Intel 8086 and 8088

How many address lines are there in a 16 to 1 multiplexer?


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2011-12-15 14:38:03
2011-12-15 14:38:03

20 address line available in 16 to 1 multiplexer 16 for input lines and 4 will be selection lines.

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Microprocessor has 16 address lines and microcontroller has 20 address lines

16 address lines can address 216 or 65536 different locations.

In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..

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.

for 16 MB memory has 24 address lines

16 bit address lines, so it address up 2^16.

16 address line are multiplexing

There are 16 address lines (8 shared by the data bus), and 8 data lines in the 8085.

It takes 16 address lines to address 64 KB of memory. (216 = 65536)

8 address lines can address 28 or 256 different memory locations.

With 8 address lines, you can choose between 256 (28) locations in memory. The 8085 has 16 address lines, so it can address 65536 (216) locations.

the 8085 microprocessor is a 8-bit microprocessor and these are bidirectional but the address lines are unidirectional.these address lines are used to address the location of the instruction in memory .these data lines are used to transfer data between processor and peripheral devices. when the address of the instruction will be recognized by the address lines the data will be send to the processor therefore the 16 address lines are not act as a data lines in 8085

you require 16 input line and 16 output line and 16 address line. Because 64 K = 26 X 210 = 216 so, 16 address lines Here N = 16, so 16 data lines will be there. .

A multiplexer will have 2n inputs, n selection lines and 1 output. A 16 input multiplexer accepts 16 inputs i. e. 24 and requires 4 selection lines. An 8 input multiplexer accepts 8 inputs i. e. 23. And it needs 3 selection lines. To realize a 16:1 multiplexer, two 8:1 multiplexers are required. They provide 16 inputs (8+8). Join the three selection lines of each MUX. Now we require 16 combinations from selection lines. i. e. 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111 ------------------------------------------------ 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 We know that 000, 001, 010, 011, 100, 101, 110,111 are common. Only the first bit differs (0 or 1). Hence, apply the fourth selection line as it is (i. e. 1) to upper 8:1 MUX and apply it complimented (i. e. 0) to lower MUX. Now it acts as 16:1 MUX.

Theoretically you use five 4-to-1 multiplexers. You use four of them to connect the 16 inputs. You then have four outputs. Take the fifth multiplexer and connect the four outputs as the inputs. The fifth multiplexer then has a single output that has multiplexed the original 16 inputs.

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.

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.

If an address bus for a given computer has 16 lines, what is the maximum amount of memory it can access?

segment is for converting physical address to logical address , here on taking 8086 microprocessor as example, we have 20 address lines but it is capable of taking only 16 address lines.... so to convert that 20 into 16 segment is used....

It should have a 16 bit data bus allowing it access from 0 - 65535 memory locations.

14 lines can address 16,384 locations. The 8085, however, has 16 lines, and can address 65,536 locations.The system design, of course, may limit that to 14, so 16,384 is the answer in that case.

00 to 0F represents 16 addresses Binary representation is 1111 therefore you need 4 address lines to connect to all addresses

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