Even though the 8085 is an 8 bit microprocessor, it can address 64K memory, because it has a 16 bit address bus.
2^14 memory locations. In general for n-bit address bus, its 2^n
microprocessor can access 2^8 points which is 256 then we have 8 bit memory = 1 bytes then 1*256 =256 bytes
2^14 memory locations. In general for n-bit address bus, its 2^n
The 8086/8088 microprocessor has a 20 bit address bus, so the number of memory locations it can address is 220 or 1,048,576.
Usually memory banks made up of SRAMs or DRAMs or EPROMs consist of the storage area provided on a microprocessor. For understanding how the address space of a 20 bit address line microprocessor is organised, read about address decoding for even and odd memory addressing through SRAMs and EPROMs.
in 8086 there is 20 bit address bus,so it can address 1,048,576 address. At each address we can store 8 bit address (1-byte)but if want to write a word(16-bit)into a memory segment to store data in byte form then we write the data in two consecutive memory address which are even(low) and odd(high) memory.
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
The memory capacity of the 8085 microprocessor is 64 kb because the address bus is 16 bits, and you can address 216, or 64kb, with a 16 bit address bus.
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The Intersil 6100 family consisted of a 12-bit microprocessor and a range of peripheral support and memory ICs.
32 bit address line can access 4GB of memory. As 2^10 -> 1KB; 2^20 -> 2MB; 2^30 -> 1GB and so on.... 32 bit gives (2^30) * (2^2) = 1GB * 4 = 4GB;