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Integrated Circuit Devices Professor Ali Javey Summer 2009 Semiconductor Fundamentals Evolution of Devices Yesterday’s Transistor (1947) Today’s Transistor (2006) Why “Semiconductors”? • Conductors – e.g Metals • Insulators – e.g. Sand (SiO2) • Semiconductors – conductivity between conductors and insulators – Generally crystalline in structure • In recent years, non-crystalline semiconductors have become commercially very important Polycrystalline amorphous crystalline What are semiconductors Elements: Si, Ge, C Binary: GaAs, InSb, SiC, CdSe, etc. Ternary+: AlGaAs, InGaAs, etc. Electrons and Holes in Semiconductors Å Structure Silicon Crystal • Unit cell of silicon crystal is cubic. • Each Si atom has 4 nearest neighbors. 5.43 Å Silicon Wafers and Crystal Planes z z z y (100) x y y x (011) (111) x (100) plane (011) flat Si (111) plane The standard notation for crystal planes is based on the cubic unit cell. Silicon wafers are usually cut along the (100) plane with a flat or notch to help orient the wafer during IC fabrication. Bond Model of Electrons and Holes (Intrinsic Si) Si Si Si Si Si Si Si Si Si Silicon crystal in a two-dimensional 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. Dopants in Silicon Si Si Si Si Si Si Si As Si Si B Si Si Si Si Si Si Si N-type Si P-type Si As (Arsenic), a Group V element, introduces conduction electrons and creates N-type silicon, and is called a donor. B (Boron), a Group III element, introduces holes and creates P-type silicon, and is called an acceptor. Donors and acceptors are known as dopants. Types of charges in semiconductors Hole Electron Ionized Donor Ionized Acceptor Mobile Charge Carriers they contribute to current flow with electric field is applied. Immobile Charges they DO NOT contribute to current flow with electric field is applied. However, they affect the local electric field EE143 – Vivek Subramanian Slide 1-9 GaAs, III-V Compound Semiconductors, and Their Dopants Ga As Ga As Ga As Ga As Ga As Ga GaAs has the same crystal structure as Si. GaAs, GaP, GaN are III-V compound semiconductors, important for optoelectronics. Which group of elements are candidates for donors? acceptors? From Atoms to Crystals conduction band Energy p Pauli exclusion principle s valence band isolated atoms lattice spacing Decreasing atomic separation Energy states of Si atom (a) expand into energy bands of Si crystal (b). The lower bands are filled and higher bands are empty in a semiconductor. The highest filled band is the valence band. The lowest empty band is the conduction band . Energy Band Diagram Conduction band Ec Eg Band gap Ev Valence band Energy band diagram shows the bottom edge of conduction band, Ec , and top edge of valence band, Ev . Ec and Ev are separated by the band gap energy, Eg . Measuring the Band Gap Energy by Light Absorption electron Ec photons Eg photon energy: h v > E g Ev hole • Eg can be determined from the minimum energy (hn) of photons that are absorbed by the semiconductor. Bandgap energies of selected semiconductors Material PbTe Ge Si GaAs GaP E g (eV) 0.31 0.67 1.12 1.42 2.25 Diamond 6.0 Semiconductors, Insulators, and Conductors Ec Top of conduction band Ec E g= 9 eV empty E g = 1.1 eV Ev Ev Si (Semiconductor) SiO (Insulator) filled Ec Conductor 2 Totally filled bands and totally empty bands do not allow current flow. (Just as there is no motion of liquid in a . totally empty bottle.) totally filled or Metal conduction band is half-filled. Semiconductors have lower E 's than insulators and can be g doped. Donor and Acceptor Levels in the Band Model Conduction Band Ed Donor Level Ec Donor ionization energy