Physics of Semiconductors. Exercises. The Evaluation of the Fermi Level in Semiconductors. B.I.Lembrikov Department of Communication Engineering Holon Academic Institute of Technology I. Problem 8. The Fermi energy level E F in the n-type semiconductor is given by E F E F i + k BT ln n p () where E F i is the Fermi energy level in the intrinsic semiconductor, T is a temperature in Kelvin, k B is the Boltzmann constant, n; p are the electron and hole concentrations. In our case (x) > (0) n i! n (x) ; p n i n n i (x) () E F E F i + k BT Substitute into (3) ln n p E F i + k BT ln N D (x) n i (x) (0) x ; E F i L where is the band gap energy. We obtain. Let x 0 E F + k BT ln (0) + x n i L E F i + ln (x) n i (3) E F (0) + k BT ln (0) : n i + 0:06 ln 0 4 0: 794 73eV : 0 0 (6). Let x L E F (L) E F (0) + 0: 794 73 + 0:06 8 : 00 7eV (7) (4) (5)
II. Problem 8.3 The impurity of Fe (iron) is not e ective as a donor in a semiconductor since its donor levels are deep (0:55eV and 0:7eV from the minimum energy of the conduction band ), and it is di cult for electrons to pass from these deep donor levels into the conduction band. The large dimension of the Fe atom results in an additional electron scattering, trapping of electrons and deteriorate the quality of a semiconductor crystal. For these reasons, Fe is not used as a donor impurity. III. Problem 8.4 The equation (8.5) gives the probability f (E D ) of the donor level E D f (E D ) + 0:5 [(E D E F ) ] (8) The probability P e of the electron to be in the conduction band is f (E D ) + 0:5 [(E D E F ) ] 0:5 [(E D E F ) ] + 0:5 [(E D E F ) ] The concentration n of the electrons in the conduction band is (the result of ex. 8.4): n P e 0:5 [(E D E F ) ] + 0:5 [(E D E F ) ] + [ (E D E F ) ] N h D + On the other hand n EF i h i (9) EF EF! n (0) where is minimum energy of the conduction band and E F is the Fermi energy level. Substitute (0) into (9) n n + h n + n N c ED N c 4 ED i n! n + n s Nc 4 ED + 0 0 ()
Only the positive root is appropriate. Then 4s Nc n 4 ED + 8 3 5 () E D 0:044eV ; :8 0 9 cm 3 ; :8 0 6 cm 3 n 0:5 "s 0:044 0 6 + 8 0 3 0:044 0 3 # hp y 0:5 ( 0:044x) 06 + 8 0 3 (0:044x) 0 3i ; x (40 50) p06 0:5 ( 0:044x) + 8 0 3 (0:044x) 0 3 p06 0:5 ( 0:044 40) + 8 0 3 (0:044 40) 0 3 : 988 64 p06 0:5 ( 0:044 80) + 8 0 3 (0:044 80) 0 3 : 940 6 p06 0:5 ( 0:044 00) + 8 0 3 (0:044 00) 0 3 : 875 p06 0:5 ( 0:044 40) + 8 0 3 (0:044 40) 0 3 : 67 35 p06 0:5 ( 0:044 00) + 8 0 3 (0:044 00) 0 3 : 09 54 p06 0:5 ( 0:044 3) + 8 0 3 (0:044 3) 0 3 : 9 5 IV. Problem 8.5 In a dielectric material the Fermi level is situated in the middle of the band gap. In metals the Fermi level is situated inside the conduction band populated with electrons. In a degenerate p type semiconductor the Fermi level is situated in the valence band. In an intrinsic semiconductor the Fermi level is situated in the middle of the band gap. V. Problem 8.6 N 8 0 4 cm 3 E F E F i + ln N 0:56 + 0:06 ln 8 04 0: 848 79eV n i : 0 0 3
