Continue to Site

W value is too big than the minimum value

Status
Not open for further replies.

deepsetan

Advanced Member level 4
Full Member level 1
Joined
May 8, 2013
Messages
119
Helped
6
Reputation
12
Reaction score
5
Trophy points
1,298
Location
Malaysia
Activity points
2,137
Hi guys

I got a problem in calculating the value of W for my transistor. During the Id-Vgs simulation, I got Id=22uA.This is my parameters

Vdd=3.3V
Vds=1.65V
Vgs=1
Vth=0.546V
Un=0.050542
tox=5.141 x 10^-9
eox=3.5 x 10^-17 F/um
Cox=3.392 x 10^-12
L=0.56u

When I calculate the value of W using saturation formula, it gives me value of 0.35 which I think too big since the minimum value of W is 0.0001. Is it normal? I think maybe I got confuse in calculating value of Cox. During the Id-Vgs simulation in LT SPICE, I dont assign the value of W and L. I just left in empty.
 

Cox = eox/tox = 3.5 x 10^-17 F/um / 5.141 x 10^-9 m = 6.8 fF/(µm)2

Probably a 0.25 or 0.18µm technology, so Wmin should be in this order.
 
Cox = eox/tox = 3.5 x 10^-17 F/um / 5.141 x 10^-9 m = 6.8 fF/(µm)2

Probably a 0.25 or 0.18µm technology, so Wmin should be in this order.

What do you mean by that?, do you mean that my W should be 0.25 or 0.18um? I'm using 0.13um technology.

- - - Updated - - -

Id=1/2UnCox(W/L)(Vgs-Vth)^2

22u=1/2(0.050542)(6.8f)(W/0.56u)(1-0.546)^2

when i calculate W,it gives big value
 

It seems Un is too big... Usually UnCox is in the order of 100uA/V^2, maybe you forget to add Un's unit?

Click "helped me" button if you agree. Thanks
 

Thanks about that but I think that value should be correct since I took it from 0.13um technology itself.

- - - Updated - - -

What do you mean by that?, do you mean that my W should be 0.25 or 0.18um? I'm using 0.13um technology.

- - - Updated - - -

Id=1/2UnCox(W/L)(Vgs-Vth)^2

22u=1/2(0.050542)(6.8f)(W/0.56u)(1-0.546)^2

when i calculate W,it gives big value

It seems Un is too big... Usually UnCox is in the order of 100uA/V^2, maybe you forget to add Un's unit?

Click "helped me" button if you agree. Thanks

I did found that the unit value for U0 is in cm^2/V-s. Do you mean that my Un=0.050542 cm^2/V-s? What is actually the correct way to convert it in order to calculate the W?
 

What do you mean by that?, do you mean that my W should be 0.25 or 0.18um? I'm using 0.13um technology.
No. Wmin is the same size or a bit bigger than your technology size value, i.e. Wmin ≧ 0.13µm in your case.


Id=1/2UnCox(W/L)(Vgs-Vth)^2

22u=1/2(0.050542)(6.8f)(W/0.56u)(1-0.546)^2

when i calculate W,it gives big value

When you get your equation right ...

Id=(1/2)*UnCox(W/L)(Vgs-Vth)2
or
W = 2Id*L / (UnCox*(Vgs-Vth)2)

... and add correct units to your calculation

W = 2*22µA*0.56µm / ((0.050542m2/Vs)*6.8fF/(µm)2*(1-0.546)2 V2)

W = 2*22µA*0.56µm / ((344µA/V2)*0.206V2) = 24.64µm/70.864 = 0.348µm

... you get a reasonable value for W in your 0.13µm technology. Here W is even smaller than L, which is ok.
 

No. Wmin is the same size or a bit bigger than your technology size value, i.e. Wmin ≧ 0.13µm in your case.




When you get your equation right ...

Id=(1/2)*UnCox(W/L)(Vgs-Vth)2
or
W = 2Id*L / (UnCox*(Vgs-Vth)2)

... and add correct units to your calculation

W = 2*22µA*0.56µm / ((0.050542m2/Vs)*6.8fF/(µm)2*(1-0.546)2 V2)

W = 2*22µA*0.56µm / ((344µA/V2)*0.206V2) = 24.64µm/70.864 = 0.348µm

... you get a reasonable value for W in your 0.13µm technology. Here W is even smaller than L, which is ok.

thanks erikl. I did found this value during my first calculation but I thought that I was wrong because most of the design W>L. Is it ok that the value of W is smaller than L. This mean that my gm will be smaller.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top