l01 01/15/021 ee 4345 - semiconductor electronics design project spring 2002 - lecture 01 professor...
TRANSCRIPT
![Page 1: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/1.jpg)
L01 01/15/02 1
EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01
Professor Ronald L. [email protected]
http://www.uta.edu/ronc/
![Page 2: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/2.jpg)
L01 01/15/02 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
![Page 3: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/3.jpg)
L01 01/15/02 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
![Page 4: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/4.jpg)
L01 01/15/02 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)
![Page 5: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/5.jpg)
L01 01/15/02 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
![Page 6: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/6.jpg)
L01 01/15/02 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
![Page 7: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/7.jpg)
L01 01/15/02 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
![Page 8: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/8.jpg)
L01 01/15/02 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
![Page 9: L01 01/15/021 EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 01 Professor Ronald L. Carter ronc@uta.edu](https://reader036.vdocuments.us/reader036/viewer/2022082711/56649f045503460f94c179e8/html5/thumbnails/9.jpg)
L01 01/15/02 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