1. 1 UCONN ECE 4243/6243 Fall 2014 Nanoscience and Nanotechnology-I L2 Energy levels in potential wells and density of states 1.Bonds and energy bands (page 38)

2. 2 Bonds (page 43) Fig. 15. Schematic representation of s, p y, sp 3, and tetrahedral orbitals Tetrahedral primitive cell of Si lattice

3. 3 Infinite Potential Well p.47 Solve Schrodinger Equation In a region where V(x) = 0, using boundary condition that  (x=0)=0,  (x=L)=0. Plot the  (x) in the 0-L region.

4. 4 In the region x = 0 to x = L, V(x) = 0 Let The solution depends on boundary conditions. Equation 2 can be written as D cosk x x + C sink x x. At x=0,, so D= 0 (2) (3) Using the boundary condition  (x=L) = 0, we get from Eq. 3 sin k x L = 0, L = L x n x = 1,2,3… (4)

5. 5 The solution depends on boundary conditions. It satisfies boundary condition when Eq.6 is satisfied. Page 52-53: Using periodic boundary condition  (x+L) =  (x), we get a different solution: We need to have k x = ---(6) If we write a three-dimensional potential well, the problem is not that much different k y = n y = 1,2,3, … k Z =, n z = 1,2,3, …. The allowed k x, k y, k z values form a grid. The cell size for each allowed state in k-space is …(7) …(8)

6. Density of States in 3D semiconductor film (pages 53, 54) 6 Density of states between k and k+dk including spin N(k)dk = …(10) k k+dk Density of states between E and E+dE N(E)dE=2 …11 E=E c + (12) …(13) N(E)dE = …14 VB E N(E) N(E)dE = …15

7. Time dependent Schrodinger Equation p.68-69 7

8. 8

9. 9 Finite Potential Well p.70

10. 10 In the barrier region

11. 11 Wavefunction matching at well-barrier boundary

12. 12 Wavefunction matching at well-barrier boundary

13. 13

14. 14 ElectronsE C1 = 72.5 meV Heavy HolesE HH1 = 22.3 meV Light HolesE LH1 = 52.9 meV

15. 15 αL/2 kL /2 Where, radius is ElectronsE C1 = 72.5 meV Heavy HolesE HH1 = 22.3 meV Light HolesE LH1 = 52.9 meV Output: Equation (15) and (12), give k, α ’ (or α). Once k is known, using Equation (4) we get: (4) The effective width, L eff can be found by: E 1 is the first energy level.

16. 16 Summary of Schrodinger Equation in finite quantum well region. In the well region where V(x) = 0, and in the barrier region V(x) = Vo = ∆E c in conduction band, and ∆E v in the valence band. The boundary conditions are: 0 Al x Ga 1-x As GaAs ∆E V ∆E C 0 V(z) z -E G ∆E V -E G +∆E V z ∆E c = 0.6∆E g ∆E v = 0.4∆E g (5) Continuity of the wavefunction (6) Continuity of the slope Eigenvalue equation: (1)

17. C. 2D density of state (quantum wells) page 84 17 Carrier density number of states (1) Without f(E) we get density of states Quantization due to carrier confinement along the z-axis.

18. 18 Fig.2. Schematic cross-section of a floating Fig.4 energy band model to program a

19. Examples of Quantum dot lasers 19

20. 20 Why quantum well, wire and dot lasers, modulators and solar cells? Quantum Dot Lasers: Low threshold current density and improved modulation rate. Temperature insensitive threshold current density in quantum dot lasers. Quantum Dot Modulators: High field dependent Absorption coefficient (α ~160,000 cm-1) : Ultra-compact intensity modulator Large electric field-dependent index of refraction change (Δn/n~ 0.1-0.2): Phase or Mach-Zhender Modulators Radiative lifetime τr ~ 14.5 fs (a significant reduction from 100-200fs). Quantum Dot Solar Cells: High absorption coefficent enables very thin films as absorbers. Excitonic effects require use of pseudomorphic cladded nanocrystals (quantum dots ZnCdSe-ZnMgSSe, InGaN-AlGaN) Table I Computed threshold current density (Jth) as a function of dot size in for InGaN/AlGaN Quantum Dot Lasers (Ref. F. Jan and W. Huang, J. Appl. Phys. 85, pp. 2706-2712, March 1999).

