The Instruction Pointer (IP) in an 8086 microprocessor contains the address of the next instruction to be executed. The processor uses IP to request memory data from the Bus Interface Unit, and then increments it by the size of the instruction.
What is the need of segments in 8086 micro-processor? Explain how the address of an instruction is calculated in 8086 using segment register
To increase the speed of the 8086, you need to increase the clock speed, reduce the number of wait states, or both. You could also optimize your code so that it runs faster. Since the 8086 is a segmented memory architecture, it is more efficient to use operands in one segment and to make near references to them.
The need for a timing diagram for a microprocessor is, primarily, to allow you to properly design the interface logic that will support the microprocessor. You need to know what lines are used to execute various data transfers, what are the timing of those lines with respect to each other, and how the microprocessor expects the external logic to behave. You can also use the timing diagram to understand how the microprocessor functions and, particularly, to know how long each instruction will take.
first connect the 8255 to the 8086 microprocessor and configure it using the CWR then connect the key board to the 8255
There are many instructions in the 8085. To find the hex code for a particular instruction, you need to look at the data sheet. For more information, please see the Related Link below.
The NOP instruction is a no-operation instruction. It does nothing to the state of the machine, except to use some time. In the case of the 8085, it uses four clock cycles plus however many wait states are need to access the NOP instruction from memory.
You need an 8086 assembly language pencil.
A microprocessor doesn't usually have memory and interfaces to perpherals, so these need to be added.
To exchange two registers, say the BX and CX registers, in the 8086 using the stack, you can use...PUSH BXPUSH CXPOP BXPOP CX... Of course, this is for 16 bit operation. If you want 8 bit operation, you will need to do more than that, because stack operations are always 16-bit operations.
the principle function of memory interfacing is to enable the microprocessor to read or write into a register of the memory chip
There are ground pins on a microprocessor chip for the same reason there are ground pins on any kind of chip - to provide a current sink path for gates that need to pull to ground. If you are asking why there are two ground pins on some processors such as the 8086/8088, the answer is that one ground pin is not enough - that if all gates pulled to ground at the same time, the current transient would destabilize the processor - so two were provided.
No, you need a real teacher, and a textbook.