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How many address lines are necessary to address two megabytes of memory?
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11
How many address lines are needed to address each machine location in a 2048 X 4 memory chip?
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.
What is the size of RAM that 8086 can address?
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).
If a memory had 16 bit address bus and a 32 bit data bus what is the largest size of the memory?
A memory with a 16 bit address bus can address 216 or 65536 distinct items. If each item is 32 bits in size, then the item is 4 bytes. The size of this memory is then 262144 bytes. (256Kb)
How many address lines are necessary to access 64K x 8 memory system using 8K x 8 memory chips?
14 lines can address 16,384 locations. The 8085, however, has 16 lines, and can address 65,536 locations.The system design, of course, may limit that to 14, so 16,384 is the answer in that case.
How do you find the memory size of n bit address line processor or controller?
2N
What size memory card for GameCube needed?
You don't need a specific size memory card to play the GameCube; any GameCube memory card will do. Different games take up a different amount of blocks on a memory card, so what size you'll need depends on what games you want to play.
Diagram to show the address mapping of RAM and Cache?
consider a RAM of 64 words with a size of 16 bits.Assume that this memory have a cache memory of 8 Blocks with block size of 32 bits.Draw a diagram to show the address mapping of RAM and Cache, if 4-way set associative memory scheme is used.
How much maximum memory size that 8085 can access?
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.
How memory space is occupied by union variables in C programming language?
A union is an aggregate of members that share the same memory address. The size of a union is determined by the largest member.
How are elements of an array stored in memory?
Computer memory is linear so a one dimensional array can be mapped on to the memory cells in rather straight forward manner.To find the actual address of an element one needs to subtract one from the position of the desired entry and then add the result to the address of the the first cell in the sequence.Having said that therefore it is necessary to know the starting address of the space allocated to the array and the size of the each element, which is same for all the elements of an array.The the location of the Ith element would be B+I*S where B is the base address(Starting address of the array) and S is the size of each element of the array.
How many address bits would be required for a memory that stores 64 4-bit words?
It requires 6 bits to address 64 words. It does not matter what the word size is.
