Generation of Free Electrons and Holes In an intrinsic semiconductor, the number of free electrons equals the number of holes. Thermal : The concentration of free electrons and holes increases with increasing temperature. Thermal : At a fixed temperature, an intrinsic semiconductor with a large energy gap has smaller free electron and hole concentrations than a semiconductor with a small energy gap. Optical: Light can also generate free electrons and holes in a semiconductor. Optical: The energy of the photons (hν) must equal or exceed the energy gap of the semiconductor (Eg) . If hν > Eg , a photon can be absorbed, creating a free electron and a free hole. This absorption process underlies the operation of photoconductive light detectors, photodiodes, photovoltaic (solar) cells, and solid state camera “chips”. Electrons and Holes in Intrinsic Semiconductor Electron energy Ec+χ CB Ec FREE e– hυ > E hυ g E hole g – HOLE e Ev VB 0 (a) (b) Fig. 5.3: (a) A photon with an energy greater than Eg can excite an electron from the VB to the CB. (b) When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Si bond is created. Electrons and Holes in Semiconductors under Electric field Conduction band — free electron Valence band - holes Energy band diagram in the presence of a uniform electric field. Shown are the upper almost-empty band and the lower almost-filled band. The tilt of the bands is caused by an externally applied electric field Intrinsic Semiconductor • Semiconductor contain no impurities. •Electron density equals to hole density. • Resulting from thermal activation or photon excitation. • Intrinsic carrier density Schematic representation of Carrier densities (a) (b) (c) (d) 1/2 g(E) ∝ (E-Ec) E E E Ec+χ [1-f(E)] CB For Area =∫ nE ( ) = ndEE electrons n (E) Ec Ec E EF EF E v Ev pE(E) For holes Area = p VB 0 g(E) f(E) nE(E) or pE(E) Fig. 5.7: (a) Energy band diagram. (b) Density of states (number of states per unit energy per unit volume). (c) Fermi-Dirac probability function (probability of occupancy of a state). (d) The product of g(E) and f(E) is the energy density of electrons in the CB (number of electrons per unit energy per unit volume). The area under nE(E) vs. E is the electron concentration in the conduction band. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Non-degenerate and Degenerate semiconductors E CB n > Nc Impurities forming CB E a band Fn Ec Ec g(E) Ev Ev EFp VB (a) (b) Fig. 5.21: (a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB. (b) Degenerate p-type semiconductor. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002) http://Materials.Usask.Ca Extrinsic Semiconductors Extrinsic semiconductors : impurity atoms dictate the properties Almost all commercial semiconductors are extrinsic Impurity concentrations of 1 atom in 1012 is enough to make silicon extrinsic at room T! Impurity atoms can create states that are in the band gap. In most cases, the doping of a semiconductor leads either to the creation of donor or acceptor levels n-type semiconductors n-Type In these, the charge carriers are negative p-type semiconductors. In these, the charge carriers are positive p-Type Band Diagram: Acceptor Dopant in Semiconductor For Si, add a group III element to “accept” an electron and make p-type Si (more positive “holes”). EC “Missing” electron results in an extra “hole”, with an acceptor energy level E just above E A F E the valence band E . A V EV p-type Si Holes easily formed in valence band, greatly increasing the electrical conductivity. Fermi level EF moves down towards EV. Typical acceptor elements are Boron, Aluminum, Gallium, Indium. Band Diagram: Donor Dopant in Semiconductor Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) For group IV Si, add a group V element to “donate” an electron and make n-type Si n-type Si (more negative electrons!). EC ED “Extra” electron is weakly bound, with EF donor energy level ED just below conduction band EC. Egap~ 1 eV EV Dopant electrons easily promoted to conduction band, increasing electrical conductivity by increasing carrier density n. Fermi level EF moves up towards EC. Typical donor elements that are added to Si or Ge are phosphorus, arsenic, antimonium. Silicon n-type semiconductors: Si Si Si Si Bonding model description: Element with 5 bonding electrons. Only 4 electrons Si Si Si Si - participate in bonding the extra e can easily Si P Si Si become a