Microprocessor Articles - Page 15 of 42

These instructions are used to transfer/branch the instructions during an execution. There are two types of branching instructions. The unconditional branch and conditional branch.The Unconditional Program execution transfer instructions are as follows.OpcodeOperandDescriptionCALLaddressUsed to call a procedure and save their return address to the stack.RET----Used to return from the procedure to the main program.JMPaddressUsed to jump to the provided address to proceed to the next instruction.LOOPaddressUsed to loop a group of instructions until the condition satisfies, i.e., CX = 0 Now let us see the Conditional Program execution transfer instructions.OpcodeOperandDescriptionJCaddressUsed to jump if carry flag CY = 1JNCaddressUsed to jump if no ... Read More

String is a group of bytes/words and their memory is always allocated in a sequential order. String is either referred as byte string or word string. Here we will see some instructions which are used to manipulate the string related operations.The String manipulation instructions are as follows.OpcodeOperandDescriptionREPInstructionUsed to repeat the given instruction till CX ≠ 0.REPE/REPZInstructionUsed to repeat the given instruction until CX = 0 or zero flag ZF = 1.REPNE/REPNZInstructionUsed to repeat the given instruction until CX = 0 or zero flag ZF = 1.MOVS/MOVSB/MOVSW----Used to move the byte/word from one string to another.COMS/COMPSB/COMPSW----Used to compare two string bytes/words.INS/INSB/INSW----Used ... Read More

The general purpose registers are used to store temporary data in the time of different operations in microprocessor. 8086 has eight general purpose registers. The description of these general purpose registersRegisterFunctionAXThis is the accumulator. It is 16-bit registers, but it is divided into two 8-bit registers. These registers are AH and AL. AX generally used for arithmetic or logical instructions, but it is not mandatory in 8086.BXBX is another register pair consisting of BH and BL. This register is used to store the offset values.CXCX is generally used as control register. It has two parts CH and CL. For different looping ... Read More

In this program we will see how to perform addition by using ports to take data and send the result into the port.Problem StatementWrite 8085 Assembly language program for interfacing between 8085 and 8255. Here Port A and Port B are holding two values, take the numbers from port A and B, add them, and send the result at port C.DiscussionThe task is very simple. At first we have to setup the control word register of 8255 chip. After that we will take the input from port A and B, add the content, and send it to port C.The control ... Read More

In this program we will see how to transfer the switch values from one port to another using 8085 and 8255 chip.Problem Statement:Write 8085 Assembly language program for interfacing between 8085 and 8255. Here eight switches are connected at port A. Transfer the status of these switches into port B. In port B the LEDs are connected.Discussion:The task is very simple. At first we have to setup the control word register of 8255 chip. After that we will take the input from port A, and send it to port B.The control word register is looks like this. It is holding ... Read More

In this program we will see how to lower and upper nibbles are masked in 8085.Problem StatementWrite 8085 Assembly language program to mask the upper and lower nibble of an 8-bit number. The number is stored at location 8000H. Lower and Upper nibbles will be stored at location 8001H and 8002H.DiscussionThe masking is basically ANDing two numbers. When we want to mask the upper nibble of an 8-bit number say 2D (0010 1101), then we will AND with 0F (0000 1111), so we will get 0D (0000 1101). By masking with F0 (1111 0000), the result will be 20 (0010 ... Read More

In this program we will see how to reverse the digits of a 16-bit number using 8085.Problem StatementWrite 8085 Assembly language program to reverse a 16-bit number stored at location 8000H-8001H. Also, store the result at 8050H – 8051H.DiscussionHere the task is too simple. There are some rotating instructions in 8085. The RRC, RLC are used to rotate the Accumulator content to the right and left respectively without carry. We can use either RRC or RLC to perform this task. In the final result each digit of the H and L are reversed, and also the H and L values ... Read More

In this program we will see how to reverse the digits of an 8-bit number using 8085.Problem StatementWrite 8085 Assembly language program to reverse an 8-bit number stored at location 8000H. Also, store the result at 8050H.DiscussionHere the task is too simple. There are some rotating instructions in 8085. The RRC, RLC are used to rotate the Accumulator content to the right and left respectively without carry. We can use either RRC or RLC to perform this task.InputAddressData……80004C……Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF0003A, 00, 80 LDA 8000HTake the number from memoryF0030F RRCRotate right without carry four timesF0040F RRC F0050F RRC F0060F RRC F00732, 50, 80 STA 8050HStore the result at memoryF00A76 HLTTerminate the ... Read More

In this program we will see how to find the minimum digit from a two-digit number.Problem StatementWrite 8085 Assembly language program to find the minimum digit from a two-digit number. The number is stored at location 8000H, store the result at 8050H.DiscussionHere we are performing this task by using masking operation. Each digit takes one nibbles. We are masking the upper nibble by ANDing with 0FH (0000 1111). Store the lower nibble into another register. After that, we are taking the upper nibble. To get it, we are shifting the number to the right four times to convert lower nibble ... Read More

In this program, we will see how to find the square of an 8-bit number.Problem StatementWrite 8085 Assembly language program to find the square of a number The number is stored at location 8000H, store the result at 8050H.DiscussionIn 8085, we cannot perform the multiplication operation directly. We are performing the multiplication by using repetitive addition. To get square of a number, we have to multiply the number with itself.InputAddressData……80000C……Flow Diagram ProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 00, 80 LXI H, 8000HLoad the number from 8000HF003AF XRA AClear accumulatorF00446 MOV B, MLoad data from memory to BF00586LOOPADD MAdd memory byte with AF00605 DCR BDecrease B by 1F007C2, 05, F0 JNZ ... Read More