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- Path: lou.teclink.net!usenet
- From: rad@teclink.net (rad)
- Newsgroups: comp.sys.amiga.programmer
- Subject: Re: Processors
- Date: 24 Mar 1996 18:12:04 GMT
- Organization: TECLink Internet Services: info@TECLink.Net
- Message-ID: <1665.6656T1237T1226@teclink.net>
- References: <4iri6d$lim@columba.udac.uu.se>
- NNTP-Posting-Host: tc2_39.teclink.net
- X-Newsreader: THOR 2.2 (Amiga;TCP/IP) *UNREGISTERED*
-
- On 21-Mar-96 12:31:09, Kristofer Maad <m93kma@sabik.tdb.uu.se> wrote:
- >travis (envision@darwin.topend.com.au) wrote:
- >: You must remember that Motorola considers the processor speed to be the bus
- >: speed. The 040 actually runs at twice the bus speed.
-
- >No, it doesn't. This has been said a thousand times on this and other
- >newsgroups: The '040 uses a double internal clock _only_ for pipeline
- >synchronization purposes. No instructions are performed in an odd
- >number of 80MHz-cycles. The fastest instruction takes one
- >40MHz-cycle. On the 486, though, the processor is _really_ clock
- >doubled, so some instructions take only one 66MHz-cycle to complete.
-
- Ummm, you're only half right. The integer unit runs at the rated speed;
- however, the FPU is run at the higher clock rate. Check out the 68040 user's
- manual and you should see that several FPU instructions take fractional
- numbers of cycles in the Execution stage... (FDIV 37.5, FMOVE, 1.5 or 4.5
- FABS, FNEG 4.5) It has been confirmed by Motorola Engineers on
- comp.sys.m68k that the FPU is based on the "double" clock.
-
- It is true though that the 68040 is not clock doubled in the sense of the
- 80486s. It does not internally generate a faster clock from a slower bus
- clock. It simply uses an external clock at half speed.
-
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- - Richard Deken EMail: (personal) rad@teclink.net -
- - VLSI Design Engineer (AuE) rad@aue.com -
- - Advanced Microelectronics PGP public key available -
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