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Short-channel MOSFETs

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isaacnewton

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vds(sat)

On page 297 of Baker's book CMOS: Circuit Design, Layout, and Simulation 2nd Edition www.cmosedu.com/cmos1/book.htm

Equation 9.54
Vov = VGS - VTH ≠ VDS,sat
How to understand this? How do define VDS,sat for short-channel MOSFETs?

The short channel effects is only for very small L transistors, right? If I choose L = 2 um for 0.18 um technology , does the short-channel effect exist?

For short channel mosfet, the drain current is
Equation 9.56
iD = vsat • Cox' • W • (VGS - VTH - VDS,sat)
 

how to avoid short channel effect

Hi Newton,
Vov = VGS - VTH ≠ VDS,sat
In my opinion, it means the criterion of saturation is no longer VGS-VTH, it has changed to another one VDS,sat, which should be calculated by using an extra kind of empirical formulation.
The relationship between short channel and long channel is just like quantum mechanics and classic mechanics.
And if you choose L = 2 um for 0.18 um technology, I think the short-channel effect exists.

Regards,
Terry
 

short channel pmos iv curve

For 0.18 um CMOS technology, L = 2 um is not short-channel any more. Could you explain more? Thanks.

qiushidaren said:
Hi Newton,
Vov = VGS - VTH ≠ VDS,sat
In my opinion, it means the criterion of saturation is no longer VGS-VTH, it has changed to another one VDS,sat, which should be calculated by using an extra kind of empirical formulation.
The relationship between short channel and long channel is just like quantum mechanics and classic mechanics.
And if you choose L = 2 um for 0.18 um technology, I think the short-channel effect exists.

Regards,
Terry
 

cmos i-v curve short channel

isaacnewton said:
For 0.18 um CMOS technology, L = 2 um is not short-channel any more. Could you explain more? Thanks.

qiushidaren said:
Hi Newton,
Vov = VGS - VTH ≠ VDS,sat
In my opinion, it means the criterion of saturation is no longer VGS-VTH, it has changed to another one VDS,sat, which should be calculated by using an extra kind of empirical formulation.
The relationship between short channel and long channel is just like quantum mechanics and classic mechanics.
And if you choose L = 2 um for 0.18 um technology, I think the short-channel effect exists.

Regards,
Terry
As far as I have known, if L<3µm, the short-channel-effect exists, and the 0.18 BSIM models we have used are for short-channel MOSFETs.
 

    isaacnewton

    Points: 2
    Helpful Answer Positive Rating
short channel effects

qiushidaren said:
As far as I have known, if L<3µm, the short-channel-effect exists, and the 0.18 BSIM models we have used are for short-channel MOSFETs.

So if I choose L = 4 um for 0.18 um CMOS technology, there will be no short channel effects.

If I choose L = 2.5 um for 2 um CMOS technology, there will be short channel effects.

Is it right?
 

isaacnewton said:
qiushidaren said:
As far as I have known, if L<3µm, the short-channel-effect exists, and the 0.18 BSIM models we have used are for short-channel MOSFETs.

So if I choose L = 4 um for 0.18 um CMOS technology, there will be no short channel effects.

If I choose L = 2.5 um for 2 um CMOS technology, there will be short channel effects.

Is it right?
I think it all depends on your models, maybe you can refer to your models to gain more information.
Yes, if L = 4µm, there will be no short-channel-effect.
 

Hi
this is njeem there will be no short channel effects if u use L=2um or little effect since for L<3um short channel effect will be present.

Bye
 

hello,
to avoide short channel effect we take 3-4 times to the min. channel length transistors. so in my opinion there won't be significant effect above 0.6-0.7u channel length.

Added after 9 minutes:

isaacnewton said:
On page 297 of Baker's book CMOS: Circuit Design, Layout, and Simulation 2nd Edition www.cmosedu.com/cmos1/book.htm

Equation 9.54
Vov = VGS - VTH ≠ VDS,sat
How to understand this? How do define VDS,sat for short-channel MOSFETs?

