Electronic – Memory interfacing with 8086

I put on my teacher/professor hat.

• your text mentions an 8Kb RAM, your diagram shows 16Kb.

• your ROM has 32 locations, hence it needs more than the 2 address lines you connect to it. Even if you intend to connect some lines to a fixed value, you should how that.

• the question explicitly asks for the PROM content, which you don't give. Yes, it is a 32x8 table of 1's and 0's, but the point is which value you put at each location! You won't need to specify all of the 32x8 locations.

I suggest you draw yourself a memory map. The smallest item you must map is 8Kb, so how many blocks will it have to cover 64Kb? How many address lines are involved in selecting one block? If less than the number required by the ROM, what are you going to do with the others? For each block, note the value of those address lines, and the value of the two select output bits. Now it is almost trivial to write out the content of the PROM. It will contain a lot of "don't care" bits.

Side note: only archaeologists make such systems from separate chips these days, and even micro*controllers* are switching from 8 bit to 32 bit. But the design of something like this is still a good test of your general knowledge of digital logic.

The 8085 has different instructions for accessing main memory and I/O 'memory'. In addition to the standard memory interface pins the 8085 also provides a pin that identifies whether a memory access cycle is accessing main or I/O. This extra line is used in the select logic of both main and I/O 'memory'.

But there is no law that I/O can only be accessed by I/O instructions: in a small system the highest address line could be used to distinguish between memory (a15=0) and I/O (a15=1), so we get 32Kb for real memory (ROM and RAM) and 32 Kb for I/O.

Note that it is even possible to use the I/O addresses to access RAM, but that is less useful because there are only 256 I/O addresses and the addressing modes available for these addresses are very limited.