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8085 Articles
Found 357 articles
Concept of Direct Memory Access (DMA)
The microcomputer system basically consists of three blocksThe microprocessorThe memories of microprocessor like EPROM and RAMThe I/O ports by which they are connected.The possible data transfers are indicated below.Between the memory and microprocessor data transfer occurs by using the LDA and STA instructions.Between microprocessor and I/O ports also data transfer occurs by the help of two instructions IN and OUT.Through DMA data transfer, data is transferred between the Input Output ports and the memory.Fig: The figure shows possible ways of data transfers in a microcomputer system.For performing the data transfer either between the microprocessor and memory or between I/O ports ...
Read MoreTimers of 8051
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 MoreAddressing modes of 8051
In this section, we will see different addressing modes of the 8051 microcontrollers. In 8051 there are 1-byte, 2-byte instructions and very few 3-byte instructions are present. The opcodes are 8-bit long. As the opcodes are 8-bit data, there are 256 possibilities. Among 256, 255 opcodes are implemented.The clock frequency is12MHz, so 64 instruction types are executed in just 1 µs, and rest are just 2 µs. The Multiplication and Division operations take 4 µsto to execute.In 8051 There are six types of addressing modes. Immediate AddressingModeRegister AddressingModeDirect AddressingModeRegister IndirectAddressing ModeIndexed AddressingModeImplied AddressingModeImmediate addressing modeIn this Immediate Addressing Mode, the data ...
Read More8051 Program to Add two 8 Bit numbers
Intel 8051 is an 8-bit microcontroller. It has many powerful instructions and IO accessing techniques. In this section, we will see one of the simplest program using 8051.Here we will add two8-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 number 5FH and D8H at location 20H and 21H, After adding them, the result will be stored at location 30H and 31H. AddressValue...20H5FH21HD8H...30H00H31H00H...ProgramMOVR0, #20H;set source address 20H to R0 ...
Read More8085 program to multiply two 8 bit numbers
In this program we will see how to multiply two 8-bit numbers using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to multiply two 8-bit numbers stored in memory location and store the 16-bit results into the memory.DiscussionThe 8085 has no multiplication operation. To get the result of multiplication, we should use the repetitive addition method.After multiplying two 8-bit numbers it may generate 1-byte or 2-byte numbers, so we are using two registers to hold the result.We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.InputAddressData......8000DC8001AC......Flow DiagramProgramAddressHEX CodesLabelsMnemonicsCommentsF00021, 00, 80LXI H, 8000HLoad ...
Read More8085 Program to Add two 8 Bit numbers
In this program, we will see how to add two 8-bit numbers using 8085 microprocessor.Problem StatementWrite 8085 Assembly language program to add two 8-bit numbers and store the result at locations 8050H and 8051H.DiscussionTo perform this task, we are using the ADD operation of 8085 Microprocessor. When the result of the addition is the 1-byte result, then the carry flag will not be enabled. When the result is exceeding the 1-byte range, then the carry flag will be 1We are using two numbers at location 8000H and 8001H. When the numbers are 6CH and 24H, then the result will be ...
Read MoreStack and the stack pointer in 8085 Microprocessor
The stack is a LIFO (last in, first out) data structure implemented in the RAM area and is used to store addresses and data when the microprocessor branches to a subroutine. Then the return address used to get pushed on this stack. Also to swap values of two registers and register pairs we use the stack as well.In the programmer‘s view of 8085, only the general purpose registers A, B, C, D, E, H, and L, and the Flags registers were discussed so far. But in the complete programmer’s view of 8085, there are two more special purpose registers, each ...
Read MoreInternal Data Memory Organization of Intel 8051
The internal data memory of 8051 is divided into two groups. These are a set of eight registers and a scratch pad memory. These eight registers are R0 toR7. The address range 00H to 07H is used to access the registers, and the rest are scratch pad memory. 8051 Provides four register bank, but only one register bank can be used at any point in time. To select the register bank, two bits of PSW (Program Status Word) are used.So the following addressing can be used to select register banks.Address RangeRegister Bank00H to 07HRegister Bank 008H to 0FHRegister Bank 110H to ...
Read MoreFlags register in 8085 Microprocessor
In 8085 microprocessor, the flags register can have a total of eight flags. Thus a flag can be represented by 1 bit of information. But only five flags are implemented in 8085. And they are:Carry flag (Cy), Auxiliary carry flag (AC), Sign flag (S), Parity flag (P), andZero flag (Z).The respective position of these flag bits in flag register has been show the below figure. The positions marked by “x” are to be considered as don't care bits in the flags register. The user is not required to memorize the positions of these flags in the flags register.Fig. Flags registerNow ...
Read MoreAdvantage of multiple chip select lines
Let’s consider that the EPROMs we have are having the starting addresses as 4000H, 4400H, …, 5C00H. 4000H in binary is 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0. In such a case, the 74138 has to be selected for the condition A15 A14 A13 = 0 1 0. This can be achieved by the connection shown in below figure which uses two invert gates.The procedure of selecting the chip74138 is denoted by the addresses A15 A14 A13 = 0 1 0 which can be implemented by any gates. We have taken ...
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