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Why does sub-threshold leakage decrease in Vt?

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cheenu2002

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Hi,
I have read in some articles that subthreshold leakage current is a major problem faced by designers in lower process nodes like 65nm and 45nm. I would like to understand the physics behind this phenomena.. Why does it increase with decrease in Vt? Can anyone give me an intuitive or mathematical answer....
Also, I would lile to know how this is handled by analog design engineers.
 

Somashekhar

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Hi,

The subthreshold region is often referred to as the weak inversion region.
If Vt of particular device is less that means Inversion layer is formed with less voltage and also tat weak inversion region is also formed at low voltage hence there will be lot of leakage.

Hence Sub-threshold voltage is the major limitation in the advancement of the technology.
 

dick_freebird

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You have a "subthreshold slope", mV per decade of current.

Processes with less total working voltage, want to push VT
lower so drive current { (Vgs-VT)^2 } is maximized for speed.
The counter-effect is leakage.

Say it's 150mV/decade, and your FET has a VT(1uA) of 0.5V.
That means at Vgs=0, you will have ideally a ~1nA per
transistor leakage. Now consider that processing and temperature
might put your VT(1uA) as low as 0.3V, and you're at 10nA
(only 2 decades down) per 1-square device.
 

eecs4ever

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When Vgs < Vt.

Subthreshold current is proportional to e^((Vgs-Vt)*q/KT)

T is temperature, K is boltzman's constant, q is 1.6e-19.

As Vt decreases, the substreshiold current increases exponentially.
 

Prashanthanilm

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When Vgs < Vt.

Subthreshold current is proportional to e^((Vgs-Vt)*q/KT)
y.


Hello,
Do you have any documents or proof regarding the quoted statement. I am confused about the current equation.Sorry for the inconvinence.Thank you.
 

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