4 Resistor MOSFET P-MOS

[forbi]

Hi,

How do i go about finding the Q-Point of the above Circuit.
Vtp = -1v, lamda is 0.01V^-1
Kp = 0.5mA/V^2

btw the Veq should be 1V.

i do by Vdd-IsRs+Vgs-IgReq-Veq=0
and Ig =0A so Is=Id

equation can be broken down into
Vdd-IdRs+Vgs-Veq=0

What i get for Vgs is 5.317v and -1.317v. it seems that my Vgs is wrong..
Vds for -1.317 is -9.87V, Id = 25.12uA

cos for both value of Vgs, |Vgs|>= |Vtp|
and both answer are in saturation region |Vds|>= |Vgs|-|Vtp|

Btw to find Vds can i do by using this equation
Vdd-IdRs-Vsd-IdRd-Vss=0

 

[kmegamind]

I solved this starting from the loop equation from GND to VDD through VGS, u get VGS = 2ID - 4 , substitute in the ID - VGS relation, get ID and VGS , that should be 0.608 and -2.784 , get VDS from the remaining loop (-6.9) the other value yields VDS < VGS-VTP

 

[forbi]

I solved this starting from the loop equation from GND to VDD through VGS, u get VGS = 2ID - 4 , substitute in the ID - VGS relation, get ID and VGS , that should be 0.608 and -2.784 , get VDS from the remaining loop (-6.9) the other value yields VDS < VGS-VTP

How you get the loop equation?

The loop equation i get from GND to VDD is
-1 +IgReq + Vgs + IdRs + 5 = 0?
It will become -Vgs = 2ID - 4

Btw, issit true for all Enhanced PMos, Vgs , Vds and Vtp are all negative voltage?

 

[jeffrey samuel]

Yup for all PMOS the values are always negative

 

[forbi]

anyone can shed some light on the equation for GND to Vdd?
Cos it seems that i'm unable to get what kmegamind equation of VGS = 2ID - 4

what i get is -1 +IgReq + Vgs + IdRs + 5 = 0
which will derive into -Vgs = 2ID - 4

Unless im suppose to treat it as "-Vgs" instead of "+Vgs" which i dun understand why.

 

[jeffrey samuel]

Use these two eqns and substitute all your values in them

Vdd+IdRs+Vds+IdRd-Vss=0

VddRb1/(Rb1+Rb2) + Vgs IdRd-Vss=0

compute Vgs from the voltage at gate Vg which is the Thevinin voltage of the Gate network then the Source voltage whish is the IdRs value the difference between them gives Vgs

 

[kmegamind]

anyone can shed some light on the equation for GND to Vdd?
Cos it seems that i'm unable to get what kmegamind equation of VGS = 2ID - 4

what i get is -1 +IgReq + Vgs + IdRs + 5 = 0
which will derive into -Vgs = 2ID - 4

Unless im suppose to treat it as "-Vgs" instead of "+Vgs" which i dun understand why.

Vgs = Vg - Vs
so it's the voltage rise from the source to gate
you can say it's like an arrow pointing from S to G
so in the equation you wrote it's Vsg not Vgs , as u pass by G then S on your way from GND to VDD (voltage rise is from gate to source).

 

[fredd]

Hi,

How do i go about finding the Q-Point of the above Circuit.
Vtp = -1v, lamda is 0.01V^-1
Kp = 0.5mA/V^2

btw the Veq should be 1V.

i do by Vdd-IsRs+Vgs-IgReq-Veq=0
and Ig =0A so Is=Id

equation can be broken down into
Vdd-IdRs+Vgs-Veq=0

What i get for Vgs is 5.317v and -1.317v. it seems that my Vgs is wrong..
Vds for -1.317 is -9.87V, Id = 25.12uA

cos for both value of Vgs, |Vgs|>= |Vtp|
and both answer are in saturation region |Vds|>= |Vgs|-|Vtp|

Btw to find Vds can i do by using this equation
Vdd-IdRs-Vsd-IdRd-Vss=0

Veq is indeed 1V. Here's the solution:
KCL at node G (gate), (5V - Veq)/80Kohms = (Veq - -5V)/120Kohms
1.5(5V - Veq) = Veq + 5V
7.5V - 1.5Veq = Veq + 5V
2.5Veq = 2.5V
Veq = 1V



- - - Updated - - -

anyone can shed some light on the equation for GND to Vdd?
Cos it seems that i'm unable to get what kmegamind equation of VGS = 2ID - 4

what i get is -1 +IgReq + Vgs + IdRs + 5 = 0
which will derive into -Vgs = 2ID - 4

Unless im suppose to treat it as "-Vgs" instead of "+Vgs" which i dun understand why.

Let's do this.
KVL from +5V to Veq then ground: 5V - 2Kohms*Id(or Is) - Vsg - Veq = 0
5V - 2Kohms*Id + Vgs -1V = 0 (take note that Vgs = -Vsg)
Vgs = 2Kohms*Id -4V
Vgs = 2KId - 4 (eqn.1)

then let's substitute eqn.1 to the general formula for Id: Id = Kp(Vgs - Vth)^2 (assuming that the effect of lambda is negligible)
Id = Kp(2KId - 4 - Vth)^2
Id = 0.5m(2KId - 4 + 1)^2
Id = [(0.5m*2K)Id - 3(0.5m)]^2
Id = (Id -1.5m)^2
Id = Id^2 - 3mId + 2.25u
0 = Id^2 - 1.003Id +2.25u
by quadratic formula,
Id = 2.243275205uA or 1.002997757A
Id = 2.243275205uA

substituting Id to eqn.1,
Vgs = (2Kohms)(2.243275205uA) - 4V
Vgs = -3.99551345V or Vsg = 3.99551345V