Mark Lundstrom 2/10/2013. SOLUTIONS: ECE 606 Homework Week 5 Mark Lundstrom Purdue University (corrected 3/26/13)

Transcription

1 SOLUIONS: ECE 606 Homework Week 5 Mark Lundstrom Purdue University corrected 6/13) Some of the problems below are taken/adapted from Chapter 4 in Advanced Semiconductor Fundamentals, nd. Ed. By R.F. Pierret. 1) he conduction band minima in GaP occurs at the boundary of the first Brillouin zone along [100] directions. he constant energy surfaces are ellipsoidal with m * 1.1m 0 and m t * 0.m 0. Determine the density- of- states effective mass for electrons in GaP. he situation here is much like silicon except that the minima occur at the BZ boundary. he result is that we can only include one- half of each ellipsoid inside the BZ. he 6 in enq. 4.5a) of ASF p. 95) is replaced by 3. We conclude: m n *) 3 /3 m * * DOS m t ) 1/3 m n *) DOS 0.79m 0 ) Assuming that the Si valence bands are spherical with m * hh 0.537m 0 and m * lh 0.153m 0, determine the fraction of holes that are in the heavy hole band. You should assume non- degenerate conditions at 300 K. Also assume that the effective masses quoted above, which were measured by cyclotron resonance at 4K, can be used at room temperature. Begin with: p N V F 1/ ) m - 3 N V 1 4 m * π m - 3 where: E V ) his expression assumes a single, parabolic energy band i.e. valley degeneracy is 1). For the heavy and light hole bands: p hh N V hh F 1/ p lh N V lh F 1/ ECE

2 he fraction of holes in the heavy hole band becomes: p hh p p hh p hh p lh N V hh N V hh N V lh Inserting numbers, we find: * m hh ) * m hh ) * m lh p hh 0.153) p p hh p Most of the holes are in the heavy hole band. his should be expected, because the heavy hole band has a much higher density of states. N hh * V m hh ) >> N lh * V m lh ) Note that we did NO need to assume MB non- degenerate) statistics. 3) In a Si MOSFE, the source and drain regions are heavily doped. Assuming that the dopants are fully ionized and that 10 0 cm - 3, compute the location of the Fermi level. Compare the answer you get assuming Maxwell- Boltzmann statistics with the answer from Fermi- Dirac statistics. You may assume 300K and that the effective density of states is N C 1 m * D 4 π cm -3 You will need to compute an inverse Fermi- Dirac integral. A FD integral tool is available on nanohub.org at: nanohub.org/resources/fdical You can also find an iphone app that does Fermi- Dirac integrals. n N C F 1/ F 1/ ECE- 606

3 F 1/ F 1/ Using the nanohub tool or the iphone app, we find: 1 E F 1/ F So the Fermi level is.184 kb above EC for FD statistics. Now repeat for Maxwell Boltzmann statistics: F 1/ 3.10 e ln 3.10) 1.13 E C L So the Fermi level is 1.13 kb above EC for MB statistics. In general, we find that the Fermi level is higher when we use FD statistics. his occurs because as states are filled, we need to go higher in the conduction band to find empty state. he implicit assumption for MB statistics is that all states are nearly empty. 4) In 3D, the distribution of electrons in the conduction band, n E), is peaked at an energy, Ê that is very near the bottom of the conduction band,. Assume Maxwell- Boltzmann carrier statistics and derive an expression for Ê for the following cases: 4a) 3D semiconductor with parabolic energy bands 4b) D semiconductor with parabolic energy bands 4a) D E) f E) D E n E e E ) dn E) de dd E de k B D E e E ) e E ECE

4 dn Ê) de 0 k dd E ) D B de Ê ) Ê in 3D, D E) K E K Ê 1/ K is a constant), so 1/ K Ê ) 1/ Ê 4b) Begin with: dd E ) D de Ê ) Ê in D, D D E) K, so 0 K nonsense. here is no peak to n E). 5) Sometimes we need to know the average kinetic energy for electrons in the conduction band or for holes in the valence band). Answer the follow questions: 5a) Derive a general expressions for the average kinetic energy per electron in the conduction band. u E E n D E) f E D E) f E)dE de 5b) Assume parabolic energy bands and evaluate the general expression for a 1D semiconductor. Simplify your answer for Maxwell- Boltzmann carrier statistics. ECE

