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The instruction opcode is a type of data contained in memory, pointed to by the PC (Program Counter) register.
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What is opcode fetch?
The process of transferring instruction codes from memory location to instruction queue register is called opcode fetch.
What is the necessity to use opcode fetch cycle for every instruction?
The microprocessor uses an opcode fetch cycle for every instruction because it has to know the opcode in order to execute it, and that is located in memory.
Difference between LDA and STA instruction in 8085?
In the 8085, the LDA instruction loads the accumulator from memory, while the STA instruction stores the accumulator to memory. LDA is a read, while STA is a write. LDA is opcode 3AH, while STA is opcode 32H.
How does microprocessor differentiate between data and instruction code?
i) Instruction code deals only with mnemonics and its corresponding opcode but data code refers to your data like 10h which is always of 8 bits or a particular address say 8080h which is of 16 bits. ii) Data is your input to the instruction but an opcode is native to your machine. iii) Data is user specific instruction while opcode is machine specific instruction iv) You can alter data code but you cannot modify an instruction opcode.
What is the difference between LDA and STA instructions?
In the 8085, the LDA instruction loads the accumulator from memory, while the STA instructionstores the accumulator to memory. LDA is a read, while STA is a write. LDA is opcode 3AH, while STA is opcode 32H.
What happens to ip after completion of fetch of instruction code?
IP is incremented after fetch of instruction opcode. Specifically, IP is incremented by the number of opcode bytes.
Is INTR is a non vectored interrupt?
Yes and no. INTR response requires an opcode fetch sequence from the interrupting hardware device, often a CALL instruction, so there is no vector table in memory for it, because you can CALL any location. On the other hand, some implementations provide an RST instruction as the opcode, making it a vectored interrupt.
What are operands?
Every instruction contains to parts: operation code[opcode],and operand. The first part of an instruction which specifies the task to be performed by the computer is called opcode. The second part of the instruction is the data to be operated on.,and it is called operand. The operand[or data]given in the instruction may be in various forms such as 8-bit or 16-bit data, 8-bit or 16-bit address, internal register or a register or memory location.
What are the difference between opcode and operand?
What is difference between oppress code and operend
What is instruction decode and operand fetch?
The opcode fetched from the memory is being decoded for the next steps and moved to the appropriate registers. Fetch operands from memory if necessary: If any operands are memory addresses, initiate memory read cycles to read them into CPU registers.
How 8085 timing diagram works based on each instruction?
Each instruction requires specific time for the execution of instruction and this time is called instruction cycle. Each instruction cycle consists 1 to 5 machine cycle -- opcode fetch, memory read, memory write, IO read, IO write and each machine cycle consist 3 to 6 T - states. Time required to execute 1 T-state = 1/ operating frequency of 8085 Microprocessor for example operating frequency = 2MHz then time required to execute 1 T-state = 0.5 uSec example: Calculate time required to execute instruction MOV C, A sol: This instruction has one machine cycle i.e. opcode fetch (In any instruction 1st cycle is always opcode fetch and opcode fetch consists 4 to 6 T state depend on the operation of particular instruction) so to execute MOV C, A required 4T states so time required to execute this instruction is 4*0.5usec = 2usec any other queries pls contect: nileshbahadure2000@yahoo.co.in example:Calculate the time required to execute LXI H,2000H sol:Here we have to draw opcode fetch and two memory reads as two bytes 00H and 20H have to be read from memory. i.e, opcode fetch+Memory reads *2(bytes address) =4+3+3 so to execute LXI H,2000H,the required T-states is 10T and time is 10*0.5usec=5usec
How an assembler converts assembly mnemonics to machine language?
Each mnemonic maps directly to a machine instruction code, known as an opcode. Some mnemonics map to more than one opcode, however the instruction's operand types will determine which specific opcode will be generated.
