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Microelectronics HW2: Calculating Concentrations & Atomic Distances in Boron-Doped Silicon, Assignments of Electrical and Electronics Engineering

Arizona State University (ASU) - TempeElectrical and Electronics Engineering

Solutions to homework problems related to microelectronics, specifically calculating concentrations and atomic distances in silicon doped with boron. It includes steps to find the concentration and average distance between boron atoms in a silicon wafer, as well as the number of silicon and boron atoms contained in the wafer.

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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EEE435/591 Microelectronics - Homework 2
1) Chapter 2, Q11 on p36 of the textbook (10 marks)
Using the normal melt expression CS = C0 k0(1-X) ko-1 the questions tells us that CS = 3x1015cm-3
at X=0 ⇒ C0k0 = 3x1015 cm-3.
Therefore at X=0.9 CS = 3x1015 (1-0.9)0.8-1 = 4.75 x 1015 cm-3
2) In Lecture 2 there is a plot showing how Cs/Co varies as a function of fraction of melt
solidified for boron and phosphorus. Use Excel, or equivalent, to plot similar curves for arsenic
and antimony.
(10 marks)
pf3

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EEE435/591 Microelectronics - Homework 2

  1. Chapter 2, Q11 on p36 of the textbook (10 marks) Using the normal melt expression CS = C 0 k0 (1-X) ko-1^ the questions tells us that CS = 3x10^15 cm- at X=0 ⇒ C 0 k0 = 3x10^15 cm-^. Therefore at X=0.9 CS = 3x10^15 (1-0.9)0.8-1^ = 4.75 x 10 15 cm-

  2. In Lecture 2 there is a plot showing how Cs/Co varies as a function of fraction of meltsolidified for boron and phosphorus. Use Excel, or equivalent, to plot similar curves for arsenic and antimony. ( 10 marks)

  1. Repeat the exercise we did in Lecture 2 but this time consider what is the weight of phosphorus that must be added to a 10 kg. charge of pure silicon to get the phosphorus-doped silicon to have a resistivity ofyou will need the density of silicon and atomic weight of phosphorus to complete this problem. 0.01Ohm.cm when one-half of the ingot is grown? Remember that

( 10 marks) From graph supplied 0.01Ω.cm n-Si is doped at ~ 4x10 18 cm-^. CS = C 0 k0 (1-X) ko-1^ and using k 0 = 0.35 for phosphorus and X=0.5 we get C 0 = 7.26 x 10^18 cm-^. But the volume of 10 kg of silicon is given by (10kg x 1000)/2.328g/cm^3 = 4296 cm^3 Therefore 4296 cm^3 of this silicon contains 3.11 10^22 phosphorus atoms But the atomic weight of phosphorus is 30.97 gm/mole i.e. 6.023 x 10 23 phosphorus atoms weighs 30.97 gm therefore 3.11 x 10^22 phosphorus atoms weighs 30.97 x 3.11 x 10 22 / 6.023 x10^23 = 1.60 gm

  1. A 150 mm diameter silicon wafer with a thickness of 650 microns contains 10mg of boron uniformly distributed in substitutional sites. Find a) the boron concentration in atoms/cm^3 and b) the average distance between boron atoms. ( 10 marks) Boron has an atomic weight of 10.81 gm/mol Therefore, 10 mg of boron contains 6.023 x10^23 x 10 -2^ /10.81 = 5.57x10 20 atoms Wafer volume = π x (7.5cm) 2 x 650x10 -4^ cm = 11.49 cm^3 a) ⇒ boron concentration ND = 5.57x10 20 / 11.49 = 4.85 x 10^19 cm -^3 ( 4 marks ) b) Each boron atom occupies (4.85x10 19 ) -1^ = 2.06 x 10-20^ cm^3 Side of each cube = cube root (2.06 x 10-20^ ) = 2.74 x 10-7^ cm = 2.74 nm ( 6 marks )