Re: address decoders

Eric Young (eyoung@interchange.ubc.ca) Thu, 7 Dec 2000 14:25:44 -0800 (PST) 

Date: Thu, 7 Dec 2000 14:25:44 -0800 (PST)
From: Eric Young <eyoung@interchange.ubc.ca>
Subject: Re: address decoders



They'll still be used to give the 2 LSB portion ofthe address.  It's just
that the decoder doesn't need to know them, since they don't give any
significant information about which device needs to be turned on. 

Eric

On Thu, 7 Dec 2000, Azuma Hashimoto wrote:

> 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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