could someone solve for me 1436 please thank you step by ste

could someone solve for me 14.36 please, thank you! step by step please!

Solution

Given Kn^I = 500uA/V²

(W/L)_{n =} 1.5 ; V_DD = 1.3V; V_tn = 0.4V

The Peak current is i_DN = Kn^I (W/L)_n [(V_DD - V_tn ) V_DD/2 - (1/2)(V_DD/2)²]

= 500X10^-6X 1.5 X [(1.3-0.4)1.3/2 - 1/2(1.3/2)²]

= 750x10^-6[0.585-0.211] = .280mA