Microcontroller Articles

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8085 Program to simulate a real-time clock

George John
George John
Updated on 30-Jul-2019 2K+ Views

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to generate real time clock using 8085.Problem Statement:Write 8085 Assembly language program to simulate real-time clock.Discussion:In this program we are creating a real time clock using 8085MPU. Here we are generating 1s delay to update seconds. This clock is 24Hrs clock. We are initializing the clock from 00:00:00. To display the values into 7-segment display we have to use some Port ICs and correct configurations. In each 60 seconds the minute field is updated, and in each 60 minutes the hour field is ...

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8085 Program to convert a 16-bit binary number to BCD

Arjun Thakur
Arjun Thakur
Updated on 30-Jul-2019 2K+ Views

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to convert 16-bit binary data to BCD data.Problem Statement:Write 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 ...

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8085 Program to add the contents of N word locations

Arjun Thakur
Arjun Thakur
Updated on 30-Jul-2019 336 Views

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to add the contents of N word locations.Problem Statement:Write 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 ...

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8085 Program to multiply two 8-bit numbers (shift and add method)

Chandu yadav
Chandu yadav
Updated on 30-Jul-2019 5K+ Views

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to multiply two numbers using shift and add methods.Problem Statement:Write 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 00H. The second ...

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Generation of .OBJ file using a cross-assembler

Ankith Reddy
Ankith Reddy
Updated on 30-Jul-2019 490 Views

Generation of .OBJ file by using a cross-assembler:The file assembly language program file, e.g. MULT.ASM which is created by using an editor is simply a text file. We cannot execute this file directly. At first we have to assemble the file, and then we have to link it. The step of assembly the translation of the program of assembly language to machine code requires the generation of a .OBJ file.We have used some examples in the text of ‘2500 A.D i.e. the 8085 cross-assembler of version 4.01.In the prompt mode the translation is:We type ‘X8085’ just for performing the translation ...

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Data file mode in 8085 Microprocessor

Ankith Reddy
Ankith Reddy
Updated on 30-Jul-2019 536 Views

DATA FILE MODE in8085 MicroprocessorWe can refer to ‘X8085 cross-assembler’ manual for details.The format of Intel Hex: We have shown the MULT.HEX file to give a brief description at a glance.:01 C100 00 04 3A:01 C200 00 05 38:02 C300 00 00 00 3B:10 C000 00 21 00 C1 5E 16 00 21 00 C2 7E 21 00 00FE 00 CA 90:10 C010 00 17 C0 19 3D C2 12 C0 22 F7 FF 22 00C3 CD BC 06 D3:01 C020 00 76 A9:00 0000 01 FFThere are several lines which the file consists of which are termed as records. ...

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Running the entire 8085 program in a single operation

Arjun Thakur
Arjun Thakur
Updated on 30-Jul-2019 426 Views

In a single operation, we can run the entire program of 8085. We run the entire program in this mode so that we can get the accurate results. If the process is unsuccessful a single-stepping by the entire program is attempted.We type ‘G’ at the prompt ‘>’. By noticing the absence of change of address after G. The prompting of the system are as follows:The address of starting is: xxxx - yy/where the memory address is xxxx and the content of the memory location is yy. Which allows the user for responding with the desired address of starting. If ...

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Use of 74138 in ALS-SDA-85M kit

Ankith Reddy
Ankith Reddy
Updated on 30-Jul-2019 233 Views

In the ALS-SDA-85M kit, we have 74138 EPROM of minimum size 16K x 8 and size of RAM IS 2K x 8, there are empty sockets for the EPROM to get expanded. The selection of these four chips is done by the integrated circuit 74138. Every time the selection of 74138 is done when IO/M* turns to 0, the election of 27138 is done when A15A14 turns to 0. 27138 have the lowest address range likeLowest address: 00 00000000000000 is 0000H.Highest address: 00 11111111111111 is 3FFFH.Whereas in 74138 the address ranges are as follows -Lowest address: 11 xxx 00000000000Highest Address: 11 xxx 11111111111.The ...

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Running 8085 program in single-step mode

Chandu yadav
Chandu yadav
Updated on 30-Jul-2019 584 Views

Running the program in Single-Step Mode:The single step mode is very useful for finding our program which has not yielded the desired outcomes after being executed in a single burst. We use it for program debugging. Similarly, like the operation to be executed singly ‘S’ is typed at the ‘>’ prompt. We can notice clearly that the change of address occurs after S. The given system clearly prompts which is stated below:The address of starting is: xxxx - yy/where the memory address is xxxx and the content of the located memory is yy which allows for responding the user ...

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Multiple memory address range in 8085 Microprocessor

Arjun Thakur
Arjun Thakur
Updated on 30-Jul-2019 1K+ Views

Suppose we have the chip select circuit as shown in the below figure, then what is the address range for the RAM?Fig. Chip selection that results in multiple address range.A15 and A14 pins are used for the selection of Random Access Memory, and pins ranging from A10 to A0 are used for the selection of location in RAM. But the value of the pins A13, A12 and A11 have non-dependent value. They play no role for selection of chip or location on RANDOM ACCESS MEMORY. Hence they are termed as don't cares and called x for more signification.We have eight ...

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