From: firstname.lastname@example.org (Peter)
Subject: Re: power consumption
Date: 11 Oct 2002 18:10:46 -0700
NNTP-Posting-Date: 12 Oct 2002 01:10:46 GMT
thanks, the info you provided was very useful. And thanks to
all others for helping out.
"Tom Faloon" wrote in message news:...
> Hi Peter,
> In most modern logic gates there are two elements contributing to power
> Total power = DC power dissipation + AC power dissipation.
> The DC power dissipation is the amount of power dissipated if the inputs and
> outputs are static, in defined logic states.
> It is calculated from Supply Voltage * quiescent current. In modern 'xMOS'
> devices the quiescent current is usually very small. In the case of the
> 74AC00, it is 20uA max. So maximum DC power dissipation @ 5V supply,
> Pdiss-dc = 100uW
> The AC power dissipation is a result of switching.
> If there is any change of logic states, internal switching transistors have
> to charge or discharge stray capacitance within the device, and this draws
> extra current from the supply. If they are switched repeatedly can result in
> appreciable current being drawn from the supply. The faster the rate of
> switching, the greater the current.
> If one element of a 74AC00 is clocked, the AC power dissipation can be
> calculated from
> Pdiss-ac = Cpd * V(squared) * frequency. (See Fairchild application note
> AN-303 for derivation of this formula.)
> Where Cpd is the manufacturers estimate of the effective 'switching
> capacitance' , V is the supply voltage, and f is the frequency in Hertz.
> The Fairchild data sheet specifies Cpd as 20pF, so if you run it at 10MHz,
> with a 5V supply, you can expect the AC power dissipation, for one element,
> to be Pdiss-ac = 20E-12 * 5^2 * 10E6 = 5mW.
> So if all four elements were driven at 10MHz, the total dissipation
> Ptotal = Pdiss-dc + 4 * Pdiss-ac = 100uW + 4 * 5mW = 20.1mW
> This is much less than the value which you calculated. Perhaps you can see
> where you (or I ?) went wrong.
> You would have to run all four elements of the 74AC00 at 100 MHz to achieve
> 200mW dissipation, but this is way beyond it's guaranteed operating speed,
> so it won't happen.)
> Of course many complex chips, dissipate much more than 200mW !
> You will find a useful application note 'AN-303 HC-MOS power dissipation'
> and a Fairchild data sheet for the 74AC00 at
> Tom Faloon
> Peter wrote in message
> > Hi you all!
> > i would like to ask how to calculate the power
> > consumption of IC's from the info in datasheet provided, say 74ac00. I
> > am aware that there are some equations that take in to account the no
> > . of input output pins, the freq of these pins working, Vcc, Icc and
> > the loading capacitance. However the power calculated from this is
> > quite large ,(200mW) and if such IC is operated ,it seems a bit
> > unrealistic if it is to be supported using batteries.
> > Do forgive me if this seems a stupid question, but help me out by
> > pointing out where i am wrong.
> > Thanks