SEMICONDUCTOR I: Doping, semiconductor statistics (REF: Sze, McKelvey, and Kittel)


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1 SEMICONDUCTOR I: Doping, semiconductor statistics (REF: Sze, McKelvey, and Kittel) Introduction Based on known band structures of Si, Ge, and GaAs, we will begin to focus on specific properties of semiconductors, including the doping problem, electron and hole population calculation, and other problems related to statistics. These are necessary tools for understanding semiconductor devices. I expect you to get familiar with the equations in the first chapter of Sze s bible (Physics of semiconductor devices, nd Ed.): knowing how to derive each equation and their physical meanings. Following this section, we will begin to explore the semiconductor transport properties, optical properties, and magnetic properties. A Bohr atom in vacuum and an impurity atom in semiconductors For a typical hydrogen atom, an electron with charge (e) and mass (m o ) is moving in free space in the presence of an attractive potential e /r at the nuclear, with r the spacing between the proton and the electron. The Bohr radius a o = h /(me ). Putting a positive ion in Si will give a very similar situation, except that now the electric field is screened to be e /(ǫr), and the electron will sense that it is in silicon and thus has an effective mass m. Here, ǫ is the dielectric constant of the host material. As a result, the radius of the ground state of the ionized impurityelectron atom will be a o ǫ/m, that is, the size is scaled by ǫ/m. The ground state binding energy (the energy it takes to ionize the impurity) also becomes 3.6eV m /ǫ. (E ionization of a hydrogen atom is m o e 4 /3π ǫ h = 3.6 ev in vacuum.) For common values of ǫ (Si:.9, Ge: 6, and GaAs: 3.) and m (in unit of m o, Si: 0.98 and 0.6, Ge: 0.04 and.64, and GaAs: 0.067), the radius of the impurity atom (when the electron is still bound to the ion) will be around 00 Å. The ionization energy is also scaled to be 3.6eV 0./0, about 00 times smaller. This is a consistent story: the size of the impurity atom must be several times larger than the lattice constant to make the screening argument justified. Let s use an example: As in Si. Initially, an electron is bound to As +, and the As atom is neutral. The ionization energy is found to be 54 mev. (Page of Sze lists many ionization energies for commonly observed elements in Si, Ge, and GaAs.) That is to say, if the electron can get 54 mev, it is able to travel free in Si and become, in the band structure language, an electron in the conduction band. When the electron is set free, the As will carry one positive charge, and there is a possibility for this ion to capture an electron and become neutral again. Since, in this example, 54 mev is the ionization energy, we therefore plot the donor energy at 54 mev below the conduction band minimum on the band structure. The position of the impurity states is only to follow the convention of Fermi statistics: the states below ǫ F is filled, and, if above, empty. By the same token, we define acceptor levels. Usually, column V (III) elements in column IV materials will become donors (acceptors). However, column IV elements in GaAs (a IIIV compound) is amphoteric. For example, Si in GaAs can be:
2 . a donor: Si Si + Ga + one electron. an acceptor : Si Si As + one hole 3. a neutral impurity: interstitial, or through compensation. Doping is of great importance to semiconductors. The impurity level is usually close to the band edges, thus by controlling the impurity concentration carrier concentration and therefore conductivity of a semiconductor sample can be tuned. This property turns out to be useful in making transistors. Bandgap change as a function of temperature Experimentally observed temperaturedependence of bandgap can be expressed as: E g (T) = E g (T = 0) [αt /(T + β)]. For Si, Ge, and GaAs, bandgap increases at low temperature. (Ref: Sze, page 5.) Carrier concentration at thermal equilibrium: INTRINSIC semiconductor (n = p = n i ) For an intrinsic semiconductor at zero temperature, all valence bands are filled and all conduction bands are empty. Free electron concentration and free hole concentration are both zero. (What is the total electron concentration?) By definition, at the conduction band, n = Ec ded(e)f(e), () where n is free electron concentration in the conduction band, Ec is the energy at conduction band minimum, D(E) is density of states, and f(e) is FermiDirac distribution function. Taken into account the anisotropic Fermi surface (ellipsoid for Si and Ge), D(E) = Mc π (E Ec) / h 3 (m m m 3 )/, () where Mc is the valley degeneracy (6 for Si, 8 for Ge, and for GaAs). Sometimes we define the density of states effective mass m de by (m m m 3 )/3. Spin degeneracy () is included in the above D(E). The effective masses (m m m 3 )/3 are along the principle axes of the ellipsoidal constant energy surface. For example, in Si, we can define m to be the longitudinal effective mass and m and m 3 to be the transverse effective mass. The FermiDirac distribution function (Ref: Kittel and many other statistical mechanics textbooks) is defined as: f(e) = + exp( E Ef kt ), (3)