For short channel mosfet, the drain current is
Equation 9.56
iD = vsat • Cox' • W • (VGS - VTH - VDS,sat)

for short channel MOSFET's velocity saturation occur before the Vds=Vgs-Vth.
This happen because of the high electric field in channel for short channel length transistors. so VDS,sat is the voltage at which carrier mobility got saturated.
 

vinodjn said:
hello,
to avoide short channel effect we take 3-4 times to the min. channel length transistors. so in my opinion there won't be significant effect above 0.6-0.7u channel length.

Added after 9 minutes:

isaacnewton said:
On page 297 of Baker's book CMOS: Circuit Design, Layout, and Simulation 2nd Edition www.cmosedu.com/cmos1/book.htm

Equation 9.54
Vov = VGS - VTH ≠ VDS,sat
How to understand this? How do define VDS,sat for short-channel MOSFETs?

For short channel mosfet, the drain current is
Equation 9.56
iD = vsat • Cox' • W • (VGS - VTH - VDS,sat)

for short channel MOSFET's velocity saturation occur before the Vds=Vgs-Vth.
This happen because of the high electric field in channel for short channel length transistors. so VDS,sat is the voltage at which carrier mobility got saturated.

i dont think that if we use L 3 - 4 times large than minimum size we can avoid short channel effect. when i simulate one transistor and plot it's IV curv, vds still affecting the drain current although the transistor is in saturation mode. this is due to channel length modulation (CLM) if i'm not mistaken. CLM is one of short channel effect?
 

The short channel effect exists all the time whatever L is. The importance of SCE depends upon the applications.
 

the short chnnel effect will be reduced if we take 3-4 times minimum.but it will not be totally removed.

we can take 5-6 times of minimum chaneel length.
also.
 

najmuus said:
the short chnnel effect will be reduced if we take 3-4 times minimum.but it will not be totally removed.

we can take 5-6 times of minimum chaneel length.
also.

agreed with that. i believe if we take L > 10µ then SCE effect is almost negligible. however it a large L. so our design should be large and that not really desirable
 

the equation for short channel MOS is very complex (example BSIM3), so how we can design with these equations. How can we use these equation, they have many parameters, we can't use them directly to design.
Anyone can give me some ideal about this problem?
Thanks
 

tran said:
the equation for short channel MOS is very complex (example BSIM3), so how we can design with these equations. How can we use these equation, they have many parameters, we can't use them directly to design.
Anyone can give me some ideal about this problem?
Thanks
no you mustn't use complex bsim3 equation to do manual analysis. Just use simple equation such as mos level 1.
 

But the results I have when I design with level 1 is not proper i.e if I use those results to simulation (use Bsim 3), I will get the results different from my caculation, those results is only right when I simulation with level1. But level1 is only the simplest model, it is not a practical model so it'results will not use for practical design, it'results only use for you understand what happened and you will have to find a way to change your design (change L W ...) so that you get the results satisfy your problem. But I want to design by use Bsim 3 model.
I want to find a way that I can design with Bsim 3 or complex model (EKV,...) directly, possibly the errors between calculation (by hand) and simulation is about 20 percentage or less than.
Have you got any ideals about that ?
Thanks
 

qiushidaren said:
isaacnewton said:
qiushidaren said:
As far as I have known, if L<3µm, the short-channel-effect exists, and the 0.18 BSIM models we have used are for short-channel MOSFETs.

So if I choose L = 4 um for 0.18 um CMOS technology, there will be no short channel effects.

If I choose L = 2.5 um for 2 um CMOS technology, there will be short channel effects.

Is it right?
I think it all depends on your models, maybe you can refer to your models to gain more information.
Yes, if L = 4µm, there will be no short-channel-effect.

Actually short channel effect is very much depend on the foudry process.... there is no exact number of the L for the short channel, but roughly if you take L=5Xmin L then the effect is negligible...... You can always request IV curve from your foundry to see where is the L turning point to consider as short channel....
 

Short Channel Effects are physical constraints on MOSFET performance and mostly considered under L<1u .
The most obvious effect is velocity saturation of carriers I think and this fact changes all the equations we use so far.

I think, It would be better If we hadn't defined a boundary value like Vth, because the underlying physics is of continious nature...
 

Vdsat of course not equals to Vgs-Vth

2/Vdsat=gm/Id
 

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