I am currently working on a reverse engineering project. I am checking out an SDRAM data sheet and it says that the memory chip is organized as 2,097,152 Words × 4 banks × 16 bits
When these values are multiplied the result is: 134217728.
Does the result mean that the SDRAM can store 134217728 bytes?
Another way I thought of it is that, the Chip contains 2,097,152 Bytes x 4 banks, which are 16 Bit Aligned.
Does the result mean that the SDRAM can store 134217728 bytes?
No, that's bits, not bytes. And your brain will find it much easier to work in 'Kilo', or 'Mega'.
134217728 bits = 128 Megabits (Mb) = 16 Megabytes (MB)
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This is just basic theoretical question. so I read that a bit consist of 0 or 1. and a byte consists of 8 bits. and in 8 bit we can store 2^8 nos.
similarly in 10 bits we store 2^10 (1024). but then why do we say that 1024 is 1 kilo bytes, its actually 10 bits which just 1.25 byte to be exact.
please share some knowledge on it
just a concrete explanation.
Bit means like there are 8 bits in 1 byte, bit is the smallest unit of any storage or you can say the system and 8 bits sums up to 1 byte.
A bit, short for binary digit, is the smallest unit of measurement used in computers for information storage. A bit is represented by a 1 or a 0 with the value true or false, also known as on or off. A single byte of information, also known as an octet, is made up of eight bits. The size, or amount of information stored, distinguishes a bit from a byte.
A kilobit is 1,000 bits, but it is designated as 1024 bits in the binary system due to the amount of space required to store a kilobit using common operating systems and storage schemes. Most people, however, think of kilo as referring to 1,000 in order to remember what a kilobit is. A kilobyte then, would be 1,000 bytes.
Assume the 16 bit no. to be 256.
So,
byte 1 = Some binary no.
byte 2 = Some binary no.
But byte 1 also represents a 8 bit no.(Which could be an independent decimal number) and so does byte 2..
So how does the processor know that bytes 1,2 represent a single no. 256 and not two separate numbers
The processor would need to have another long type for that. I guess you could implement a software equivalent, but for the processor, these two bytes would still have individual values.
The processor could also have a special integer representation and machine instructions that handle these numbers. For example, most modern machines nowadays use twos-complement integers to represent negative numbers. In twos-complement, the most significant bit is used to differentiate negative numbers. So a twos-complement 8-bit integer can have a range of -128 (1000 0000) to 127 (0111 111).
You could easily have the most significant bit mean something else, so for example, when MSB is 0 we have integers from 0 (0000 0000) to 127 (0111 1111); when MSB is 1 we have integers from 256 (1000 0000) to 256 + 127 (1111 1111). Whether this is efficient or good architecture is another history.
In 32-bit systems, 2^32 different address can be generate. Each address' size is 32 bit. So 2^32 x 32 gives us maximum memory size (bit). But it's equal with 16 GB. I don't understand. This must be 4 GB but not?
Because 2^32 = 4294967296 = 4 GiB.
Suppose that I have a computer with an address register of size 16 bits (MAR, for example). The smallest addressable unit in this computer is a word and each word is of size 2 bytes. What is the maximum memory size (in bytes) this system can support?
I thought it would be 2^16 = 65536 bytes, but the part about the smallest addressable unit implies that this is not the way to solve it.
Thanks in advance
There is no direct correlation to the maximum amount of memory a system can support, and the size of address registers.
16bit computers 30 years ago could very well support more than 64 kilobytes. On the other hand, modern 64bit processors typilcally only have lanes for 52 bits (or less), but even so a typical computer cannot nearly support 2^52 bytes of memory.
Typical 64bit computers today could in theory address 16 exibytes, but present-time CPUs only support 4 petabytes of phyisical and 256 terabytes of per-process virtual memory. Typical desktop mainboards support 128GiB maximum, if you buy extra expensive DIMMS. With affordable DIMMS, you're limited to about half as much (there are only so and so many slots).
Operating systems typically allow for main memory sizes in the hundreds of gigabytes only (e.g. 512 GiB for Windows 8 enterprise/professional, and 128GiB otherwise, or as little as 16GiB for Windows 7 Home Premium)
Generally the smallest addressable size is one byte, as you have calculated it, if it were one byte it would be 2^16*1 = 65536 bytes. However, because on this system there are two bytes per address, it is actually 2^16*2 = 131072 bytes.
Does anyone know what aspect ratio mean in memories? and how it's different in block ram and distributed ram in Xilinx FPGA?
Thanks
"Aspect ratio" refers to the number of address bits and data bits when accessing memory.
For example, say you have memory that holds 65,536 bits. If you use 8 data bits per address, you have 8,192 addresses (65,536 / 8 = 8,192), which means 13 address bits (2^13 = 8,192). So one aspect ratio for accessing 65,536 bits is 8 data bits and 13 address bits.
But say you want 16 data bits per address. Then you can only have 4,096 addresses since 4,096 * 16 is 65,536, and that's all the memory you have. In that case, you can only have 12 address bits. So another aspect ratio for accessing 65,536 bits of memory is 16 data bits and 12 address bits.
Aspect Ratio is another term that is used for the data organization of a RAM block.
In Xilinx FPGAs, a Block RAM is a dedicated two-port memory containing several kilobits of RAM. The FPGA contains several (or many) of these blocks.
Inside of each small logic block is a configurable lookup table. It is normally used for logic functions, but you can reconfigure it as a few bits of RAM. You can combine several (or many) of them into a larger RAM. This is distributed RAM.
Both types of RAM can be initialized with data, or used as ROM.
More Information can be found here:
http://www.xilinx.com/support/index.htm#nav=sd-nav-link-182711&tab=tab-sd