Question

6. (15 points) Hall measurement was performed on an n-type semiconductor and a Hall coefficient R = -8.5 cm/C was obtained. (a) (pt) What is the carrier concentration of the semiconductor? Ans. (1) 7.35 x 1029m 5.35 x 1025m (11) 6.5 x 10 cm (iv) 8.35 x 1015 cm? (v) 2.45 x 1023m- (6) 3pt) Given that the resistivity of the semiconductor is 0.015. - cm at room temperature (300K), what is the electron mobility of the material? Ans. (1) 567 m/s 750 cm2/sm 55 cm/s (iv) 567 cm?/Vs 750 cm?/Vs (c) A p-type layer of the same semiconductor with a hole concentration of 3x10 cm is joined with this n-type semiconductor, forming a p-n junction at T-300K (Assuming E-1.4 eV, n = 3.6 x 102m) I (4 pt)Calculate the built-in potential of this p-njunction. Ans (1) 5.2 v 1.26 V (iii) 0.24 V (iv) 2.2 v 0.88 V cur II. (5 pt) What is the depletion width on the n side of the junction? (Assuming the permittivity e = 8.0 x 10-13 Ans. (1) 0.2 m 0.05 pm V (iii) 82 nm 8.2 nm 0.08 cm (iv)

please answer as soon as possible

Answer 1

-camiers Concentration of charge RA al age of single carrier -I Ru = Hall coefficient n. 9 9 = char na Concentration. for 9 = + 1.6x10'sc magnitude) 1 RH 9 (-8.5x154) (+126816"C) 23 for ha tyre semjonkurencing n = -1 را -ند This is the concentration of Corrier. b Resistivity = f and mobility = 4 They are related by - 4 = Rr. ſ 3 -Ans- - D 9 M= 8.5 cm le = 567 cm?/vs 0.015 2-cm This is the mobility: built in potential (vo) KT IN NANO h; 2 Where KT = Thermal vo voltage of temperature NA = acceptor density hole concentration & 3X122 / m² No = Bonar den density electron concetration. = 7.35x16 7.35x1373 nia Intrinsic concentration. - V @quilivalent -

V= (1-38 x10"") (300) 26xlov or 26mV 1.6X10ls V. - 26x1o? In / 3x102x 7.35x135 (3:6x10'?) V=26x10's In/1.7014х 10² This is the built in potential = 1.26 v Ans-(ii) 5) The depletion width is :- 2 EV q (Nn+No) No n A Xn 2x 8X16" X 3x1622 (1.6x16 "(7.35 +0.3)x 10 35 +0+3)x 17x 7.35x1333 Xn -8 0.82 X10 depletion width = An= 8.2 X10 8.2 nm Ans. (iv)