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From: dana@locus.com (Dana H. Myers)
Newsgroups: sci.electronics
Subject: Re: 80386.
Message-ID: <1991Jan14.205104.2829526@locus.com>
Date: 14 Jan 91 20:51:04 GMT
References: <1990Dec18.234020.2491@uoft02.utoledo.edu> <4400001@hpsgwp.sgp.hp.com>
Organization: Locus Computing Corporation, Inglewood, CA
Lines: 83

In article <4400001@hpsgwp.sgp.hp.com> plim@hpsgwp.sgp.hp.com (Peter Lim) writes:
>/ dana@locus.com (Dana H. Myers) /  2:14 pm  Jan 10, 1991 /
>
>$   The 25Mhz chips DO run hotter than the the 16Mhz parts. CMOS power
>$ consumption occurs during transitions. There are two reasons why;(1)
>$ current must flow through the gate capacitances (standard 1/(2*pi*f*C)
>$ stuff)
>$ Resultantly, the power consumption of CMOS parts
>$ is roughly proportional to frequency of operation. A 25Mhz will consume
>$ about 25/16 (or 150%) of the power a 16Mhz part consumes.

  I should phrase this better. A 25Mhz part will consume about 25/16 (or
about 150%) of the power the same part at 16Mhz.

>If I remember correctly, the energy dissipated in a CMOS chip is
>proportional to the square of the frequency. So, 25 MHz vs. 16 MHz power
>supply would be 25^2 / 16^2 = 244 %. Quite a hell lot more power.
>Assuming voltage and impedance stay about constant, heating rate of
>the chip will be about proportional to power dissipation. But since
>higher temperature will give rise to higher rate of heat loss; may be
>a part running at 25 MHz will be about twice as hot as a part running
>at 16 MHz.
>
>(Somehow, can't quite remember the equation that says that power is
>proportional to freq^2).

Well, let's derive something close.

Power is given by:

	P=IV		

    Where:
	P = power in Watts
	I = current in Amperes
	V = voltage in Volts

Since we are assuming the power is disspated in capacitance, we'll
use the simple form of capacitive reactance:

	Xc = 1 / (2*Pi*f*C)

    Where:
	Xc = capacitive reactance, in Ohms
	Pi = pi, 3.14159...
	f = frequency, in Hz
	C = capacitance, in Farads

Using Ohms Law for the current:

	Idynamic = 2*Pi*f*C*V

Resulting in a dynamic power dissipation of:

	Pdynamic = V^2 * 2 * Pi * f * C

You'll see that the dynamic component of the power dissipation is
directly proportional to the frequency of operation, and proportional
to the square of the applied voltage.

 In reality, the power consume is the sum of the static and dynamic
components:

	Ptotal = Pstatic + Pdynamic

which can also be expressed:

	Ptotal = V * (Istatic + Idynamic)
or:
	Ptotal = V * (Istatic + 2 * Pi * f * C * V).

    This is a very simplified picture. I might have left lots of
stuff out. However, graphs of current vs. frequency for popular
CMOS parts (386 included) have a linear slope.

Dana


-- 
/*
 * Dana H. Myers KK6JQ 		| Views expressed here are	*
 * (213) 337-5136 		| mine and do not necessarily	*
 * dana@locus.com		| reflect those of my employer	*