Semiconductors

SemiconductorsThe importance of the band theory is that is provides us with a clear means of understanding how solids may be insulators, semiconductors or metals. Full/empty bands give no conduction At O K, if the filled bands are separated from empty bands, we have insulating properties. If the separation is > 5 eV the substance remains an insulator above O K, whereas semiconducting properties arise if the filled and empty levels lie within 0 - 2 eV of one another. In semiconductor physics we use a distinct terminology. The filled energy band is called the valence band (V.B.) and the empty band immediately above it the conduction band (C.B.). The zero energy level is taken as the top of the valence band. The energy gap separating the two bands is denoted by Eg. Any process that excites an electron from the valence to the conduction band produces two charge carriers, electron in the conduction band and the positive hole left in the valence band for conduction. (Thermal and optical excitation processes are two important ways of inducing semiconductor behaviour, which will normally involve both electrons and holes). E Metal Semiconductor Insulator C.B EF V.B. At 300 K Si: Eg = 1.17 eV, Ge: Eg = 0.75 eV, C(diamond): Eg = 1.17 eV Charge carriers in semiconductors σ =e⎡ nµ + p µ p ⎤ ⎣ n ⎦ µ= eτ m* n = # electrons/volume p = # holes/volume Both electrons and holes give conduction in the same direction Metals → σ not strongly dependent on T (just through τ), N(EF) ≈ constant. Semiconductors → σ strongly dependent on T because n, p are strong functions of T. Intrinsic semiconductors Semiconductors with “free” electrons/holes created by electronic excitations from V.B. → C.B. Occupation obeys Fermi statistics. Let's call the thermal equilibrium concentration for an intrinsic semiconductor no and po. n0 = 3 * 2 n 2 3 3 ∞ 1 e ( E − EF ) / k B T +1  e − ( E − EF ) / k B T 1 (2m ) e EF / kBT ∫ ( E − Ec ) 2 e − E / kBT dE 2π = Ec ∞ * 2 1 (2mn ) n0 = e − ( Ec − EF ) / kBT ∫ ( E − Ec ) 2 e − ( E − Ec ) / kBT dE 2 3 2π = Ec Let x = ( E − Ec ) k BT 3 ∞ * 2 3 1 (2mn ) Therefore n0 = (k BT ) 2 e − ( Ec − EF ) / kBT ∫ x 2 e − x dx 2 3 2π = 0 * ⎛ 2π mn k BT ⎞ 2 − ( Ec − EF ) / kBT n0 = 2 ⎜ ⎟ e 2 h ⎝ ⎠ 3 . * 2 1 (2mn ) 2 − Dc ( E ) = ( E E ) c 2π 2 =3 E > Ec Dv ( E ) = (2m* p) 2π = 3 2 2 3 ( Ev − E ) 3 ∞ 1 2 Ev > E 1 e ( E − EF ) / k B T * 2 (2mn ) Thus n0 = 2 3 2π = Ec ∫ (E − E ) c 1 2 +1 dE Since E-EF>>kBT. For an intrinsic semiconductor the Fermi energy is in the middle of the band gap and n0=p0 ∞ Thus n0 = ∫ Dc ( E ) f ( E . T )]dE v 3 where Dc(E) and Dv(E) are the density of states in the conduction band and the valence band respectively.A schematic of the density of states. T )dE and Ec p0 = Ev −∞ ∫ D ( E )[1 − f ( E. c. ⎛ m* 3 p EF = EFi = Emidgap + k BT ln ⎜ * ⎜ 4 ⎝ mn ⎞ ⎟ ⎟ ⎠ .Similarly ⎛ 2π m* p k BT p0 = 2 ⎜ ⎜ h2 ⎝ ⎞ 2 − ( EF − Ev ) / kBT e ⎟ ⎟ ⎠ 3 The parameters n0 and p0 are normally referred to as intrinsic electron and hole concentrations respectively (ni and pi) * ⎛ 2π mn k BT ⎞ 2 − ( Ec − EF ) / kBT n0 = ni = 2 ⎜ ⎟ e h2 ⎝ ⎠ 3 ⎛ 2π m* p k BT p0 = pi = ni = 2 ⎜ ⎜ h2 ⎝ 3 2 i ⎞ 2 − ( EF − Ev ) / kBT e ⎟ ⎟ ⎠ 3 3 2 ⎛ 2π k BT ⎞ * * n0 p0 = n = 4 ⎜ m m e − ( Ec − Ev ) / kBT ( ) n p ⎟ 2 ⎝ h ⎠ 2 ⎛ 2π k BT ⎞ −( E ) / k T * * n0 p0 = n = 4 ⎜ m m e g B .Law of Mass Action ( ) n p ⎟ 2 ⎝ h ⎠ 3 3 2 i Fermi level of intrinsic semiconductors For intrinsic semiconductors n0 = p0 implies * ⎛ 2π m* ⎛ 2π mn k BT ⎞ 2 − ( Ec − EF ) / kBT p k BT 2⎜ e 2 = ⎜ ⎟ 2 ⎜ h h2 ⎝ ⎠ ⎝ 3 ⎞ 2 − ( EF − Ev ) / kBT e ⎟ ⎟ ⎠ 3 e − ( Ec − EF ) / kBT ⎛ m* p =⎜ * ⎜m ⎝ n ⎞ 2 − ( EF − Ev ) / kBT e ⎟ ⎟ ⎠ 3 EF = ⎛ m* Ec + Ev 3 p + k BT ln ⎜ * ⎜ 2 4 ⎝ mn ⎞ ⎟ ⎟ ⎠ We normally denote EF by EFi for an intrinsic s. 07 Arsenide 0. By adding dopants.5 x 1010 2.4 x 1013 1. .56 0.08 Germanium 0. EFi is right in the middle of the gap. semiconductors can be given extra carriers.37 0.8 x 106 These carrier concentrations are not enough to carry reasonable current for devices. EFi > Emid gap EFi < Emid gap that is slightly above the centre that is slightly below the centre Effective masses and intrinsic carrier concentrations for some commonly used semiconductors (at 300K) mn*/mo mp*/mo Silicon 1. If mp* < mn*.48 ni(cm-3) 1.55 Gallium 0.For mp* = mn*. If mp* > mn* . . defects or departures from exact stoichiometry. As. The influence of added impurities or defects gives rise to extrinsic properties. Sb) Acceptors: Group III elements (B. Al.Extrinsic Semiconductors The properties of semiconductors are strongly affected by the presence of impurities. Electrically active impurities (donors or acceptors) give rise to ‘free’ electrons or holes.6eV 8ε 02h2 • For the loosely bound electron in the semiconductor m0 → me* and ε0 → εrε0 where εr is the dielectric constant in the semiconductor. • The binding energy of the electron in the hydrogen atom is given by E =− m0e4 = − 13. Can be estimated using Bohr model of Hydrogen atom. Ga. In) Impurity dopants → substitutionals Energy required for donor electrons to be excited to conduction band is small compared to thermal electrons. Donors: Group V elements (P. n-type p-type .025 m0 → and εr ≈ 12 ∴ Binding energy of the loosely bound electron is given by E = − (13.] At fairly moderate temperatures.026 eV 2 (12) • The energy states of these electrons lie very close to the bottom of the conduction band.08 = − .6 eV ) 1.∗ 4 m ee E =− 2 2 2 8ε r ε 0 h • If for silicon me*= . [Note that the thermal energy kBT at room temperature (300 K) is 0.025 eV. Similar situation occurs when dopant atoms are acceptors which will be ionized creating extra holes in the valence band. all the donor atoms are ionized giving electrons to conduction band. Thus n0 and p0 can be written as * * ⎛ 2π mn ⎛ 2π mn k BT ⎞ 2 − ( Ec + EFi − EFi − EF ) / kBT k BT ⎞ 2 − ( Ec − EFi ) / kBT − ( EFi − EF ) / kBT n0 = 2 ⎜ e 2 e = ⎟ ⎜ ⎟ e h2 h2 ⎝ ⎠ ⎝ ⎠ 3 3 3 3 ⎛ 2π m* p k BT p0 = 2 ⎜ ⎜ h2 ⎝ ⎞ 2 − ( EF + EFi − EFi − Ev ) / kBT ⎛ 2π m* p k BT e = 2⎜ ⎟ ⎟ ⎜ h2 ⎠ ⎝ ⎞ 2 − ( EFi − Ev ) / kBT − ( EF − EFi ) / kBT e e ⎟ ⎟ ⎠ n0 = ni e − ( EF − EF )/ kBT i p0 = ni e − ( EF − EF )/ kBT i and n0 p0 = ni2 . which will then change the values of the no and po.Carrier densities in doped semiconductors Law of mass action still applies i. n0 p0 = 4 ⎛ ⎜ 2 2π k BT ⎞ − ( E g ) / k BT * * ⎟ ( mn m p ) e 2 ⎝ h ⎠ 3 3 * ⎛ 2π mn k BT ⎞ 2 − ( Ec − EF ) / kBT = N c e− ( Ec − EF ) / kBT n0 = 2 ⎜ ⎟ e 2 h ⎝ ⎠ where * ⎛ 2π mn k BT ⎞ Nc = 2 ⎜ ⎟ 2 h ⎝ ⎠ 3 2 3 and ⎛ 2π m* p k BT p0 = 2 ⎜ ⎜ h2 ⎝ where ⎛ 2π m* p k BT Nv = 2 ⎜ ⎜ h2 ⎝ ⎞ 2 − ( EF − Ev ) / kBT = N v e− ( EF − Ev ) / kBT e ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ 3 2 3 The difference is that the Fermi Energy may vary through the band gap.e. n0 > ni and po < ni → an n-type conductor po > ni and no < ni.can be formed by diffusing acceptor impurities into an n-type material or by diffusing donor impurities into a p-type material Nd – donor impurity concentration Na – acceptor impurity concentration When Nd > Na → n-type compensated semiconductor Na > Nd → p-type compensated semiconductor Na = Nd semiconductor shows intrinsic characteristics