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.
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The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
Since DRAM uses address multiplexing(essentially each line accesses both columns and rows in memory) you need to take 2 to the 24th power to find your total capacity(basically doubling the number of address lines). In this case it would be 16,777,216.
18
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
To calculate the number of address lines required for a 64 kB memory, first convert 64 kB into bytes: 64 kB = 64 × 1024 bytes = 65,536 bytes. The number of address lines needed can be determined using the formula (2^n = \text{total number of addresses}), where (n) is the number of address lines. Since 65,536 is (2^{16}), you need 16 address lines to address a 64 kB memory.
14*8=112 bits
Microprocessor has 16 address lines and microcontroller has 20 address lines
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
a return address should have 2 lines
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.
20 address lines are required
In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.