Question

No: Date Exercise Calculate the VI, for an n-transistor at 300k for a with a si subtracts with NA= 1.80x1016 cm, a sioa Oxide with thickness 200 À ni = 1:45 * 100 cm process gate Lot @ 300k) -0.02 fev Esi= 11.7 Esioa: 3.9 E. - 8.858 10 14 cm - tox - 2004 = 0.2x10-5 cm. KT :0.026 ev kt = KT = 0.0268 -0.026v (9 e e | Uri : 2KT en INAL +(2 Esi to q NA (2kT ln (NA! Cox

how to calculate Vti? and what value for q?

Answer 1

Solution: Given, (for n-type transistor) *74)(- To find Car = 3.9 0.2x10-5 =1.95 x106_F *** cm =1.95 ME KT To find the term kt (1.38x10-2) (300) a 1.6x10-19 = 0.026

To find Va 0.026)In (1.45 x 1010 8x10-17 ( 1.8x1026) VA =2(0.026)In 1983.2016 1 2(11.7)(8.85x10-14) (1.6x10-19)(1.8x1016) 1.95x10 = 0.7296+(3.059)} 0.7296)] = 0.7296+1.4939 V = 2.224 V The charge q value is 1.6x10- C.