3 where k is Boltzmann s constant, T the temperature, and E f the Fermi energy. (Whenever a temperature is given or FermiDirac statistics is used, it implies that the electrons are in equilibrium among themselves. If electrons and the lattice are also in equilibrium, then the lattice temperature should be the same as the electron temperature.) Homework: Derive the form of D(E) and f(e). For a simple example, volume = 4 3 πk3, energy= h k /m, i.e., k = (m E/ h ) /. Since = π dk, D(E) is then (from spin degeneracy) [ (4πk 3 /3)/ E] = 4π m 3/ E / /(π h 3 ). Finally, multiply the above result by valley degeneracy. For an ellipsoid (anisotropic effective mass), E f = h (k xf m x + k yf m y + k zf m z ). Or, = kxf (a + m x ) kyf (a + k zf m y ) (a, where a = E m z ) f / h. That is, a change in k x, k y, and k z are linked, and the dk relation is: mx = dky x m = dkz y m. Now, the Fermi ellipsoid has a volume of 4 z 3 π(k xf +kyf +kzf). When energy is increased a little bit, the change in k will be scaled by the m. So, when energy is increased, the small volume increase is 4π 3 (k xfk yf dk z + k yf k zf dk x + k zf k xf dk y ) = 4π[k m 3 yfk x zf m dk x + k xf k m m z y yf m dk x + k zf k xf x x m dk x ] = 4π x 3 d( Ef h ) [ E f h m x m y m z 3 ] = the expression in equation (). Now, putting D(E) and f(e) together, and define several new terms: dk x m x [ E f h m x m y m z 3] = 4π 3 3 n = N c π F / (E f Ec) kt, (4) where N c is the effective density of states in the conduction band and is defined as N c = ( πm dekt h ) 3/ Mc, (5) and F / is called FermiDirac integral, defined by F / (η f ) = 0 η / dη + e (η eta f ). (6) For a plot of calculated FermiDirac integral, see page 8 of Sze. Homework: Use computer to calculate this integral and plot or table the results like what shown in Fig. 0 of Sze. Numerical calculation of this FermiDirac integral is often simplified by using fitted equations. (Ref: Selberherr: Analysis and simulation of semiconductor devices, page 57.) Two asymptotic behaviors of F / are of interest to our discussion:. F / (x) π ex, for x ; 3
4 . F / (x) 3 x3/, for x. So, for nondegerate semiconductors, where in practice either the temperature is high, or the doping level is low, as an end result the Fermi level is more than several kt below the conduction band minimum, we can simplified the calculation of n (from the asymptotic value of FermiDirac integral) and get: n = N c e (Ec E f ) kt. (7) Degenerate electron gas just means that the Fermi level is close enough (about several kt) to the conduction band minimum or valence band top, therefore, we have to use the full definition in equation (?) to calculate the electron or hole concentrations. Notice that given the relatively large bandgap of Si, Ge, and GaAs, if the sample is degenerate for electrons, it will be nondegenerate for holes, and vice versa. Homework: Calculate N c for Si, and compare your result with what listed in Sze (page 850, appendix H). Similar equations holds for holes (in Si, Ge, and GaAs):. p = N v π F / (Ev E f ) kt,. N v = ( πm dhkt h ) 3/, (Mv = ), 3. m dh = (m 3/ lh + m 3/ hh ) /3, 4. p = N v e (E f Ev) kt (if nondegenerate). (lh: light hole, hh: heavy hole) For Si, N c = /cm 3 N v = /cm 3, and N i = /cm 3. For GaAs, N c = /cm 3 N v = /cm 3, and N i = /cm 3. The definition of intrinsic semiconductor is that n = p = n i. This result is coming from that all electrons in the conduction band must be excited from valence band. Therefore, there is equal number of electrons and holes in the semiconductor, and we just further define their concentration to be n i. (By the way, intrinsic semiconductors are, by definition, nondegenerate.) The Fermi level of an intrinsic semiconductor is called E i. We here list some basic relations for intrinsic semiconductors:. n = N c e (Ec E f ) kt ;. p = N v e (E f Ev) kt ; 4
5 3. n = p = n i np = n i n i = N c N v e Eg/kT ; 4. n = p E i = E f intrinsic = (Ec+Ev) + kt ln(nv N c ) = (Ec+Ev) + 3kT 4 ln(m dh m de ). Homework: Calculate n i as a function of temperature for Si, Ge, and GaAs, and plot the results like what shown in Fig. of Sze. Homework: For intrinsic Si, at what temperature would n = 0 7 /cm 3? Homework: Does E i have temperature dependence? If so, where is it from? (What is the physics involved?) DOPED SEMICONDUCTOR: calculation of n, p, and E f for all cases (to be more specific, both degenerate and nondegenerate) (Ref: Page 70 of McKelvey) Charge neutrality We will derive relations for a very generalized case, where in a semiconductor there are free electrons, free holes, ionized donors, unionized donors, ionized acceptors, and unionized acceptors. At thermal equilibrium, the net charge density should be zero, and the Fermi level is flat everywhere. The charge neutrality rule is based on the physical consideration that air molecules can neutralize the sample. And, the Fermi level must be flat in the sample, otherwise there is current flowing which indicates a nonequilibrium state. So, we have contributions from various terms, and each one deserves a detailed look. n + n d + N a = p + p a + N d, (8) where the definition of the terms are listed below:. n is free electron concentration in the conduction band, n = up E c ded e (E)f(E); 5
6 . n d is (bound) electron concentration at the donor cites, n d =, g= (due to spin, please refer to page 70 in J. McKelvey for a derivation); N d + g ee d E f /kt 3. N a is the concentration of acceptors, a known value when the sample is prepred; 4. p is free hole concentration in the valence band, p = down E v ded h (E)f(E); 5. p a is (bound) hole concentration at the acceptor cites; p a = N a N a = N a + g e(e f Ea)/kT, g=4 (due to spin and the additional hh/lh degeneracy). 6. N d is the concentration of donors, a known value when the sample is prepred. Also, N a is the concentration of (access) electrons at acceptors. So, N a = N a For the same reason, the concentration of ionized donors will be N + d + e(e d E f )/kt ). +4e (Ea E f )/kt. = N d n d = N d ( Using the graphic solution shown in page 4 of Sze, we can then calculate n(t), p(t), and E f (T). (Ref: page 4 of Sze) We are also interested to get analytical equations. In the following, we will discuss several useful limits. 6
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