Here we will see 8085 Microprocessor program, that will find the multiplication result of two BCD numbers.Problem Statement −Write an 8085 Assembly language program to find two BCD number multiplication. The numbers are stored at location 8000H and 8001H.Discussion −In this program the data are taken from 8000H and 8001H. The result is stored at location 8050H and 8051H.As we know that 8085 has no multiply instruction so we have to use repetitive addition method. In this process after each addition we are adjusting the accumulator value to get decimal equivalent. When carry is present, we are incrementing the value of MS-Byte. ... Read More

Let us see one 8085 Microprocessor problem. In this problem we will see how to multiply two numbers using shift and add methods, not by using additive approach.Problem Statement −Write an 8085 Assembly language program to multiply two 8-bit numbers using shift and add method.Discussion −The shift and add method is an efficient process. In this program, we are taking the numbers from memory location 8000H and 8001H. The 16 bit results are storing into location 8050H onwards.In this method we are putting the first number into DE register pair. The actual number is placed at E register, and D is holding ... Read More

Here we will see one 8085 Microprocessor program. This program will add the contents of N word locations.Problem Statement −Write an 8085 Assembly language program to add N 16-bit numbers stored into memoryDiscussion −The 16-bit numbers are stored into memory location 8001H onwards. The value of N is stored at location 8000H. After addition, the result will be stored at location 8050H onwards.In 8085 we have few number of registers. So we are storing the count into memory, when we need to update it, we will fetch it from memory, update it and then again store it into memory.Here the 16-bit numbers ... Read More

Here we will see one 8085 Microprocessor program. This program will sort a sequence of numbers in reverse order using selection sort technique.Problem Statement −Write an 8085 Assembly language program to sort a given sequence using selection sort in descending order. The numbers are stored at 8001H onwards. 8000H is holding the block size.Discussion −In the selection sorting technique, we will choose the minimum or the maximum term from a set of numbers. In this case we are considering the sorting in descending order, so we are choosing the maximum number. By taking the maximum number, we are swapping it with the ... Read More

Here we will see one 8085 Microprocessor program. In this program we will see how to sort a sequence of numbers using selection sort.Problem Statement −Write an 8085 Assembly language program to sort a given sequence using selection sort in ascending order. The numbers are stored at 8001H onwards. 8000H is holding the block size.Discussion −In the selection sorting technique, we will choose the minimum or the maximum term from a set of numbers. In this case we are considering the sorting in ascending order, so we are choosing the minimum number. By taking the minimum number, we are swapping it with ... Read More

Here we will see one 8085 microprocessor program that will help to arrange the numbers in ascending or descending order based on our choice. Sort using bubble sort technique.Problem Statement −Write an 8085 Assembly language program to perform bubble sorting operation on a set of data, and arrange them into ascending or descending order based on choice.Discussion −In this program we are arranging some numbers into ascending or descending order based on some choice. We are storing the choice at location A000H. If the choice value is 00H, then data will be sorted in ascending order, otherwise it will be sorted in ... Read More

Here we will see one 8085 program to arrange a block of numbers in ascending order.Problem Statement −Write an 8085 program to sort numbers in ascending order where n number of numbers are stored in consecutive memory locations starting from 8041H and the value of n is available in memory location 8040H (Using BUBBLE sort).Discussion −In this program we will arrange the numbers in bubble sorting technique. In this sorting technique, it will be executed in different pass. In each pass the largest number is stored at the end of the list. Here we are taking the numbers from location 8041H to ... Read More

Here we will see one 8085 program. This program will perform different operations on BCD numbers based on choice.Problem Statement −Write an 8085 Assembly language program to perform some operations on two 8-bit BCD numbers base on our choice.Discussion −In this program we are taking a choice. The choice value is stored at memory location 8000H (named as X). And the BCD numbers are stored at location 8001H and 8002H. We are storing the result at location 8050H and 8051H.Here if the choice is 00H, then it will perform addition, for 01H, it will perform subtraction, and for 02H, it will do ... Read More

Here we will see one 8085 Microprocessor program. This program will be used to convert 16-bit binary data to BCD data.Problem Statement −Write an 8085 Assembly language program to convert 16-bit binary data to BCD data. The binary data is stored at location 8000H and 8001H.Discussion −This problem is solved by implementing 16-bit counter. We are storing the 16-bit number at first, then decreasing the numbers one by one, and increasing the decimal value by adjusting the decimal value. To increase the value, we can use the INR instruction, but INR instruction does not affect the carry flag. So here we are ... Read More

Here we will see one program that can perform subtraction for multi-byte BCD numbers using 8085 microprocessor.Problem Statement −Write an 8085 Assembly language program to subtract two multi-byte BCD numbers.Discussion −The numbers are stored into memory, and one additional information is stored. It will show us the byte count of the multi-byte BCD number. Here we are choosing 3-byte BCD numbers. They are stored at location 8001H to 8003H, and another number is stored at location 8004H to 8006H. The location 8000H is holding the byte count. In this case the byte count is 03H.For the subtraction we are using the 10’s ... Read More