of depletion region, so capacitance model is Cox in series with the depletion region capacitance Cb gate Si/SiO2 surface bulk ε ox C ox = -------t ox εs C b = -----Xd Note that Xd is a function of VGB C = C ox C b ■ Inversion: bulk charge is no longer changing with VGB --> an increment in gate charge is mirrored in the inversion layer under the gate. The capacitance is therefore the same as in accumulation --> C = Cox EE 105 Spring 1997 Lecture 7 Understanding MOS Capacitors ■ Step 1: identify the flatband voltage from the gate and bulk potentials in equilibrium ■ Step 2: determine whether VGB > VFB leads to accumulation or to depletion substrate is n-type --> accumulation substrate is p-type --> depletion Why? positive charge on gate ( since VGB - VFB > 0 V) must be mirrored by a negative charge in the substrate. n-type substrate: negatively charged electrons are accumulated under the gate p-type substrate: negatively charged ionized acceptors are left, after holes are repelled away from positive charge on gate ■ Step 3: construct C(VGB) plot, using the knowledge that the substrate is depleted on the other side of VFB from accumulation in Step 2 and that inversion occurs after depletion. Calculation of VT and Cmin is necessary to quantify the plot ■ Additional data point: determine state of MOS structure in thermal equilibrium (VGB = 0 V) ... accumulation or [depletion/inversion] Example: gate: p+ polysilicon (where φp+ = - 550 mV); gate oxide thickness = 200 Å , substrate: n type silicon, φn = 480 mV (Nd = 1018 cm-3) VFB = - (-550 mV - 480 mV) = + 1.03 V VGB - VFB > 0 V --> accumulated; substrate is depleted for VGB < 1.03 V Check: VGB = 0 --> negative charge on gate; positive in bulk (since gate is at - 0.55 V and substrate is at + 0.48 V in thermal equilibrium) --> positive donors in depletion region under gate ... and possibly holes due to inversion EE 105 Spring 1997 Lecture 7 MOS Capacitance-Voltage Curve ■ Evaluate threshold voltage VTp 2qε s N d ( 2φ n ) V Tp = V FB – 2φ n – ------------------------------------- = 1.03 – 2 ( 0.48 ) – 3.28 = – 3.21 V C ox ■ ■ Minimum capacitance occurs just prior to inversion and is the series combination of the oxide capacitance and the maximum depletion capacitance: 3.45 ×10– 13 1.04 ×10– 12 ε ox ε s 2 C min = -------- ------------------ = ---------------------------- ---------------------------- = 1.16 fF/cm – 6 – 6 2 ×10 2.9 ×10 t ox X d, max Maximum capacitance is Cox = 1.72 fF/cm2. C/Cox 1 0.75 1.16/1.72 = 0.67 0.5 0.25 1.03 V -3.21 V 0 -4 -3 -2 -1 0 1 2 VGB EE 105 Spring 1997 Lecture 7 EE 105 Spring 1997 Lecture 7