Integrated Circuit Devices

1. Integrated Circuit Devices Professor Ali Javey Summer 2009 Semiconductor Fundamentals

2. Evolution of Devices Yesterday’s Transistor (1947) Today’s Transistor (2006)

3. 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

4. What are semiconductors Elements: Si, Ge, C Binary: GaAs, InSb, SiC, CdSe, etc. Ternary+: AlGaAs, InGaAs, etc.

5. Electrons and Holes in Semiconductors Silicon Crystal Structure • Unit cell of silicon crystal is cubic. • Each Si atom has 4 nearest neighbors. Å

6. (011) (100) Silicon Wafers and Crystal Planes z z z · The standard notation for crystal planes is based on the cubic unit cell. y y y x x x (111) · Silicon wafers are usually cut along the (100) plane with a flat or notch to help orient the wafer during IC fabrication. Si (111) plane

7. Bond Model of Electrons and Holes (Intrinsic 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.

8. Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Dopants in Silicon As B 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. ·

9. Immobile Charges they DO NOT contribute to current flow with electric field is applied. However, they affect the local electric field Ionized Donor Ionized Acceptor Hole Mobile Charge Carriers they contribute to current flow with electric field is applied. Electron Types of charges in semiconductors EE143 – Vivek Subramanian

10. GaAs, III-V Compound Semiconductors, and Their Dopants 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?

11. conduction band p Energy s valence band isolated atoms lattice spacing Decreasing atomic separation From Atoms to Crystals Pauli exclusion principle · 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 .

12. Energy Band Diagram Conduction band E c Band gap E g E v 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 .

13. Material PbTe Ge Si GaAs GaP Diamond E (eV) 0.31 0.67 1.12 1.42 2.25 6.0 g Measuring the Band Gap Energy by Light Absorption electron E c photons E g photon energy: h v > E g E v hole • Eg can be determined from the minimum energy (hn) of photons that are absorbed by the semiconductor. Bandgap energies of selected semiconductors

14. Semiconductors, Insulators, and Conductors E c Top of conduction band = E 9 eV g empty E c = E 1.1 eV g filled E E E v v c Conductor SiO (Insulator) Si (Semiconductor) 2 · 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 half-filled. · 's than insulators and can be Semiconductors have lower E g doped.

15. m0q4 Hydrogen: E = = 13.6 eV ion 8e02h2 Donor and Acceptor Levels in the Band Model Conduction Band E c E Donor Level d Donor ionization energy Acceptor ionization energy Acceptor Level E a E v Valence Band Ionization energy of selected donors and acceptors in silicon

16. Dopants and Free Carriers Donors n-type Acceptors p-type Dopant ionization energy ~50meV (very low).

17. + + - - - - n n N N p p N N a a d d General Effects of Doping on n and p _ + = 0 Charge neutrality: _ N : number of ionized acceptors /cm3 a + N : number of ionized donors /cm3 d Assuming total ionization of acceptors and donors: = 0 N : number of acceptors /cm3 a N : number of donors /cm3 d

18. Density of States E g c DE E E c c g(E) E E v v g v

19. Thermal Equilibrium

20. 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.)

21. At E=EF, f(E)=1/2

22. 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?

23. Nc is called the effective density of states (of the conduction band) .

24. Nv is called the effective density of states of the valence band.

25. 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

26. Remember: the closer E moves up to E , the larger n is; f c the closer E moves down to E , the larger p is. f v ´ ´ For Si, N = 2.8 10 19 cm -3 and N = 1.04 10 19 cm -3 . c v Ec Ec Ef Ef Ev Ev

27. Example: The Fermi Level and Carrier Concentrations Where is Ef for n =1017 cm-3? Solution: 0.146 eV E c E f E v

28. The np Product and the Intrinsic Carrier Concentration and Multiply • In an intrinsic (undoped) semiconductor, n = p = ni .

