Found 475 Articles for 8085

In 8085 Instruction set, and specially in its logical group of instructions, we have AND, OR, XOR, NOT type of instructions. 8085 does not have instructions to perform NAND, NOR, XNOR operations directly. Now let us discuss the instructions to perform OR operations only.To perform ORing of two numbers, 8085 imposes the restriction that one of the operands must be kept in the Accumulator. The other operand can be at any one of the following possible locations −ClassificationsExamplesThe other operand can be kept in 8-bit immediate data in the instruction.ORI 43HORI FFHThe other 8-bit operand can be kept in a ... Read More

In 8085 Instruction set, and specially in its logical group of instructions, we have AND, OR, XOR, NOT type of instructions. 8085 does not have instructions to perform NAND, NOR, XNOR operations directly. Now let us discuss the instructions to perform AND operations only.To perform ANDing of two numbers, 8085 imposes the restriction that one of the operands must be kept in the Accumulator. The other operand can be at any one of the following possible locations −ClassificationsExamplesThe other operand can be kept in 8-bit immediate data in the instruction.ANI 43HANI FFHThe other 8-bit operand can be kept in a ... Read More

Let us consider we want to add two decimal numbers 38 and 45. They will be represented in BCD as 0011 1000 and 0100 0101. The addition results in 0111 1101. But the answer will be incorrect if we want to interpret this result as a BCD number. The result will be not only incorrect but also illegal as 1101, which we obtained as the last nibble in the answer is not a valid BCD number. Here, in such situations, we can use DAA to have the BCD sum as outcome. All that is required to be done is to ... Read More

In 8085 Instruction set, and specially in its arithmetic group of instructions, we have only add and subtract instructions. 8085 does not have instructions to perform multiplication or division numbers. Now let us discuss the instructions to perform addition operations only.To perform addition of two numbers, 8085 imposes the restriction that one of the operands must be kept in the Accumulator. The other operand can be at any one of the following possible locationsClassificationsExamplesThe other operand can be kept in 8-bit immediate data in the instruction.ADI 43HADI FFHThe other 8-bit operand can be kept in a memory location and whose ... Read More

INTEL 8085 has a very enriched Instruction Set. Varieties of instructions it can execute. Instructions will have different Byte counts, ranging from 1-Byte to 3-Bytes. Opcode always occupies 1-Byte in the memory. As we know that, with 8 bits for the opcode, 28 = 256 distinct opcodes are possible. In hexadecimal notation, the opcodes can range from 00H to FFH. Each opcode will correspond to an instruction. Thus from the calculation, it is possible to have 256 instructions in the instruction set of 8085. However, only 246 opcodes are implemented in 8085. They can be classified under 66 types, which ... Read More

INTEL 8085 has a very enriched Instruction Set. Varieties of instructions it can execute. All these instructions can be broadly classified as follows −DescriptionNo. of opcodesNo. of instruction typesExampleData transfer Instructions8313MOV, MVI etc.Arithmetic Instructions6214ADD, SUB etc.Logical Instructions4315AND, XOR etc.Stack Instructions159PUSH, POP etc.Branch Instructions368JMP, JZ etc.I/O Instructions22IN, OUT etc.Interrupt Instructions55RST 0, RST 1 etc.Total24666Following is the table showing the list of Control instructions with their meanings.OpcodeOperandMeaningExplanationNOPNoneNo operationNo operation is performed, i.e., the instruction is fetched and decoded only.HLTNoneHalt and enter wait stateThe CPU finishes executing the current instruction and stops further execution. An interrupt or reset is necessary to exit from ... Read More

In this section, we will see how we can use the Zilog Z-80 Microprocessor to move a block of data to another location. There is one assumption, there is sufficient distance between source and destination. So blocks are non-overlapping. Basically the block movement is not exact moving, it is copying the data to other locations.The number of items in the block is given at location 5000H, and the block is located at position 5050H.So before movement, the items in the memory is looking like this.AddressValue5000H04H...5050H89H5051H7AH5052H2FH5053H56H...Now, we are writing a program at location 8000H to move the block contents to other ... Read More

In this section we will see how we can use the Zilog Z-80 Microprocessor to Exchange the contents of each element from two different blocks.The number of items in each block are given at location 5000H, and the blocks are at position 5050H and 5070H.So before swapping the items in the memory is looking like thisAddressValue5000H04H...5050H89H5051H7AH5052H2FH5053H56H...5070HAFH5071HA9H5072HFBH5073H21H...Now, we are writing a program at location 8000H to exchange the block contents.ProgramAddressHex CodesLabelsMnemonicsComments800021 00 50LD HL, 5000HLoad the HL pair with 5000H8003DD 21 50 50LD IX, 5050HSet the index register with 5050H8007DD 7E 00LD A, (IX+00H)Load Acc with IX + 00H800ADD 46 20LD ... Read More

Now, in this section we will see how we can use the Zilog Z-80 Microprocessor to add Multi-byte numbers.In this example, we are using 4-byte numbers (56 2F 7A 89)16 and (21 FB A9 AF)16In the memory, at first, we are storing the byte counts, and then the numbers (from least significant bytes to Most significant bytes) in different segments. So after storing the data, the memory structure will be look like thisAddressValue5000H04H...5050H89H5051H7AH5052H2FH5053H56H...5070HAFH5071HA9H5072HFBH5073H21H...Now, we are writing a program at location 8000H to add these two 4-byte number and store the result at location 5090H onwards.ProgramAddressHex CodesLabelsMnemonicsComments800037SCFSet the carry flag80013FCCFComplement the ... Read More

As we know that the Intel 8085 has five interrupt pins (TRAP, RST7.5, RST6.5, RST6.5 and INTR), but the Zilog Z-80 has only two interrupt pin. The NMI and INT . But it has superior interrupt structure compared to 8085.The INT InterruptIt is an active low, level triggered input interrupt. This is maskable and using DI instruction this can be disabled. When the interrupt pin is disabled, the Z-80 will not be interrupted if the IO devices enables the INT pin. Even after the reset, it will be disabled. So if we want that the MPU will be interrupted by the pin, there must be ... Read More