Wn/Wp in digital circuit

[taranom1]

Hi all!

How can I determine Wn/Wp ratio exactly by having Vth in digital circuit?

In full swing digital circuits the input and output are equal to vdd or 0, so are the Vgs and Vds equal to vdd or 0 in equation below?
Do I have to use triode and saturation equations for each transistor?
Ids(triode) = kw/l(Vgs-Vth-Vds/2)×Vds

One more question:
Is eatch of Wn or Wp important in effecting on Vth? or determining Wn/Wp ratio is enough?

 

[taranom1]

Is each of Wn or Wp important in effecting on Vth? or determining Wn/Wp ratio is enough?

I mean that for wn/wp, 60/120 is same as 180/360?

 

[Dominik Przyborowski]

For narrow width devices Vth is different than for wider ones.

 

[dick_freebird]

If Vth is your inverter threshold then there is a unique case (width
ratio) where NMOS ID(Vth) = PMOS ID(VDD-VTH). This will differ as
VDD changes. You probably care most about VDDmin as this is your
most challenging timing case. It will change with process and temp
as well. It will also change when you start stacking and paralleling
transistors in more interesting (than inverter) logic configurations.
And standard cell area optimization emphasis may preclude you
making so, in those cases.

 

[analogLow]

Hi,
> I mean that for wn/wp, 60/120 is same as 180/360?
Yes and no. First, the ratio is important to ensure that the drive strength of the NMOS and PMOS are similar. You want the inverter to equal low-to-high and high-to-low transitions. So yes, in terms of matching the low-to-high and high-to-low delay, 60/120 and 180/360 are very similar. But remember that 180/360 adds more capacitance ... so the delay is more than with 60/120 (but still has matched transition times).

One other note ...changing the transistor size has very minimal impact on Vth and I wouldn't worry about it unless you are at high frequencies or very low voltages.

My advice for these situations is always: sweep , sweep, sweep. Run simulations and sweep variables and see what happens. The more graphs you have ... the more insight you have into your technology and circuit behaviour.

Hope that helps,
analogLow :thumbsup: