Lecture 17 - p-n Junction. October 11, Ideal p-n junction in equilibrium 2. Ideal p-n junction out of equilibrium

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6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-1 Lecture 17 - p-n Junction October 11, 22 Contents: 1. Ideal p-n junction in equilibrium 2. Ideal p-n junction out of equilibrium eading assignment: del Alamo, Ch. 7, 7.1-7.2 (7.2.1-7.2.3)

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-2 Key questions What happens if you bring into intimate contact an n-type region and a p-type region? What happens to the electrostatics of a p-n junction if one applies a bias accross? What are the dominant physics of current flow in a p-n junction under bias? What underlies the rectifying behavior of the p-n junction?

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-3 Motivation: pn junctions everywhere! Eample: CMOS PMOS NMOS n + p + n p+ n + n + p p + n

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-4 1. Ideal p-n junction in equilibrium p n E o E o χ W Sp W Sn χ E F E F Eo χ qφ bi Ec WSp Eo WSn χ EF Ev Ec Ev qφ bi = W Sp W Sn =(E C E F ) (E C E F ) = kt ln n o( ) n o ( ) Then: φ bi = kt q ln N DN A n 2 i

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-5 -p ρ qnd n depletion approimation eact -qna metallurgical junction -p ε n p - - -- - - - - - - n + + + + + + + + + + + + + + εma φ dipole of charge φbi -p n log po, no NA po no ND ni 2 NA ni 2 ND

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-6 Do depletion approimation: Volume charge density: ρ() ρ() qn A in p-qn: < p in SC: p << Electric field: ρ() qn D in SC: << n ρ() in n-qn: n < E() in p-qn: p E() qn A ɛ ( + p) in SC: p E() qn D ( n ) in SC: n ɛ E() in n-qn: n Electrostatic potential [select φ( = ) = ]: φ() qn A 2 p 2ɛ in p-qn: p φ() qn A 2ɛ (2 +2 p ) in SC: p φ() qn D 2ɛ (2 2 n ) in SC: n φ() qn D 2 n 2ɛ in n-qn: n

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-7 Two unknowns: n and p. demand overall charge neutrality: qn A p = qn D n potential difference across structure must be φ bi : Solve for n and p : φ( n ) φ( p )= qn D 2 n 2ɛ qn A 2 p 2ɛ = φ bi n = 2ɛN A φ bi qn D (N D + N A ) p = 2ɛN D φ bi qn A (N D + N A ) Total SC width: SC = 2ɛ(N D + N A )φ bi qn A N D Maimum electric field: E ma = 2qN A N D φ bi ɛ(n D + N A )

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-8 symmetric junction: N A = N D : p = n φ( p ) = φ( n ) asymmetric junction: i.e. N A >N D : p < n φ( p ) <φ( n ) strongly asymmetric junction: i.e. p + -n junction N A N D : p n SC 2ɛφ bi qn D E ma 2qN Dφ bi ɛ the lowly-doped side controls everything φ( p ) φ( n ) φ bi

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-9 ρ qnd SC ε SC εma φ φbi SC EF Ec Ev

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-1 2. Ideal p-n junction out of equilibrium Electrostatics emember, battery grabs on majority carrier Fermi levels and splits them: qφ bi E F p n q(φ bi -V) V E fp qv E fn E fp q(φ bi -V) qv E fn In forward and reverse bias: φ bi φ bi V

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-11 Qualitatively, electrostatics unchanged out of equilibrium use equilibrium equations: SC (V )= 2ɛ(N D + N A )(φ bi V ) V = SC (V =) 1 qn A N D φ bi E ma (V ) = 2qN A N D (φ bi V ) ɛ(n D + N A ) V = E ma (V =) 1 φ bi ρ qnd -p(v) n(v) -qna V> V= V< ε ε ma (V) φ φbi-v φbi φbi-v

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-12 Depletion capacitance Think differentially: ρ p - - - - n - - - - + + - - + + + + + + - - - + + + + - - + + + - + + + qn D -p(v) Q -Q ρ + Q +Q -qna Q n(v) V V+ V -p(v) n(v) Q C(V )= ɛ SC (V ) Then: C(V )= ɛqn A N D 2(N D + N A )(φ bi V ) = C(V =) 1 V φ bi

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-13 C(V )= ɛqn A N D 2(N D + N A )(φ bi V ) C(V =) = 1 V φ bi For p + -n junction: C(V )= ɛqn D 2(φ bi V ) Capacitance dominated by lowly-doped side Technique to etract φ bi and N low : 1 C 2 = 2(φ bi V ) ɛqn D Q C 1 C 2-2 εqn D φ bi V φ bi V φ bi V C

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-14 Eperimental verification:

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-15 I-V characteristics F e = a) equilibrium F h = Fe b) forward bias Fh Fe c) reverse bias Fh

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-16 In TE: balance between electron and hole flows across SC balance between and in QN s I = In forward bias: energy barrier to minority carriers reduced minority carrier injection >in QN s I e qv/kt In reverse bias: energy barrier to minority carriers increased minority carrier etraction >in QN s I saturates to a small value

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-17 Key thinking to construct model for I-V characteristics: junction voltage sets concentration of carriers with enough energy to get injected; rate of carrier injection set by minority carrier transport and / rates in quasi-neutral regions; SC is in quasi-equilibrium. Fe Fh

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-18 F e F h -p n Strategy for deriving first-order model for I-V characteristics: Compute diode current density as follows: J = J e ( p )+J h ( n ) Compute each minority carrier current contribution as follows: J e ( p ) = qn ( p )v e ( p ) J h ( n ) = qp ( n )v h ( n ) Use epressions of v e ( p ) and v h ( n ) derived for similar minority carrier type problems in Ch. 5. Derive epressions for n ( p ) and p ( n ) assuming quasi-equilibrium across the space-charge region. Unified result for forward and reverse bias.

6.72J/3.43J - Integrated Microelectronic Devices - Fall 22 Lecture 17-19 Key conclusions Built-in potential of p-n junction: φ bi = kt q ln N DN A n 2 i Depletion approimation: two quasi-neutral regions separated by a space-charge region. In strongly asymmetric junction electrostatics dominated by region with lowest doping level; i.e. for p + -n junction: SC 2ɛφ bi E ma qn D 2qN Dφ bi ɛ Electrostatics out of equilibrium same as in TE if φ bi φ bi V. Depletion capacitance due to SC width modulation: C(V )= ɛ SC (V ) Forward bias: junction barrier carrier injection recombination in QN s I e qv/kt everse bias: junction barrier carrier etraction generation in QN s I saturates with reverse V