Acceptor ionization energy Acceptor Level Ea Valence Band Ev Ionization energy of selected donors and acceptors in silicon Donors Dopant Sb Ionization energy, E c –E d or E a –E v (meV) 39 Hydrogen: E ion = P 44 m0 q 4 8e0 2h2 Acceptors As 54 B 45 = 13.6 eV Al 57 In 160 Dopants and Free Carriers Donors n-type Acceptors p-type Dopant ionization energy ~50meV (very low). General Effects of Doping on n and p _ Charge neutrality: + + n Na p Nd = 0 _ Na Nd : number of ionized acceptors /cm3 + : number of ionized donors /cm3 Assuming total ionization of acceptors and donors: n + Na - p - Nd Na : number of acceptors /cm3 Nd : number of donors /cm3 =0 Density of States E gc DE Ec Ec g(E) Ev Ev gv gc ( E ) number of states in DE 1 3 DE volume eV cm mn* 2mn* E - Ec gc ( E) 2 h3 gv ( E ) m*p 2m*p Ev - E 2 h3 Thermal Equilibrium 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 high-energy states under the agitation of thermal energy (vibrating atoms, etc.) At E=EF, f(E)=1/2 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? Nc is called the effective density of states (of the conduction band) . Nv is called the effective density of states of the valence band. Intrinsic Semiconductor • Extremely pure semiconductor sample containing an insignificant amount of impurity atoms. n = p = ni Ef lies in the middle of the band gap Material Ge Si GaAs Eg (eV) 0.67 1.12 1.42 ni (1/cm3) 2 x 1013 1 x 1010 2 x 106 Remember: the closer Ef moves up to E c , the larger n is; the closer Ef moves down to Ev , the larger p is. For Si, Nc = 2.8 ´1019 cm-3 and Nv = 1.04 ´1019 cm-3 . Ec Ev Ef Ec Ev Ef Example: The Fermi Level and Carrier Concentrations Where is Ef for n =1017 cm-3? Solution: n Nc e - ( Ec - E f ) / kT Ec - E f kT ln Nc n 0.026 ln 2.8 1019 / 1017 0.146 eV 0.146 eV E E E c f v The np Product and the Intrinsic Carrier Concentration Multiply n Nc e - ( Ec - E f ) / kT and p Nv e np Nc Nv e-( Ec -Ev ) / kT Nc Nv e np ni - ( E f - Ev ) / kT - Eg / kT 2 ni Nc Nv e - E g / 2 kT • In an intrinsic (undoped) semiconductor, n = p = ni . EXAMPLE: Carrier Concentrations Question: What is the hole concentration in an N-type semiconductor with 1015 cm-3 of donors? Solution: n = 1015 cm-3. 2 ni 10 20 cm -3 p 15 -3 105 cm -3 n 10 cm After increasing T by 60C, n remains the same at 1015 cm-3 while p - E / kT increases by about a factor of 2300 because ni 2 e g . Question: What is n if p = 1017cm-3 in a P-type silicon wafer? Solution: 2 ni 1020 cm-3 n 17 -3 103 cm-3 p 10 cm General Effects of Doping on n and p I. N d - N a ni (i.e., N-type) n Nd - Na p ni n 2 If N d N a , n Nd II. N a - N d ni (i.e., P-type) If N a N d , p Na p ni Nd 2 and p Na - Nd n ni and 2 p n ni Na 2 EXAMPLE: Dopant Compensation What are n and p in Si with (a) Nd = 61016 cm-3 and Na = 21016 cm-3 and (b) additional 61016 cm-3 of Na? (a) n N d - N a 4 1016 cm -3 p ni / n 1020 / 4 1016 2.5 103 cm-3 2 (b) Na = 21016 + 61016 = 81016 cm-3 > Nd! p N a - N d 8 1016 - 6 1016 2 1016 cm -3 n ni / p 1020 / 2 1016 5 103 cm-3 2 n = 41016 cm-3 ++++++ ...... ++++++ ...... Nd = 61016 cm-3 Nd = 61016 cm-3 Na = 21016 cm-3 Na = 81016 cm-3 ........... - - - - - - - -. . . . . . p = 21016 cm-3 Chapter Summary Energy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef. n Nc e - ( Ec - E f ) / kT p Nv e - ( E f - Ev ) / kT n Nd - Na p Na - Nd np ni 2 Thermal Motion • Zig-zag motion is due to collisions or scattering with imperfections in the crystal. • Net thermal velocity is zero. • Mean time between collisions (mean free time) is m ~ 0.1ps Thermal Energy and Thermal Velocity 3 2 1 2 electron or hole kinetic energy kT mvth2 vth 3kT meff 3 1.38 10 -23 JK -1 300K 0.26 9.1 10 -31 kg 2.3 105 m/s 2.3 107 cm/s ~8.3 