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Microcontroller Articles
Page 8 of 33
Absolute addressing mode in 8085 Microprocessor
In this mode, the data is directly copied from the given address to the register. This absolute addressing mode is also called a direct addressing mode. For example LDA 3000H: means the data at address 3000H is copied to register A.LDA 4050HLet us consider LDA 4050 Has an example instruction of this type. It is a 3-Byte instruction. The initial content of memory address 4050H is ABH. initial accumulator content is CDH. As after execution A will be initialized with value ABH. Memory location 4050H will still remain with the content ABH. The results of the execution of this instruction ...
Read MoreIntel 8259A Programmable Interrupt Controller
8085 microprocessor, consists of five interrupt input pins named as RST 5.5, RST 6.5, RST7.5, INTR, and TRAP respectively. When there are a maximum of five I/O devices they want to perform driven interrupt data transfer, which is connected to the five interrupt input pins. Now considering the case where there are more than five I/O devices which would like to perform an interrupt driven data transfer scheme. Here for some pins termed as an interrupt, we use more than one I/O device to the process. Most of the microprocessors nowadays have the configuration of these interrupt input pins. There ...
Read MoreInterrupt mask register in 8259
It stores the levels of interrupts to be masked by means of storing the bits of the interrupt level already masked. It differs from other registers by means of only masking of the bits. Other processes remained intact. Let’s take for the assumption that the requests to the IR4 AND IR6 should not be an interrupt to the processor which can be well achieved by setting the bits of IMR to 1. The IMR is written by OCW1 command. The processor here also has the capability to read the contents of the IMRregister. To complete this task, the processor has ...
Read MoreIn-service register in 8259
Also, an 8-bit register which keeps track records of the interrupt requests that are currently being executed. If the request IR6 is currently being served, whose contents of ISR will be 01000000. If by any means the request to IR3 becomes active during the service process of IR6, 8259 sets bit 3 of ISR to 1 and activates the output INT. But bit 6 of ISR always remains set at 1 asIR6 request which is not fully serviced. Hence the contents ofISR become 01001000. The following assumptions stated below helps this to happen.Until 8259 operates in a complete nested mode, without ...
Read MoreSlave register in 8259
It is also an 8-bit register. The processor here writes to SLR but cannot read. The content of this register here implies a different meaning for Master8259 and Slave 8259. Through Master 8259 information is carried through the IR inputs to which Slave 8259s are connected. If the SLR of Master 8259 is loaded with the value 00001111, then it signifies that:The Slave exists on 8259 namely called as IR0, IR1, IR2 and IR3.No Slave exists on 8259 on this registers IR4, IR5, IR6 and IR7.A Slave 8259 serves information about the IR input of Master 8259 to which the ...
Read MoreProgramming the 8259 with no slaves
Now in this topic we assume that 8085 is the processor which is used in this microcomputer system. In this slave, no 8259 slaves are used. We should examine properly before 8259 PIC is used in the microcomputer system for performing the interrupt control application. 8259 is configured in such a fantastic way that is found that a variety of information are provided like for IR0 request IV, interrupts like level or edge-triggered, whether 8259s are used single or many, if ICW4 is in need or not and whether for the interrupt requests masking should be done or not. This ...
Read MoreProgram to Divide two 8 Bit numbers in 8051 Microprocessor
Here we will see the division operation. This operation will be used 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 number 0EH and 03H at location 20H and 21H, After dividing the result will be stored at location 30H and 31H.AddressValue …20H0EH21H03H …30H00H31H00H …Program MOV R0, #20H ; set source address 20H ...
Read MoreProgram to Multiply two 8 Bit numbers in 8051 Microprocessor
Here we will see how 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 MoreProgram to Subtract two 8 Bit numbers in 8051 Microprocessor
Now, in this section we will see how to subtract two 8-bit numbers using 8051 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 second operand.We are taking two number 73H and BDH at location 20H and 21H, After subtracting the result will be stored at location 30H and 31H.AddressValue …20H73H21HBDH …30H00H31H00H …Program MOV R0, #20H ; set source address 20H to R0 MOV R1, #30H ; set destination address 30H to R1 ...
Read MoreProgram to Divide two 8 Bit numbers in 8085 Microprocessor
Here we will see 8085 program. This program will divide two 8-bit numbers using 8085 microprocessor.Problem Statement −Write an 8085 Assembly language program to divide two 8-bit numbers and store the result at locations 8020H and 8021H.Discussion −The 8085 has no division operation. To get the result of division, we should use the repetitive subtraction method.By using this program, we will get the quotient and the remainder. 8020H will hold the quotient, and 8021H will hold remainder.We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.InputThe Dividend: 0EHThe Divisor 04HThe Quotient will be ...
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