Equilibrium electron and hole concentrations in a compensated semiconductor n o . Similarly. when EF < EFi.When EF > EFi. → a p-type conductor Compensated semiconductors Semiconductor that contains both donor and acceptor impurity atoms in the same region .t o t a l e le c t r o n c o n c e n t r a t io n therm al e le c t r o n s d o n o r e le c t r o n s EC + total N d + + + Ed io n iz e d N d + io n iz e d N a to tal N a Ea + + + + + + + EV therm al a c c e p t o r h o le s p o t o t a l h o le c o n c e n t r a t io n . ni is a very strong function of temperature.= negatively charged acceptor states Total no. additional electron-hole pairs will be generated so that the ni2 term in the equation may begin to dominate.= Na – pa Therefore no + Na .nd If the material is completely ionized nd = pa = 0 and thus no + Na = po + Nd combined with n0 p0 = ni2 n0 = ( Nd − Na ) ⎛ N − Na ⎞ 2 + ⎜ d ⎟ + ni . So as the temperature is raised.for a p-type semiconductor ⎟ 2 2 ⎝ ⎠ 2 . of negative charges = no + Nawhere Na.for an n-type semiconductor 2 2 ⎝ ⎠ 2 p0 = ( Na − Nd ) ⎛ N − Nd ⎞ + ⎜ a + ni2 .pa = po + Nd .no = thermal + donor electrons po = thermal + acceptor holes Total no. The semiconductor will eventually loose . of positive charges = po + Nd+ where Nd+ = positively charged donor states If nd and pa are the electron and hole concentrations of donor and acceptor states (we have to consider uncharged states) then Nd+ = Nd – nd Na. n0 extrinsic intrinsic partial ionization 100 200 300 400 500 T(K) Position of the Fermi level We can then determine the position of the Fermi level as a function of doping concentration.its extrinsic characteristics.EV = k BT ln ( NV / N a ) As we increase Na. EF moves closer to Ev . EF . EF moves closer to Ec p-type semiconductor for which Na >> ni and Nd = 0 po ≈ Na Therefore.EF = k BT ln ( N c / N d ) As we increase Nd. As a function of temperature. n0 = N c e− ( Ec − EF ) / kBT and p0 = N v e − ( EF − Ev ) / kBT n-type semiconductor for which Nd >> ni and Na = 0 no ≈ Nd Therefore Ec . the electron concentration in the semiconductor shows the following variation. Variation of EF with impurity concentration Ec n-type EF p-type Ev 1012 Na(Nd) 1018 . n p E . . As a result diffusion of majority carriers takes place across the metallurgical junction.field Depletion layer Diffusion of majority carriers creates a depletion layer around the junction where an electric field exists resulting a potential barrier across the junction.The Semiconductor Diode Semiconductor diode is made by joining a p-type semiconductor and an n-type semiconductor metallurgically. The potential barrier is known as the built-in potential barrier. Diffusion of majority carriers will be balanced by the opposing electric field at some point so that the depletion layer will attain a certain width. D = ρ ).field Ep= 0 Ep= -eVbi Ep= eVbi Ep= 0 The built-in electric field and the potential can be found using one JJJ K of the Maxwell’s equations ( ∇. .E . field p n x = −xp 0 x = xn If the impurity concentrations in the p and n sides are Na and Nd respectively the charge densities in the depletion region can be written as ρ ( x) = eN d . dE ( x) dx In one dimension. = ρ ( x) ε sε 0 . electric field E and D JG JG are related through D = ε sε 0 E ∇.E . 