In 8085 Instruction set, Read Interrupt Mask. It is a 1-Byte multi-purpose instruction. It is used for the following purposes.To check whether RST7.5, RST6.5, and RST5.5 are masked or not;To check whether interrupts are enabled or not;To check whether RST7.5, RST6.5, or RST5.5 interrupts are pending or not;To perform serial input of data.Mnemonics, OperandOpcode (in HEX)BytesRIM201To get the status information about the interrupt system, Read Interrupt Mask instruction provides status information about interrupt system and this instruction can be used for serial input of data. Through this RIM instruction, 8085 can know which interrupt is masked or unmasked, etc. The ... Read More

In 8085 Instruction set, SIM stands for “SetInterrupt Mask”. It is 1-Byte instruction and it is a multi-purpose instruction. The main uses of SIM instruction are –Masking/unmasking of RST7.5, RST6.5, and RST5.5Reset to 0 RST7.5 flip-flopPerform serial output of dataMnemonics, OperandOpcode(in HEX)BytesSIM301When SIM instruction is executed then the content of theAccumulator decides the action to be taken. So before executing the SIM instruction, it is mandatory to initialize Accumulator with the required value. The meaning and purpose of the various bits of the accumulator when SIM is executed has been depicted below – Note that except bit 5, which is a ... Read More

In 8085 Instruction set, SIM (Set Interrupt Mask) and RIM (Read Interrupt Mask) instructions can perform mask and unmask RST7.5, RST6.5, and RST5.5 interrupt pins and can also read their status. In 8085 Instruction set, SIM stands for “Set Interrupt Mask”. It is 1-Byte instruction and it is a multi-purpose instruction. The main uses of SIM instruction are –Masking/unmasking of RST7.5, RST6.5, and RST5.5Reset to 0 RST7.5 flip-flopPerform serial output of dataMnemonics, OperandOpcode(in HEX)BytesSIM301When SIM instruction is executed then the content of the Accumulator decides the action to be taken. So before executing the SIM instruction, it is mandatory to initialize ... Read More

In 8085 is having five internal interrupt signals that correspond to the five external interrupt pins. And these interrupts are only recognized only when the internal interrupt signal is activated, the 8085 gets interrupted, provided higher priority internal interrupt signals are not active at the same time. The 8085 checks all these internal interrupt signals in the penultimate clock cycle of the last machine cycle of instruction.With in 8085, there is a flip-flop called IE flip-flop. Here IE stands for Interrupt Enable. 8085 interrupt system is disabled whenever this flip-flop is reset to the 0 states. That is, even if ... Read More

In Intel 8051, there are two 16-bit timer registers. These registers are known as Timer0 andTimer1. The timer registers can be used in two modes. These modes areTimer mode and the Counter mode. The only difference between these two modes is the source for incrementing the timer registers. Timer ModeIn the timer mode, the internal machine cycles are counted. So this register is incremented in each machine cycle. So when the clock frequency is 12MHz, then the timer register is incremented in each millisecond. In this mode it ignores the external timer input pin.Counter ModeIn the counter mode, the external events ... Read More

As we have seen there are five different interrupts in 8051. These interrupts are INT0, INT1, TO, T1, TI/RI.There are six states in each machine cycle. These states are S1 to S6. All of the interrupts are sampled at the end of state S5 of each machine cycle. When the instruction takes more than one machine cycle, then the samples are polled during the next machine cycle. When an interrupt flag is set at the S5 of the first machine cycle, then the polling cycle will find it. The interrupt system generates LCALL instruction to call appropriate ISS.There are some ... Read More

Now in this section, we will see the interrupt structure of Intel 8051 microcontroller.Interrupts are basically the events that temporarily suspend the main program, pass the control to the external sources and execute their task. It then passes the control to the main program where it had left off.8051 has five interrupts. These interrupts are INT0,  INT1, TO ,T1 , TI/RI. All of the interrupts can be enabled or disabled by using the IE (interrupt enable) register.The interrupt addresses of these interrupts are like below −InterruptAddressINT00003HINT1000BHT00013HT1001BHTI/RI0023HInterrupt Enable (IE)RegisterThis register can be used to enable or disable interrupts programmatically. This register ... Read More

Now we will see another arithmetic operation. The divide operation to divide two 8-bit numbers using this 8051 microcontroller. The register A and B will be used in this operation. No other registers can be used for division. The result of the division has two parts. The quotient part and the remainder part. Register A will hold Quotient, and register B will hold Remainder.We are taking two number0EH and 03H at location 20H and 21H, After dividing the result will be stored at location 30H and 31H.  AddressValue...20H0EH21H03H...30H00H31H00H...ProgramMOV R0, #20H;set source address 20H to R0 MOV R1, #30H;set destination address ... Read More

Now we will try to multiply two 8-bit numbers using this 8051 microcontroller. The register A and B will be used for multiplication. No other registers can be used for multiplication. The result of the multiplication may exceed the 8-bit size. So the higher order byte is stored at register B, and lower order byte will be in the Accumulator A after multiplication.We are taking two number FFH and FFH at location 20H and 21H, After multiplying the result will be stored at location 30H and 31H.  AddressValue...20HFFH21HFFH...30H00H31H00H...Program        MOV R0, #20H;set source address 20H to R0   ... Read More

Here we will see how to subtract two 8-bit numbers using this microcontroller. The register A(Accumulator) is used as one operand in the operations. There are seven registers R0 – R7 in different register banks. We can use any of them as the second operand.We are taking two number73H and BDH at location 20H and 21H, After subtracting the result will be stored at location 30H and 31H.  AddressValue...20H73H21HBDH...30H00H31H00H...ProgramMOVR0, #20H;set source address 20H to R0 MOVR1, #30H;set destination address 30H to R1 MOVA, @R0;take the value from source to register A MOVR5, A; Move the value from A to ... Read More