29. 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. After increasing T by 60C, n remains the same at 1015 cm-3 while p increases by about a factor of 2300 because . Question:What is n if p = 1017cm-3 in a P-type silicon wafer? Solution:

30. General Effects of Doping on n and p I. (i.e., N-type) If , and II. (i.e., P-type) , and If

31. EXAMPLE: Dopant Compensation • What are n and p in Si with (a) Nd = 61016 cm-3 and Na = 21016 cm-3 and (b) additional 61016 cm-3 of Na? • (a) • (b) Na = 21016 + 61016 = 81016 cm-3 > Nd! n = 41016 cm-3 . . . . . . . . . . . . + + + + + + + + + + + + Nd = 61016 cm-3 Nd = 61016 cm-3 Na = 81016 cm-3 Na = 21016 cm-3 . . . . . . . . . . . . . . . . . - - - - - - - - p = 21016 cm-3

32. Chapter SummaryEnergy band diagram. Acceptor. Donor. mn, mp. Fermi function. Ef.

33. Thermal Motion • Zig-zag motion is due to collisions or scattering • with imperfections in the crystal. • Netthermal velocity is zero. • Mean time between collisions (mean free time) is m ~ 0.1ps

34. Thermal Energy and Thermal Velocity electron or hole kinetic energy ~8.3 X 105 km/hr

36. Drift Electron and Hole Mobilities • Drift is the motion caused by an electric field.

38. Effective Mass In an electric field, E, an electron or a hole accelerates. Electron and hole effective masses electrons Remember : F=ma=-qE holes

39. Remember : F=ma=mV/t = -qE

40. Electron and Hole Mobilities = t m v q E mp p q t E mp = v m p = m = - m v E v E p n • p is the hole mobility and n is the electron mobility

41. Electron and Hole Mobilities v = E ;  has the dimensions of v/E Electron and hole mobilities of selected semiconductors Based on the above table alone, which semiconductor and which carriers (electrons or holes) are attractive for applications in high-speed devices?

42. Drift Velocity, Mean Free Time, Mean Free Path EXAMPLE: Given mp = 470 cm2/V·s,what is the hole drift velocity at E= 103 V/cm? What is tmp 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 = mpE = 470 cm2/V·s  103 V/cm =4.7 105cm/s tmp = mpmp/q =470 cm2/V ·s  0.39  9.110-31 kg/1.610-19 C = 0.047 m2/V ·s  2.210-12 kg/C = 110-13s = 0.1 ps mean free path = tmhnth ~ 1 10-13 s  2.2107 cm/s = 2.210-6 cm = 220 Å = 22 nm This is smaller than the typical dimensions of devices, but getting close.

43. Mechanisms of Carrier Scattering • There are two main causes of carrier scattering: • 1. Phonon Scattering • 2. Impurity (Dopant) Ion Scattering Phonon scatteringmobility decreases when temperature rises:  = q/m  T vth  T1/2

44. 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.

45. Total Mobility Na + Nd (cm-3)

46. Temperature Effect on Mobility Question: What Nd will make dmn/dT = 0 at room temperature?

47. E J p unit + area n + Drift Current and Conductivity Jp = qpv A/cm2 or C/cm2·sec If p = 1015cm-3 and v = 104 cm/s, then Jp= 1.610-19C  1015cm-3 104cm/s = Current density EXAMPLE:

48. E J p unit + area n + Drift Current and Conductivity E + - • Remember: • Holes travel in the direction of the Electric field • Electrons travel in the direction opposite to that of the E-field

49. Drift Current and Conductivity Jp,drift = qpv = qppE Jn,drift = –qnv = qnnE Jdrift = Jn,drift + Jp,drift = (qnn+qpp)E =  E conductivity of a semiconductor is  = qnn + qpp resistivity of a semiconductor is  = 1/

50. DOPANT DENSITY cm-3 P-type N-type RESISTIVITY (cm) = 1/ Relationship between Resistivity and Dopant Density