Messages in this thread | | | From | Alexander.vanLuijpen@nym ... | Subject | Re: [OFFTOPIC] Re: disk head scheduling | Date | Tue, 23 Mar 1999 12:22:27 +0100 |
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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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