21. Nanophotonics Si nanophotonics Surface enhanced Raman effect via plasmon formation in thin metal films or gold nanoparticles. Plasmons are modified by functionalized nanoparticles enabling biosensing of proteins etc. 21

22. 22 Semiconductor Background Review (ECE 4211 Chapters 1 and 2) Energy bands in semiconductors: Direct and indirect energy gap N- and p-type doping, Carrier concentrations: np=n i 2 Fermi-Dirac Statistics & Fermi level Drift and diffusion currents P-n junctions: Forward/Reverse biased Heterojunctions

23. 23 Conductivity σ, Resistivity ρ= 1/ σ Current density J in terms of conductivity  and electric field E: J =  E =  (-  V) = -   V I = J A =  E (W d), In n-type Si,  n  q  n n no + q  p p no

24. 24 Carrier Transport: Drift and Diffusion Drift Current: I n = J n A = - (q  n n) E A Diffusion Current density: J n = +q D n  n, [Fick's Law] Total current = Diffusion Current + Drift Current Einstein’s Relationship: D n /μ n =kT/q P no n no =ni 2 n-Si N D =N n =n no P po n po =ni 2 p-Si N A =P p =P po

25. 25 Drift and Diffusion of holes in p-Si In p-type Si, The conductivity is:  n  q  p p po + q  n n po Drift Current: I p drift = J p A = (q  p p) E A Diffusion Current density: J p = - q D p  p, [Fick's Law] Diffusion current: I p diff = - q A D p  p Einstein’s Relationship: D n /μ n =kT/q Total hole current = Diffusion Current + Drift Current I p = - q A D p  p + (q  p p) E A

26. 26 Carrier concentration When a semiconductor is pure and without impurities and defects, the carrier concentration is called intrinsic concentration and it is denoted by n i. i.e. n=p=n i. n i as a function of Temperature, see Figure 17 (page 28) and Fig. 11 (page 69). Also, n i can be obtained by multiplying n and p expressions (apge 68 of notes)

27. 27 Extrinsic Semiconductors: Doped n- and p-type Si, GaAs, InGaAs, ZnMgSSe IIIrd or Vth group elements in Si and Ge are used to dope them to increase their hole and electron concentrations, respectively. Vth group elements, such as Phosphorus, Arsenic, and Antimony, have one more electron in their outer shell, as a result when we replace one of the Si atoms by any one of the donor, we introduce an extra electron in Si. These Vth group atoms are called as donors. Once a donor has given an electron to the Si semiconductor, it becomes positively charged and remains so. Whether a donor atom will donate its electron depends on its ionization energy E D. If there are N D donor atoms per unit cm 3, the number of the ionized donors per unit volume is given by

28. 28 Fermi-Dirac Statistics We have used a statistical distribution function, which tells the probability of finding an electron at a certain level E. This statistics is called Fermi-Dirac statistics, and it expresses the probability of finding an electron at E as E f is the energy at which the probability of finding an electron is ½ or 50%. In brief, donor doped semiconductors have more electrons than holes. E

29. 29 Acceptors and p-type semiconductors: We can add IIIrd group elements such as Boron, Indium and Gallium in Si. When they replace a Si atom, they cause a deficiency of electron, as they have three electrons in their outer shell (as compared to 4 for Si atom). These are called acceptor atoms as they accept electrons from the Si lattice which have energy near the valence band edge E v. Eq. 12 expresses the concentration of ionized acceptor atoms (on page 71). N - A is the concentration of the ionized acceptor atoms that have accepted electrons. E A is the empty energy level in the acceptor atom. Hole conduction in the valence band: The electron, which has been accepted by an acceptor atom, is taken out of the Si lattice, and it leaves an empty energy state behind. This energy state in turn is made available to other electrons. It is occupied by other electrons like an empty seat in the game of musical chairs. This constitutes hole conduction.

