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Design >> Analog Design >> VDSAT more than VGS-VTh
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Message started by cmosa on Mar 27^th, 2018, 1:55pm

Title: VDSAT more than VGS-VTh
Post by cmosa on Mar 27^th, 2018, 1:55pm

Simulating sub 28nm planner technology node IO 1.8V devices.
Device Sizes: W=1u; L=1.5u
Operating points from Spectre sims.

VGS= 730mV
Vth=455mV
VDS= 400mV
VDsat= 330mV

Clearly, VGS-Vth is 275mV whereas Vdsat is 330mV
I have a very little margin on VDS/VDsat.
If my Vdsat is lower (~275mV as expected), I have enough margin.

I have read and heard that due to channel length modulation effect and/or velocity saturation effect Vdsat is lower than expected VGS-Vth.
however I see VDSAT more than Vgs-Vth !!!

Any explanation for Vdsat > than Vgs-Vth?
Could it be due to metal gate technology?
Could it be some error in the technology file setup since its a relatively new setup for us?

Thanks

Title: Re: VDSAT more than VGS-VTh
Post by AnalogDE on May 9^th, 2018, 12:42am

Saturation condition for short channel devices isn't going to match square law. I suggest you look at whatever small signal parameters are of interest to you and make sure those are still OK at your biasing point.

Title: Re: VDSAT more than VGS-VTh
Post by Nyquist on Jul 10^th, 2018, 5:11pm

Hi,

I think of VDSAT and VGS-TH as two different parameters. VGS-VTH is the overdrive voltage available to me while VDSAT is the voltage that signals the onset of saturation.

Having VGS-VTH < VDSAT means you are operating your transistor in weak inversion/moderate inversion region.

As a rule of thumb, you could also annotate your gmoverid to see if it is > 15. This value might differ for your technology, but it gives a feel for you on where you are operating your transistor.

Can you post the picture of the circuit, you are trying to simulate ?
Also, May I know why the W > L ?

Title: Re: VDSAT more than VGS-VTh
Post by A Kumar R on Jul 23^rd, 2018, 11:07pm

I want to understand this topic a little more deeply.

Why is it that a potential difference between Gate-to-Drain(VGD) < Vth of the device ALONE can't saturate the current through the device?

My point is that once we set VGD < Vth, we have created a pinch-off region near the drain-end and this is the on-set of saturation. So, >= 275mV of VDS is enough to saturate then why should we be concerned about a VDSAT of 330mV at all?

are we saying till we hit a VDS of 330mV device won't saturate, why?

Title: Re: VDSAT more than VGS-VTh
Post by Nyquist on Jul 24^th, 2018, 12:34pm

Hi Kumar,

I get your point. Even I felt the same when I started. VGS-VTH = VDSAT is based on the square law. For short channel devices that do not follow square law, VDSAT is no longer equal to VGS-VTH.

you can refer to the discussion here https://www.edaboard.com/showthread.php?321080-What-are-betaeff-and-vdsat-parameters-found-from-Cadence-simulation.

Title: Re: VDSAT more than VGS-VTh
Post by A Kumar R on Jul 25^th, 2018, 8:53am

Can I take it as VDSAT being something independent of VTH of the device?

Title: Re: VDSAT more than VGS-VTh
Post by Nyquist on Aug 6^th, 2018, 11:48am

Hi Kumar,

I am not an expert in device technology. But I would take VTH and VDSAT as independent parameters, just like you mentioned for any technology <0.5um