Question

For the TSMC 0.18 μm process NMOS transistor, what are the following? TOX = 4.1 nm and the dielectric is SiO₂.

a. C_ox', fF/μm²

b. q∅_s1, use the substrate doping Boron, 2.35e17 cm^-3 (Hint: #10 of 09 Slides)

c. . q∅_M, if the gate is aluminium (you will need to look this up)

d. q∅_M, if the gate is poly, degeneratively doped with phosphorus, 5.02e20 cm^-3

Answer 1

C_{​​​​​​Ox} = (e_{​​​​​​r} * e_{​​​​​​o}​​​​​)/tox. ............(1)

Where e_​​r = 3.9 for SiO_​​​2

e_{​​​​​o} = 8.854 * 10^-3

Substitute above values in the equation (1)

C_{​​​​​​ox} = (8.854 * 10^-3 * 3.9)/(4.1*10^-9)

= 8.422 *10^-3

  = 0.008422 F/micro m^{​​​​​​2}