ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

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

Linearly graded junction: Why: Why: get a more uniform E (x) over a bigger x region VM Ayres, ECE875, S14

Linearly graded junction: How: Q and  E (x)  i (x) VM Ayres, ECE875, S14

Q and  ~ a x E (x) ~ ax 2 + B  i (x) ~ ax 2 + Bx +C Linearly graded junction: How:

VM Ayres, ECE875, S14

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

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

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 cm -3 n cm -3 p cm -3

VM Ayres, ECE875, S14

1st2nd3rd

VM Ayres, ECE875, S14 q q

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

@ 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

Different nature of a metal Lots of e- and NO E gap E C = at E F Individual energy band diagrams:

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

When in physical contact E Fm and E Fs align:

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)

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)

2. metal with big work function/n-type semiconductor

-- 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

Schottky Barrier: N D + on n-side -- N d + N d + n -- N d + N d +

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 +

3. metal with small work function/p-type semiconductor

++ 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

++ N a - N a - p ++ N a - N a - Schottky Barrier: N A - on p-side

++ 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

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

Example from Exam:

Answer: Ei

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

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

Example: (a) Evaluate the energy barrier qV 0 = q  bi for previous problem (b) Draw the band-bending diagram

Answer: -- N d + N d + n -- N d + N d + q  bi =qV 0 = eV (a) qV 0 = (a) Band-bending diagram: Find W:

Equilibrium: metal contact to n-type Si when work functions q  m > q  s EFEF metal Junction Depletion region W n 0 = 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

= 1.14 x cm = 0.14  m

Answer: -- N d + N d + n -- N d + N d + q  bi =qV 0 = eV (a) qV 0 = (a) Band-bending diagram: W = 0.14  m Also: q  B = eV = 0.2 eV