isaacnewton
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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 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
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.
I think it all depends on your models, maybe you can refer to your models to gain more information.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?
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)
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.
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.
no you mustn't use complex bsim3 equation to do manual analysis. Just use simple equation such as mos level 1.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
qiushidaren said:I think it all depends on your models, maybe you can refer to your models to gain more information.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?
Yes, if L = 4µm, there will be no short-channel-effect.
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