Solution 3:

To determine ;

(i) Cox :

Eox = ( Kox ) ( Eo ) = 3.9 x 8.854 x 10 ^ -14

= 3.45 x 10^ -13 F/cm

Cox = Eox / tox = 3.45 x 10^-13 / 0.07 x 10^-4

= 49.28 x 10^-9

(ii) K' :

K' = (un )( Cox ) = 580 x 49.28 x 10^-9

= 2.86 x 10^-5 F/v.s

(iii) K :

K = K' x (W/L) = 2.86 x 10^-5 x 2/1 = 5.72 x 10^-5 A/v^2

= 57.2 uA/v^2

(iv) IDS:

Since VDS > VGS - VT the transistor is in saturation mode :

IDS = (K/2) [ ( VGS -VT ) ^ 2 ]

= (57.2/2) [ ( 2.3 -1 )^ 2]

= 48.334 uA