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Microprocessor Articles
Page 13 of 42
Programming the 8259 with no slaves
Now in this topic we assume that 8085 is the processor which is used in this microcomputer system. In this slave, no 8259 slaves are used. We should examine properly before 8259 PIC is used in the microcomputer system for performing the interrupt control application. 8259 is configured in such a fantastic way that is found that a variety of information are provided like for IR0 request IV, interrupts like level or edge-triggered, whether 8259s are used single or many, if ICW4 is in need or not and whether for the interrupt requests masking should be done or not. This ...
Read More8085 Program to convert ASCII to HEX
Now let us see a program of Intel 8085 Microprocessor. This program will convert ASCII to HEXvalues.Problem StatementWrite 8085 Assembly language program to convert ASCII to Hexadecimal character values. DiscussionWe know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is39H (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 less ...
Read More8085 Program to convert HEX to ASCII
Now let us see a program of Intel 8085 Microprocessor. This program will convert HEX to ASCII values.Problem StatementWrite 8085 Assembly language programs to convert Hexadecimal characters to ASCII values.DiscussionWe know that the ASCII of number 00H is 30H (48D), and ASCII of 09H is39H (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 More8085 Program to find Square of a number using look up table
Now let us see a program of Intel 8085 Microprocessor. This program is mainly for finding a square of a number.Problem StatementWrite 8085 Assembly language program to find the square of a number using Look-Up Table. DiscussionIn this example, we are using Look-Up table to find the square of a number. The main limitation of this program is that it can find a square in range 0-F. When the input is above F, it will fail. We are storing the square result into the memory. When the number is greater than F, we are storing FFH into memory to indicate an ...
Read More8085 Program to Subtract two 8 Bit numbers
In this program, we will see how to subtract two 8-bit numbers using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to subtract two 8-bit numbers and store the result at locations 8050H and 8051H.DiscussionIn 8085, the SUB instruction is used 2’s complemented method for subtraction. When the first operand is larger, the result will be positive. It will not enable the carry flag after completing the subtraction. When the result is negative, then the result will be in 2’s complemented form and carry flag will be enabled. We are using two numbers at location 8000H and 8001H. When the numbers are ...
Read More8085 Program to Multiply two 8 bits numbers
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 two8-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, 80LXIH, 8000HLoad first ...
Read More8085 Program to Divide two 8 Bit numbers
In this program, we will see how to divide two 8-bit numbers using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to divide two 8-bit numbers and store the result at locations 8020H and 8021H.DiscussionThe 8085 has no division operation. To get the result of the division, we should use the repetitive subtraction method. By using this program, we will get the quotient and the remainder. 8020H will hold the quotient, and 8021H will hold the remainder.We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.InputThe Dividend: 0EHThe Divisor 04HThe Quotient will be ...
Read MoreWhat is the human brain made of?
In the Central Nervous System, Human Brain is the main organ. It comprises of the cerebrum, the brain stem, and the cerebellum. It controls the vast majority of the body activities, processing, incorporating, and coordinating the data it gets from the sensory organs and deciding the instructions sent to the rest of the body. The mind is contained in and secured by the skull bones of our head.The human cerebrum is mainly made up of neurons, glial cells, and veins. Interneurons, pyramidal cells including Betz cells, motor neurons, and cerebellar Purkinje cells are the various types of neurons. Betz cells are ...
Read MoreProgram to Divide two 8 Bit numbers in 8051 Microprocessor
Here we will see the division operation. This operation will be used to divide two 8-bit numbers using this 8051 microcontroller. The register A and B will be used in this operation. No other registers can be used for division. The result of the division has two parts. The quotient part and the remainder part. Register A will hold Quotient, and register B will hold Remainder.We are taking two number 0EH and 03H at location 20H and 21H, After dividing the result will be stored at location 30H and 31H.AddressValue …20H0EH21H03H …30H00H31H00H …Program MOV R0, #20H ; set source address 20H ...
Read MoreProgram to Multiply two 8 Bit numbers in 8051 Microprocessor
Here we will see how to multiply two 8-bit numbers using this 8051 microcontroller. The register A and B will be used for multiplication. No other registers can be used for multiplication. The result of the multiplication may exceed the 8-bit size. So the higher order byte is stored at register B, and lower order byte will be in the Accumulator A after multiplication.We are taking two number FFH and FFH at location 20H and 21H, After multiplying the result will be stored at location 30H and 31H.AddressValue …20HFFH21HFFH …30H00H31H00H …Program MOV R0, #20H ; set source address 20H to R0 ...
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