Microcomputer Circuits and Processes

Look again at figure 2.4. You will see that the buffers' enable
terminals are connected to wires labelled 'control', which must ultimately
be connected to the CPU. The CPU will decide when either
device will be enabled on to the bus. As you continue reading this
chapter, you will realize that the CPU has a lot of controlling to do, and
its control connections to the various devices on the bus are therefore
sent down the control bus. You will meet the control signals on this bus
one by one.
The situation so far is shown in figure 2.5. We have not given a name
to the top bus; think of it for the moment as an 'information' bus used to
shunt numbers to and fro. So how many wires should be used in this
bus, or put another way, how 'wide' should it be? Well, you know from
binary arithmetic that to write the numbers 0 to 10 in binary, you need
four binary digits or bits. For example 2 10 = 0010 2 and 9 10 is 1001 2 .* So
if we only wanted to send numbers from 0 to 10 down the bus, 4 wires
would do. So it was in the days of the 4-bit 4004. But if you want to send
English characters and words down the bus then you need a wider bus.
Taking 8 bits gives at once 2 8 = 256 possible symbols. That sounds a bit
more like it, but there is a bit more to it than that, as we will now see
when we look at computer memory.
,...--------,
information
bus
CPU
control
bus
D
television
screen
printer
Figure 2.5
Addition of control bus provides signals to enable the buffers.
A FIRST LOOK AT MEMORY
The basic requirement of memory is to be able to write some
information into it, leave it there, and come back later to read it. This
can be done with floppy disks, tape, or pencil and paper, but here we are
concerned with the type of memory all computers have, semiconductor
memory. The basic unit of memory is a single storage cell which may
hold binary 0 or binary 1. We will not worry about what electronics
could be in the cell just yet - you probably know one arrangement of
gates or flip-flops that could do the job. It must be possible to read and
* 2 10 (read as 'two base ten') means 2 in our common base-IO counting system. 0010 2 (read
as 'one zero base two') = (l X 2 1 ) + (0 x 2°) = 2 10 • And 1001 2 is (1 X 2 3 ) + (0 X 2 2 ) +
(Ox2 1 )+(l X 2°)=9 10 •
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