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EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01

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

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EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01

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

  2. 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. Si Energy BandStructure 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. Si Bond ModelAbove 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. Band Model forthermal carriers • Thermal energy ~kT generates electron-hole pairs • At 300K Eg(Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> ni = 1E10/cm3

  6. 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. Band Model fordonor electrons • Ionization energy of donor Ei = Ec-Ed ~ 40 meV • Since Ec-Ed ~ kT, all donors are ionized, so ND ~ n • Electron “freeze-out” when kT is too small

  8. Acceptor: Holedue 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. Classes ofsemiconductors • Intrinsic: no = po = ni, since Na&Nd << ni =[NcNvexp(Eg/kT)]1/2,(not easy to get) • n-type: no > po, since Nd > Na • p-type: no < po, since Nd < Na • Compensated: no=po=ni, w/ Na- = Nd+ > 0 • Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants

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