an opcode (operation code) is the portion of a machine language instruction that specifies the operation to be performed. Their specification and format are laid out in the instruction set architecture of the processor in question (which may be a general CPU or a more specialized processing unit). Apart from the opcode itself, an instruction normally also has one or more specifiers foroperands (i.e. data) on which the operation should act, although some operations may have implicit operands, or none at all. There are instruction sets with nearly uniform fields for opcode and operand specifiers, as well as others (the x86architecture for instance) with a more complicated, varied length structure.by: HerLoyd
A code that leaves a spare bit to indicate that if that bit is set, consider this byte and the next byte to be defining the entire code.
· 1)Syntax errors like missing commas· 2)Invalid opcode· 3)Duplicate definition of symbol· 4)Undefined symbol· 5)Missing START· 6)Missing END· 7)Symbol defined
Symbols to represent instructions Each instruction (opcode) is given some mnemonic name Various addressing modes might select different opcodes for same mnemonic Symbols to represent labels and variables The addresses where variables are stored can be named Addresses in the code can be labelled
1. FI (fetch instruction) - get the next instruction 2. DI (decode instruction) - decode the opcode and operands 3. CO (calculate operands) - calculate EA of the operands 4. FO (fetch operands) - fetch operands from memory (not necessary for register data) 5. EI (execute instruction) - execute instruction storing result if necessary 6. WO (write operand) - write the result in MEM
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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.
3 for opcode fetch, 1 for opcode decode, 3 for operand fetch, and 3 for opcode store, for a total of 10, not including wait states.
Summary − So this instruction XCHG requires 1-Byte, 4-Machine Cycles (Opcode Fetch) and 4 T-States for execution as shown in the timing diagram.
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The STA instruction in the 8085 has 13 T states, 4 for opcode fetch, 3 for low half immediate address fetch, 3 for high half immediate address fetch, and 3 for data write.
The OUT instruction on the 8085 uses 10 T cycles, 3 for opcode fetch, 1 for opcode decode, 3 for port address fetch, and 3 for port data store. Any wait states encountered are above and beyond that.
It will require 4 machine cycles, 1.opcode fetch 2.memory read 3. memory read 4. memory write
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.
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
An opcode is an instruction. An operand is information used by the opcode. Not all opcodes require operands.
The MOV A,B instruction requires 1 machine cycle and 4 T-states, 3 to fetch the opcode, and 1 to decode/execute it.