Path: utzoo!attcan!uunet!husc6!rutgers!rochester!pt.cs.cmu.edu!b.gp.cs.cmu.edu!Ralf.Brown@B.GP.CS.CMU.EDU
From: Ralf.Brown@B.GP.CS.CMU.EDU
Newsgroups: comp.sys.ibm.pc
Subject: Re: RAM speed per clock rate
Summary: don't follow that table, please!
Message-ID: <22faf142@ralf>
Date: 6 Aug 88 12:45:06 GMT
Sender: netnews@pt.cs.cmu.edu
Lines: 33
In-Reply-To: <3075@tekig4.TEK.COM>

In article <3075@tekig4.TEK.COM>, mstevens@tekig4.TEK.COM (Michael Stevens) writes:
}There have been numerous requests for info on how fast a RAM chip is
}needed for a clock rate of X.  Generally, the equation is (assuming 0
}wait state operation is desired):
}
}        1 / (Clock Speed in Hertz) = RAM speed in nanoSeconds
Only on 0ws 80286 or 80386 machines.  The actual formula is
        1/clock speed * memory_access_clocks / 2 = RAM speed in ns
where memory_access_clock is 4 for 8088, and 2+ws for 80286/386.

}     4.77Mhz clock:     1 /  4,770,000 = 210 nS or faster parts.
Actually, a 4.77MHz 8088 would let you use 400 ns parts! (if any existed)
}     6Mhz clock:        1 /  6,000,000 = 167 nS or faster parts.
}     8Mhz clock:        1 /  8,000,000 = 125 nS or faster.
}     10Mhz:             1 / 10,000,000 = 100 nS or faster.
}     12Mhz:             1 / 12,000,000 =  83 nS or faster.
}     16Mhz:             1 / 16,000,000 =  62 nS or faster.
[...]
}If your system runs with 1 wait state, it means that 2 clock cycles
}are used to access the RAM (among other things), so a 12Mhz 1 Wait state
}system can use 150ns RAMS, but to run a 12Mhz system at 0 wait states

Your system will die a very quick death trying to use 150ns chips at 12MHz,
even with one wait state!  8088s use four clocks per memory access, 80286/386
machines use two clocks plus wait states.  Dynamic RAMs require a memory
cycle time of twice the access time (due to setup and recovery times).  It's
just coincidence that the two-clock memory cycle on 286/386 machines cancels
that factor of two.
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