2kb=2*1024=2048
2^11=2048
therefore 11 address lines are required
2kb of memory requires 11 address lines. (211 = 2048)
The time required by a processor to access data or to write data from and to memory chip is referred as access time.
The time required by a processor to access data or to write data from and to memory chip is referred as access time.
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.
Max. memory address space= 216 X 2 bytes = 128 Kbytes
32 bit processor can access 4294967296 bit memory adderss.
Memory is microchip; address are processor board slots
8088 processor accessed 1MB
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
microprocessor can access 2^8 points which is 256 then we have 8 bit memory = 1 bytes then 1*256 =256 bytes
ransom access memory which is permenant
1 The processor puts the required addresses on to the address bus 2 Any addresses that invoke chip select are decoded 3 Chip select is generated 4 The processor waits for memory to settle 5 The processor generates a memory write control bit (MEMW) 6 The processor puts the data on to the data bus 7 The contents are written to a specific location on memory.
RAM (Random-access memory)