An NMOS transistor with kn = 1mA/V2 and Vt = 1V is operated with VGS = 2.5V. At what value of VDS does the transistor enter the saturation region? What value of ID is obtained in saturation?

a) the point at which the transistor enters the saturation region is 1.5 v

b) the value of ID is obtained in saturation is 1.125 mA

Explanation:

Given the data in the question;

for an NMOS, the condition for the saturation is;

V_DS ≥ V_GS - V_t

V_GS is 2.5 v and V_t is 1 v

so we substitute

V_DS ≥ 2.5 - 1

V_DS = 1.5 v

so the point at which the transistor enters the saturation region is 1.5 v

The drain current I_d in the saturation region;

I_d = 1/2×μₙ×Cₐₓ×W/L×( V_GS - V_t)²

= 1/2Kₙ (  V_GS - V_t)²

our Kₙ is 1 mA/V², V_GS is 2.5v and V_t is 1 v

so we substitute

I_d = 1/2(1 mA/V²)( 2.5 - 1 )²

= 1/2(1 mA/V²)( 2.25)

= 1.125 mA

therefore, the value of ID is obtained in saturation is 1.125 mA

answer:

make you upset

explanation: