Lecture 6 PN Junction and MOS Electrostatics(III) MetalOxideSemiconductor Structure


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1 Lecture 6 PN Junction and MOS Electrostatics(III) MetalOxideSemiconductor 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 Spring 2007 Lecture 6 1
2 1. Introduction MetalOxideSemiconductor structure MOS at the heart of the electronics revolution: Digital and analog functions MetalOxideSemiconductor FieldEffect Transistor (MOSFET) is key element of Complementary Metal OxideSemiconductor (CMOS) circuit family Memory function Dynamic Random Access Memory (DRAM) Static Random Access Memory (SRAM) NonVolatile Random Access Memory (NVRAM) Imaging Charge Coupled Device (CCD) and CMOS cameras Displays Active Matrix Liquid Crystal Displays (AMLCD) Spring 2007 Lecture 6 2
3 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, V GB = 0V) Spring 2007 Lecture 6 3
4 For most metals on psi, equilibrium achieved by electrons flowing from metal to semiconductor and holes from semiconductor to metal: Remember: n o p o =n i 2 Fewer holes near Si / SiO 2 interface ionized acceptors exposed (volume charge) Spring 2007 Lecture 6 4
5 Space Charge Density In semiconductor: spacecharge region close Si /SiO 2 interface can use depletion approximation In metal: sheet of charge at metal /SiO 2 interface Overall charge neutrality x = t ox ; σ = Q G t ox < x < 0; ρ o (x) = 0 0 < x < x do ; ρ o (x) = qn a x > x do ; ρ o (x) = Spring 2007 Lecture 6 5
6 Electric Field Integrate Poisson s equation E o (x 2 ) E o (x 1 ) = 1 ε ρ( x ) d x At interface between oxide and semiconductor, there is a change in permittivity change in electric field x 2 x 1 ε ox E ox = ε s E s E ox E s = ε s ε ox Spring 2007 Lecture 6 6
7 Start integrating from deep inside semiconductor: x > x do ; E o (x) = 0 0 < x <x do ; E o (x) E o (x do ) = 1 qn a d x ε s x x do = qn a ε s ( x x do ) t ox < x < 0; x < t ox ; E(x) = 0 E o (x) = ε s ε ox E o (x = 0 + ) = qn ax do ε ox Spring 2007 Lecture 6 7
8 Electrostatic Potential (with φ = n o = p o = n i ) φ = kt q ln n o n i φ = kt q ln p o n i In QNRs, n o and p o are known can determine φ in pqnr: p o = N a in n + gate: n o = N d+ φ p = kt q ln N a n i φ g = φ n + Builtin potential: φ B = φ g φ p = φ n + + kt q ln N a n i Spring 2007 Lecture 6 8
9 To obtain φ o (x), integrate E o (x); start from deep inside semiconductor bulk: x 2 φ o (x 2 ) φ o (x 1 ) = E o ( x ) d x x 1 x > x do ; φ o (x) = φ p 0 < x < x do ; φ o (x) φ o (x do ) = qn a ( x x do )d x ε s φ o (x) = φ p + qn a 2ε s x x do ( x x do ) 2 t ox < x < 0; AT x = 0 φ o (x) = φ p + qn a 2ε s φ o (x) = φ p + qn x 2 a do 2ε s + qn a x do ε ox ( x do ) 2 ( x) x < t ox ; φ o (x) = φ n + Almost done Spring 2007 Lecture 6 9
10 Still do not know x do need one more equation Potential difference across structure has to add up to φ B : 2 φ B = V B,o + V ox,o = qn a x do 2ε s Solve quadratic equation: + qn a x do t ox ε ox x do = ε s t ε ox 1 + 2ε 2 φ ox B 2 ox qε s N a t 1 ox = ε s C ox 1+ 2C 2 φ ox B qε s N a 1 where C ox is the capacitance per unit area of oxide C ox = ε ox t ox Now problem is completely solved! Spring 2007 Lecture 6 10
11 There are also contact potentials total potential difference from contact to contact is zero! Spring 2007 Lecture 6 11
12 3. MOS with applied bias V GB 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? Spring 2007 Lecture 6 12
13 Potential difference shows up across oxide and SCR in semiconductor Oxide is an insulator no current anywhere in structure In SCR, quasiequilibrium situation prevails New balance between drift and diffusion Electrostatics qualitatively identical to thermal equilibrium (but amount of charge redistribution is different) np = n i Spring 2007 Lecture 6 13
14 Apply V GB >0: potential difference across structure increases need larger charge dipole SCR expands into semiconductor substrate: Simple way to remember: with V GB >0, gate attracts electrons and repels holes Spring 2007 Lecture 6 14
15 Qualitatively, physics unaffected by application of V GB >0. Use mathematical formulation in thermal equilibrium, but: φ B φ B + V GB For example, to determine x d (V BG ): φ B + V GB = V B (V GB ) + V ox (V GB ) = qn a x d 2 (V GB ) 2ε s + qn a x d (V GB )t ox ε ox x d (V GB ) = ε s C ox 1 ε s qn a 1+ 2C 2 ( ox φb + V GB ) φ(0) = φ s = φ p + qn a x d 2 (V GB ) 2ε s φ s gives n & p concentration at the surface Spring 2007 Lecture 6 15
16 What did we learn today? Summary of Key Concepts Charge redistribution in MOS structure in thermal equilibrium SCR in semiconductor builtin 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 Spring 2007 Lecture 6 16
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