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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 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 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 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 program for bubble sort
In this program we will see how to sort a block of bytes using bubble sorting technique.Problem StatementWrite 8085 Assembly language 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).DiscussionIn 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 ...
Read More8085 Program to convert two-digit hex to two ASCII values
Now let us see a program of Intel 8085 Microprocessor. This program will convert 8-bit numbers to two digit ASCII values.Problem StatementWrite 8085 Assembly language program where an 8-bit binary number is stored in memory location 8050H. Separate each nibbles and convert it to corresponding ASCII code and store it to the memory location 8060H and 8061H.DiscussionIn this problem we are using a subroutine to convert one hexa-decimal digit (nibble) to its equivalent ASCII values. As the 8-bit number contains two nibbles, so we can execute this subroutine to find ASCII values of them. We can get the lower nibble ...
Read More8085 Program to perform bubble sort in ascending order
In this program we will see how to sort a block of bytes in ascending order using bubble sorting technique.Problem StatementWrite8085 Assembly language 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).DiscussionIn 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 ...
Read More8085 Executing the program and checking result
In this section we will see how to use 8085 to write a program in 8085 kit. We will also see how to debug the program and check the result after successful execution. Let us see a typical keypad structure of 8085 kit. (This keyboard pattern may vary indifferent kits of different manufacturers)The following table will show the functionalities of different control keys. There are 16alphanumeric keys (0-9, A-F) to provide data and address −KeysFunctionalitiesRESETReset the systemVCT INTVector Interrupt. It generates hardware interruptRST 7.5 via keypadSHIFTProvides second level commands to all keysGOExecute the programSIExecute in Single Step ModeEXREGExamine Register. It allows ...
Read More8085 Program to Exchange 10 bytes
In this program we will see how to exchange a block of 10-byte data using 8085.Problem StatementWrite 8085 Assembly language program to exchange a block of data, where block size is 10.DiscussionThe data are stored at location 8010H to 8019H and 9010H to 9019H. The location 8000H is holding the number of bytes to exchange. In this case number of bytes are 10D so it will be 0AH.The logic is very simple, The HL and DE register pair is pointing the first and second data block respectively. By taking the data we are just swapping the values of each memory ...
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