[comp.arch] 040 vs 030 SPEC nos.

abaum (Allen Baum) (05/21/91)

In article <1991May20.145332.19406@murdoch.acc.Virginia.EDU> clc5q@hemlock.cs.Virginia.EDU (Clark L. Coleman) writes:

>In article <177@armltd.uucp> abaum (Allen Baum) writes:
>>
>>A...there are several reasons why the '040 SPEC numbers are better-
>>1- faster clock, resulting from better process technology,
>
>This does not make any sense to me. The only SPEC numbers for
>the 68040 are for 25MHz chips. There are 68030 chips running at 25MHz
>for several years now, and likewise for the 68882 FPU that went with
>the 68030. Any improvement in SPEC numbers over the 68030 did't come
>from the 68040 having a faster clock rate.

I'm sorry to disappoint you, but the 68040 runs effectively a half
speed bus. A 25Mhz '040 is running at 50Mhz internally. There are,
in fact, two clock pins, one at 25Mhz, and one at 50. For comparison,
you should be comparing 50Mhz '030s and 25Mhz '040s.

billl@oakhill.sps.mot.com (Bill Ledbetter) (05/29/91)

abaum (Allen Baum) writes:

>I'm sorry to disappoint you, but the 68040 runs effectively a half
>speed bus. A 25Mhz '040 is running at 50Mhz internally. There are,
>in fact, two clock pins, one at 25Mhz, and one at 50. For comparison,
>you should be comparing 50Mhz '030s and 25Mhz '040s.

Well, in general, an 040 at 25MHz runs both the internals (caches, ALUs,
etc.) and the external bus at 25MHz: the 50MHz input is used to get timing
edges.  In general, a 25MHz 030 runs the internals at 12.5MHz and the
external bus at 25Mhz.

---------------------------------------------------------------------------
Bill Ledbetter
Motorola Inc.    MD OE37                 email: cs.utexas.edu!oakhill!billl
6501 William Cannon Dr. West
Austin, TX 78735-8598


-- 
---------------------------------------------------------------------------
Bill Ledbetter
Motorola Inc.    MD OE37                 email: cs.utexas.edu!oakhill!billl
6501 William Cannon Dr. West

crisp@mips.com (Richard Crisp) (06/07/91)

In article <1991May28.183453.13821@oakhill.sps.mot.com> billl@oakhill.sps.mot.com (Bill Ledbetter) writes:
>abaum (Allen Baum) writes:
>
>>I'm sorry to disappoint you, but the 68040 runs effectively a half
>>speed bus. A 25Mhz '040 is running at 50Mhz internally. There are,
>>in fact, two clock pins, one at 25Mhz, and one at 50. For comparison,
>>you should be comparing 50Mhz '030s and 25Mhz '040s.
>
>Well, in general, an 040 at 25MHz runs both the internals (caches, ALUs,
>etc.) and the external bus at 25MHz: the 50MHz input is used to get timing
>edges.  In general, a 25MHz 030 runs the internals at 12.5MHz and the
>external bus at 25Mhz.
>


Attempting to state things more succintly:
The execution unit of an 030 cycles at frequency f/2 with an input clock
frequency of f. The execution unit of an 040 cycles at frequency f with
an input clock of frequency f.

---new material---
An interesting metric is the ratio of the external bus cycle time to
the cycle time of the execution unit. I will define the external bus
cycle time as the minimum amount of time required for an external bus
cycle. I will define the execution unit cycle time at the amount of
time it takes for two operands to be read from the register file, perform
an ALU operation and then write back the result.

When we designed the 68000 back in the late 70's it worked like this:
With an input clock of frequency f, the execution unit of the 68000 ran 
at frequency f/2. The minimum bus cycle was f/4. The ratio of minimum 
bus cycle time to execution unit cycle time was therefore 2.

For the 68020:

With an input clock of frequency f, the execution unit of the 68020 ran
at frequency f/2. The minimum bus cycle was f/3. The ratio of minimum
bus cycle time to execution unit cycle time was therefore 3/2.

For the 68030:

With an input clock of frequency f, the execution unit of the 68030 ran
at frequency f/2. the minimum bus cycle was f/2. The ratio of minimum
bus cycle time to execution unit cycle time was therefore 1.

For the 68040:

With an input clock (the one that it is sold as running at) of frequency f,
the execution unit runs at frequency f. The minimum bus cycle is f/2. the
ratio of minimum bus cycle time to execution unit cycle time is therefore 2.

In summary:

processor		year		bus cycle time/EU cycle time

68000			1980		2

68020			1984		1.5

68030			1988		1

68040			1990		2

It looks like those guys at Moto have really been making some progress!



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