W value is too big than the minimum value

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 W_min should be in this order.

erikl said:

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 W_min should be in this order.

Click to expand...

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.

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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.

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deepsetan said:

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.

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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

Click to expand...

KingOfHearts said:

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

Click to expand...

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?

deepsetan said:

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.

Click to expand...

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

deepsetan said:

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

Click to expand...

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.

erikl said:

No. W_min is the same size or a bit bigger than your technology size value, i.e. W_min ≧ 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.

Click to expand...

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.

deepsetan said:

Is it ok that the value of W is smaller than L.

Click to expand...

Yes, as I told you above.

deepsetan said:

This mean that my gm will be smaller.

Click to expand...

You chose an Id of 22µA, so you can't expect higher gm. If you need greater gm, you must use higher Id.