The transistor in the circuit shown below has parametes IDSS

The transistor in the circuit shown below has parametes I_DSS = 8 mA and V_P = -4 V. Determine the Q-point

Solution

as the gate current is negligible we can assume that resistor R1 and resistor R2 form a voltage divider circuit with the supply voltage.

hence voltage V2 across R2 resistor,

V₂=V_DD R₂/(R₁+R₂)= 20X60/(140+60) = 20X60/200= 6 V

V₂= V_GS+I_DR_S

V_GS= V₂-I_DR_S

To get VGS we need to know the ID

we know that

ID= IDSS(1-VGS/VP)²

or ID= IDSS(1- (V2-IDRS) / VP)²

ID= 8(1- (6-ID x 2)/-4)²

ID= 8((4+6-IDx2)/4)²

ID= 8((10-IDx2)/4)²

ID= (10-ID x2 )²/2

2ID= (10-2ID)²

2I_D= 100- 40ID + 4ID²

4ID²-42I_D +100 = 0

2ID^2-21I_D+50=0

OR 2ID²-25ID+4ID+50=0

ID= 3.64 mA or 6.8mA

value of ID 6.8mA will not be taken as it is greater than Vp

so, ID= 3.64mA

now VGS=v2-I_D R_s

_{= 6-3.64x2}

_{= -1.28v}

_{applying KVL at the output ,}

V_DS= V_DD- ID(R_S+R_D)

=20- 3.64(2.7+2)

20-3.64 X (4.7)

=2.892V

HENCE ID= 3.64mA V_DS= 2.892 V

this is the Q-point