The 8051 Microcontroller is designed by Intel in 1981. 8051 is 40 pin DIP IC. This is an 8-bit microcontroller.It is a very famous Microcontroller of Intel family. As we know Microprocessor like 8085, Z-80 or M6800 contains the ALU, control unit and some registers, but there were no internal storages. The Microcontroller contains the RAM and ROM, IO ports inside it. Some microcontroller also contains the ADC in it. As most of the required components are inside the chip, so the cost of application is also reduced.These kinds of microcontrollers are used in different appliances like televisions, washing machines, ... Read More

Now, in this section, we will see how we can use the Motorola M6800 Microprocessor to add multi-Byte numbers.AddingMulti-Byte NumberIn this example, we are using 4-Byte numbers (56 2F 7A 89)16 and (21 FB A9 AF)16In the memory, at first, we are storing the Byte counts, and then the numbers (from least significant Bytes to Most significant Bytes) in different segments. So after storing the data, the memory structure will be looking like thisAddressValue5000H04H...5050H89H5051H7AH5052H2FH5053H56H...5070HAFH5071HA9H5072HFBH5073H21H...Now, we are writing a program to add these two 4-Byte number and store the result at location 5090H onwards.Program      CLC       LDX#$5050 ... Read More

In Motorola M6800, there are two hardware interrupt pins. These pins are NMI and. IRQ  These pins are active low input pins. The first one is non-maskable and the second one is maskable and lower priority interrupt. When the IM flag is 1, or CCR register is set, the interrupt is masked or disabled.When the Micro processor enters into some Interrupt Service Subroutine (ISS), it uses SEI instruction to mask the interrupt even if the IRQ activated. The reverse action can be performed using the CLI instruction. It can unmask the interrupt.When in the interrupt is occurred, the M6800 follows ... Read More

In this section, we will see the different types of instructions of Motorola M6800 microprocessor. There are 72 different types of instructions and 197 different opcodes. So there are 51 1-Byte instruction, 103 2-Byte instruction and 43 3-Byte instruction.As we know that the Intel 8085 has 246 opcodes, though 6800 is more powerful than 8085. The Z-80 has 700 instructions but M6800 has some more advanced branching instructions.The different instruction groups are like these −Data Transfer GroupArithmetic GroupLogical GroupBranch GroupMiscellaneous InstructionsData transfer GroupIn this group, there are 14 instructions. We can find 38 opcodes for these 14 instructions. These instructions ... Read More

We have seen the Internal structure and registers of Motorola M6800 Microprocessor. In this section we will see the addressing modes of M6800.There are six addressing modes in M6800 MPU. These modes are −Immediate Addressing ModeImplied Addressing ModeDirect Addressing ModeExtended Addressing ModeIndexed Addressing ModeRelative Addressing ModeNow Let us see some basic syntax of M6800 Assembly Language Programming.If a number is 8CH in hexadecimal, then we have to use $ sign before it. So it will be $8C. A number without $ is treated as decimal number. Similarly, an immediate data is denoted by # symbol. #50FF is a data, but ... Read More

In this section, we will see the basic architecture of the Motorola M6800 Microprocessor, and different registers to write programs into it.To write programs we have to care about the registers and some instructions to access them during program execution. So from this diagram, we can see that there are two 8-bit Accumulators A and B, some 16-bit registers (Program Counter PC, Index Register IX, Stack Pointer SP), and 8-bit flag register CCR.Both of the Accumulators Aand B have the same prominence in the Instruction set. There are very few instructions in 6800, where the Accumulator A is used but B ... Read More

The Motorola M6800 is 40pin DIP Microprocessor. Here we will see the actual pin level diagram of M6800 and also the functional pin diagram of it.The M6800 requires some additional chips to provide the required functions. These chips are:6870 (clock generator)6830 (ROM) or 68708(EPROM)6810 (RAM)6820 (PeripheralInterface Adapter)6850 (AsynchronousCommunications Interface Adapter)6828 (PriorityInterrupt Controller)Now see the Pin Level diagram of Motorola M6800This is the actual pin diagram of the M6800 Microprocessor. Now we will see the functional pin diagram of it.Now let us see the Pin functions of the M6800 microprocessor.PinsTypeFunctionA15– A0Output16-bit address bus, which provides the addresses for memory (up to ... Read More

The Motorola M6800 is 40pin DIP IC. It is an 8-bit Microprocessor. It was introduced in 1974 by Motorola. It was designed nearly at the same time when the 8080 was introduced. Like 8080, it also had 16-bit address bus to handle 64KB of data, and 8-bit data bus to read and write data. The M6800 has very few registers. It has two Accumulators A and B, and the Index Register. This Microprocessor only supports the memory mapped IO because there is no instruction regarding IO operations in its instruction set.The M6800 has six addressing modes, 72 instructions and at least ... Read More

In 8085 Instruction set, IN is a mnemonic that stands for INput the Byte from input port’s content to the accumulator. Input port’s 8-bit address is indicated in the instructions as a8. It occupies 2 Bytes in the memory. First Byte specifies the opcode, and the next Byte provides the 8-bit input port address. Mnemonics, OperandOpcode(in HEX)BytesIN Port-addressDB2IN F0H is an example instruction of this type. The result of execution of this instruction is shown below with an example.BeforeAfter(A)Any ValueABHInput Port F0HABHABHIN instruction is the only instruction using which read the input port content to the Accumulator. A possible chip select ... Read More

In 8085 Instruction set, HLT is the mnemonic which stands for ‘Halt the microprocessor’ instruction. It is having a size of 1-Byte instruction. Using these particular instructions, as 8085 enters into the halt state, so we can put the8085 from further processing of next instructions. This is indicated by S1 and S0 control signals. During the halt, S1 and S0 output signals will become 0 0. Mnemonics, OperandOpcode(in HEX)BytesHLT761The 8085 comes out of the Halt state when a valid interrupt occurs. In such a case, it executes the corresponding interrupt service subroutine depending upon the interrupt number and then it continues ... Read More