It depends on the type of architecture and controller u use. It can be found in the instruction set documentation. It requires 18 cycles on the Intel 8085.
How_many_machine_cycles_require_for_call_instruction_in_8085
The CALL instruction takes 18 T states; 3 for opcode fetch, 1 for decode, 6 for address fetch, 2 for stack pointer manipulation, and 6 for writing on the stack.
These 18 T states are divided up into 5 machine cycles, opcode fetch, address fetch, address fetch, stack write, and stack write.
when conditional jump instruction is executed it has 10 m/c cycles bt when nt executed it has 7 m/c cycles....while unconditional jump instruction has 10 m/c cycles...
The Intel 8086/8088 is not a 32 bit machine. It is a 16 bit machine. It has 72 instruction types.
In 8086 the instruction queue is 6 byte long. This is because even the longest 8086 instruction is 6 byte long. Thus it is possible to prefetch even the longest instruction in the instruction set.
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It is not possible to calculate the number of machine cycles from first principles without going into design details of the CPU in question. You will need the reference book or card for the specific microprocessor model you are using. That will give you the actual number of cycles that are required for each instruction family. Note that many instructions take a variable number of cycles based on where they fetch their operands from. In the 8088 and 8086, in many cases the number of machine cycles is given in the form "4+EA". This means that you have to look up the number of cycles for a specific Effective Address, which is part of the op code, and add it to the number of cycles to execute the op code. There will be a table of machine stated for each type of Effective Address determination as well. Older machines like the 8085, 8086, and 8088 will actually have a few instructions, notably rotate and shift instructions, where the time is given as something like "4+2s". In these cases, the value "s" is the number of positions you are shifting the operand; to shift it 7 places takes 14 machine states, over and above the initial 2.
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.
6 bytes
6 bytes
To speed up the program execution, the B.I.U fetches six instructions byte ahead of time from memory and stores in FIFO registers called "queue". Since EU and B.I.U are independent, the B.I.U fetches additional instruction, while EU decodes and executes a previously fetched instruction. When the execution unit is ready for its next instruction, it simply reads the instruction from the queue. The main point here is that, fetching of the next instruction is overlapped with the execution of current instruction. This is called pipe lining pipe lining saves time required for executing instructions. For example, if a multiplication operation is to be performed with its operands and registers, the microprocessor takes 100 clock-cycles, however, without pipe lining and with pipe lining, the number of clock cycles required to multiply two operands get relatively reduced. There are many instructions in 8086 microprocessor that require huge number of clock cycles. Thus, the prefetch queue becomes a circular part in enhancing the performance of the system.
in 8086, there is instruction queue of 6 byte. It is one of the reason behind giving name. 8086 was introducing pipeline architecture.
Yes.
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