Question

Is it possible for 2 segment to overlap in 8086?

Answer 1

\nIn 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. \nbut 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.. \nIn 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. \nbut 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 size is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.