6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-1 Contents: Lecture 4 - PN Junction and MOS Electrostatics (I) Semiconductor Electrostatics in Thermal Equilibrium September 20, 2005 1. Non-uniformly doped semiconductor in thermal equilibrium 2. Quasi-neutral situation 3. Relationships between φ() and equilibrium carrier concentrations (Boltzmann relations), 60 mv Rule Reading assignment: Howe and Sodini, Ch. 3, 3.1-3.2
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-2 Key questions Is it possible to have an electric field inside a semiconductor in thermal equilibrium? If there is a doping gradient in a semiconductor, what is the resulting majority carrier concentration in thermal equilibrium?
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-3 1. Non-uniformly doped semiconductor in thermal equilibrium Consider first uniformly doped n-type Si in thermal equilibrium: N d N d ()=N d n-type lots of electrons, few holes focus on electrons n o = N d independent of Volume charge density [C/cm 3 ]: ρ = q(n d n o )= 0
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-4 Net, consider piece of n-type Si in thermal equilibrium with non-uniform dopant distribution: N d N d () What is the resulting electron concentration in thermal equilibrium?
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-5 Option 1: Every donor gives out one electron n o () = N d () n o, N d N d () n o ()=N d ()? Gradient of electron concentration: net electron diffusion not thermal equilibrium!
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-6 Option 2: Electron concentration uniform in space: n o = n ave = f () n o, N d N d () n o = f()? Think about space charge density: ρ() = q[n d () n o ()] = n o () ρ() If N d () =0 electric field net electron drift not thermal equilibrium!
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-7 Option 3: Demand J e = 0 in thermal equilibrium (and J h = 0 too) at every Diffusion precisely balances drift: J e () = Je drif t ()+ Je dif f () = 0 What is n o () that satisfies this condition? n o, N d partially uncompensated donor charge - N d () + n o () net electron charge In general, then: n o () = N d () What are the implications of this?
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-8 Space charge density: ρ() = q[n d () n o ()] n o, N d partially uncompensated donor charge - N d () + n o () net electron charge ρ +
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-9 Electric field: Gauss equation: de d ρ = ɛ s Integrate from = 0 to : n o, N d E() E(0) = 1 ρ()d ɛ s 0 - N d () + n o () ρ + E
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-10 Electrostatic potential: Integrate from = 0 to : dφ = E d φ() φ(0) = E()d 0 Need to select reference (physics is in potential difference, not in absolute value!); select φ( = 0) = φ ref : n o, N d - N d () + n o () ρ + E φ φ ref
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-11 Given N d (), want to know n o (), ρ(), E(), and φ(). Equations that describe problem: dn o J e = qµ n n o E + qd n =0 d de d q = (N d n o ) ɛ s Epress them in terms of φ: dφ dn o qµ n n o + qd n = 0 (1) d d d 2 φ d 2 q = (n o N d ) (2) ɛ s Plug [1] into [2]: d 2 (ln n o ) q 2 = (n o N d ) (3) d 2 ɛ s kt One equation with one unknown. Given N d (), can solve for n o () and all the rest, but...... no analytical solution for most situations!
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-12 2. Quasi-neutral situation d 2 (ln n o ) q 2 = (n o N d ) d 2 ɛ s kt If N d () changes slowly with : n o () also changes slowly with d 2 (ln n o ) d 2 small = n o () N d () n o () tracks N d () well minimum space charge semiconductor is quasi-neutral n o, N d N d () n o () = N d () Quasi-neutrality good if: n o N d n o N d 1 or 1 n o N d
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-13 3. Relationships between φ() and equilibrium carrier concentrations (Boltzmann relations) From [1]: µ n dφ 1 dn o = D n d n o d Using Einstein relation: q dφ d(ln n o ) = kt d d Integrate: q (φ φ ref )=ln n o ln n o (ref) = ln kt n o (ref) Then: n o = n o (ref)e q(φ φ ref )/kt n o
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-14 Any reference is good. In 6.012, φ ref =0 at n o (ref) = n i. Then: n o = n i e qφ/kt If do same with holes (starting with J h = equilibrium, or simply using n o p o = n 2 i ): 0 in thermal p o = n i e qφ/kt Can rewrite as: φ = kt q ln n o n i and kt φ = q ln p o n i
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-15 2 60 mv Rule: At room temperature for Si: φ = (25 mv )ln n o n i = (25 mv ) ln(10) log n o n i Or φ (60 mv ) log n o 10 10 For every decade of increase in n o, φ increases by 60 mv at 300K. Eample 1: n o =10 18 cm φ = (60 mv ) 8 = 480 mv
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-16 With holes: φ = (25 mv )ln p o n i = (25 mv ) ln(10) log p o n i Or φ (60 mv ) log p o 10 10 Eample 2: n o =10 18 cm p o =10 2 cm φ = (60 mv ) ( 8) = 480 mv
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-17 Relationship between φ and n o and p o : φ (mv) φ ma =550 mv 480 360 240 120 0-120 -240-360 -480 φ min =-550 mv n o =p o =n i 100 102 104 106 108 101010121014101610181020 n o (cm-3) 102010181016101410121010 108 106 104 102 100 p o (cm-3) Note: φ cannot eceed 550 mv or be smaller than 550 mv (beyond these points, different physics come into play).
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-18 Eample 3: Compute potential difference in thermal equilibrium between region where n o =10 17 cm and region where p o =10 15 cm : φ(n o =10 17 cm )= 60 7 = 420 mv φ(p o =10 15 cm )= 60 5 = 00 mv φ(n o =10 17 cm ) φ(p o =10 15 cm ) = 720 mv Eample 4: Compute potential difference in thermal equilibrium between region where n o =10 20 cm and region where p o =10 16 cm : φ(n o =10 20 cm )= φ ma = 550 mv φ(p o =10 16 cm )= 60 6 = 60 mv φ(n o =10 20 cm ) φ(p o =10 16 cm ) = 910 mv
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-19 Boltzmann relations readily seen in device behavior! 2 pn diode current-voltage characteristics: 2 Bipolar transistor transfer characteristics:
6.012 - Microelectronic Devices and Circuits - Fall 2005 Lecture 4-20 Key conclusions It is possible to have an electric field inside a semiconductor in thermal equilibrium non-uniform doping distribution. In a slowly varying doping profile, majority carrier concentration tracks well doping concentration. In thermal equilibrium, there are fundamental relationships between φ() and the equilibrium carrier concentrations Boltzmann relations (or 60 mv Rule ).