The 8086/8088 can address a maximum of 220, or 1,048,576, or 1 MB of 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.
16-bit memory
Max. memory address space= 216 X 2 bytes = 128 Kbytes
The 8086/8088 processor is a 16 bit processor. In a 16 bit two's complement notation, the maximum number is 0x7FFF, or 32767, while the minimum number is 0x8000, or -32768.
65536 bytes, because the 8051 family has a 16 bit external address buss.
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
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85 is a 8 bit processor,number of flags are 5 and memory capacity is 64KB while 86 is a 16 bit processor ,number of flags are 9 and memory capacity is 1 MB.The main difference between 8085 and 8086 is that 8086 uses pipelining.
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