5 u E n L E ) D 1D E) f E)dE D 1D E) f E)dE D 1D E) g V π m D * E You should be able to show: n L D 1D E) f E)dE N 1D F 1/ *), where: N 1D m D * π. Now, lets work on the numerator: E E ) 1 π m D * E 1 1 e E ) de 1 π 1/ de m E E * C D 1 e E ) η E ) de η ) E 1 π 1/ dη m k * B η D 1 e η E m * k D B π 0 η 1/ dη 1 e η η 1/ dη π 1 e η F η 1/ F 0 E m * k D B π F 1/ See Notes on FD Integrals ) N 1D F 1/ **) ECE

6 Combine *) and **) to find: u 1D E k B n L F 1/ F 1/ ) For a nondegenerate semiconductor, all FD integrals reduce to exponentials, so we find: u 1D k B 5c) Assume circular, parabolic energy bands and evaluate the general expression for a D semiconductor. Simplify your answer for Maxwell- Boltzmann carrier statistics. We begin the same way u D E n S E ) D D E) f E)dE D D E) f E)dE Use the D DOS: D D E) m * D π and eventually find: u D E n S F η 1 F ) F 0 For a nondegenerate semiconductor, all FD integrals reduce to exponentials and we find: u D ECE

7 5d) Assume spherical, parabolic energy bands and evaluate the general expression for a 3D semiconductor. Simplify your answer for Maxwell- Boltzmann carrier statistics. We begin the same way u 3D E E n Use the 3D DOS: D 3D E) m D D 3D E) f E D 3D E π 3 E) f E)dE de and eventually find: u 3D E n 3k B F F 1/ ) For a nondegenerate semiconductor, all FD integrals reduce to exponentials and we find: u 3D 3 6) Assume Si at 300 K, doped with arsenic at cm - 3. Make reasonable assumptions and answer the following questions. 6a) Compute the density of electrons in the conduction band. 6b) Compute the location of the Fermi level. 6c) Compute the fraction of the dopants that are ionized. 6d) Compute the density of holes in the valence band. ECE

8 For problems 6) - 10), let s assume at 300 K from Pierret, ASF, nd Ed., p. 113) N C 1 m * n 4 π N V 1 m * p 4 π cm cm -3 It should be understood that the effective masses in these expression are the density- of- states effective masses. For problems 6) 10), the starting point is space charge neutrality: p n N A 0 N V F 1/ E V N C F 1/ 1 g D e E D ) N A 1 g A e E A ) 0 For this problem, we can assume that N A 0. Since the doping is moderate, we assume non- degenerate carrier statistics. Because Arsenic is a shallow dopant and we expect the Fermi level to be well below ED, we can assume that the dopants are fully ionized. Because the semiconductor is extrinsic, n >> p. We conclude that n to a very good approximation. All these assumptions can be checked as we proceed. 6a) n cm -3 6b) Determine EF from the electron density: n N C e ) k B ln n N C ) ln ) So the assumptions that the semiconductor is non- degenerate is good. 6c) Begin with: ECE

9 1 1 g D e E D ) 1 1 g D e E D According to Fig. 4.14, p. 111 of Pierret, ASF: E D ev.08 and we also know that for donors, g D. 1 1 e ) 1 1 e So our assumptions that the dopants are fully ionized is good. 6d) Use the Law of Mass action: np n i p n i n p 10 4 cm So our assumption that n >> p is good. 7) Assume Si at 640 K, doped with arsenic at cm - 3. Make reasonable assumptions and answer the following questions. 7a) Compute the density of electrons in the conduction band. 7b) Compute the location of the Fermi level. 7c) Compute the fraction of the dopants that are ionized. 7d) Compute the density of holes in the valence band. Dopants should be fully ionized and semiconductor should be nondegenerate, but n i may be large, which means that p may not be negligible. From Fig. 4.17, p. 117 of Pierret, ASF we find: n i 640 K) cm -3 he space charge neutrality condition becomes: ECE

10 p n N D N A 0 p n n i n n 0 which can be solved for n 7a) Solve for n: n n i n i 1/ 1/ ) 1/ n n cm -3 7b) Determine EF from the electron density: n N C e ) k B ln n N C ) but NC is temperature dependent. N C 1 m * n 4 π N C 300K 300 ln ) cm but realize that kb is bigger now). 7c) Proceeding as in 6c), we find: E D ev 0.98 and 1 1 e ) 1 Dopants are fully ionized as assumed. 7d) Use the Law of Mass Action: np n i 1 e ECE

