ee 5340 semiconductor device theory lecture 03 – spring 2011
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EE 5340 Semiconductor Device Theory Lecture 03 – Spring 2011. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc. Review the Following. R. L. Carter’s web page: www.uta.edu/ronc/ - PowerPoint PPT PresentationTRANSCRIPT
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EE 5340Semiconductor Device TheoryLecture 03 – Spring 2011
Professor Ronald L. [email protected]
http://www.uta.edu/ronc
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Review the Following• R. L. Carter’s web page:
– www.uta.edu/ronc/• EE 5340 web page and syllabus. (Refresh
all EE 5340 pages when downloading to assure the latest version.) All links at:– www.uta.edu/ronc/5340/syllabus.htm
• University and College Ethics Policies– www.uta.edu/studentaffairs/conduct/
• Makeup lecture at noon Friday (1/28) in 108 Nedderman Hall. This will be available on the web.
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First Assignment• Send e-mail to [email protected]
– On the subject line, put “5340 e-mail”– In the body of message include
• email address: ______________________• Your Name*: _______________________• Last four digits of your Student ID: _____
* Your name as it appears in the UTA Record - no more, no less
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Second Assignment• Submit a signed copy of the
document posted at
www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf
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Kronig-Penney ModelA simple one-dimensional model of a
crystalline solid• V = 0, 0 < x < a, the ionic region• V = Vo, a < x < (a + b) = L,
between ions• V(x+nL) = V(x), n = 0, +1, +2, +3,
…, representing the symmetry of the assemblage of ions and requiring that y(x+L) = y(x) exp(jkL), Bloch’s Thm
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K-P Potential Function*
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K-P Impulse Solution• Limiting case of Vo-> inf. and b ->
0, while a2b = 2P/a is finite• In this way a2b2 = 2Pb/a < 1,
giving sinh(ab) ~ ab and cosh(ab) ~ 1
• The solution is expressed byP sin(ba)/(ba) + cos(ba) =
cos(ka)• Allowed valued of LHS bounded by
+1• k = free electron wave # = 2p/l
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K-P Solutions*
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K-P E(k) Relationship*
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Analogy: a nearly-free electr. model• Solutions can be displaced by ka =
2np• Allowed and forbidden energies• Infinite well approximation by
replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of 1
22
22
4
kEhm
p
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Generalizationsand Conclusions• The symm. of the crystal struct.
gives “allowed” and “forbidden” energies (sim to pass- and stop-band)
• The curvature at band-edge (where k = (n+1)p) gives an “effective” mass.
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Silicon BandStructure**• Indirect Bandgap• Curvature (hence
m*) is function of direction and band. [100] is x-dir, [111] is cube diagonal
• Eg = 1.17-aT2/(T+b) a = 4.73E-4 eV/K b = 636K
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Generalizationsand Conclusions• The symm. of the crystal struct.
gives “allowed” and “forbidden” energies (sim to pass- and stop-band)
• The curvature at band-edge (where k = (n+1)p) gives an “effective” mass.
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Analogy: a nearly-free electr. model• Solutions can be displaced by ka =
2np• Allowed and forbidden energies• Infinite well approximation by
replacing the free electron mass with an “effective” mass (noting E = p2/2m = h2k2/2m) of 1
22
22
4
kEhm
p14
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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
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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
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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)
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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 = 1.45E10/cm3
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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 orbits19
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Bohr model H atom-like orbits at donor• Electron (-q) rev. around proton
(+q)• Coulomb force,
F=q2/4peSieo,q=1.6E-19 Coul, eSi=11.7, eo=8.854E-14 Fd/cm
• Quantization L = mvr = nh/2p• En= -(Z2m*q4)/[8(eoeSi)2h2n2] ~-
40meV• rn= [n2(eoeSi)h2]/[Zpm*q2] ~ 2 nm
for Z=1, m*~mo/2, n=1, ground state
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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
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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, has one -q ion
• #B atoms = #holes, so neutral
• H atom-like orbits22
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Hole orbits andacceptor states• Similar to free electrons and donor
sites, there are hole orbits at acceptor sites
• The ionization energy of these states is EA - EV ~ 40 meV, so NA ~ p and there is a hole “freeze-out” at low temperatures
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Impurity Levels in Si: EG = 1,124 meV• Phosphorous, P: EC - ED = 44 meV• Arsenic, As: EC - ED = 49 meV• Boron, B: EA - EV = 45 meV• Aluminum, Al: EA - EV = 57 meV• Gallium, Ga: EA - EV = 65meV• Gold, Au: EA - EV = 584
meV EC - ED = 774 meV 24
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References *Fundamentals of Semiconductor
Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.
**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.
M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.