Does any1 know what is the maximum theoretical hard disk size that can be connected to a 32 Bit microprocessor?
And there are two mechanisms by which the micrprocessor accesses the addresses of the hard disk what are those two mechanisms..keeping in mind that we are to discuss the schemes from the microprocessor's end and not from hard disks end.....
Please Reply as Early as Possible...
I'm not quite sure that I understand your question very well, but the size of the hard drive is dependent on the architecture of the computer (e.g. motherboard, BIOS) and it is also OS and FS dependent.
Please provide more details if you want more detailed answer.
im studying the microprocessor systems course...
answer me keeping in view the architecture of a 32 bit Microprocessor e.g intel 80486....it has a data bus of 32 bit etc etc....i know that the MAX ram this processor can address will be 2^32 bytes....i.e...4GB...thats fine...My question is what is the MAX theoretical Hard Disk size that can be attached with this 32 bit Processor....and what is the mechanism by which it can address such a big hard disk .... my teacher told me that there are two mechanisms ......so i m searching for them...which ones are they?........i want an answer from a microprocessors view point like how does it deal with such a big harddisk and lengthy addresses of the hard disk memory blocks....like there storage on its registers may be..etc etc...i hope its detailed enough now.....
thanks for the articles..but i got a query in ma mind that ..if theoretically unlimited amount of hard disk space can b attached with a microprocessor..... then LBA and CHS are techniques to manage the addresses of the hard disk itself on its end???
these schemes have nothing to do with the microprocessor???
and yes then would a microprocessor pick up such long hexadecimal harddisk addresses (even larger than 32 bits long) when the registers of the microprocessors are like 32 bit ( as in this particular topic) ????
Well, where to start?
The developers are trying to answer your questions by creating new schemes for addressing and accessing the hard drives as their sizes increase every year.
The limitations are coming mostly with the requirements for compatibility with the older systems and for other historical reasons.
Let’s take a look at the simplest one C/H/S. First of all the microprocessor system never addresses a single byte/word on the hard drive. This is extremely inefficient mostly because of the speed of the hard drives.
A unit red at once is normally a sector, which size is in most cases 512 bytes. Theoretically it could be other.
The read/write operations are controlled by the disk controller which has some standard interface and provides set of commands with parameters for accessing the underlying device. So in order to read a sector from the drive you have to provide the controller with the right command and to supply the parameters to it i.e. in our case the cylinder #, the head # and the sector #. These are three numbers which theoretically could be of any size (of course in practice it won’t be the case)
Having a 32-bit microprocessor doesn’t mean that you cannot work with numbers larger than 32 bits. I’m not talking about the fact that some processors are equipped with the capability to combine two registers in one of double size to use it for some calculations and for addressing purposes. No, they are algorithms to process numbers of any bit size even with the 8-bit microprocessor. Think about it.