30. 30 Donors and acceptors in compound semiconductors (see Problem set before chapter 1, p. 26) Semiconductors such as GaAs and InGaAs or ZnMgSSe are binary, ternary, and quaternary, respectively. They represent III-V and II-VI group elements. For example, the doping of GaAs needs addition of group II or VI elements if we replace Ga and As for p and n-type doping. In addition, we can replace Ga by Si for n-type doping. Similarly, if As is replaced by Si, it will result in p-type doping. So Si can act as both n and p-type dopant depending which atom it replaces. Whether Si is donor or acceptor depends on doping temperature.

31. 31 Calculation of electron and hole concentrations in n-type and p- type semiconductors Method #1: (simplest) Simple expressions for electron and hole concentrations in n-Si having N D concentration of donors (all ionized). Electron concentration is n = n n or n no =N D, (here, the subscript n means on the n-side or in n-Si; additional subscript ‘o’ refers to equilibrium). Hole concentration is p no =n i 2 /N D. For p-Si having N A acceptor concentration (all ionized), we have p= p p =N A, and electron concentration n po = (n i 2 )/N A, Method#2 (simpler) Here, we start with the charge neutrality condition. Applying charge neutrality, we get : total negative charge density = total positive charge density i.e. qN D + + qp no = qn no, here p no and n no are the hole concentrations in the n-type Si at equilibrium. But p no or hole concentration = n i 2 /N D. Substituting p no in the charge neutrality equation, we get electron concentration by solving a quadratic equation [Eq 8, page 71]. Its solution is:

32. 32 Method#3 (Precise but requires E f calculations) qN D + + qp no = qn no, Charge neutrality condition in n-type semiconductor can be written as:[Eq. 6 on page 70] Here, we have ignored the factor of ½ from the denominator of the first term. In this equation, we know all parameters except E f. One can write a short program and evaluate E f or assume values of E f and see which values makes left hand side equal to the right hand side.

33. 33 Effect of Temperature on Carrier Concentration The intrinsic and extrinsic concentrations depend on the temperature. For example, in Si the intrinsic concentrations at room temperature (~300K) is n i =1.5x10 10 cm -3. If you raise the temperature, its value increases exponentially (see relation for n i ).

34. 34 Another way of looking at the carrier concentration expression [pages 52-59] The electron concentration in conduction band between E and E+dE energy states is given by dn = f(E) N(E) dE. To find all the electrons occupying the conduction band, we need to integrate the dn expression from the bottom of the conduction band to the highest lying level or energy width of the conduction band. That is, [see page 56 notes] This leads to electron concentration (see page 57 and page 68): This equation assumes that the bottom of the conduction band E c = 0. An alternate expression results, if E c is not assumed to be zero.

35. 35  Direct and Indirect Energy Gap Semiconductors Fig. 10b. Energy-wavevector (E-k) diagrams for indirect and direct semiconductors(page 22). Here, wavevector k represnts momentum of the particle (electron in the conduction band and holes in the valence band). Actually momentum is = (h/2  )k = k

36. 36 Electrons & HolesPhotonsPhonons StatisticsF-D & M-BBose-Einstein Velocity v th,v n 1/2 mv th 2 =3/2 kT Light c or v = c/n r n r = index of refraction Sound v s = 2,865 meters/s in GaAs Effective Mass m n, m p (material dependent) No mass Energy E-k diagram E elec =25meV to 1.5eV ω-k diagram (E=hω) ω~10 15 /s at E~1eV E photons = 1-3eV ω-k diagram (E= ω) ω~5x10 13 /s at E~30meV E phonons = 20-200 meV Momentum P= k k=2π/λ λ=2πv elec /ω momentum: 1000 times smaller than phonons and electrons P= k k=2π/λ λ=2πv s /ω

37. 37 P-n Junctions (See Overview, Pp. 7-11] Before Going to Chapter 2) Use (V bi – V f ) for forward-biased junctions, and (V bi + V r )for reverse-biased junctions.

