Find the MOSFET bias point

[Daniel San]

1. The problem
   
   Given the following circuit find the bias point (VDS,IDS).
   
   
   
   
   
   
2. The attempt at a solution:
   
   The first thing I did was find the value of VG, the voltage from the gate of the mosfet to the -VDD source.
   
   VG = 2VDD * R2/(R1+R2)
   
   To find the current IDS I needed the voltage from the gate to the source, VGS, which is:
   
   VGS = VG - RsID
   
   Since we're operating in the saturation region I can replace ID by:
   VGS = VG - Rsk(VGS-VT)^2 (1+λVDS)
   
   where VDS is the voltage between the drain and the source.
   I also know that:
   VDS = 2VDD - (Rd+Rs)ID
   
   This is as far as I can get. I'm a little confused regarding what I should do next.
   
   If anyone could help I'd appreciate.
   
   Thanks.
   Daniel.
   
   
   

 

[Debdut]

I_D = 0.5K(V_GS - V_T)²(1+λV_DS) ... (1)

Now,
V_DS = V_D - V_S
= (V_DD - I_DR_d) - (I_DR_s)
= V_DD - I_D(R_d + R_s)
= 10 - 31k×I_D ... (2)

V_GS = V_G - V_S
= V_DD×R₂/(R₁ + R₂) - I_DR_s
= 1.25 - 1k×I_D ... (3)

Putting (2) and (3) in (1) you shall get,
I_D = 250µ(0.25 - 1k×I_D)(1.2 - 620×I_D)
I_D = 250µ(0.3 - 1355×I_D + 620k×I_D²)

The above eqⁿ gives 2 values of I_D = 8.58 mA and 56.4 µA

For I_D = 56.4 µA and V_DS = (10 - 31k×56.4µ) V = 8.25 V
For I_D = 8.58 mA, V_DS becomes negative
Hence, I_D = 56.4 µA and V_DS = 8.25 V

 

[Daniel San]

Putting (2) and (3) in (1) you shall get,
I_D = 250µ(0.25 - 1k×I_D)(1.2 - 620×I_D)
I_D = 250µ(0.3 - 1355×I_D + 620k×I_D²)

The problem is that there's a square missing in the expression, which complicates the expression.

I_D = 250µ(0.25 - 1k×I_D)²(1.2 - 620×I_D)

With that square we get a cubic expressionl, which I can't solve!

 

[Debdut]

I'm very sorry for the mistake.
Can you post what the cubic equation becomes?

 

[Debdut]

OK, I've found the equation -

I_D = 250µ(0.25 - 1k×I_D)²(1.2 - 620×I_D)
155k×I_D³ - 377.5×I_D² + 1.1596875×I_D - 18.75µ = 0
I_D = 16.253 µA
The other two values of I_D are imaginary

Putting value of I_D in equation (2),
V_DS = 9.496 V

Therefore, I_D = 16.253 µA, V_DS = 9.5 V
NOTE : Equation solved using Calculator CASIO fx - 991MS