Path: utzoo!utgpu!water!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!mailrus!cornell!batcomputer!itsgw!steinmetz!uunet!pilchuck!del
From: del@Data-IO.COM (Erik Lindberg)
Newsgroups: comp.sys.ibm.pc
Subject: Re: Speeding up old PC's
Message-ID: <1017@pilchuck.Data-IO.COM>
Date: 20 Oct 88 01:24:37 GMT
References: <235494b5@ralf> <1109@bucket.UUCP>
Reply-To: del@pilchuck.Data-IO.COM (Erik Lindberg)
Organization: Data I/O Corporation; Redmond, WA
Lines: 48

In article <1109@bucket.UUCP> leonard@bucket.UUCP (Leonard Erickson) writes:
>In article <235494b5@ralf> Ralf.Brown@B.GP.CS.CMU.EDU writes:

><Your experience must be exclusively with systems using extremely slow logic
><chips.  My 9.8/1 AT clone works just fine with 150ns chips....
>
>You should have paid more attention to what I posted. *my* example was a 10 
>Mhz *XT*, not an AT. And II've checked with some others. 10Mhz XT motherboards
>WILL NOT work with 150 ns RAM. 
>

I have a 10 Mhz XT MB that works just fine with 150 ns RAM. I didn't check
out the math to figure out where the error in caclulation is, but I know
from years of experience that 10 Mhz (AT *OR* XT) should require < 120ns
access times. Those same years of experience bring to mind the obvious
solution to the question of how some machines can work at 10 Mhz using
150ns chips.
      It's called "wait states" folks! For those of you new to the term,
it involves tugging on the "ready" line to the CPU which tells it to wait
for slow memory to catch up. That is why my 10 Mhz clone runs with 150ns
chips. It is also why it only gets a Norton SI of 1.9 when obviously:
10Mhz > 4.77Mhz * 2

      Back in the good old days when you bought a memory board it had a
jumper on it which allowed you to select 0, 1, 2, or sometimes more, wait
states for that board. This allowed you to select the ram to match your
budget, while still driving the CPU as fast as possible. PC clone board
manufacturers seem to have dropped that, the probably figured we weren't
smart enough to select the correct jumper.

PS. I don't have the original article referred to, but I dug this up:

>In article <235494b5@ralf> Ralf.Brown@B.GP.CS.CMU.EDU writes:
><}<and for a 10/1 AT
><}<
><}<        1000/10 * 3 / 2 = 100 * 3 / 2 = 150ns
><

Certainly the arithmetic is correct here, but I don't see how this bears
any resemblance at all to the timing data found in the Intel data book.
Access time is not a simple function of the clock cycle time.
There are such things as set up times and data sample times that must be
observed. It is left as an exercise for the interested reader to calculate
the actual access time required.

-- 
del (Erik Lindberg) 
uw-beaver!tikal!pilchuck!del