Question

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?

Answer 1

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

Step-by-step 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