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ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu
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VM Ayres, ECE875, S14 Chp. 02: pn junction: Info: Linearly graded junction Multiple charge layers Example Lecture 20, 24 Feb 14 Chp. 03: metal-semiconductor junction: Schottky barrier Review Examples New
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Linearly graded junction: Why: Why: get a more uniform E (x) over a bigger x region VM Ayres, ECE875, S14
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Linearly graded junction: How: Q and E (x) i (x) VM Ayres, ECE875, S14
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Q and ~ a x E (x) ~ ax 2 + B i (x) ~ ax 2 + Bx +C Linearly graded junction: How:
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VM Ayres, ECE875, S14
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The curvature of the initial C-V curve is different from that for an abrupt junction The slope gives the grading constant a The intercept gives the equilibrium built in potential bi Practical: 1/ VM Ayres, ECE875, S14
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Chp. 02: pn junction: Info: Linearly graded junction Multiple charge layers Example Lecture 20, 24 Feb 14 Chp. 03: metal-semiconductor junction: Schottky barrier Review Examples New
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VM Ayres, ECE875, S14 Example: Set up the answer: (a) Find bi at equilibrium fro the following doping profile in si at 300 K (b) Draw the energy band-bending diagram p 10 17 cm -3 n 10 16 cm -3 p 10 15 cm -3
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VM Ayres, ECE875, S14
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1st2nd3rd
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VM Ayres, ECE875, S14 q q
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Chp. 02: pn junction: Info: Linearly graded junction Multiple charge layers Example Lecture 20, 24 Feb 14 Chp. 03: metal-semiconductor junction: Schottky barrier Review Examples New
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@ Interconnects: Use energy band diagrams to describe what is happening One question to answer: is it an Ohmic contact or a Schottky barrier contact? Interconnect contacts are key for nanotechnology: –MOSFET: Ohmic contact = good –NanoFET: SB contact = good
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Different nature of a metal Lots of e- and NO E gap E C = at E F Individual energy band diagrams:
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Need 1 description: Work Function of the metal q m : where is E F = E C relative to E vac Need 2 descriptions: Electron affinity q s = where is E C relative to E vac Work Function q s = where is E F relative to E vac
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When in physical contact E Fm and E Fs align:
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Four cases = the same approach: 1. metal with small work function/n-type semiconductor: Ohmic (barrier) 2. metal with big work function/n-type semiconductor: Schottky barrier 3. metal with small work function/p-type semiconductor: Schottky barrier 4. metal with big work function/p-type semiconductor: Ohmic (barrier) In every case, use logic: do I need to make the metal more n-type (add e- from semiconductor) or less n-type (e- move into semiconductor)
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Four cases = the same approach: 1. metal with small work function/n-type semiconductor: Ohmic (barrier) 2. metal with big work function/n-type semiconductor: Schottky barrier 3. metal with small work function/p-type semiconductor: Schottky barrier 4. metal with big work function/p-type semiconductor: Ohmic (barrier) In every case, use logic: do I need to make the metal more n-type (add e- from semiconductor) or less n-type (e- move into semiconductor)
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2. metal with big work function/n-type semiconductor
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-- N d + N d + n -- N d + N d + electrons move to metal side leaving N d + behind Size of n-side strip is set by doping concentration and can be large To bring the Fermi energy level of the metal up: make the metal more n-type
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Schottky Barrier: N D + on n-side -- N d + N d + n -- N d + N d +
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Schottky Barrier: Very narrow region with high concentration of e- similar to ionized N A = large -- N d + N d + n -- N d + N d +
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3. metal with small work function/p-type semiconductor
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++ N a - N a - p ++ N a - N a - electrons move to p-side and recombine with its large hole population. This leaves N a - strip Size of p-side strip is set by doping concentration and can be large To bring the Fermi energy level of the metal down: make the metal less n-type
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++ N a - N a - p ++ N a - N a - Schottky Barrier: N A - on p-side
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++ N a - N a - p ++ N a - N a - Schottky Barrier: N A - on p-side Very narrow region = high concentration exposed + nuclei similar to ionized N D = large
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VM Ayres, ECE875, S14 Chp. 02: pn junction: Info: Linearly graded junction Multiple charge layers Example Lecture 20, 24 Feb 14 Chp. 03: metal-semiconductor junction: Schottky barrier Review Examples New
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Example from Exam:
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Answer: Ei
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E C – E F = Egap/2 – (E F – E i ) = E F – E i = kT ln(N D /n i ) E C – E F = Streetman n i
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Made the metal more n- type to bring E Fm up to E Fs Electrons left the semiconductor and went into the metal. The semiconductor is n- type: N d + left behind. -- N d + N d + n -- N d + N d + Size W D of n-side depletion region is set by doping concentration and can be large
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Example: (a) Evaluate the energy barrier qV 0 = q bi for previous problem (b) Draw the band-bending diagram
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Answer: -- N d + N d + n -- N d + N d + q bi =qV 0 = 0.057 eV (a) qV 0 = (a) Band-bending diagram: Find W:
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Equilibrium: metal contact to n-type Si when work functions q m > q s EFEF metal Junction Depletion region W n 0 = 10 17 cm -3 - EFEF P+ P+ P P P P P P P qV 0 Neutral region n-side EiEi E (x) Although the charges are balanced, the layer on the metal side is very thin, similar to ionized N A = large
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= 1.14 x 10 -5 cm = 0.14 m
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Answer: -- N d + N d + n -- N d + N d + q bi =qV 0 = 0.057 eV (a) qV 0 = (a) Band-bending diagram: W = 0.14 m Also: q B = 4.0 -3.8 eV = 0.2 eV
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