MOSFET Transistor in Saturation Design the circuit below so

MOSFET - Transistor in Saturation

Design the circuit below so that the transistor operates in saturation with
I_D = 0.5mA and V_D = 3V.

Given V_TP = -1V, K_p = 1mA/V², Assume ? = 0.

What is the largest value that R_D can have for which the transistor will remain in saturation?

Solution

Notice that VDS = VGS Therefore: VDS >> VGS– Vt is true, and the transistor is in the saturation region thus ID = ½ [n Cox (W/L)] ( VGS – Vt ) 2 solving for VGS – Vt ( VGS – Vt ) 2 = ( 2 ID ) / [ n Cox (W/L) ] Substituting values, we obtain: ( VGS – Vt ) 2 = 0.16 V2 ; VGS – Vt = 0.4 V ; VGS = 0.4 + Vt = 1 V VGS = 1 V, VDS = VD = 1 V RD = (VDD – VD ) / ID = (3 V – 1V) / 0.16 mA = 12.5 kNotice that: VDS = 0.05 V < VGS – Vt = 5V – 1V = 4 V Therefore that transistor is in the Triode Region ID = kn’ (W/L) [ ( VGS – Vt )VDS – ½ VDS 2 ] substituting values we obtain: ID = 0.2 mA RD = ( VDD – VD ) / ID = ( 5 V – 0.05V ) / 0.2 mA = 24.75 k The effective Drain to Source resistance is: rDS = VDS/ ID = 0.05V/0.2mA = 250 Because IG = 0, we can apply voltage divider to the gate: VG= RG2/(RG1+RG2) VDD = 5 V Because VG > Vt the transistor is on Assume saturation region, solve the problem, and then verify Notice that: VS = (ID )(RS ) = 6 ID (ID in mA) Then: VGS = VG – VS = 5 V – 6 ID ID = ( ½ kn’)(W/L)( VGS – Vt ) 2 = ( 1/2 mA/V2 )(5 V – 6 ID – 1V ) 2 18 ID 2 - 25 ID + 8 = 0 Solving for ID we obtain: ID = 0.89 mA and ID = 0.5 mA