11 p n 16 i n p cm -3 So in this case, the hole concentration is not negligible. he semiconductor is nearly intrinsic. 8) Assume Si at 77 K, doped with arsenic at cm - 3. Make reasonable assumptions and answer the following questions. 8a) Compute the density of electrons in the conduction band. 8b) Compute the location of the Fermi level. 8c) Compute the fraction of the dopants that are ionized. 8d) Compute the density of holes in the valence band. In this case, we worry that there may not be enough thermal energy to ionize the donors. he intrinsic concentration should be very small, so holes can be ignored. he space charge neutrality condition becomes: p n N A 0 n N C F 1/ 1 g D e E D ) 0 Assume non- degenerate carrier statistics check later): N C exp 1 g D e E D ) 0 his equation can be solved. he answer is eqn. 4.88) in Pierret, ASF where ln N ζ N C 1 4 N ζ ) 1/ 1 { } N ζ N C g D )e E D ) Now lets put numbers in: ECE

12 N C N C 300K ) cm ev N ζ N C g D )e E D ) k B )e 54) ) 1/ 1 { } ln ln { 1/ 1 } ) 65.8 Note that E D 8.13 So the Fermi level is above the donor level, but far enough below EC that we can use non- degenerate carrier statistics. 8a) Since we know the Fermi level, we can get the electron density: n N C exp exp b) We already solved this problem n cm -3 8c) Since n , we deduce 0. We could also get this directly from 1 N D 1 g D e E D ) 8d) he number of holes is essentially zero. We could get this from the law of mass action, but then we need to figure out ni at 77 K. Another way is: ECE

13 p N V exp E V N V exp E V N V exp E G exp N V 1 m * p 4 π N V 300K cm -3 Also need the bandgap of Si at 77 K. From pp of Pierret, ASF, we have E G ) E G 0 α β For Si: E G ) ev so p exp exp ) p cm -3 his is essentially zero hardly worth the effort of computing it. Since we have n and p, we can deduce n i cm -3 9) Assume Si at 300 K, doped with arsenic at cm - 3. Make reasonable assumptions and answer the following questions. 9a) Compute the density of electrons in the conduction band. 9b) Compute the location of the Fermi level. 9c) Compute the fraction of the dopants that are ionized. 9d) Compute the density of holes in the valence band. ECE

14 his requires a little thought the temperature is high enough so the donors may be fully ionized, but the number of dopants is high, so the Fermi level might be too close to the donor level to assume complete ionization. It might also be too close to the conduction band to assume non- degenerate carrier statistics. Let assume full ionization and nondegenerate carrier statistics and see where it leads. 9a) Assume complete ionization: n cm -3 9b) Determine EF from the electron density: n N C e ) k B ln n N C ) ln ) So the assumptions that the semiconductor is non- degenerate is OK. 9c) As in problem 6c): 1 1 e ) 1 1 e So our assumptions that the dopants are fully ionized is not justified. We need to start over and solve the problem using the procedure of problem 8). he rest of this problem is left as an exercise for the interested student! 10) Assume Si at 300 K, doped with arsenic at 10 0 cm - 3. Make reasonable assumptions and answer the following questions. In this case, you should assume that heavy doping effects cause the dopants to be fully ionized and the bandgap to shrink by 0.1 ev.) 10a) Compute the density of electrons in the conduction band. 10b) Compute the location of the Fermi level. 10c) Compute the fraction of the dopants that are ionized. 10d) Compute the density of holes in the valence band. ECE

15 We expect the Fermi level to be inside the conduction band. We can assume full ionization and ignore holes, but we cannot use nondegenerate statistics. n to a very good approximation. 10a) n 10 0 cm -3 10b) Determine EF from the electron density: n N C F 1/ N C cm -3 1 n 10 0 F 1/ N C F 1/ F 1/ 3.10).184 from Fermi Dirac Integral Calculator on nanohub.org So the semiconductor is, indeed degenerate with EF inside the conduction band. 10c) By assumption, we are above the metal- insulator transition and 1 10d) Use the Law of Mass Action: np n i But bandgap narrowing increases ni. n i 10 0 exp ΔE G / ) 10 0 exp 100 / 6) n i p n i n p cm So our assumption that n >> p is good. ECE

16 11) For the energy band sketched below, provide sketches of the following: 11a) the carrier densities, nx), and px) vs. position. 11b) the electrostatic potential, ψ x), vs. position. ECE

17 11c) the electric field E vs. position. 11d) the space charge density, ρ x) vs. position ECE