Lecture 6 PN Junction and MOS Electrostatics(III) Metal-Oxide-Semiconductor Structure Outline 1. Introduction to MOS structure 2. Electrostatics of MOS in thermal equilibrium 3. Electrostatics of MOS with applied bias Reading Assignment: Howe and Sodini, Chapter 3, Sections 3.7-3.8 6.012 Spring 2009 Lecture 6 11. Introduction Metal-Oxide-Semiconductor structure MOS at the heart of the electronics revolution: • Digital and analog functions – Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET) is key element of Complementary Metal-Oxide-Semiconductor (CMOS) circuit family • Memory function – Dynamic Random Access Memory (DRAM) – Static Random Access Memory (SRAM) – Non-Volatile Random Access Memory (NVRAM) • Imaging – Charge Coupled Device (CCD) and CMOS cameras • Displays – Active Matrix Liquid Crystal Displays (AMLCD) 6.012 Spring 2009 Lecture 6 2 Metal interconnectto gateMetal interconnect to bulk0xGate oxideεox = 3.9 εop-typeεs = 11.7 εon+ polysilicon gateFigure by MIT OpenCourseWare.2. MOS Electrostatics in equilibrium Idealized 1D structure: • Metal: does not tolerate volume charge – ⇒ charge can only exist at its surface • Oxide: insulator and does not have volume charge – ⇒ no free carriers, no dopants • Semiconductor: can have volume charge – ⇒ Space charge region (SCR) In thermal equilibrium we assume Gate contact is shorted to Bulk contact. (i. e, VGB = 0V) 6.012 Spring 2009 Lecture 6 3For most metals on p-Si, equilibrium achieved by electrons flowing from metal to semiconductor and holes from semiconductor to metal: Remember: n opo=ni2 6.012 Spring 2009 Lecture 6 4 Fewer holes near Si / SiO2interface⇒ionized acceptors exposed (volume charge)Space Charge Density • In semiconductor: space-charge region close Si /SiO2 interface – can use depletion approximation • In metal: sheet of charge at metal /SiO2 interface • Overall charge neutrality x = −tox; σσσσ= QG −tox < x < 0; ρρρρo(x) = 0 0 < x < xdo ; ρρρρo (x) = −qNa x > xdo; ρρρρo (x) = 0 6.012 Spring 2009 Lecture 6 5Electric Field Integrate Poisson’s equation 1 x2 Eo(x2) − Eo(x1) = ∫ρρρρ(x′) dx′ εεεεx1 At interface between oxide and semiconductor, there is a change in permittivity ⇒ change in electric field εεεεoxEox =εεεεsEs Eox =εεεεs ≈ 3 Eεεεεs ox 6.012 Spring 2009 Lecture 6 6Start integrating from deep inside semiconductor: x > xdo; Eo (x) = 0 0 < x <xdo; Eo(x) − Eo (xdo) = 1 εεεεs −qNa d ′x xdo x ∫ = − qNa εεεεs x − xdo( ) −tox< x < 0; Eo (x) = εεεεs εεεεox Eo(x = 0+ ) = qNaxdo εεεεox x < −tox ; E(x) = 0 6.012 Spring 2009 Lecture 6 7Electrostatic Potential (with φφφφ = 0 @ no = po = ni) φφφφ= kT q • ln no ni φφφφ= − kT q • ln po ni In QNRs, no and po are known ⇒ can determine φ in p-QNR: po = N a ⇒ in n+-gate: no = Nd + ⇒ φφφφp = − kT q • ln Na ni φφφφg = φφφφn + Built-in potential: φφφφB = φφφφg − φφφφp = φφφφn + + kT q • ln Na ni 6.012 Spring 2009 Lecture 6 8To obtain φφφφo(x), integrate Eo(x); start from deep inside semiconductor bulk: φφφφo (x2 ) − φφφφo (x1) = − Eo ( ′x ) d′x x1 x2 ∫ x > xdo; φ o (x) = φ p 0 < x < xdo; φ o (x) − φ o (xdo ) = − − qNa ε s ′x − xdo( )d′xxdo x ∫ φ o (x) = φ p + qNa 2ε s x − xdo( )2 AT x = 0 φ o (0) = φ p + qNa 2ε s xdo( )2 −t ox < x < 0; φ o (x) = φ p + qNa xdo 2 2ε s + qNa xdo ε ox −x( ) x < −t ox ; φ o (x) = φ n + Almost done …. φB = φ n + − φ p 6.012 Spring 2009 Lecture 6 9Still do not know xdo ⇒⇒⇒⇒ need one more equation Potential difference across structure has to add up to φB: Solve quadratic equation: where C ox is the capacitance per unit area of oxide φφφφB = VB,o + Vox,o = qNaxdo 2 2εεεεs + qNa xdotox εεεεox xdo = εεεεs εεεεox tox 1 + 2εεεεox 2 φφφφB qεεεεs Natox 2 − 1 = εεεεs Cox 1+ 2C ox 2 φφφφB qεεεεsNa −1 Now problem is completely solved! Cox = εεεεox tox 6.012 Spring 2009 Lecture 6 10There are also contact potentials ⇒⇒⇒⇒ total potential difference from contact to contact is zero! 6.012 Spring 2009 Lecture 6 113. MOS with applied bias VGB Apply voltage to gate with respect to semiconductor: Electrostatics of MOS structure affected ⇒ potential difference across entire structure now ≠ 0 How is potential difference accommodated? 6.012 Spring 2009 Lecture 6 12Potential difference shows up across oxide and SCR in semiconductor Oxide is an insulator ⇒ no current anywhere in structure In SCR, quasi-equilibrium situation prevails ⇒ New balance between drift and diffusion • Electrostatics qualitatively identical to thermal equilibrium (but amount of charge redistribution is different) 2• np = ni 6.012 Spring 2009 Lecture 6 13Apply VGB>0: potential difference across structure increases ⇒ need larger charge dipole ⇒ SCR expands into semiconductor substrate: Simple way to remember: with VGB>0, gate attracts electrons and repels holes. 6.012 Spring 2009 Lecture 6 14Qualitatively, physics unaffected by application of VGB >0. Use mathematical formulation in thermal equilibrium, but: φφφφB →φφφφB + VGB For example, to determine xd(VBG): φφφφB + VGB = VB (VGB ) + Vox (VGB) 2qNaxd (VGB ) qNaxd (VGB)tox= + 2εεεεs εεεεox 2C ox 2 (φφφφB + VGB) εεεεs xd (VGB ) = 1+εεεεsqNa − 1 Cox 2 φφφφ(0) =φφφφs =φφφφp + qNaxd (VGB) 2εεεεs 6.012 Spring 2009 Lecture 6 15 φsgives n & p concentration at the surfaceWhat did we learn today? Summary of Key Concepts • Charge redistribution in MOS structure in thermal equilibrium – SCR in semiconductor – ⇒⇒⇒⇒ built-in potential across MOS structure. • In most cases, we can use depletion approximation in semiconductor SCR • Application of voltage modulates depletion region width in semiconductor – No current flows 6.012 Spring 2009 Lecture 6 16MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Spring 2009 For information about citing these materials or our Terms of Use, visit:
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