By: David Kanter (dkanter.delete@this.realworldtech.com), December 15, 2011 4:20 pm

Room: Moderated Discussions

Rohit (@.) on 12/15/11 wrote:

---------------------------

>Konrad Schwarz (no.spam@no.spam) on 12/15/11 wrote:

>---------------------------

>>You write:

>>

>>As the equation makes clear, subthreshold leakage increases exponentially with temperature.

>>

>>The exponent in the equation is -B * V / T

>>

>>If T tends to infinity, the quotient tends toward minus zero,

>>and the exponential tends toward one. Isn't the dominant factor in the equation then the T squared?

>>

>>This would mean that subthreshold leakage increases >>quadratically with temperature.

>

>That dependence is the asymptotic limit. Normally, you are in the regime where

>V is MUCH larger than the thermal voltage, which is exponential matters.

So ironically one of my friends who has a PhD in math made this same point. As Rohit mentioned, it's not the asymptotic behavior that is really a concern.

Tongue in cheek:

Asymptotically as T-->infinity, the leakage of the chip goes to 0 since it burns up : )

David

---------------------------

>Konrad Schwarz (no.spam@no.spam) on 12/15/11 wrote:

>---------------------------

>>You write:

>>

>>As the equation makes clear, subthreshold leakage increases exponentially with temperature.

>>

>>The exponent in the equation is -B * V / T

>>

>>If T tends to infinity, the quotient tends toward minus zero,

>>and the exponential tends toward one. Isn't the dominant factor in the equation then the T squared?

>>

>>This would mean that subthreshold leakage increases >>quadratically with temperature.

>

>That dependence is the asymptotic limit. Normally, you are in the regime where

>V is MUCH larger than the thermal voltage, which is exponential matters.

So ironically one of my friends who has a PhD in math made this same point. As Rohit mentioned, it's not the asymptotic behavior that is really a concern.

Tongue in cheek:

Asymptotically as T-->infinity, the leakage of the chip goes to 0 since it burns up : )

David

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