conduction electron Si Si Si Si p-type semiconductors: Si Si Si Si Bonding model description: Element with 3 bonding electrons. Since 4 Si Si Si Si electrons participate in bonding and only 3 are Si Si B Si available the left over “hole” can carry charge Si Si Si Si Question: How many electrons and holes are there in an intrinsic semiconductor in thermal equilibrium? Define: -3 no equilibrium (free) electron concentration in conduction band [cm ] -3 po equilibrium hole concentration in valence band [cm ] Certainly in intrinsic semiconductor: no = po = ni -3 ni intrinsic carrier concentration [cm ] As T ↑ then ni ↑ nO= p O= n i As Eg ↑ then ni ↓ What is the detailed form of these dependencies? We will use analogies to chemical reactions. The electron-hole formation can be though of as a chemical reaction…….. Similar to the chemical reaction……… − + bond⇔ e + h + − H O2 ⇔ H() + OH The Law-of-Mass-Action relates concentration of reactants and reaction products. For water…… Where E is the energy released or consumed during the reaction…………. H[+ ][ OH − ] ⎛ E ⎞ K = ≈exp⎜ − ⎟ HO[]2 ⎝ kT ⎠ This is a thermally activated process, where the rate of the reaction is limited by the need to overcome an energy barrier (activation energy). By analogy, for electron-hole formation: [no ][ p o ] ⎛ Eg ⎞ =K ≈exp⎜ − ⎟ []bonds ⎝ kT ⎠ Where [bonds] is the concentration of unbroken bonds and Eg is the activation energy [bonds]>> n ,p In general, relative few bonds are broken to o o form an electron-hole and therefore the [bonds] =constan t number of bonds are approximately constant. ⎛ E ⎞ n p≈exp⎜ − g ⎟ Two important results: o o ⎜ kT ⎟ ⎛ E ⎞ ⎝ ⎠ 1)…………. ⎜ g ⎟ n≈i exp⎜ − ⎟ ⎝ 2kT ⎠ 2)……………. 2 nO× p O = n i The equilibrium np product in a semiconductor at a certain temperature is a constant, specific to the semiconductor. Effect of Temperature on Intrinsic Semiconductivity The concentration of electrons with sufficient thermal energy to enter the conduction band (and thus creating the same concentration of holes in the valence band) ni is given by ⎛− ΔE ⎞ ⎜ ⎟ n i ≈ exp ⎜ ⎟ ⎝kB T ⎠ For intrinsic semiconductor, the energy is half way across the gap, so that ⎛ − E ⎞ ⎜ g ⎟ ni ≈ exp⎜ ⎟ ⎝ 2kB T⎠ Since the electrical conductivity σ is proportional to the concentration of electrical charge carriers, then ⎛ − E ⎞ ⎜ g ⎟ σ= σO exp⎜ ⎟ ⎝ k2 B T⎠ Thermal Stimulation P = Ratio of the number of electrons ⎛ Δ− E ⎞ P = exp⎜ ⎟ promoted to conduction band and the ⎜ Tk ⎟ ⎝ B ⎠ number of electrons in the system Suppose the band gap is Eg = 1.0 eV T(°K) kBT(eV) E/kBT ⎛ ΔE ⎞ Δ exp ⎜ − ⎟ ⎝ k BT⎠ 00∞ 0 100 0.0086 58 0.06x10-24 200 0.0172 29 0.25x10-12 300 0.0258 19.4 3.7 x10-9 400 0.0344 14.5 0.5x10-6 Example Calculate the number of Si atoms per cubic meter. The density of silicon is 2.33g.cm-3 and its atomic mass is 28.03g.mol-1. Then, calculate the electrical resistivity of intrinsic silicon at 300K. For Si at 300K 16 -3 -19 2 -1 2 -1 ni=1.5x10 carriers.m , q=1.60x10 C, μe=0.135m (V.s) and μh=0.048m .(V.s) Solution NA× ρ Si 28 −3 22 −3 nSi = 5= . 00Si × 10 . atoms − 5= .Si m 00 × 10 atoms − . cm ASi =σ n ×i qμ ×()e μ +h 16 3 −19 2 −1 2 −1 σ1=/ .() 5 × 1carriers 10. 6)( 10× mC 0 .() 135 m() V .+ s.m 0 () V . s 048= . 0 .=σ 4385 ×−3 10 Ωm − (1 ) 1 ρ =resistivit=2 = . y 28 ×3 10 Ωm − σ Example The electrical resistivity of pure silicon is 2.3x103Ω-m at room temperature (27oC ~ 300K). Calculate its electrical conductivity at 200oC (473K). -5 Assume that the Eg of Si is 1.1eV ; kB =8.62x10 eV/K ⎛ − E ⎞ ⎛ − E ⎞ g g ⎛ − Eg ⎞ σ =.C exp⎜ ⎟ σ 473 =.C exp⎜ ⎟ σ =.C exp⎜ ⎟ ⎜ k2 T⎟ 2⎜ k () 473⎟ 300 ⎜ ⎟ ⎝ B ⎠ ⎝ B ⎠ 2⎝ kB() 300⎠ σ 473 ⎛ − Eg − Eg ⎞ = exp⎜ − ⎟ σ 3002⎝ 473k()()B 2kB ⎠ 300 ⎛σ ⎞ − Eg Eg Eg ⎛ 1 1 ⎞ 1. eV 1 ⎛ 1 1 ⎞ ln⎜ 473 ⎟ = + = − = − ⎜ ⎟ ⎜ ⎟ −5 ⎜ ⎟ 2⎝σ 300 ⎠ 473k()()B 2kB 300kB ⎝ 300 2 4732⎠ 8(.) 62× 10⎝ 300 473 ⎠ ⎛σ 473 ⎞ ln⎜ 7⎟ = . 777 ⎝σ 300 ⎠ 1 σ= σ ()2385 = ().(.)2385= 1 Ωm 04−1 473 300 2 3. × 103 Example: For germanium at 25oC estimate (a) the number of charge carriers, (b) the fraction of total electrons in the valence band that are excited into the conduction band and ⎛ − E ⎞ (c) the constant A in the expression g when E=E /2 n= Aexp ⎜ ⎟ g ⎝ 2kB T⎠ Data: Ge has a diamond cubic structure with 8 atoms per cell and valence of 4 ; 2 2 a=0.56575nm ; Eg for Ge = 0.67eV ; μe = 3900cm /V.s ; μh = 1900cm /V.s ; -5 ρ = 43Ω-cm ; kB=8.63x10 eV/K (a) Number of carriers TC= 25o −5 2kB 2= T( 8 )( 63 .