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SubjectRe: [OFFTOPIC] Re: disk head scheduling
Date

David Wragg <dpw@doc.ic.ac.uk> wrote:
> "Richard B. Johnson" <root@chaos.analogic.com> writes:
> > [snip]
> > Expensive disk drives now do full track buffering. This costs
> > money because RAM costs money. To buffer one full track on a
> > Disc drive requires CAPACITY / (HEADS * CYLINDERS) which can
> > be upwards of 100 megabytes of high-speed SRAM. Sector buffering
> > is always necessary. It is part of the de-serializer and is
> > required because the Disc internals are never synchronous with
> > the outside world.
>
> Why SRAM? The nice thing about SRAM is its low latencies. If you're
> slurping a whole track off a disk, latencies shouldn't be an issue;
> you might need lots of bandwidth, but you can get high bandwidth from
> DRAM.
>
> Besides, 100 megabytes of SRAM is 100*(8 bits in a byte)*(6
> transistors for an SRAM cell) = 2.4 billion transistors! Ignoring
> issues of cost, where exactly do they put all those chips? I don't
> know much about the semiconductor state-of-the-art, but I do know that
> at current levels of integration it would be a squeeze to fit that lot
> on the back of a disk.

Actually: 100*8*6 is 4.8 billion transistors. With an estimate of 5 um2
per cell in our CMOS18 process, this would result in a die size of
approximatelly 41 cm2.


Vriendelijke Groeten / Kind Regards,

Alexander van Luijpen

Philips Semiconductors Nederland
Test and Product Engineering
MOS4YOU - C075 OTP / Consumer Systems Nijmegen - BL Video

email: Alexander.vanLuijpen@nym.sc.philips.com email:
vluijpen@cistron.nl
tel: (+31)-24-353 4639
tel: (+31)-24-378 9475


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