Lecture 4 PN Junction and MOS Electrostatics(I) Semiconductor Electrostatics in Thermal Equilibrium Outline • Non­uniformly doped semiconductor in thermal equilibrium • Relationships between potential, φ(x) and equilibrium carrier concentrations, po(x), no(x) –Boltzmann relations & “60 mV Rule” • Quasi­neutral situation Reading Assignment: Howe and Sodini; Chapter 3, Sections 3.1­3.2 6.012 Spring 2009 Lecture 4 1 1. Non­uniformly doped semiconductor in thermal equilibrium Consider a piece of n­type Si in thermal equilibrium with non­uniform dopant distribution: Nd Nd(x) x What is the resulting electron concentration in thermal equilibrium? n­type ⇒ lots of electrons, few holes ⇒ focus on electrons 6.012 Spring 2009 Lecture 4 2 OPTION 1: electron concentration follows doping concentration EXACTLY ⇒ n o (x) = N d (x) no, Nd no(x)=Nd(x)? Nd(x) x Gradient of electron concentration ⇒ net electron diffusion ⇒ not in thermal equilibrium! 6.012 Spring 2009 Lecture 4 3 OPTION 2: electron concentration uniform in space n o (x) = n ave ≠ f(x) no, Nd Nd(x) no = f(x)? x Think about space charge density: ρ(x) ≈ q[N d (x) − n o (x)] If Nd(x) ≠ no(x) ⇒ ρ(x) ≠ 0 ⇒ electric field ⇒ net electron drift ⇒ not in thermal equilibrium! 6.012 Spring 2009 Lecture 4 4 OPTION 3: Demand that Jn = 0 in thermal equilibrium at every x (Jp = 0 too) Diffusion precisely balances Drift J n (x) drift = J n (x) + diff Jn (x) = 0 What is no(x) that satisfies this condition? partially uncompensated � donor charge no, Nd Nd(x) + no(x) - net electron charge x Let us examine the electrostatics implications of no(x) ≠ Nd(x) 6.012 Spring 2009 Lecture 4 5 Space charge density ρ(x) = q[Nd (x) − n o (x)] partially uncompensated � donor charge no, Nd Nd(x) + no(x) - net electron charge x ρ + x − 6.012 Spring 2009 Lecture 4 6 Electric Field Poisson’s equation: dE ρ (x) = dx εs Integrate from x = 0: x E(x) − E(0) = 1 εs ′ x ′ ∫ ρ(x )d 0 no, Nd Nd(x) + no(x) x ρ + x − E x 6.012 Spring 2009 Lecture 4 7 Electrostatic Potential dφ = −E(x) dx x Integrate from x=0: ∫ φ(x) − φ(0) = − E (x )d ′ x ′ 0 Need to select reference (physics is in the potential difference, not in absolute value!); Select φ(x = 0) = φref no, Nd Nd(x) + no(x) x ρ + x − E x φ φref x 6.012 Spring 2009 Lecture 4 8 2. Relationships between potential, φ(x) and equilibrium carrier concentrations, po(x), no(x) (Boltzmann relations) dno Jn = 0 = qnoµ n E + qDn dx µ n dφ 1 dno • = • Dn dx no dx Using Einstein relation: q dφ d(ln no ) • = kT dx dx Integrate: ( ) q no φ − φref = ln no − ln no,ref = ln kT no,ref ( q φ − φ ref no = no,ref exp kT Then: 6.012 Spring 2009 Lecture 4 ) 9 Any reference is good In 6.012, φref = 0 at no,ref = ni Then: no = ni e qφ kT If we do same with holes (starting with Jp = 0 in thermal equilibrium, or simply using nopo = ni2 ); po = n i e −qφ kT We can re­write as: kT no φ= • ln q ni and kT po φ=− • ln q ni 6.012 Spring 2009 Lecture 4 10 “60 mV” Rule At room temperature for Si: no n o φ = (25m ) • ln = (25mV) • ln (10) • log ni n i Or φ ≈ (60m )• log n o n i EXAMPLE 1: 18 n o = 10 cm 6.012 Spring 2009 −3 ⇒ φ = (60m ) × 8 = 480 mV Lecture 4 11 “60 mV” Rule: contd. With holes: po φ = − (25m )• ln = − (25m ni po )• ln (10) • log ni Or φ ≈ − (60 m ) • log p o n i EXAMPLE 2: 18 n o = 10 cm −3 ⇒ φ = −(60m 6.012 Spring 2009 2 ⇒ p o = 10 cm −3 ) × −8 = 480mV Lecture 4 12 Relationship between φ, no and po: po, equilibrium hole concentration (cm−3) p-type 1019 1018 1016 1014 1012 φp+� � n-type intrinsic 1010 108 106 104 102 101 φn+� φ (mV) −550 −480 −360 −240 −120 0 120 240 360 � 480 550 −550 −480 φp+� −360 −240 −120 0 120 240 360 480 550 φn+� φ (mV) � 101 102 � 104 106 p-type 108 1010 1012 1014 intrinsic 1016 1018 1019 n-type no, equilibrium electron concentration (cm−3) Note: φ cannot exceed 550 mV or be smaller than -550 mV. (Beyond this point different physics come into play.) 6.012 Spring 2009 Lecture 4 13 Example 3: Compute potential difference in thermal equilibrium between region where no = 1017 cm­3 and no = 1015 cm­3. φ(no = 1017 cm− 3 ) = 60 × 7 = 420 mV φ(no = 1015 cm− 3 ) = 60 × 5 = 300 mV φ(no = 1017 cm− 3 ) − φ(n o = 1015 cm −3 ) = 120 mV Example 4: Compute potential difference in thermal equilibrium between region where po = 1020 cm­3 and po = 1016 cm­3. φ( po = 10 20 cm− 3 ) = φmax = − 550mV φ( po = 10 16 cm −3 ) = − 60 × 6 = − 360mV φ( po = 10 20 cm− 3 ) − φ(po = 1016 cm− 3 ) = −190 mV 6.012 Spring 2009 Lecture 4 14 3. Quasi­neutral situation If Nd(x) changes slowly with x ⇒ no(x) also changes slowly with x. WHY? Small dno/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) ≈ N d (x) no(x) tracks Nd(x) well ⇒ minimum space charge ⇒ semiconductor is quasi­neutral no, Nd Nd(x) no(x) = Nd(x) x 6.012 Spring 2009 Lecture 4 15 What did we learn today? Summary of Key Concepts • It is possible to have an electric field inside a semiconductor in thermal equilibrium – ⇒ Non­uniform doping distribution. • In thermal equilibrium, there is a fundamental relationship between the φ(x) and the equilibrium carrier concentrations no(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 16 MIT OpenCourseWare http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.