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Id Versus Vds

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12 Threads found on edaboard.com: Id Versus Vds
Thanks for holddreams's reply. If vth is not equal to vth0, how should I set Vgs in spectre? is there any ideas?
Phenomena like DIBL will lower the effective VT of FETs in absolute terms. None of the lengths you show are likely to see much of this in submicron technologies. Edge effects may also be a factor, a small region at the edge is a somewhat effective MOS structure but with some degraded attributes from strain and oxide / interface quality.
How to find the I-V characteristics Id versus vds? of an NMOS transistor Also.. How do I sweep the VGS and vds from 0 to 3V? I am having a bit of trouble using spice, if anyone knows a good tutorial "for dummies site" please let me know...
hey,moonnightingale here's my answer , I hope it can be helpful. As the nmos is in the "Saturation Region" and the current source is a dc source, it means: Id=0.5 * Un * Cox * (w/L) * (Vgs-Vth)^2 * (1+λvds) and here ,Vgs=vds : Id=0.5 * Un * Cox * (w/L) * (vds-Vth)^2 * (1+λvds) the right part of the equation (...)
plot id vs. vgs in cadence then take the derivative of the curve
there is not a sharp transistion from sub-threshold operation to moderate inversion. Usually, when you have the Id ~ Vg curve with the fixed vds, you will get a somehow constant slop curver after some Vg, just extend that curve toward you x-axis and the intercept point will be quite accute of your Vth
Hi all, If I have a diode connected NFET being biased by an ideal current source. As the temperature goes up, based square law, the vds now equal to Vgs should go up too. But my vds versus temp simulation result is actually opposite. Does anyone know why? Also, in this configuration, how does the vdsat change over (...)
I think he is confused. Vsat is velocity saturation in short(0.5 to 0.25 micron) and narrow(0.18 to 0.13 micron) channel transistors. It has nothing to do with vdsat. Vsat is due to the E-field from drain acting on the electrons in the channel and source when VDD is applied to the drain. Therefore constant electron mobility no longer holds true!
If u see the output curve between vds and Id u can easily see that in st region , for a very small change in i/p, there will be a large change in o/p(Id) ,,this is what we require in we bias it in saturation region
It seems to assume a quadratic behaviour, so probably that is the ID vs. VGS curve, i.e. the drain current versus gate voltage curve
There's obviously a misunderstanding of technical terms. In MOS literature, triode region is defined by the boundary condition vds ≤ VGS - VTH, while in saturation (or in traditional vacuum electronics terms pentode) region vds > VGS - VTH applies. The region near the origin with resistive vds versus ID behavio
Trying to follow the example in Allen's book for a two stage op-amp design (6.3) Could someone explain the constraints placed on the common mode input range? Vin_max = VDD - sqrt(I5/beta3) - |VT03max| + VT1min Vin_min = VSS + sqrt(I5/beta1) + |VT1max| +vds5(sat) I don't understand his terminology, what is VT03max, and what is VT1min and