Found 409 Articles for Microcontroller

In 8085 Instruction set, there is one mnemonic XCHG, which stands for eXCHanGe. This is an instruction to exchange contents of HL register pair with DE register pair. This instruction uses implied addressing mode. In the instruction, we don’t mention as “XCHG HL, DE”. It is implied that it will deal with HL and DEregister pairs. So we write only XCHG as mnemonic. That’s why it is called an implied addressing mode. As it is 1-Byte instruction, so It occupies only 1-Byte in the memory. After execution of this instruction, the content between H and D registers and L and ... Read More

In this mode, the data is transferred from one register to another by using the address pointed by the register. Register indirect addressing mode also used to call as indirect addressing mode. For example MOV A, M: means data is transferred from the memory address pointed by the register pair HLto the register A.MOV E, MIt occupies only 1-Byte in memory. MOV E, M is an example instruction of this type. It is a 1-Byte instruction. Suppose E register content is DBH, H register content is 40H, and L register content is 50H. Letus say location 4050H has the data ... Read More

In this mode, the data is directly copied from the given address to the register. This absolute addressing mode is also called a direct addressing mode. For example LDA 3000H: means the data at address 3000H is copied to register A.LDA 4050HLet us consider LDA 4050 Has an example instruction of this type. It is a 3-Byte instruction. The initial content of memory address 4050H is ABH. initial accumulator content is CDH. As after execution A will be initialized with value ABH. Memory location 4050H will still remain with the content ABH. The results of the execution of this instruction ... Read More

In this mode, the data is copied from one register to another. For example, MOV A, B: means data in register B is copied to register A.MOV E, HIt occupies only 1-Byte in memory. MOV E, H is an example instruction of this type. It is a 1-Byte instruction. Suppose E register content is AB H, and H register content is 9C H. When the 8085 executes this instruction, the contents of E register will change to 9C H.This is shown as follows.BeforeAfter(E)ABH9CH(H)9CH9CHAddressHex CodesMnemonicComment20045CMOV E, HE ← HNote that H register’s content has not been changed at all. Although Intel has called ... Read More

In this mode, the 8/16-bit data is specified in the instruction itself as one of its operand. For example MVI E, ABH: means ABH is copied into register A. Here the operand is immediately available in the instruction.MVI E ABHBeforeAfter(A)Any valueABHAs example, if we consider instruction MVI E, ABH then it means that ABH will be moved or copied to the register E. And, as a result, the previous value of E will get overwritten.AddressHex CodesMnemonicComment20001EMVI E, ABHE ← ABH2001ABABH as operandThis instruction will have seven T-states as shown below.Summary  − So this instruction MVI E, ABH requires 2-Bytes, 2-Machine ... Read More

Using 16-bit address, 8085 can access one of the 216= 64K locations. As a single hexadecimal digit can be expressed in4-bit notation so, in 8085, memory address can be expressed using four hexadecimal digits. Similarly, for convenience, we can represent all 8085 CPU registers as A, B, C etc. using binary notations. Internally 8085 specifies these registers using 0s and 1s only. So3-bits are just enough to represent a register. The 3-bit register codes for the registers of 8085 are shown in the following tableWith 3-bit register code, eight registers can be specified in maximum as 23= 8. On the ... Read More

Using mnemonics without any alteration in the content, data can be transferred in three different cases – From one register to another registerFrom the memory to the register andFrom the register to the memoryThese can be guided by addressing modes. Addressing modes in 8085 can be classified into 5 groups −Immediate addressing modeRegister addressing modeDirect addressing modeIndirect addressing modeImplied addressing modeImmediate addressing modeIn this mode, the 8/16-bit data is specified in the instruction itself as one of its operands. For example MVI E, ABH means ABH is copied into register A.MVI E ABHBeforeAfter(A)Any valueABHAs an example, if we consider instruction MVI E, ... Read More

In this program, we will see how to multiply two 8-bit numbers using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to multiply two 8-bit numbers stored in a memory location and store the 16-bit results into the memory.DiscussionThe 8085 has no multiplication operation. To get the result of multiplication, we should use the repetitive addition method. After multiplying two 8-bit numbers it may generate 1-byte or 2-byte numbers, so we are using two registers to hold the result.We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.InputAddressData......8000DC8001AC......Flow DiagramProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 00, 80LXI H, ... Read More

In this program, we will see how to check the 4th bit of an 8-bit number.Problem StatementWrite 8085 Assembly language program to check whether the fourth bit of a byte is 0 or 1.When it is 0, store 00H at any specified location, and when it is 1, store FFH at the specified location.DiscussionWe are considering the 8-bit number, and storing 00H or FFH by checking the 4th bit on the number from left. The logic behind it is very simple. We are just performing bit-wise and operation on the given data with 08H. If the result is non-zero, then the 4th ... Read More

In this program, we will see how to add a block of data using the 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to add N 1-byte numbers. The value of N is provided.DiscussionIn this problem, we are using location 8000H to hold the length of the block. The main block is stored from address 8010H. We are storing the result at location 9000H and 9001H. The 9000H holding the lower byte, and 9001H is holding the upper byte.Repeatedly we are taking the number from the memory, then adding it with the accumulator and increase the register E content when carry ... Read More