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....
mujhe nahi aata
There is insufficient information in the question to properly answer it. Which Motorola microprocessor are you talking about? Please restate the question.
Because the processor is a 16 bit processor, and 64k is what you can address with a 16 bit processor.
There are four segment registers in the 8086/8088, Code Segment (CS), Stack Segment (SS), Data Segment (DS), and Extra Segment (ES). As a result, there are four segments that can be directly addressed at a particular time, i.e. without an extra instruction to reload a segment register.
A 7 segment display is an I/O device, and it is not necessarily used in microprocessors. The choice of I/O devices is a function of system design, not microprocessor design.
In the 8086/8088 microprocessor, the code segment is used to fetch the opcode and any additional instruction bytes that might be part of the instruction, while the data segment is used to fetch and/or store any operand bytes that the instruction requires to be manipulated.This is in the case of no segment override prefix.
In the 8086 microprocessor, an intra-segment jump refers to a jump instruction that occurs within the same segment of memory, allowing the program to branch to a different address within that segment. Conversely, an inter-segment jump involves jumping to a different segment, which requires specifying both the segment and offset addresses to access the target location. Intra-segment jumps are typically faster since they do not require segment register changes, while inter-segment jumps can introduce additional overhead due to the need for segment management.
A seven segment[E1]consist of seven light -emitting diode segments and one segment for the decimal point.The LEDs are physically arranged.To display a number, the necesssary segments are lit by sending an appropriate signals for current flow to the diodes
Physical address in the 8086/8088 is {Selected Segment Register} * 16 + {Effective Offset Address}. It is a 20-bit address .
the data path between the microprocessor RAM and the peripherals along which communication travels is called the data bus.
How does a microprocessor decode?
The plural of microprocessor is microprocessors.