X 105 km/hr Drift Electron and Hole Mobilities • Drift is the motion caused by an electric field. Effective Mass In an electric field, E, an electron or a hole accelerates. electrons Remember : F=ma=-qE holes Electron and hole effective masses m n /m 0 m p /m 0 Si 0.26 0.39 Ge 0.12 0.30 GaAs 0.068 0.50 GaP 0.82 0.60 Remember : F=ma=mV/t = -qE Electron and Hole Mobilities mp v qE mp qE mp v mp v pE q mp p mp v - nE q n mn mn • p is the hole mobility and n is the electron mobility Electron and Hole Mobilities v = E ; has the dimensions of v/E cm/s cm 2 . V/cm V s Electron and hole mobilities of selected semiconductors n (cm2/V∙s) p (cm2/V∙s) Si 1400 Ge 3900 GaAs 8500 InAs 30000 470 1900 400 500 Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices? Drift Velocity, Mean Free Time, Mean Free Path EXAMPLE: Given p = 470 cm2/V·s, what is the hole drift velocity at E = 103 V/cm? What is mp and what is the distance traveled between collisions (called the mean free path)? Hint: When in doubt, use the MKS system of units. Solution: n = pE = 470 cm2/V·s 103 V/cm = 4.7 105 cm/s mp = pmp/q =470 cm2/V ·s 0.39 9.110-31 kg/1.610-19 C = 0.047 m2/V ·s 2.210-12 kg/C = 110-13s = 0.1 ps mean free path = mhnth ~ 1 10-13 s 2.2107 cm/s = 2.210-6 cm = 220 Å = 22 nm This is smaller than the typical dimensions of devices, but getting close. Mechanisms of Carrier Scattering There are two main causes of carrier scattering: 1. Phonon Scattering 2. Impurity (Dopant) Ion Scattering Phonon scattering mobility decreases when temperature rises: phonon phonon 1 1 -3 / 2 T phonon density carrier thermal velocity T T 1 / 2 = q/m T vth T1/2 Impurity (Dopant)-Ion Scattering or Coulombic Scattering Boron _ Ion - Electron - + Electron Arsenic Ion There is less change in the direction of travel if the electron zips by the ion at a higher speed. 3 th 3/ 2 v T impurity Na + Nd Na + Nd Total Mobility 1600 1 1400 Electrons 1 1000 2 -1 -1 Mobility (cm V s ) 1200 1 + 1 phonon impurity 1 phonon + 1 impurity 800 600 400 Holes 200 0 1E14 1E15 1E16 1E17 1E18 1E19 -3 Na +Concenration Nd (cm-3) Total Impurity (atoms cm ) 1E20 Temperature Effect on Mobility 10 15 Question: What Nd will make dn/dT = 0 at room temperature? Drift Current and Conductivity E Jp unit area + + n Current density EXAMPLE: Jp = qpv A/cm2 or C/cm2·sec If p = 1015cm-3 and v = 104 cm/s, then Jp= 1.610-19C 1015cm-3 104cm/s = A/cm 2 1.6 C/s cm 2 1.6 Drift Current and Conductivity E Jp unit area + + n + E - Remember: • Holes travel in the direction of the Electric field • Electrons travel in the direction opposite to that of the E-field Drift Current and Conductivity Jp,drift = qpv = qppE Jn,drift = –qnv = qnnE Jdrift = Jn,drift + Jp,drift = (qnn+qpp)E = E conductivity of a semiconductor is = qnn + qpp resistivity of a semiconductor is = 1/ DOPANT DENSITY cm-3 Relationship between Resistivity and Dopant Density P-type N-type RESISTIVITY (cm) = 1/ Ec and Ev vary in the opposite direction from the voltage. That is, Ec and Ev are higher where the voltage is lower. V x Ec E Ec-Ereference = -qV Ev x Variation in Ec with position is called band bending Diffusion Current Particles diffuse from a higher-concentration location to a lower-concentration location. Diffusion Current dn J n ,diffusion qDn dx dp J p ,diffusion - qD p dx D is called the diffusion constant. Signs explained: p n x x Total Current – Review of Four Current Components JTOTAL = Jn + Jp dn Jn = Jn,drift + Jn,diff = qnnE + qDn dx dp Jp = Jp,drift + Jp,diff = qppE – qD p dx JTOTAL = Jn,drift + Jn,diff + Jp,drift + Jp,diff Chapter Summary vp pE vn - nE J p ,drift qp pE Jn ,drift qn nE dn J n ,diffusion qDn dx dp J p ,diffusion - qD p dx kT Dn n q kT Dp p q