Semiconductor Devices 26 Atsufumi Hirohata Department of Electronics

Semiconductor Devices 26 Atsufumi Hirohata Department of Electronics 11: 00 Tuesday, 2/December/2014 (P/T 006)

Exercise 5 State the following metals in contact with Si to form either Schottky or Ohmic contacts based on their energy diagram. Assume the following parameters: Si electron affinity: c = 4. 05 e. V and Si bandgap: Eg = 1. 11 e. V. Metal Work function n-type f. M [e. V] Si Pt 6. 30 Au 4. 80 Cu 4. 18 Ni 4. 01 p-type Si

Answer to Exercise 5 The electron affinity is defined as For an n-type contact, : Ohmic contact : Schottky contact with the barrier height of For an p-type contact, : Ohmic contact : Schottky contact with the barrier height of Metal Work n-type Si function f. M [e. V] p-type Si Pt 6. 30 Schottky contact Ohmic contact Au 4. 80 Schottky contact Cu 4. 18 Schottky contact Ni 4. 01 Ohmic contact Schottky contact f. B = 6. 30 – 4. 05 = 2. 25 e. V f. B = 4. 80 – 4. 05 = 0. 75 e. V f. B = 4. 18 – 4. 05 = 0. 13 e. V f. M (= 4. 01) < c (=4. 05) f. M (= 6. 30) > c + Eg (= 4. 05 + 1. 11) f. B = 4. 05 + 1. 11 – 4. 80 = 0. 36 e. V f. B = 4. 05 + 1. 11 – 4. 18 = 0. 98 e. V f. B = 4. 05 + 1. 11 – 4. 01 = 1. 15 e. V

Answer to Exercise 5 Metal Work n-type Si function f. M [e. V] p-type Si Pt 6. 30 Schottky contact Ohmic contact Au 4. 80 Schottky contact Cu 4. 18 Schottky contact Ni 4. 01 Ohmic contact Schottky contact f. B = 6. 30 – 4. 05 = 2. 25 e. V f. B = 4. 80 – 4. 05 = 0. 75 e. V f. B = 4. 18 – 4. 05 = 0. 13 e. V f. M (= 4. 01) < c (=4. 05) f. M (= 6. 30) > c + Eg (= 4. 05 + 1. 11) f. B = 4. 05 + 1. 11 – 4. 80 = 0. 36 e. V f. B = 4. 05 + 1. 11 – 4. 18 = 0. 98 e. V f. B = 4. 05 + 1. 11 – 4. 01 = 1. 15 e. V n-type semiconductor Metal

26 Schottky Junction • • • Image force Schottky effect Depletion layer capacity • Ohmic contact

Schottky Barrier Definition of energy at the Schottly barrier : c * S. M. Sze, Physics of Semiconductor Devices (Wiley, New York, 2006).

Image Force Origin of image force : Metal (M) Vacuum Force between the two electrons : * S. Kishino, Physics of Semiconductor Devices (Maruzen, Tokyo, 1995).

Image Force at a Metal Semiconductor Interface Potential energy for an electron : Image force at a metal semiconductor interface : Vacuum level Vacuum * S. Kishino, Physics of Semiconductor Devices (Maruzen, Tokyo, 1995).

External Electric Field Application Under an electric field E : Schottky barrier at a metal semiconductor interface : Vacuum level Vacuum Schottky effect * S. Kishino, Physics of Semiconductor Devices (Maruzen, Tokyo, 1995).

Metal - Semiconductor Junction Realistic energy diagram of a Schottky junction : Junction Under a forward bias application : Under a reverse bias application : * S. Kishino, Physics of Semiconductor Devices (Maruzen, Tokyo, 1995).

Depletion Layer Capacity Under reverse bias, Poisson’s equation is defined as q(Vbi + VR ) where the donor density is assumed to be constant. Here, the boundary conditions are at at q. VR EC EF Depletion layer EV Charge distributions Therefore, the depletion layer width w can be determined by q. ND w x Electric field distributions Depletion layer capacity C is Potential distributions Vbi + VR w x * www. tc. knct. ac. jp/~hayama/denshi/chapter 3. ppt

Ohmic Contact By highly doping an interfacial region : Vacuum level Metal Vacuum level Highly doped semiconductor Semiconductor Junction * S. Kishino, Physics of Semiconductor Devices (Maruzen, Tokyo, 1995).

Realistic Schottky Barrier Image force and Shottky barrier : * S. M. Sze, Physics of Semiconductor Devices (Wiley, New York, 2006).

Exercise 6 Calculate the depletion layer capacity at a reverse bias VR = 0. 5 V in a Au/n-Si Schottky diode. Assume the following parameters: Au work function: f. M = 4. 80 e. V n-region: doping density of ND = 1 1021 m-3 Si electron affinity: c = 4. 05 e. V Si Fermi level: EF = EC – 0. 15 e. V permittivity: e = e e 0 = 12. 0 8. 854 10 -12 F/m and q = 1. 6 10 -19 C. q(Vbi + VR ) q. VR Depletion layer EC EF EV