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

Now let us see a program of Intel 8085 Microprocessor. This program will convert ASCII to binary values.Problem StatementWrite 8085 Assembly language program to check whether a bit pattern is palindrome or not.DiscussionIn this program we are taking the number from location 8000H. The program will return 00H if the number is not palindrome, otherwise it will return FFH.Let the Input is 18H so the binary value is (0001 1000) this is a palindrome. The number 52H(0101 0010) it is not a palindrome.In this problem we are taking the first number into accumulator, then shifting it to the left. When ... Read More

In this program we will see how to convert binary numbers to its BCD equivalent.Problem StatementA binary number is store dat location 800H. Convert the number into its BCD equivalent and store it to the memory location 8050H.DiscussionHere we are taking a number from the memory, and initializing it as a counter. Now in each step of this counter we are incrementing the number by 1, and adjust the decimal value. By this process we are finding the BCD value of binary number or hexadecimal number.We can use INR instruction to increment the counter in this case but this instruction ... Read More

In this program we will see how to convert BCD numbers to binary equivalent.Problem StatementA BCD number is stored at location 802BH. Convert the number into its binary equivalent andstore it to the memory location 802CH.DiscussionIn this problem we are taking a BCD number from the memory and converting it to its binaryequivalent. At first we are cutting each nibble of the input. So ifthe input is 52 (0101 0010) then we can simply cut it by masking the number by 0FH and F0H. When the Higher order nibble is cut, thenrotate it to the left four times to transfer ... Read More