In this program we will see how to add two 16-bit numbers.Problem StatementWrite8085 Assembly language program to add two 16-bit number stored in memory location 8000H – 8001H and 8002H – 8003H.DiscussionIn this program we are pointing the operand addresses using HL and DE register pair. Then adding LSBytes by ADD operator, and after that adding MSBytes using ADC operator to consider the carry flag result. The 16-bit result will be stored at BC register, and by checking the carry bit after addition we can simply put 1 into memory.We are taking two numbersBCAD + FE2D = 1BADAInputAddressData......8000AD8001BC80022D8003FE......Flow DiagramProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, ... Read More

In this program we will see how to find the smallest number from a block of bytes using 8085.Problem StatementWrite 8085 Assembly language program to find the smallest number from a block of bytes.DiscussionIn this program the data are stored at location 8001H onwards. The 8000H is containing the size of the block. After executing this program, it will return the smallest number and store it at location 9000H.Logic is simple, we are taking the first number at register B to start the job. In each iteration we are getting the number from memory and storing it into register A. ... Read More

In this program we will see how to find the largest number from a block of bytes using 8085.Problem StatementWrite 8085 Assembly language program to find the largest number from a block of bytes.DiscussionIn this program the data are stored at location 8001H onwards. The 8000H is containing the size of the block. After executing this program, it will return the largest number and store it at location 9000H.Logic is simple, we are taking the first number at register B to start the job. In each iteration we are getting the number from memory and storing it into register A. ... Read More

According to the microprocessor point of view, the 8253 is designed and has some specialty port chip I/O. We don't use it for interfering the I/O devices. For performing the application of time it is used. 8253 has the addressed A1 and A0 input pins.The counters have width of 16 bits. If they were 8-bits wide, the delay in time that would be generated is very small. The Least Significant Byte and the Most Significant Byte of a counter is selected by using the same address of the port.The processor here writes to the control port to configure the working of ... Read More

As a DIP package Intel 8253 is a 24-pin programmable IC available. IChas three counters which work independently and whose width is of16-bits. In addition, we have a control port to decide what is the mode of working of the three counters. The physical and functional pin diagrams of them are indicated below.Fig.Diagram of 8253 pin basedFig.Pin diagram functionalVcc and GnThese are the Power supply and ground pins which 8253 uses +5V as power supplyD7-0For the communication of the processor there are eight functional pinsRD*This reads counter information it is active low pinWR*Writes control informationCS*It selects the chip which is ... Read More

The interrupt requests are accepted by 8259 from eight interrupting devices on the pin ranging fromIR0 toIR7. After that, it identifies the priority interrupt having the highest request from the inputs which are active. It is possible for us to configure the 8259 for the mode of operation of "fixed priority" mode.Here, among the priorities, IR0 has the highest and IR7 has the lowest. If the three inputs IR2, IR4, and IR6 are in active state, then IR2 will have is the highest priority interrupt request than the other active requests. We can mask the requests of the interrupts by installing ... Read More

Many situations may arise, where a microcomputer system requires accurate time delays. For example, when we implement a real time clock, the time needs to get updated at least for once in every second.We generate accurate time delays by using a few instructions in a loop.It is completely software based, where 8085 does not perform any work which is beneficial except the generation of time delays.The delay in time or Time delay scan also be generated by hardware method also. As an example, a 555 timer chip can also be used for the generation of time delays or delay in ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will calculate the multiplication of two 16-bit numbers.Problem StatementWrite 8085 Assembly language program to multiply two 16-bit numbers stored at 8000H -8001H and 8002H - 8003H.DiscussionThis program takes the 16 bit data from memory location 8000H – 8001Hand 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 input AddressData......80002480011C800252800307......Flow DiagramProgramAddressHEXCodesLabelsMnemonicsCommentsF00031, 00, 20LXI SP, 2000H   InitializeStack pointerF0032A, 00, ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will find the multiplication result of two BCD numbers.Problem StatementWrite 8085 Assembly language program to find two BCD number multiplication. The numbers are stored at location 8000H and 8001H.DiscussionIn 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 ... Read More

Now let us see a program of Intel 8085 Microprocessor. This program will find the LCM of two8-bit numbers.Problem StatementWrite 8085 Assembly language program to find LCM of two 8-bit numbers stored at location8000H and 8001HDiscussionIn this program we are reading the data from 8000H and 8001H. By loading the number, we are storing it at C register, and clear the B register. The second number is loaded into Accumulator. Set DE as the 2's complement of BCregister. This DE is used to subtract BC from HL pair.The method is like this:let us say the numbers are 25 and 15. ... Read More