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Microcontroller Articles
Page 27 of 33
Generation of rectangular wave using DAC interface
We write a program for the generation of rectangular interface of Digital to Analog Converter (DAC) interference: Let us consider a problem solution in this domain. The problem states that: To get unipolar output, J1 is shorted to J2 on the interface. To display the waveform on a CRO connect pin 1 of connector P1 to CRO signal pin, and pin 2 of connector P1 to CRO ground pin.The program is stated as below.; FILE NAME DAC_TO_RECT.ASM ORG C100H X DW 00FFH ; ‘OFF’ time is proportional to this value Y DW 00C0H ; ‘ON’ time is proportional to this value ...
Read MoreGeneration of triangular wave using DAC interface
We write an 8085 assembly language program for the generation of triangular waveform using the Digital to Analog Converter (DAC) interface. The display of the waveform is seen on the CRO.Let us consider a problem solution in this domain. The problem states that: To get unipolar output, J1 is shorted to J2 on the interface. To display the waveform on a CRO, connect pin 1 of connector P1 to CRO signal pin, and pin 2 of connector P1 to CRO ground pin.Program; FILE NAME DAC_TO_TRIANG.ASM ORG C100H X DW 00FFH ; the fall of rise and time I proportional directly to ...
Read MoreRotation of stepper motor in forward and reverse directions
Let us consider ALS-NIFC-01, which is a stepper motor interface. Using 26-core flat cable, it is connected to ALS kit. It will be used for interfacing two stepper motors. In our current experiment, we use only one stepper motor. The motor has a step size of 1.8°. The stepper motor works on a power supply of +12V. Power supply of +5V (white wire), GND (black), and +12V (red) is provided to the interface. Note that -12V supply is not used by the interface. We shall have to make sure that the +12V supply has adequate current rating to drive the ...
Read MoreIn 8085 Microprocessor, compare I/O port chips and memory chips
Information is also stored in an Input Output port chip similar to a memory chip. Information of 1 byte are stored in an Input Output port chip on the other hand information of few bytes are stored in the Input Output port chips. An example to be cited as only 1 byte of information is stored in Intel 8212 I/O port chip, but 3 bytes of information are stored in Intel 8255 chip. Moreover, a large number of memory locations like 1K, 4K, 8K etc. are contained in the memory chips. We select memory chip location by the address pins ...
Read MoreMerits of I/O-mapped I/O and demerits of memory-mapped I/O
Before having a discussion regarding the merits of I/O mapped I/O and demerits of memorymapped I/O, let us have a generic discussion regarding the difference between I/O mapped I/O and memory mapped I/O.In Memory Mapped Input Output −We allocate a memory address to an Input Output device.Any instructions related to memory can be accessed by this Input Output device.The Input Output device data are also given to the Arithmetic Logical Unit.Input Output Mapped Input Output −We give an Input Output address to an Input Output device.Only IN and OUT instructions are accessed by such devices.The ALU operations are not directly ...
Read More8085 Block movement without overlap
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 MoreDescription of the pins of 8257
The 8257 pins are described is given in the table below.Fig. Physical pin diagram of Intel 8257Fig: Functional pin diagram of Intel 82578257 is using a power of 5V.D7-0/A15-8For communicating with the processor there are 8 bidirectional data pins, when the processor is in active and the 8257 s active state it is in slave mode. When the processor remains in the HOLD state and 8257 behaves as the master, they are used to send out the Most Significant 8 bits of memory address.A3-0When the processor remains in active state and are used as address input pins of 8257. Hence ...
Read More8085 program to subtract two 8-bit numbers with or without borrow
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 with or without borrow 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. ...
Read More8085 program to find the sum of a series
In this program we will see how to add a blocks of data using 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 accumulator and increase the register E content when carry flag is ...
Read More8085 program to add numbers in an array
In this program we will see how to add a blocks of data using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to add numbers in an array, where the size of the array is N. 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 accumulator and ...
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