PHYS208 PN Junction. Olav Torheim. May 30, 2007


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1 1 PHYS208 PN Junction Olav Torheim May 30, 2007
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3 1 Intrinsic semiconductors The lower end of the conduction band is a parabola, just like in the quadratic free electron case (E = h2 k 2 2m ). The density of states for free particles in a box is: g(e) = V 3/2 2m 2π 2 h 2 E 1/2 We assume that the electrons in the conduction band behave like free electrons, and therefore use the the same density of states as for the free electron model. We only correct for the effective mass and that we have to count the energy from the bottom of the conduction band and up: g c (E) = V 3/2 2m 2π 2 h 2 (E E c ) 1/2 The number of free electrons in the conduction band is then given by the following integral:
4 N c = g c (E) f F D (E)dE = 2V ( m kt 3/2 E c 2π h 2 ) e µ Ec We can make the same calculation for the number of holes in the valence band if we use the effective mass for holes and integrate from the top of the valence band and down to inifinity. Here we also have to take into consideration that f h (E) = 1 f F D (E). N v = E v g v (E) (1 f F D )(E)dE = 2V ( m 3/2 h kt 2π h 2 ) e Ev µ The chemical potential of the intrinsic semiconductor is found be setting N c = N v : µ = E c + E v kt ln(m h m e ) Which shows that the chemical potential must be in the middle of the energy gap:
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6 In the same way, we can find the intrinsic carrier concentration by multiplying N c /V and N v /V : n 2 i = n c n v = 4 m e m h 3/4 (kt/2π h 2 ) 3 e Eg/kT The intrinsic carrier concentration is dependent only on the energi gap and not on the Fermi level (which means that the law of mass action, n c n v = n 2 i, is valid also for doped semiconductors). 2 Doped semiconductors Impurity atoms are added to provide loosely bound electrons (ndoping with group 5 material) or holes (pdoping with group 3 material) that are thermally excited to the conduction or valence band. The impurity atoms are usually not close enough for the wavefunctions to overlap, therefore the impurity states are localized. Their energy levels are either right below the conduction band or right above the valence band. How close they are can be approximated with a modified hydrogen atom model:
7 E ion = 13.6eV m m ɛ 0 2 ɛ Here the electron mass m is replaced by the effective mass m and the dielectricity constant refers to the crystal and not to free space. Example: Phoshorus ndoping of silicon. ɛ r = 11.7 and m is 0.19 near the bottom of the conduction band for Si, which gives us E ion = 0.019eV. The donor m impurity therefore requires 0.019eV to become ionized and the position in the band gap of the donor impurity level is just ev below the conduction band edge. It can be shown using similar arguments as in the intrinsic case that the Fermi level in the ndoped case is in the middle between the band egdes and the impurity states: E F = E c + E d 2
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9 And in the pdoped case: E F = E v + E a 2 Since the electrons in the semiconductors are modeled as Bloch electrons, they all have a nonvanishing group velocity. On the band edges this group velocity is v g = 1 h de = 0. This implies that the kinetic energy on the band egde is dk zero, and the position of the band egde as a function of position is therefore changing in the same way as the potential energy is changing.
10 When pdoped and ndoped materials are joined, there is an inbalance between the ntype and ptype carrier concentrations on each side because of the different Fermi levels. Ntype and ptype majority carriers will therefore diffuse into each others regions and recombine until a potential is built up that opposes the diffusion currents. The potential is called the builtinpotential, φ BI, and it raises the band edges on the pside so the Fermi levels on each side are aligned:
11 After equilibrium is established, there is balance between the diffusion current caused by the carrier inhomogenity (given by Ficks law) and the drift current caused by the electric field (given by simple kinetic free electron theory): which leads to n(x)µe( x x) = D dn dx E x (x) = D µ 1 n dn dx The builtin potential can now be calculated: φ BI = φ( ) φ( ) = E x dx = D µ 1 n(x) dn dx dx = D n( ) ln µ n( ) Now we have the condition for the energy difference, but D, n( ) and n( ) µ are still unknown. It is therefore time to introduce EinsteinNernst. We start by writing the electric field as the gradient of the potential:
12 E x (x) = dφ dx Therefore µ n e dφ dx = ed dn dx So dn dφ = µ n D The different equation has the solution n = n 0 e µ D φ But because this nonuniform electron distribution has to be maintaned against the potential φ using thermal energy, we must also have
13 n = n 0 e eφ kt φ Thus we finally arrive at the EinsteinNernst relation: µ D = Now we continue to evaluate ln n( ). In the nregion the density of free n( ) electrons is approximately n( ) = N D. In the pregion, the negative carrier concentration is taken from the law of mass action, n( ) = n2 i N A. From this we obtain the builtin potential: e kt φ BI = D µ lnn( ) = kt e lnn AN D n 2 i After establishing equilibrum between drift and diffusion currents, we are left with a depletion region in the middle of the junction where the holes and
14 electrons have recombined and we have approximately no free carriers. Starting out with the Poisson equation, we can make a model of this depletion region characterized by the parameters kt e, N A and N D : d2 V dx 2 = Cρ(x) The net charge density, ρ, is approximated as zero outside the depletion region. Inside the depletion region, we have approximately ρ = qn A on the p side ( d p x 0) and ρ = qn D on the nside (0 x d n ). Starting on the p side of the depletion region, integration of the Poisson equation yields Another integration yields ɛ = dv dx = en AC(d p + x) V = en A C( x2 2 + d px + d2 p 2 ) = 1 2 en AC(d p + x) 2
15 for d p x 0, so that V (x = 0) = V 1 = en A C d2 p 2. In the same way we have for the n side that V (x = d p ) V (x = 0) = V 2 = en D C d2 n 2. And we have an equation for the total potential across the junction: V C = φ BI + V R = e C 2 (N Ad 2 p + N D d 2 n). We now define d = d p + d n. And since N D d n = N A d p (charge balance), we finally arrive at an equation that combines the potential across the junction with the geometry and the doping profile of the depletion region: V C = 1 N D N A 2 Ced2. N D + N A
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17 We say that it is impossible to stop the drift current, since there always will be minority carriers on each side of the junction that are thermally excited and swept across the electric field (which is on the order of V/cm). When it comes to the diffusing majority carriers, there will be a Boltzmann distribution on how many carriers that have energy enough to break through the potential barrier. This recombination current is therefore proportional to the Boltzmann factor e qv C/kT. When no external potential is applied, the generation and recombination currents are in equilibrium: i tot = i rec i gen = i gen i gen = 0 (1) When an external potential, φ ext, is applied, the recombination current is increased with the Boltzmann factor e qφext/kt, while the generation current remains the same: i tot = i gen e qφext/kt i gen = i gen (e qφext/kt 1) (2) which shows us that the pn junction acts as a rectifier.
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