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L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter

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Presentation on theme: "L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter"— Presentation transcript:

1 L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2 L01 01/15/022 Silicon Covalent Bond (2D Repr) Each Si atom has 4 nearest neighbors Si atom: 4 valence elec and 4+ ion core 8 bond sites / atom All bond sites filled Bonding electrons shared 50/50 _ = Bonding electron

3 L01 01/15/023 Si Energy Band Structure at 0 K Every valence site is occupied by an electron No electrons allowed in band gap No electrons with enough energy to populate the conduction band

4 L01 01/15/024 Si Bond Model Above Zero Kelvin Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds Free electron and broken bond separate One electron for every “hole” (absent electron of broken bond)

5 L01 01/15/025 Band Model for thermal carriers Thermal energy ~kT generates electron-hole pairs At 300K E g (Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> n i = 1E10/cm3

6 L01 01/15/026 Donor: cond. electr. due to phosphorous P atom: 5 valence elec and 5+ ion core 5th valence electr has no avail bond Each extra free el, -q, has one +q ion # P atoms = # free elect, so neutral H atom-like orbits

7 L01 01/15/027 Band Model for donor electrons Ionization energy of donor E i = E c -E d ~ 40 meV Since E c -E d ~ kT, all donors are ionized, so N D ~ n Electron “freeze- out” when kT is too small

8 L01 01/15/028 Acceptor: Hole due to boron B atom: 3 valence elec and 3+ ion core 4th bond site has no avail el (=> hole) Each hole adds -(- q) and has one -q ion #B atoms = #holes, so neutral H atom-like orbits

9 L01 01/15/029 Classes of semiconductors Intrinsic: n o = p o = n i, since N a &N d << n i =[N c N v exp(E g /kT)] 1/2,(not easy to get) n-type: n o > p o, since N d > N a p-type: n o < p o, since N d < N a Compensated: n o =p o =n i, w/ N a - = N d + > 0 Note: n-type and p-type are usually partially compensated since there are usually some opposite-type dopants


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