80186 and now AMY chip

From: Eric J. Korpela <korpela_at_ellie.ssl.berkeley.edu>
Date: Tue Nov 21 13:28:20 2000

> There are different measurements around, like data path size,
> register size or _marketing_ size.
>
> Now let us look at the 68K family
> 68000:
> Data path: 16
> Register: 32
> ISA: 32
> Marketing: 16

You missed the important one, and the only one that counts
when determining the bitness of the processor, the width of
the (integer) ALU.

Clearly, regardless of the width of the registers and the
address bus size, the Z80 is an 8 bit processor, as is the
6502. The 8088 and 8086 are both 16 bitters. The 68000 and
68010 are 16 bitters. The 68020 is a 32 bitter as is the
x386.

> So if this is a 16 Bit CPU, what's a 68008 ? 8 Bit ?
> Still the same CPU core - and further more, is a 68020
> now 32 Bit ? Again, wheres the 'new' 32 Bit part?

All obvious answers when you use the ALU size. The 68008 is
a 16 bit processor. The 68020 is a 32 bit processor.

> Well, this part is marketing, you still may argue that a
> external data bus is a valid scheme for naming - nice, but
> then every x86 chip since the original Pentium is a true
> 64 Bit CPU. Sounds wrong, huh ? But it's true! (according
> to said scheme)

No, every x86 chip since the original 386 is a 32 bit CPU since
the width of the integer ALU is 32 bits.

Eric

> /360&up-ish CPUs where available with ALUs from 8 to 32 bits,
> internal data pathes from 8 to 256 bits, external data pathers
> from 8 to 1024 bits (128 bytes per memory cycle), but nobody
> will doubt that they are all 32 Bit CPUs.

I will. Anything with an 8 bit ALU is an 8 bit processor, regardless
of compatibility and wishful thinking. :)

Eric
Received on Tue Nov 21 2000 - 13:28:20 GMT