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ENE 311 Lecture 6 Donors and Acceptors • We have learned how to find new position of Fermi level for extrinsic semiconductors. • Now let us consider the new electron density in case of both donors ND and acceptors NA are present simultaneously. • The Fermi level will adjust itself to preserve overall charge neutrality as n N A p ND (1) Donors and Acceptors • By solving (1) with n. p ni2 , the equilibrium electron and hole concentrations in an n-type semiconductors yield 1 nn N D N A 2 ni2 pn nn N D N 2 A 4n 2 i Donors and Acceptors • Similarly to p-type semiconductors, the electron and hole concentrations are expressed as 1 p p N A N D 2 ni2 np pp N N D 2 A 4ni2 Donors and Acceptors • Generally, in case of all impurities are ionized, the net impurity concentration ND – NA is larger than the intrinsic carrier concentration ni; therefore, we may simply rewrite the above relationship as nn N D N A if N D N A p p N A N D if N A N D Donors and Acceptors • The figure shows electron density in Si as a function of temperature for a donor concentration of ND = 1015 cm-3. • At low temperature, not all donor impurities could be ionized and this is called “Freeze-out region” since some electrons are frozen at the donor level. Donors and Acceptors • As the temperature increased, all donor impurities are ionized and this remains the same for a wide range of temperature. • This region is called “Extrinsic region”. Donors and Acceptors • Until the temperature is increased even higher and it reaches a point where electrons are excited from valence band. • This makes the intrinsic carrier concentration becomes comparable to the donor concentration. • At this region, the semiconductors act like an intrinsic one. Donors and Acceptors • If the semiconductors are heavily doped for both n- or p-type, EF will be higher than EC or below EV, respectively. • The semiconductor is referred to as degenerate semiconductor. • This also results in the reduction of the bandgap. Donors and Acceptors • This also results in the reduction of the bandgap. The bandgap reduction Eg for Si at room temperature is expressed by N Eg 22 meV 18 10 where the doping is in the unit of cm-3. Donors and Acceptors • Ex. Si is doped with 1016 arsenic atoms/cm3. Find the carrier concentration and the Fermi level at room temperature (300K). Donors and Acceptors Soln At room temperature, complete ionization of impurity atoms is highly possible, then we have n = ND = 1016 cm-3. 9.65 10 n p ND 1016 2 i 9 2 9.3 103 cm-3 DonorsE and Acceptors N E kT ln N D Soln 2.86 1019 0.0259 ln 10 10 The Fermi level measured from the bottom of the conductionband 0.205iseV C C F Donors and Acceptors Soln The Fermi level measured from the intrinsic Fermi level is n EF Ei kT ln ni ND kT ln n i 1016 0.0259 ln 9 9.65 10 0.358 eV Direct Recombination • When a bond between neighboring atoms is broken, an electron-hole pair is generated. • The valence electron moves upward to the conduction band due to getting thermal energy. • This results in a hole being left in the valence band. Direct Recombination • This process is called carrier generation with the generation rate Gth (number of electron-hole pair generation per unit volume per time). • When an electron moves downward from the conduction band to the valence band to recombine with the hole, this reverse process is called recombination. • The recombination rate represents by Rth. Direct Recombination • Under thermal equilibrium, the generation rate Gth equals to the recombination rate Rth to preserve the condition of pn ni2 • The direct recombination rate R can be expressed as R np where is the proportionality constant. Direct Recombination • Therefore, for an n-type semiconductor, we have Gth Rth nn0 pn0 (3) where nn0 and pn0 represent electron and hole densities at thermal equilibrium. Direct Recombination • If the light is applied on the semiconductor, it produces electron-hole pairs at a rate GL, the carrier concentrations are above their equilibrium values. • The generation and recombination rates become G GL Gth R nn pn nn0 n pn0 p where n and p are the excess carrie concentrations Direct Recombination n nn nn 0 p pn pn 0 • n = p to maintain the overall charge neutrality. • The net rate of change of hole concentration is expressed as dpn G R GL Gth R dt (7) Direct Recombination • In steady-state, dpn/dt = 0. From (7) we have GL R Gth U (8) where U is the net recombination rate. Substituting (3) and (5) into (8), this yields U nn0 pn0 p p (9) Direct Recombination For low-level injection p, pn0 << nn0, (9) becomes U nn 0 p pn pn 0 pn pn 0 1/ nn 0 p (10) where p is called excess minority carrier lifetime . pn pn0 pGL Direct Recombination We may write pn in the function of t as pn (t ) pn 0 p GL exp t / p Direct Recombination • Ex. A Si sample with nn0 = 1014 cm-3 is illuminated with light and 1013 electron-hole pairs/cm3 are created every microsecond. If n = p = 2 s, find the change in the minority carrier concentration. Direct Recombination Soln Before illumination: pn0 ni2 / nn0 (9.65109 )2 /1014 9.31105 cm-3 After illumination: 1013 6 pn pn 0 pGL 9.3110 2 10 6 2 1013 cm-3 10 5 Continuity Equation • We shall now consider the overall effect when drift, diffusion, and recombination occur at the same time in a semiconductor material. • Consider the infinitesimal slice with a thickness dx located at x shown in the figure. Continuity Equation • The number of electrons in the slice may increase because of the net current flow and the net carrier generation in the slice. • Therefore, the overall rate of electron increase is the sum of four components: the number of electrons flowing into the slice at x, the number of electrons flowing out at x+dx, the rate of generated electrons, and the rate of recombination. Continuity Equation • This can be expressed as n J e ( x) A J e ( x dx ) A Adx Gn Rn Adx t e e • where A is the cross-section area and Adx is the volume of the slice. Continuity Equation • By expanding the expression for the current at x + dx in Taylor series yields J e J e ( x dx) J e ( x) dx ... x • Thus, we have the basic continuity equation for electrons and holes as n 1 J e Gn Rn t e x p 1 J h G p Rp t e x (14) Continuity Equation • We can substitute the total current density for holes and electrons and (10) into (14). dn dx dn J h e h nE eD p dx J e ee nE eDn Continuity Equation • For low-injection condition, we will have the continuity equation for minority carriers as n p n p 2n p n p n po E n p e e E Dn 2 Gn t x x x n pn E pn pn pn pno pn h h E Dp Gp 2 t x x x p 2 (15) The Haynes-Shockley Experiment • This experiment can be used to measure the carrier mobility μ. • The voltage source establishes an electric field in the ntype semiconductor bar. Excess carriers are produced and effectively injected into the semiconductor bar at contact (1). • Then contact (2) will collect a fraction of the excess carriers drifting through the semiconductor bar. The Haynes-Shockley Experiment • After the pulse, the transport equation given by equation (15) can be rewritten as pn pn 2 pn pn pno h E Dp 2 t x x p • If there is no applied electric field along the bar, the solution is given by x2 N t pn ( x, t ) exp pno (16) 4 D t 4 D pt p p The Haynes-Shockley Experiment • N is the number of electrons or holes generated per unit area. If an electric field is applied along the sample, an equation (16) will becomes 2 x p Et N t pn ( x, t ) exp pno 4 Dpt p 4 Dpt The Haynes-Shockley Experiment Ex. In Haynes-Shockley experiment on n-type Ge semiconductor, given the bar is 1 cm long, L = 0.95 cm, V1 = 2 V, and time for pulse arrival = 0.25 ns. Find mobility μ. The Haynes-Shockley Experiment Soln L 0.95 cm vD 3800 cm/s t 0.25 ns V1 E 2 V/cm 1 cm vD 3800 cm/s 1900 cm 2 /V.s E 2 V/cm The Haynes-Shockley Experiment Ex. In a Haynes-Shockley experiment, the maximum amplitudes of the minority carriers at t1 = 100 μs and t2 = 200 μs differ by a factor of 5. Calculate the minority carrier lifetime. Soln 2 x p Et N t pn ( x, t ) exp pno 4 D pt p 4 D pt 2 x p Et N t p pn pno exp 4 D pt p 4 D pt The maximum amplitude t N p exp 4 D pt p Therefore, t2 exp t1 / p 200 100 p (t1 ) 200 exp 5 p (t2 ) 100 p t1 exp t2 / p p 79 s Thermionic emission process • It is the phenomenon that carriers having high energy thermionically emitted into the vacuum. • In other words, electrons escapes from the hot or high temperature surface of the material. • This is called “thermionic emission process”. Thermionic emission process Vn • Electron affinity qχ is the energy difference between the conduction band edge and the vacuum. • Work function q is the energy between the Fermi level and the vacuum level in the semiconductor. Thermionic emission process • It is clearly seen that an electron can thermionically escape from the semiconductor surface into the vacuum if its energy is above qχ. • The electron density with energies above qχ can be found by q Vn nth n( E )dE NC exp kT q where Vn is the difference between the bottom of the conduction band and the Fermi level. Thermionic emission process • If escaping electrons with velocity normal to the surface and having energy greater than EF + q, the thermionic current density is equal to J qvN ( E ) F ( E )dE 4 (2m)3/ 2 ( E EF ) / kT qv dE e 3 h Thermionic emission process h2k 2 • Then we use p = mv and E 2 , and after 8 m integration, it yields q J A T exp kT 2 4 qmk A* Richardson constant 3 h 1.2 106 A/(m 2 .K 2 ) 120 A/(cm 2 .K 2 ) * 2 Thermionic emission process Ex. Calculate the thermionically emitted electron density nth at room temperature for an n-type silicon sample with an electron affinity qχ = 4.05 eV and qVn = 0.2 eV. If we reduce the effective qχ to 0.6 eV, what is nth? Thermionic emission process Soln 4.05 0.2 52 nth (4.05 eV) 2.86 10 exp 10 0 0.0259 0.6 0.2 19 6 -3 nth (0.6 eV) 2.86 10 exp 1 10 cm 0.0259 19 As we clearly see that at the room temperature, there is no thermionic emission of electrons into the vacuum. This thermionic emission process is important for metal-semiconductor contacts. Tunneling Process • The figure shows the energy band when two semiconductor samples are brought close to each other. • The distance between them (d) is sufficient small, so that the electrons in the left-side semiconductor may transport across the barrier and move to the right-side semiconductor even the electron energy is much less than the barrier height. • This process is called “quantum tunneling process”. Tunneling Process • The transmission coefficient can be expressed as 2me* (qV0 E ) T 2 exp 2d 2 A C 2 • This process is used in tunnel diodes by having a small tunneling distance d, a low potential barrier qV0, and a small effective mass.