It depends on how wide the data buses are on each chip, and how they're connected. If they're one byte wide, you could need over 256 million addresses, one for each byte. if they're wider, and connected to show an even wider combined data bus, it could be much less; around 32 million.
The No.of Address Lines for a Memory Chip depends on the Size of the memory chip.
1KB memory chip will have 10 address line(i.e.,2^10=1KB)...
If the chip is SRAM it will take 11 address lines to select a memory location.
If the chip is DRAM it will take 6 multiplexed address lines (row/column) to select a memory location.
11. 211 = 2048.
212 = 4096
11
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
The memory address space is 64 MB, which means 226. However, each word is 4 bytes, which means that you have 224 words. This means you need log2 224 or 24 bits, to address each word.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
Yes and no. All memory location from 0H to 0FFFFH are addressable, but some of them are needed for the program, interrupt vectors, and the stack, so you would need to pay attention to where things are located in memory to design an appropriate program. In addition, if your system is using memory mapped I/O, some locations will be reserved.
if we have 10 address bits then we can have a memory module of 2^10 = 1024 b = 1kb so for 16kb we need 16*2^10 = 2^14=16kb here we need 2 chips as 1 chip only provid us 2^12 memory. Address and data bus are multiplexed to reduce complexity.
To implement array data structure, memory bytes must be reserved and the accessing functions must be coded. In case of linear arrays, the declaration statements tell how many cells are needed to store the array. The following characteristics of the array are used to calculate the number of cells needed and to find the location or address of any element of the array.1. The upper bound (UB) of the index range.2. The lower bound (LB) of the index range. In C/C++, LB is zero.3. The location in memory of the first byte in the array, called base address of the array (Base)4. The number of memory bytes needed for each cell containing one data element in the array (size, denoted by W)By cell we mean a unit of memory bytes that will be assigned to hold a value of respective data element.During the compilation of the program, the information about characteristics of the array is stored in a table called DOPE VECTOR. When compiler comes across references to an array element, it uses this information that will calculate the element's location in memory at runtime.
No. Any computer needs memory - among other things, it needs RAM to store data temporarily. In the case of computers such as laptops, the memory needed is normally in a separate card - not built into the machine.