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

In this program, we will see how to move blocks of data from one place to another.Problem StatementWrite 8085 Assembly language program to move a data block. The blocks are assumed to be overlapping. The block size is given, the block is starting from Xand we have to move it to the location Y.DiscussionThe overlapping block movement is tricky; we need to use some special tricks to move a block in overlapping address locations. Here the block is starting at position X, we have to move it to position Y. The location Y is inside the block. So Y < ... Read More

In this program, we will see how to move blocks of data from one place to another.Problem StatementWrite 8085 Assembly language program to move a data block. The blocks are assumed to be non-overlapping. The block size is given, the block is starting from X and we have to move it to the location Y.DiscussionThe non-overlapping block movement is relatively an easy task. Here the block is starting at position X, we have to move it to position Y. The location Y is far away from X. So Y > X + block size.In this program, the data are stored ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program is mainly for adding multi-digit BCD (Binary Coded Decimal) numbers.Problem StatementWrite 8085 Assembly language program to add two multi-byte BCD (Binary Coded Decimal) numbers. DiscussionWe are using 4-byte BCD numbers. The numbers are stored into the memory at location 8501H and8505H. One additional information is stored at location 8500H. In this place, we are storing the byte count. The result is stored at location 85F0H.The HL pair is storing the address of first operand bytes, the DE is storing the address of second operand bytes. C is holding the ... Read More

Microprocessor 8085 Architecture which is shown in the below figure consists of various units and each unit has its respective functionality.Fig: 8085 Microprocessor ArchitectureThese units are listed below-AccumulatorIn Intel 8085microprocessor, accumulator acts an 8-bit register to store 8-bit data to perform arithmetic and logical operation on them. The final result stored in the accumulator. It is also called A register.Arithmetic Logic Unit (ALU)It basically performs 8-bit arithmetic operations like +, -, *, / and to perform logical operations like AND, OR, NOT etc. ALU gets inputs from the accumulator and temporary register on which it performs such operations. After processing ... Read More

Registers B, C, D, E, H, and L are general purpose registers in 8085 Microprocessor. All these GPRS are 8-bits wide. They are less important than the accumulator. They are used to store data temporarily during the execution of the program. For example, there is no instruction to add the contents of Band E registers. At least one of the operands has to be in A. Thus to add Band E registers, and to store the result in B register, the following have to be done.Move to A register the contents of B register.Then add A and E registers. The ... Read More

Register A is an 8-bit register used in 8085 to perform arithmetic, logical, I/O & LOAD/STORE operations. Register A is quite often called as an Accumulator. An accumulator is a register for short-term, intermediate storage of arithmetic and logic data in a computer's CPU (Central Processing Unit).In an arithmetic operation involving two operands, one operand has tobe in this register. And the result of the arithmetic operation will be stored or accumulated in this register. Similarly, in a logical operation involving two operands, one operand has to be in the accumulator. Also, some other operations, like complementing and decimal adjustment, ... Read More