Re: address decoders

Azuma Hashimoto (azuma_h@hotmail.com) Thu, 07 Dec 2000 14:05:28 -0800 

From: "Azuma Hashimoto" <azuma_h@hotmail.com>
Subject: Re: address decoders
Date: Thu, 07 Dec 2000 14:05:28 -0800



Thanks,
do you know what the two LSB are being used for?

>The two missing bits are the two LSB, which will always be active, no
>matter what device (RAM, I/O, etc.) is being used.  The I/O was 2 bits
>(the smallest of any of the devices), and the CPU 20, so 20-2=18
>
>Eric
>
>On Thu, 7 Dec 2000, Azuma Hashimoto wrote:
>
> > Hi.  I have a question about exercise 7 from the microcomputer 
>architecture
> > handout.  In it we are supposed to draw out a diagram for a 
>microcomputer.
> > But when you did it in class the address decoder only gets an 18 bit
> > address?
> > Why is that?  Is it because the I/O chip could always potentially be 
>active
> > with any other activity?  So that when you have the EPROM and I/O active 
>at
> > the same time the address decoder needs
> > 16(for EPROM) + 2(for I/O) = 18 bits?????
> > 
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