38. 38 Shockley's equation Here, the reverse saturation current Is = A J s, and reverse saturation current density J s is Here, D p is the diffusion coefficient of holes, p no is the hole concentration under equilibrium on n-Si side, L p is the diffusion length of holes (L 2 p = D p x  p ).  p is the average lifetime of injected holes. Similarly, L n is the diffusion length of electrons injected from n-side into p- side and npo is the minority electrons on p-side at equilibrium.

39. 39 Charge distribution, Field, Built-in Voltage Poisson's Equation  ·D =  Or,  2 V= -    r,  = q(N D + + p- N A - -n)]

40. 40 Fig. 4. Schematic representation of energy band diagram for a p-n junction showing electron energy.

41. 41 Energy band diagrams: Homojunction & Heterojunction Homojunction Heterojunction

42. 42 Equilibrium (VA)(VA) -xp-xp xnxn 0 lnln lplp pn N A = 10 19 cm -3 VAVA I Non- equilibrium x nene pepe -x po x no 0 lnln lplp pn oo N A = 10 19 cm -3 N D = 10 16 cm -3 n(x)on(x)o p(x)op(x)o E(x)E(x) NANA NDND Fig. 5.Carrier distribution in a p-n junction under equilibrium and under non-equilibrium (under forward biasing).

43. 43 Energy band diagrams in Heterojunctions Ei E c -E fp E i -E fn

44. 44 Energy band diagrams in Heterojunctions By definition, the built in voltage is the difference between the two Fermi levels (E fp1 and E fn2 for p-GaAs and n-AlGaAs, respectively. Here, we have not used the E i we use the difference between the Fermi level and the band edge (i.e. either E c2 -E fN2 or E fp1 -E v1 ). qV bi = - q(      q[(    c2 - E fn2 ) –{    gp1 – (E fp1 –E v1 )}] = - q[(      gp1 + (E c2 - E fn2 ) + (E fp1 –E v1 )] qV bi =  E c +  gp1 -(E c2 - E fn2 ) - (E fp1 –E v1 ) V bi =1/q[  E c +  gp1 -(E c2 - E fn2 ) - (E fp1 –E v1 )] Here, (E c2 - E fn2 ) = (kT) ln(N C /n) and, (E fp1 –E v1 ) = + (kT) ln(N v /p)

45. 45 Energy band diagram in a heterojunction P-AlGaAs/n-GaAs.

46. 46 L8, 2/10/05 Heterojunctions and Junction Fabrication Techniques Heterojunctions? General: Higher injection efficiency with lower doping levels in the wider energy gap semiconductor Laser Diodes: Carrier confinement in a narrow layer, if needed (useful in lasers to generate photons in a narrow layer (smaller d); minority carriers are not as readily injected from a narrower gap material into wider gap material. Laser Diodes:Photon confinement in a three layer sandwich of low- high-low index of refraction (e.g. AlGaAs-GaAs-AlGaAs); n r AlGaAs)<n r (GaAs). See homework #1. Quantum Well Lasers: thin low energy gap active layer permits confinement of carriers in very narrow layer (~50-70Angstroms) forming quantum wells and providing lower threshold operation. Quantum wire and quantum dot lasers: Lower threshold and temperature insensitivity

47. 47 Solar cells: wider gap semiconductor acts as the window where photons enter the device and are absorbed in the lower energy gap material Solar cells: Provides higher operating voltage for a given current; and minimize recombination of carriers at the surface. Heterojunction Bipolar transistors (HBTs): High injection efficiency even with lower emitter concentrations permits the use of a very highly doped base. This in turn reduces the base transit time 9one of the main factors limiting the unit gain cutoff frequency f T. Flexibility in designing higher current gain, reducing resistance (e.g. sub-collector). Why heterojunctions?