Solution 1 :

VDS = VGS - VT = 3.5 - 1 = 2.5 V

The drain - source voltage at the edge of saturation :

(i) using the linear expression is :

ID(LIN) = K [ ( VGS - VT ) VDS - ( VDS^2 /2 ) ]

ID(LIN) = 30 [ ( 3.5 - 1 )( 2.5 ) - (2.5^2 / 2) ]

ID(LIN) = 93.75 uA

(ii) using the saturation drain current expression :

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

ID(SAT) = (30/2) [ (3.5 - 1 )^2 ]

ID(SAT) = 93.75 uA

 

(iii) At the edge of saturation, the drain currents calculated

using the linear and saturation expressions have the same value.