E-MOSFET IDQ and VDSQ with graphical Q point

Given: VDD = 40 V, R1 = 22 MΩ, R2 = 18 MΩ, RD = 3 kΩ, RS = 0.82 kΩ, VGS(Th) = 5 V, ID(on) = 3 mA at VGS(on) = 10 V.

Step 1: Find k

ID = k (VGS − VGS(Th))²

k = ID(on) / (VGS(on) − VGS(Th))² = 3 mA / (10 − 5)² = 0.12 mA/V²

Step 2: Gate voltage (gate current is zero)

VG = VDD · R2/(R1 + R2) = 40 × 18/40 = 18 V

Step 3: Bias equation

VGS = VG − ID RS = 18 − 0.82 ID (ID in mA, RS in kΩ)

ID = 0.12 (18 − 0.82 ID − 5)² = 0.12 (13 − 0.82 ID)²

Expanding: 0.0807 ID² − 3.5584 ID + 20.28 = 0, that is

ID² − 44.10 ID + 251.3 = 0

ID = 6.72 mA or 37.4 mA

For ID = 37.4 mA, VGS = 18 − 30.7 = −12.6 V, which is below the threshold, so this root is rejected.

IDQ = 6.72 mA, VGSQ = 18 − 0.82(6.72) = 12.49 V

Step 4: VDS

VDSQ = VDD − ID (RD + RS) = 40 − 6.72 × 3.82 = 14.3 V

Check: VGS − VGS(Th) = 7.49 V < VDS = 14.3 V, so the MOSFET is in saturation and the square law is valid.

Graphical Q point

(a) Transfer characteristic. Plot ID = 0.12 (VGS − 5)²:

Draw the bias line ID = (18 − VGS)/0.82 through (VGS = 18 V, ID = 0) and (VGS = 0, ID = 21.95 mA). It cuts the curve at Q = (12.49 V, 6.72 mA).

(b) Output characteristic. DC load line: VDS = 40 − 3.82 ID, through (VDS = 40 V, ID = 0) and (VDS = 0, ID = 10.47 mA). Q lies on the VGS = 12.49 V curve at VDS = 14.3 V, ID = 6.72 mA.

 ID(mA)
 10.5 *
      |  *
  6.7 |------Q  (VGS=12.5V)
      |       \ *
      |         \  *
      +-----------14.3---*-----> VDS
                           40 V

Answer: IDQ = 6.72 mA, VDSQ = 14.3 V, VGSQ = 12.5 V (saturation region).