MOSFET Q point graphically

The circuit is a drain-feedback biased enhancement MOSFET: the 10 MΩ resistor connects the drain to the gate. Since IG = 0, there is no drop across it, so VGS = VDS.

Step 1: find k

Given ID(on) = 6 mA at VGS(on) = 8 V and VGS(Th) = 3 V:

ID = k (VGS - VT)²   →   k = ID(on)/(VGS(on) - VT)² = 6 mA/(8 - 3)² = 0.24 mA/V²

Step 2: bias equations

Drain loop (dc load line):  VDS = VDD - ID RD = 12 - 2 ID      (ID in mA)
Feedback:                    VGS = VDS
Device:                      ID = 0.24 (VGS - 3)²

Substituting VGS = 12 - 2 ID:

ID = 0.24 (12 - 2 ID - 3)² = 0.24 (9 - 2 ID)²
   = 0.24 (81 - 36 ID + 4 ID²)
0.96 ID² - 9.64 ID + 19.44 = 0
ID = [9.64 ± √(92.9 - 74.65)]/1.92 = [9.64 ± 4.27]/1.92
ID = 7.25 mA   or   2.79 mA

The root 7.25 mA gives VGS = 12 - 14.5 < 0 (impossible: MOSFET off), so it is rejected.

IDQ  = 2.79 mA
VGSQ = VDSQ = 12 - 2 × 2.79 = 6.41 V

Check saturation: VDS = 6.41 V > VGS - VT = 3.41 V. Yes.

Graphical solution

Plot the transfer curve ID = 0.24 (VGS - 3)² and the bias line ID = (12 - VGS)/2 (since VGS = VDS):

The line passes through (12 V, 0) and (0 V, 6 mA), e.g. at VGS = 6 V it is 3 mA and at 7 V it is 2.5 mA.

 ID(mA)
  6 |*                   .  
    | \               .      transfer curve
    |  \          .
  3 |   \      .
    |    \ .*Q (6.41 V, 2.79 mA)
    |   . \
    +--+---+--+-----+---- VGS
       3   6.4        12

The intersection of the curve and the load/bias line is the Q-point.

Answer: Q-point: IDQ ≈ 2.79 mA, VDSQ = VGSQ ≈ 6.41 V.