The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
There are 24 address lines required for 16 mb. That covers 12 mb. The next step down is 23 address lines, which is 8 mb. The 8085 and 8086/8088 cannot address 12 mb. Only the 80286 and higher can.
There are 20 address lines and 16 data lines in the 8086 microprocessor. The low order 16 address lines are multiplexed with the data lines. Some of the high order address lines are multiplexed with status lines.
There are eight datalines, D0 through D7, in the 8085 microprocessor. They are shared, or multiplexed with the eight low order address lines, A0 through A7, and are called AD0 through AD7 on the pinout drawing.
The highest memory address in the 8086/8088 is FFFFFH.
The address bus in the 8085 is 16 bits wide.
If you are addressing bytes, then 512K words (16 bit words) requires 20 address lines.I gave that answer because the question was categorized 8086/8088. If you are addressing words, then the answer is 19 address lines.
Microprocessor has 16 address lines and microcontroller has 20 address lines
The 8086/8088 has an internal 20-bit address bus and 16-bit data bus. Externally, the address bus is 20-bits, and the data bus is 16-bits for the 8086 and 8-bits for the 8088.The data bus in the 8086 is 16 bits in size, while the address bus is 20.
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.
In the 8085, the high order address is A8-A15. In the 8086/8088, the high order address is A8-A19. (For a 16-bit address, the answer is A8-A15, but the answer above reflects the chosen categories, 8085 and 8086/8088, with the 8086/8088 running in 20-bit mode.) In Windows XP, running in 32-bit mode, the high order address is A8-A31, a 32 bit address.