0 ≤ x ≤ xn ρ ( x) = − eN a .E = JJK ε sε 0 ρ . − x p ≤ x ≤ 0 If the dielectric constant of the material is εs. C = − 1 εsε0 n ∴ E(x) = eNd (x− xn) E ( x) = 0 at x = xn εsε0 Electric field in the p-side of the depletion region In a similar manner we can show that eN ∴ E(x) = − a (x+ xp) for − xp ≤ x ≤ 0 εsε0 Thus the variation of electric field in the depletion region shows the following behaviour. ρ ( x) = eN d dE ( x) dx = ε sε 0 dx eN d ∫ dE = ∫ ε ε E ( x) = eN d s 0 ε sε 0 eN d x + C1 Applying the boundary condition eNd x wehave.Electric field in the n-side of the depletion region For 0 ≤ x ≤ xn . . E(x) x = − xp x=0 x = xn x Emax Knowing the expression for the electric field E(x). ie. |Φ(xn)| Vbi = Φ ( xn ) = e 2 ( N d xn + Na x2 p) 2ε s . the electric potential Φ(x) can also be found. It can be shown that Φ ( x) = eN a εs ( x + x p )2 − xp ≤ x ≤ 0 and Φ ( x) = eN d εs x2 ( xn x − ) 2 0 ≤ x ≤ xn In order to obtain these expressions we use the boundary conditions (i) Φ(x) = 0 at x = -xp and (ii) Φ(x) is continuous at x = 0. The built-in potential Vbi = |Φ(x)| at x= xn. Using the expression for electric potential in the n-side of the depletion region we can express the built-in potential as follows. Depletion layer width Vbi = Φ ( xn ) = e 2 ( N d xn + Na x2 p) 2ε s Also the electric field of the depletion region is continuous at the metallurgical junction (at x = 0) ∴ N d xn = N a x p Combining the two equations we have ⎧ 1 ⎪ 2ε V ⎛ N ⎞ ⎛ xn = ⎨ s bi ⎜ a ⎟ ⎜ ⎪ ⎩ e ⎝ Nd ⎠ ⎝ Na + Nd ⎧ 2ε V ⎪ ∴ x p = ⎨ s bi ⎪ e ⎩ ⎞⎫ ⎪ ⎟⎬ ⎠⎪ ⎭ 1 2 ⎛ Nd ⎞ ⎛ 1 ⎜ ⎟⎜ ⎝ Na ⎠ ⎝ Na + Nd ⎫ ⎞⎪ ⎟⎬ ⎪ ⎠⎭ 1 2 So the width of the depletion layer can be expressed as W = xn + x p ⎧ ⎪ 2ε V ⎛ N + N d W = ⎨ s bi ⎜ a ⎪ e ⎝ Na Nd ⎩ ⎞⎫ ⎪ ⎟⎬ ⎪ ⎠⎭ 1 2 Action of a biased p-n junction As we discussed earlier. the potential barrier across the p-n junction at equilibrium can be sketched as below. . This will lower the potential across the junction. there will be an electric field that is in the direction opposite to that of the built-in electric field. Φ(x) V0 Vbi Vbi-V0 x = -xp 0 x = xn .Φ(x) Vbi x = -xp 0 x = xn When the junction is forward biased. Junction is forward biased with an externally applied voltage V0 When the potential difference across the junction is reduced. the I-V characteristics of a diode takes the form as shown in the figure. Φ(x) V0 iB Vbi+V0 x = -xp 0 x = xn Junction is reverse biased with an externally applied voltage V0 Raising/lowering of the potential barrier across the junction depending on the bias. As has been discussed in second year electronics course.). the diffusion current persists and will become large with the applied voltage. . the opposite will happen and the potential across the junction will be raised. When the junction is reverse biased. makes the diode a non-linear device (The resistance of the diode is much larger for the reverse bias. more of the conduction electrons on the n-side can diffuse across the junction. Since there is a plentiful supply of conduction electrons on the n-side. In fact the diffusion current increases exponentially with the applied voltage. I V . EcP Ecn Evn EFn EFP EvP eφsn eχn ∆Ec EcP Ecn Evn EvP ∆Ev An example is GaAs/AlGaAs heterojunction in which Eg (AlGaAs) = Eg (GaAs) + ∆Ev + ∆Ec ∆Ev and ∆Ec are known as the band offsets. the corresponding band alignment is known as a type II alignment . When it is not. When the smaller band gap lies entirely within the larger band gap it is known as a type I alignment.Semiconductor heterostructures A heterojunction is an interface between two different types of semiconductor. EcP ∆Ec Ecn Evn EFn EFP ∆Ev EvP . Double Heterostructures P-AlGaAs n-GaAs N-AlGaAs N-AlGaAs p-GaAs P-AlGaAs . These double hetero-structures have very important applications in semiconductor lasers. quantum well lasers and other devices. .