Therefore it is a 32KB SRAM chip? (8192words * 4bytes_in_a_word = 32768 bytes. 32768bytes / 1024 = 32KB.) Then the chip is it is a 32kb x 8bit SRAM, right?
Therefore it is a 32KB SRAM chip? (8192words * 4bytes_in_a_word = 32768 bytes. 32768bytes / 1024 = 32KB.) Then the chip is it is a 32kb x 8bit SRAM, right?
I think the technical definition of a word is two bytes (that's what we learned in class when working the 68000). HOWEVER it seems to vary from chip to chip what is considered a word and sometimes it is 4.
Depends on the architecture. Different processors have different word lengths. Intel x86 used 16 bit words; MIPS and PowerPC used 32 bit words, etc. Some architectures have used variable-length words.
I've never used one, but I'd read that to mean it can store 8192 8-bit words, which would also match the 256 X 256 bit storage array shown in the block diagram.
I checked wikipedia for an article on a word, but it never came up. How did you get that?
I never use Wikipedia's search engine; I just type "<search term> wikipedia" into my Google box in Firefox. If I put in "word length wikipedia" it comes right up. Just putting in "word wikipedia" isn't too helpful though. Sometimes it takes a couple of tries to find terms which work well.
Your memory chip is a 64k (actually 65,536) bit chip. It is arranged as 8K x 8. So, it should have 8 data pins and 12 address pins. If you connected it to a 16 bit processor then you would gain 4k words of memory. Connect it to a 32 bit one and you would add 2K words etc.
I looked up "Byte", "Bit", "Nibble", "Binary Number System", and the like, but none the the pages had "Word" as a "See also". I also did the disambiguation of "word", but nothing came up for "computer word"...
Oh well. I now know what i need to know for the moment...
I looked up "Byte", "Bit", "Nibble", "Binary Number System", and the like, but none the the pages had "Word" as a "See also". I also did the disambiguation of "word", but nothing came up for "computer word"...
Oh well. I now know what i need to know for the moment...
Bit (one bit)
Byte (8 bits)
Word (16 bits - two bytes)
Longword (32 bits - four bytes)
Basically it comes from the fact that microprocessors tended to be 8 bit, then 16, then 32.
But you really have to UNDERSTAND what you're dealing with, for example a 'word' in a PIC 16F chip is only 14 bits - because that's actually the width of the memory, and some of the lower (and older chips) are only 12 bit.
No, it comes from pretty well any microprocessor (not just x86, which were pretty late comers) - the majority of which were 8 bit, but even 16 and 32 bit ones use the same, because they still handle a lot of data as 8 bit.
Going back before microprocessors, Octal was common instead of binary, because of the 'word' sizes used - but with the advent of 8 bit microprocessors Octal 'disappeared' and Hexadecimal took it's place, as much more suitable for the 8 bit data bus and 16 address bus that was common.
I generally associate the word size of a machine, with a Princeton architecture, with the width of the ALU pathway. This has an effect on the width of the pathway to and from memory. Prior to the existance of semiconductor memories word sizes that were not a multiple of 8 were common.
In a machine with a Harvard architecture, like the PIC, the definition of word size has at least two obvious answers. Most technical people are comfortable with with using the width of the ALU data path.
Marketing types, given a choice, will use the largest plausible number which in the case of the PIC would be the size of the instruction word.
Yes, although I think it makes far more sense to consider it an 8k one, as you don't use bits of memory, you use bytes - you could use 8 of those to give you a full 64K on your Z80!
MicroComputer: In Z80 terms it's an 8K SRAM. Your Z80 can address a total of 64K bytes total.
MicroController: EEPROMs (SPI & I2C) are generally referred to by total bits 1M = 64K bytes or 32K words BUT most microcontroller ROM / RAM are rated by bytes still.
Blame the marketing people.