
A 1 of 4 decoder need not be the same as a 1 to 4 demultiplexer, although
many commercial SSI implementations allow you to use one as such. Strictly,
a 1 of 4 decoder need only take two lines in and make one output change
state, according to the inputs.

A demux, on the other hand, uses two control inputs to determine which
of four outputs will reflect the state of the input signal. So there are
three inputs required.

A decoder can be used as a demux if it is equipped with an output enable
input, since this can be used as the data input (e.g. when high, all
outputs are high; when low, only the selected (by control inputs) output
will be low).

An eight way decoder is created by using the high order bit (bit 2) to
select which of two four way demuxes is enabled. Thus you achieve your
aim of having only one output of eight reflecting the input bits. Note
that this method cannot be used to create a true eight way demux, since
you have no data input (the enable line on a four way decoder) left
once you commit the enable lines to their intended purpose.

A sixteen way decoder obviously requires four, four-way decoders, plus
a mechanism to enable only one of the four at a time. Therefore, use
the fifth decoder, attached to the two high order bits, to provide the
four enable lines.

Of course, the two low order bits must be connected in parallel to the
four final stage decoders.

Please give me the credit when you submit your homework.

JohnH

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    Email: johnh@mpce.mq.edu.au, Ph: +61 2 805 8959, Fax: +61 2 805 8983
