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8085 Articles
Page 13 of 36
Program to convert HEX to ASCII in 8085 Microprocessor
Here we will see one 8085 Microprocessor program. That program will convert HEX to ASCII values.Problem Statement −Write an 8085 Assembly language program to convert Hexadecimal characters to ASCII values.Discussion −We know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is 39H (57D). So all other numbers are in the range 30H to 39H. The ASCII value of 0AH is 41H (65D) and ASCII of 0FH is 46H (70D), so all other alphabets (B, C, D, E, F) are in the range 41H to 46H.Here we are providing hexadecimal digit at memory location 8000H, The ASCII equivalent ...
Read MoreProgram to convert ASCII to HEX in 8085 Microprocessor
Here we will see one 8085 Microprocessor program. This program will convert ASCII to HEX values.Problem Statement −Write an 8085 Assembly language program to convert ASCII to Hexadecimal character values.Discussion −We know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is 39H (57D). So all other numbers are in the range 30H to 39H. The ASCII value of 0AH is 41H (65D) and ASCII of 0FH is 46H (70D), so all other alphabets (B, C, D, E, F) are in the range 41H to 46H.Here the logic is simple. We are checking whether the ASCII value is ...
Read MoreProgram to convert BCD to HEX in 8085 Microprocessor
Here we will see one 8085 program, that program will convert BCD numbers to HEX equivalent.Problem Statement −A BCD number is stored at location 802BH. Convert the number into its binary equivalent and store it to the memory location 802CH.Discussion −In this problem we are taking a BCD number from the memory and converting it to its binary equivalent. At first we are cutting each nibble of the input. So if the input is 52 (0101 0010) then we can simply cut it by masking the number by 0FH and F0H. When the Higher order nibble is cut, then rotate it to ...
Read MoreProgram to multiply two 16-bit binary numbers in 8085 Microprocessor
Here we will see one program for Intel 8085 Microprocessor. This program will calculate the multiplication of two 16-bit numbers.Problem Statement −Write an 8085 Assembly language program to multiply two 16-bit numbers stored at 8000H - 8001H and 8002H - 8003H.Discussion −This program takes the 16 bit data from memory location 8000H – 8001H and 8002H – 8003H. The 32 bit results are stored at location 8050H – 8053H.Here we have tested with two 16 bit numbers. The results are as follows.1111H × 1111H = 01234321H 1C24H × 0752H = 00CDFF88HInputfirst inputAddressData……800011800111800211800311……second inputAddressData……80002480011C800252800307……Flow DiagramProgramAddressHEX CodesLabelsMnemonicsCommentsF00031, 00, 20LXI SP, 2000HInitialize Stack pointerF0032A, 00, 80LHLD ...
Read MoreProgram to multiply two 2-digit BCD numbers in 8085 Microprocessor
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 MoreProgram to multiply two 8-bit numbers (shift and add method) in 8085 Microprocessor
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 MoreProgram to do an operation on two BCD numbers based on the contents of X in 8085 Microprocessor
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 MoreProgram to convert a 16-bit binary number to BCD in 8085 Microprocessor
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 MoreProgram for subtraction of multi-byte BCD numbers in 8085 Microprocessor
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 More8085 program to alternate D0 bit with specified delay
Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to alternate the D0 bit and send as output.Problem StatementWrite 8085 Assembly language program to alternate D0 bit. And send this as output.DiscussionAlternating D0 bit and sending as output is like generating the square wave. We are adding extra delay in each phase. To generate square wave with 8085, we will rotate 10101010 (AAH) continuously, and send D0 as output. We will mask the accumulator content by 01H. If this is 0, then output will be 0, if it is 1, output will ...
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