EECS130 Integrated Circuit Devices


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1 EECS130 Integrated Circuit Devices Professor Ali Javey 8/30/2007 Semiconductor Fundamentals Lecture 2 Read: Chapters 1 and 2
2 Last Lecture: Energy Band Diagram Conduction band E c E g Band gap E v Valence band Energy band diagram shows the bottom edge of conduction band, E c, and top edge of valence band, E v. E c and E v are separated by the band gap energy, E g.
3 Temperature Effect on Band Gap p conduction band Energy s valence band isolated atoms Decreasing atomic separation lattice spacing How does the band gap change with temperature?
4 Measuring the Band Gap Energy by Light Absorption electron photons photon energy: h v > E g E g E c E v hole E g can be determined from the minimum energy (hν) of photons that are absorbed by the semiconductor. Bandgap energies of selected semiconductors Material PbTe Ge Si GaAs GaP Diamond E g (ev)
5 Semiconductors, Insulators, and Conductors E c E g=1.1 ev E c Ev E g= 9 ev E v Top of conduction band empty filled E c Si (Semiconductor) SiO 2 (Insulator) Conductor Totally filled bands and totally empty bands do not allow current flow. (Just as there is no motion of liquid in a totally filled or. totally empty bottle.) Metal conduction band is halffilled. Semiconductors have lower E g 's than insulators and can be doped.
6 Donor and Acceptor Levels in the Band Model Conduction Band Donor Level E c E d Donor ionization energy Acceptor Level Valence Band Acceptor ionization energy E a E v Ionization energy of selected donors and acceptors in silicon Donors Acceptors Dopant Sb P As B Al In Ionization energy, E c E d or E a E v (mev) m 0 q 4 Hydrogen: E ion = = 13.6 ev 8ε 02 h 2
7 Dopants and Free Carriers Donors ntype Acceptors ptype Dopant ionization energy ~50meV (very low).
8 Effective Mass In an electric field,, an electron or a hole accelerates. Remember : F=ma=qE electrons holes Electron and hole effective masses Si Ge GaAs GaP m n /m m p /m
9 Density of States E E c ΔΕ E c g c g(e) E v E v g v g c ( E) g g c v number of states in ΔE ΔE volume 1 ev cm 3 * * mn 2mn ( E) 2 3 π h * * mp 2mp ( E) 2 3 π h ( E E ) c ( E E) v
10 Thermal Equilibrium
11 Thermal Equilibrium An Analogy for Thermal Equilibrium Sand particles Dish Vibrating Table There is a certain probability for the electrons in the conduction band to occupy highenergy states under the agitation of thermal energy (vibrating atoms, etc.)
12 At E=E F, f(e)=1/2
13 Assume the two extremes: High Energy (Large E): EE f >> kt, f(e) 0 Low Energy (Small E): EEf << kt, f(e) 1
14 Effect of T on f(e) T=0K
15 Question If f(e) is the probability of a state being occupied by an electron, what is the probability of a state being occupied by a hole?
16 n = electron density : number of unbound electrons / cm 3 p = hole density : number of holes / cm 3
17 N c is called the effective density of states (of the conduction band).
18 N v is called the effective density of states of the valence band.
19 Intrinsic Semiconductor Extremely pure semiconductor sample containing an insignificant amount of impurity atoms. n = p = n i E f lies in the middle of the band gap Material Ge Si GaAs E g (ev) n i (1/cm 3 ) 2 x x x 10 6
20 ntype intrinsic
21 Remember: the closer E f moves up to E c, the larger n is; the closer E f moves down to E v, the larger p is. For Si, N c = cm 3 and N v = cm 3. E c E f E c E v E v E f
22 Example: The Fermi Level and Carrier Concentrations Where is E f for n =10 17 cm 3? Solution: n = N c e ( E E c f )/ kt E c E f = kt ln ( ) ( N n = ln /10 ) = ev c ev E c E f E v
23 The np Product and the Intrinsic Carrier Concentration Multiply n = N c e ( E E c f )/ kt and p = N v e ( E E f v )/ kt np = N c N v e ( E E kt E c v )/ = NcNve g / kt 2 np = n i n i = N c N v e E g / 2kT In an intrinsic (undoped) semiconductor, n = p = n i.
24 EXAMPLE: Carrier Concentrations Question: What is the hole concentration in an Ntype semiconductor with cm 3 of donors? Solution: n = cm 3. p ni 10 cm = 15 3 n 10 cm = 10 5 cm 3 After increasing T by 60 C, n remains the same at cm 3 while p 2 Eg / kt increases by about a factor of 2300 because n e. Question: What is n if p = cm 3 in a Ptype silicon wafer? i Solution: n ni 10 cm = 17 3 p 10 cm = 10 3 cm 3
25 EXAMPLE: Complete ionization of the dopant atoms N d = cm 3 and E c E d =45 mev. What fraction of the donors are not ionized? Solution: First assume that all the donors are ionized. n = N d = cm 3 E f = E c 146meV 45meV 146 mev E d E c E f Probability of nonionization = e 1 E v 1+ e = ( E E ) / kt ((146 45)meV) / 26meV d Therefore, it is reasonable to assume complete ionization, i.e., n = N d. f 1
26 Doped Si and Charge What is the net charge of your Si when it is electron and hole doped?
27 Bond Model of Electrons and Holes (Intrinsic Si) Si Si Si Si Si Si Si Si Si Silicon crystal in a twodimensional representation. Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si When an electron breaks loose and becomes a conduction electron, a hole is also created.
28 Dopants in Silicon Si Si Si Si Si Si Si As Si Si B Si Si Si Si Ntype Si Si Si Si Ptype Si As (Arsenic), a Group V element, introduces conduction electrons and creates Ntype silicon, and is called a donor. B (Boron), a Group III element, introduces holes and creates Ptype silicon, and is called an acceptor. Donors and acceptors are known as dopants.
29 General Effects of Doping on n and p Charge neutrality: N _ a N + d n + N a _ p N + d : number of ionized acceptors /cm3 : number of ionized donors /cm3 = 0 Assuming total ionization of acceptors and donors: N a N + d n + N p = 0 a N d : number of ionized acceptors /cm3 : number of ionized donors /cm3
30 General Effects of Doping on n and p I. N N >> n (i.e., Ntype) d a i n = N d N a p = n 2 i n If N d >> N a, n = Nd and p = 2 n i N d II. N N >> a d n i (i.e., Ptype) p = n = N a N n 2 i p d If N >>, a N d p = Na and n = 2 n i N a
31 EXAMPLE: Dopant Compensation What are n and p in Si with (a) N d = cm 3 and N a = cm 3 and (b) additional cm 3 of N a? (a) n = N d Na = cm 3 p 2 i = n / n = / = cm 3 (b) N a = = cm 3 > N d! p = N a Nd 2 i n = n / p = 8 10 = n = cm N d = cm / = 5 10 = cm cm 3 N d = cm 3 N a = cm 3 N a = cm p = cm 3
32 Carrier Concentrations at Extremely High and Low Temperatures intrinsic regime ln n n = N d freezeout regime high temp. room temperature cryogenic temperature 1/T
33 Infrared Detector Based on Freezeout To image the blackbody radiation emitted by tumors requires a photodetector that responds to hν s around 0.1 ev. In doped Si operating in the freezeout mode, conduction electrons are created when the infrared photons provide the energy to ionized the donor atoms. electron photon E c E d E v
34 Chapter Summary Energy band diagram. Acceptor. Donor. m n, m p. Fermi function. E f. n p = = N N c v e e ( E E c f ( E E f v )/ kt )/ kt n p = = N d N a N a N d np = n i 2
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