Find the MOSFET bias point

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Daniel San

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  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.


 
Last edited:

ID = 0.5K(VGS - VT)2(1+λVDS) ... (1)

Now,
VDS = VD - VS
= (VDD - IDRd) - (IDRs)
= VDD - ID(Rd + Rs)
= 10 - 31k×ID ... (2)

VGS = VG - VS
= VDD×R2/(R1 + R2) - IDRs
= 1.25 - 1k×ID ... (3)

Putting (2) and (3) in (1) you shall get,
ID = 250µ(0.25 - 1k×ID)(1.2 - 620×ID)
ID = 250µ(0.3 - 1355×ID + 620k×ID2)

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

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

Putting (2) and (3) in (1) you shall get,
ID = 250µ(0.25 - 1k×ID)(1.2 - 620×ID)
ID = 250µ(0.3 - 1355×ID + 620k×ID2)

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

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

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

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

OK, I've found the equation -

ID = 250µ(0.25 - 1k×ID)2(1.2 - 620×ID)
155k×ID3 - 377.5×ID2 + 1.1596875×ID - 18.75µ = 0
ID = 16.253 µA
The other two values of ID are imaginary

Putting value of ID in equation (2),
VDS = 9.496 V

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

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