17 address lines and 8 data lines. 2^17=128k
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.
Number lines
It lines up the data and keeps your information organized, making it easier to manipulate the data and draw conclusions.
4
12
how many interrupts in 8051
the internal RAM is for data storage and for many applications is all that is needed.
There are 16 data lines in 8086.
65536 bytes, because the 8051 family has a 16 bit external address buss.
A: Because it can only control 8 bits of data.
5 data input lines and 5 data output lines
8
In other words, how do you save temporary data such as data stored in registers r0 to r7 ? The simplest way to use on-chip RAM of the 8051 is to compile your C code with the "--model-small" small memory model option of the SDCC compiler, which tells it to put all variables in on-chip RAM of the 8051.
A 2K X 8 memory requires 11 address lines and 8 data lines
17 address lines and 8 data lines. 2^17=128k
Because its a microcontroller with an 8 bit data bus width.