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.
0FH is a literal constant for the integer value 15. In the 8086/8088, as an opcode, it is the lock prefix, which means that no other CPU or bus master can take the bus during the execution of the instruction.
The 8086/8088 instruction queue is a buffer that holds opcode bytes that have been prefetched by the bus interface unit. This speeds up operations of the processor by helping to reduce fetch latency, i.e. to improve the probability that an opcode byte fetched by the processor is already available.
The 8086 microprocessor has 40 pins.
It is mightily referring to Microprocessor 8086 . I think you saw "8086 microprocessor". The 8086 is nothing it indicates the number of microprocessor same as Digital or analog ic's . 8086 microprocessor has 20 Address buses and 8 data buses which has 1 Mb inbuilt memory for performing several type of airthmatical and logical operation.
queue of 8086 microprocessor is 6 bits
The 8086 Microprocessor operate to require frequency that is provided by clock generator to 8086 Microprocessor and also Synchronization various component of 8086.
it is a 16 bit microprocessor & it has 40 pin .
Pins 1 and 20 in the 8086 microprocessor are (both) power and signal ground (GND).
8086 main application is to evaluate the arithmetic operations in any systems that uses 8086
Two ground pins are used in the 8086 microprocessor to increase the bus pull-down current capacity.
Which microprocessor accepts the program written for 8086 without any changes?