VI. Problem 8.7 n i ( N v ) h 3 (k BT ) 3 (m em h) 34 For Si m e 0:6m 0 ; m h 0:49m 0; :ev ;For InP m e 0:077m 0 ; m h 0:64m 0 ; :35eV! n iinp :46 0 6 cm 3 : For Si we have n iinp (6:660755 0 34 ) 3 :3806568 0 3 300 3 0:077 0:64 9: 0 6 34 :35 :6 0 9 :3806568 0 3 300 n iinp : 60 6 0 4 4: 74 0 n i Si : 4 7 0 3 m 3 (6:660755 0 34 ) 3 :3806568 0 3 300 3 0:6 0:49 9: 0 6 34 : :6 0 9 :3806568 0 3 300 n i Si 5: 34 9 0 4 4: 09 5 0 0 : 5 9 0 5 m 3 : 5 90 5 : 4 70 3 73: 4 n i Si : 5 9 05 73: 4 n iinp : 4 7 03 VII. Problem 8.8 (a).in a metal, at T 0 K all electrons are placed in the conduction band at energies E < E F : The Fermi level in metals is situated in the conduction band. We choose the minimum of the conduction band 0: Then the number of electrons n is equal to the number of states and has the form n ZE F 0 4 h 3 (m ) 3 p EdE 4 h 3 (m ) 3 3 E3 F (3) 4
From (3) we obtain E F h m 3 3n 8 For n 0 3 cm 3 0 9 m 3 and m m 0 9:093897 0 3 kg we have E F 6:660755 0 68 9:093897 0 3 3 0 9 3 : 58 7 0 8 J 8 : 58 7 0 8 :6 0 9 7: 866 9eV (b). In the region of the Maxwell-Boltzmann statistics E E F and the ratio of electron concentrations n ; with di erent energies E ; is n E E n where in our case the mimimal energy E E F, E is the work function of the metal and the height of the potential barrier is E E E F 4eV and For T 300 K n n n n 4 4 : 53 7 0 67 0:06 For T 000 K n n 4 300 9: 50 0 0:06 000 For n 0 3 cm 3 at T 300 K electrons cannot pass over the potential barrier. At T 000 K the number of electrons n 9: 50 0 n 9: 50 0 cm 3 can overcome the potential barrier. VIII. Problem 8.9 At T 0 K the Fermi level for holes is situated at the bottom of the band gap.. The concentration n i (T ) is given by n i (T ) AT 3 Eg 5
3 ni (T ) T n i (T ) T + E g T T 3 Eg At T 300 K in Ge n i (T 300 K) :4 0 3 cm 3 : Hence n i (T 00 K) n i (T 300 K) :4 0 3 n i (T 400 K) :40 3 3 00 0:67 300 0:06 400 300 3 00 Eg 300 3. At T 300 K in Ge p 0 5 cm 3 T T ( 3) 9: 73cm 3 0:67 0:06 E F p (300 K) E F i + ln n i p E F i + ln n i T T 300 9: 58 70 4 cm 3 400 :4 03 E F i +0:06 ln 0 5 E F i 9: 697 0 ev 9: 697 0 0:67 9: 697 0 0: 38 03eV At T 400 K the concnetrations of acceptors and electrons are of the same order of magnitude. Hence p p n i 0! p + n + p; pn n i! n i p + p 0 s NA : 55 3 0 5 cm 3 q + n i 504 + (5 0 4 ) + (9: 58 7 0 4 ) E F p (400 K) E F i + 0:06 400 9: 58 7 04 ln 300 : 55 3 0 5 E F i : 79 4 0 0:67 : 79 4 0 0: 37 09eV E F p (00 K) E F i + 0:06 00 9: 73 ln 300 5 0 4 E F i 0: 63 93 0:67 0: 63 93 0:0 70 7eV 6
IX. Problem 8.0 (a)we use the results of the problem 8.4, equation (): n s 4 4 Nc + 8 N c 3 5 and for E D 0:05eV we get Ec E D 0:05 6: 84 0:06 0 3! n 0:5 p 06 + 8 0 6: 84 3 6: 84 0 3 0: 986 68 (b)for E D 0:eV we get Ec E D 0: 46: 83 0:06 and n 0:5 p 06 + 8 0 46: 83 3 46: 83 0 3 0: 90 64 X. Problem 8. We use the equation of the intrinsic temperature T max when the intrinsic concentration of holes p i n i is equal to the concentration of acceptors p i n i (N v ) max and max ln and we get equation (8.7) (N v ) ln (N v ) On the other hand, T max k B ln (NvNc) 0:97 3:7 0 4 (T 300) ev T max 0:97 3:7 0 4 (T max 300) k B ln (NvNc) 7
" # T max k B ln (N v ) + 3:7 0 4 0:97 + 3:7 0 4 300 (a) 0 4 cm 3 0:97 + 3:7 0 4 300 T max i hk B ln (NvNc) + 3:7 0 4 k B ln (N v ) 8:67385 0 5 ln :04 09 6 0 8 0 4 (b) 0 6 cm 3 : 943 6 0 3 ev K T max 0:97 + 3:7 0 4 300 : 943 6 0 3 + 3:7 0 4 467: 4 K k B ln (N v ) 8:67385 0 5 ln :04 09 6 0 8 0 6 T max 0:97 + 3:7 0 4 300 : 49 9 0 3 + 3:7 0 4 7: 3 K 8