1 / 41

ECE 875: Electronic Devices

ECE 875: Electronic Devices. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu. Lecture 20, 24 Feb 14. Chp. 02: pn junction: Info: Linearly graded junction Multiple charge layers Example.

lilac
Télécharger la présentation

ECE 875: Electronic Devices

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ECE 875:Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu

  2. Lecture 20, 24 Feb 14 • Chp. 02: pn junction: • Info: Linearly graded junction • Multiple charge layers • Example • Chp. 03: metal-semiconductor junction: Schottky barrier • Review • Examples • New VM Ayres, ECE875, S14

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

  4. Linearly graded junction: How: Q and r E (x) yi(x) VM Ayres, ECE875, S14

  5. Linearly graded junction: How: Q and r ~ a x E (x) ~ ax2 + B yi(x) ~ ax2 + Bx +C

  6. Linearly graded junction: How: VM Ayres, ECE875, S14

  7. VM Ayres, ECE875, S14

  8. Practical: 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 ybi 1/ VM Ayres, ECE875, S14

  9. Lecture 20, 24 Feb 14 • Chp. 02: pn junction: • Info: Linearly graded junction • Multiple charge layers • Example • Chp. 03: metal-semiconductor junction: Schottky barrier • Review • Examples • New VM Ayres, ECE875, S14

  10. Example: Set up the answer: (a) Find ybi at equilibrium fro the following doping profile in si at 300 K (b) Draw the energy band-bending diagram p 1017 cm-3 n 1016 cm-3 p 1015 cm-3 VM Ayres, ECE875, S14

  11. VM Ayres, ECE875, S14

  12. 1st 2nd 3rd VM Ayres, ECE875, S14

  13. q q VM Ayres, ECE875, S14

  14. Lecture 20, 24 Feb 14 • Chp. 02: pn junction: • Info: Linearly graded junction • Multiple charge layers • Example • Chp. 03: metal-semiconductor junction: Schottky barrier • Review • Examples • New VM Ayres, ECE875, S14

  15. @ 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

  16. Individual energy band diagrams: Different nature of a metal Lots of e- and NO Egap EC = at EF

  17. Need 2 descriptions: Electron affinity qcs = where is EC relative to Evac Work Function qFs = where is EF relative to Evac Need 1 description: Work Function of the metal qFm: where is EF = EC relative to Evac

  18. When in physical contact EFm and EFs align:

  19. 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 barrier3. metal with small work function/p-type semiconductor: Schottky barrier4. 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)

  20. 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 barrier3. metal with small work function/p-type semiconductor: Schottky barrier4. 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)

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

  22. electrons move to metal side leaving Nd+ behind • Size of n-side strip is set by doping concentration and can be large --Nd+ Nd+ n --Nd+ Nd+ To bring the Fermi energy level of the metal up: make the metal more n-type

  23. Schottky Barrier: ND+ on n-side --Nd+ Nd+ n --Nd+ Nd+

  24. Schottky Barrier: Very narrow region with high concentration of e- similar to ionized NA = large --Nd+ Nd+ n --Nd+ Nd+

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

  26. electrons move to p-side and recombine with its large hole population. This leaves Na- strip • Size of p-side strip is set by doping concentration and can be large ++Na-Na- p ++Na- Na- To bring the Fermi energy level of the metal down: make the metal less n-type

  27. Schottky Barrier: NA- on p-side ++Na-Na- p ++Na- Na-

  28. Schottky Barrier: Very narrow region = high concentration exposed + nuclei similar to ionized ND = large NA- on p-side ++Na-Na- p ++Na- Na-

  29. Lecture 20, 24 Feb 14 • Chp. 02: pn junction: • Info: Linearly graded junction • Multiple charge layers • Example • Chp. 03: metal-semiconductor junction: Schottky barrier • Review • Examples • New VM Ayres, ECE875, S14

  30. Example from Exam:

  31. Answer: Ei

  32. EC – EF = Egap/2 – (EF – Ei) = EF – Ei = kT ln(ND/ni) Streetman ni EC – EF =

  33. --Nd+ Nd+ n --Nd+ Nd+ Made the metal more n-type to bring EFm up to EFs Electrons left the semiconductor and went into the metal. The semiconductor is n-type: Nd+ left behind. • Size WD of n-side depletion region is set by doping concentration and can be large

  34. Example:(a) Evaluate the energy barrier qV0 = q ybi for previous problem(b) Draw the band-bending diagram

  35. Answer: (a) qV0 = --Nd+ Nd+ n --Nd+ Nd+ q ybi =qV0 = 0.057 eV (a) Band-bending diagram: Find W:

  36. Equilibrium: metal contact to n-type Si when work functions qFm > qFs Junction metal n0= 1017 cm-3 Although the charges are balanced, the layer on the metal side is very thin, similar to ionized NA = large qV0 EF EF - - P+ P+ P P P P P P P Ei Neutral region n-side E (x) Depletion region W

  37. = 1.14 x 10-5 cm = 0.14 mm

  38. Answer: (a) qV0 = --Nd+ Nd+ n --Nd+ Nd+ q ybi =qV0 = 0.057 eV Also: qFB = 4.0 -3.8 eV = 0.2 eV (a) Band-bending diagram: W = 0.14 mm

More Related