Lecture 4 PN Junction and MOS Electrostatics(I) Semiconductor Electrostatics in Thermal Equilibrium Outline • Non-uniformly doped semiconductor in thermal equilibrium • Relationships between potential, φ(x) and equilibrium carrier concentrations, po(x), no(x) –Boltzmann relations & “60 mV Rule” • Quasi-neutral situation Reading Assignment: Howe and Sodini; Chapter 3, Sections 3.1-3.2 6.012 Spring 2009 Lecture 4 11. Non-uniformly doped semiconductor in thermal equilibrium Consider a piece of n-type Si in thermal equilibrium with non-uniform dopant distribution: n-type ⇒ lots of electrons, few holes ⇒ focus on electrons Nd Nd(x) x 6.012 Spring 2009 Lecture 4 2 What is the resulting electron concentration in thermal equilibrium?OPTION 1: electron concentration follows doping concentration EXACTLY ⇒ no(x) = Nd(x) Gradient of electron concentration ⇒ net electron diffusion ⇒ not in thermal equilibrium! no, Nd no(x)=Nd(x)? Nd(x) x 6.012 Spring 2009 Lecture 4 3OPTION 2: electron concentration uniform in space no(x) = nave ≠ f(x) If Nd(x) ≠ n o(x) ⇒ ρ(x) ≠ 0 ⇒ electric field ⇒ net electron drift ⇒ not in thermal equilibrium! Think about space charge density: ρρρρ(x) ≈ q N d(x) − no(x)[ ] no, Nd Nd(x) x no = f(x)? 6.012 Spring 2009 Lecture 4 4OPTION 3: Demand that J n = 0 in thermal equilibrium at every x (Jp = 0 too) Let us examine the electrostatics implications of n o(x) ≠ Nd(x) J n(x) = J n drift(x) + Jn diff (x) = 0 What is n o(x) that satisfies this condition? no, Nd no(x) Nd(x) + -net electron charge partially uncompensated � donor charge x 6.012 Spring 2009 Lecture 4 5 Diffusion precisely balances DriftSpace charge density ρρρρ(x) = q Nd(x) − no(x)[ ] no, Nd no(x) Nd(x) + -net electron charge partially uncompensated � donor charge ρ − + x x 6.012 Spring 2009 Lecture 4 6Electric Field Poisson’s equation: dE dx = ρ(x) ε s Integrate from x = 0: E(x) − E(0) = 1 εεεεs ρρρρ( ′x )d′x 0 x ∫ no, Nd no(x) Nd(x) + -ρ − + x x x E 6.012 Spring 2009 Lecture 4 7Electrostatic Potential dφ dx = −E(x) Integrate from x=0: φφφφ(x) − φφφφ(0) = − E ( ′x )d′x 0 x ∫ Need to select reference (physics is in the potential difference, not in absolute value!); Select φ(x = 0) = φref no, Nd no(x) Nd(x) + -ρ φ − + x x x x φref E 6.012 Spring 2009 Lecture 4 82. Relationships between potential, φφφφ(x) and equilibrium carrier concentrations, po(x), no(x) (Boltzmann relations) dnJn = 0 = qnoµnE + qDn o dx µµµµn • dφφφφ= 1 • dno Dndx no dx Using Einstein relation: q • dφφφφ= d(ln no) kTdx dx Integrate: q no(φφφφ− φφφφref )= ln no − ln no,ref = ln kT no,ref Then: q(φφφφ− φφφφref ) no = no,ref exp kT 6.012 Spring 2009 Lecture 4 9Any reference is good In 6.012, φref = 0 at no,ref = ni Then: no = nieqφφφφkT If we do same with holes (starting with Jp = 0 in thermal equilibrium, or simply using nopo = ni 2 ); φφφφ= kT q • ln no ni po = nie −qφφφφkT We can re-write as: φφφφ= − kT q • ln po ni and 6.012 Spring 2009 Lecture 4 10“60 mV” Rule At room temperature for Si: φφφφ= (25m )• ln n= (25mV( )• n o )• ln 10 log o ni ni Or oφφφφ≈ (60m )• log n ni EXAMPLE 1: n = 1018cm −3 ⇒ φφφφ=(60m)× 8 = 480 mVo 6.012 Spring 2009 Lecture 4 11“60 mV” Rule: contd. With holes: φφφφ= −(25m )• ln po = −(25m )• ln 10)po (• log ni ni Or oφφφφ≈ −(60 m )• log p ni EXAMPLE 2: 18 −32 −3 no = 10 cm ⇒ po = 10 cm ⇒ φφφφ= −(60m )× −8 = 480mV 6.012 Spring 2009 Lecture 4 12Relationship between φφφφ, no and po : φ (mV) φ (mV) intrinsicp-type po, equilibrium hole concentration (cm−3) n o, equilibrium electron concentration (cm−3) n-type p-type n-typeintrinsic 1019 101 102 104 106 108 1010 1012 1014 1016 1018 1019 1018 1016 1014 1012 1010 108 106 104 102 101 −550 −550 −480 −480 −360 −360 −240 −240 −120 −120 0 0 120 120 240 240 360 360 480 480 550 550 φ p+� � φ p+� � φ n+� � φ n+� � 6.012 Spring 2009 Lecture 4 13 Note: φ cannot exceed 550 mV or be smaller than -550 mV.(Beyond this point different physics come into play.)Example 3: Compute potential difference in thermal equilibrium between region where no = 1017 cm-3 and n o -3= 1015 cm . φφφφ(no ==== 1017 cm −−−−3) ==== 60 ×××× 7 ==== 420 mV φφφφ(no ==== 1015 cm −−−−3) ==== 60 ×××× 5 ==== 300 mV φφφφ(no ==== 1017 cm −−−−3) −−−−φφφφ(no ==== 1015 cm −−−−3) ==== 120 mV Example 4: Compute potential difference in thermal equilibrium between region where po = 1020 cm-3 and po -3= 1016 cm . φφφφ( po ====1020 cm −−−−3) ====φφφφmax ==== −−−−550mV φφφφ( po ====1016 cm −−−−3) ==== −−−−60 ×××× 6 ==== −−−−360mV φφφφ( po ====1020 cm −−−−3) −−−−φφφφ(po ==== 1016 cm −−−−3) ==== −−−−190 mV 6.012 Spring 2009 Lecture 4 14N3. Quasi-neutral situation If Nd(x) changes slowly with x ⇒ n o(x) also changes slowly with x. WHY? Small dn o/dx implies a small diffusion current. We do not need a large drift current to balance it. Small drift current implies a small electric field and therefore a small space charge Then: no(x) ≈ Nd (x) n o(x) tracks Nd(x) well ⇒ minimum space charge ⇒ semiconductor is quasi-neutral no, Nd no(x) d(x) Nd(x) x = 6.012 Spring 2009 Lecture 4 15What did we learn today? Summary of Key Concepts • It is possible to have an electric field inside a semiconductor in thermal equilibrium – ⇒⇒⇒⇒ Non-uniform doping distribution. • In thermal equilibrium, there is a fundamental relationship between the φ(x) and the equilibrium carrier concentrations n o(x) & po(x) – Boltzmann relations (or “60 mV Rule”). • In a slowly varying doping profile, majority carrier concentration tracks well the doping concentration. 6.012 Spring 2009 Lecture 4 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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