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

In 8085 Instruction set, for 16-bit addition, there is one instruction available that is DAD rp instruction. It is a 1-Byte instruction. With this instruction, with the content of the HL register pair, the contents of the mentioned register pair will get added and the result thus produced will be stored on the HL register pair.Mnemonics, OperandOpcode(in HEX)BytesDAD B091DAD D191DAD H291DAD SP391As an example, let us consider the execution of the DADB instruction. Let us suppose, that the initial content of HL register pair is 5050H and content of BC register pair is 4050H. So now if we execute instruction ... Read More

In 8085 Instruction set, OUT is a mnemonic that stands for OUTput Accumulator contents to an output port whose8-bit address is indicated in the instruction as a8. It occupies 2 Bytes in the memory. First Byte specifies the opcode, and the next Byte provides the 8-bit port address. Mnemonics, OperandOpcode(in HEX)BytesOUT Port-AddressD32OUT F0H is an example instruction of this type. The result of execution of this instruction is shown below with an example.BeforeAfter(A)ABHABHOutput Port F0HAny ValueABHOUT instruction is the only instruction using whichAccumulator contents can be sent out to an output port. A possible chip select circuit to connect an output ... Read More

In 8085 Instruction set, there are two instructions in 8085 for communication with I/O ports. They are the IN and OUT instructions. The IN or OUT instruction mnemonics should be followed by an 8-bit port address. Thus we can have 28 = 256 input ports and 256 output ports are possible in 8085-based microcomputers. IN and OUT both are 2-Bytes instructions.Mnemonics, OperandOpcode(in HEX)BytesIN Port-addressDB2OUT Port-AddressD32In case of IN instruction, the current 8-bit content of the PORT# will be made available on to the Accumulator. Let us suppose with the PORT#, 8 DIP switches are connected. And their states are ON-ON-OFF-ON-ON-ON-OFF-ON. ... Read More

In 8085 Instruction set, RSTn is actually standing for “Restart n”. And in this case,  n has a value from 0 to 7 only. Thus the eight possible RST instructions are there, e.g. RST 0, RST 1, …, RST 7. They are 1-Byte call instructions. Functionally RST n instruction is similar with:RST n = CALL n*8For example, let us consider RST 4 is functionally equivalent to CALL 4*8, i.e. CALL 32 = CALL 0020H. The advantage of RST 2 is that it is only 1 Byte, whereas CALL 0010H is 3-Byte long. Thus RST instructions are useful for branching to frequently used subroutines.Mnemonics, OperandOpcode(in ... Read More

In 8085 Instruction set, RM is a mnemonic, which stands for “Return if Minus”. This instruction is used to return to the main program, only if S flag value is 1. If S flag value is 0, program flow continues in the subroutine sequentially. It is a 1-Byte instruction.Mnemonics, OperandOpcode(in HEX)BytesRMF81Let us consider the following sample code for a better explanation –AddressHex CodesMnemonicComment200031LXI SP, 5000HSP ← 5000H.Initializing the SP200100Low order Byte of the address200250High order Byte of the address200321LXI H, 4050HHL ← 4050H, Initializing the HL register pair200450Low order Byte of the address200540High order Byte of the address2006CDCALL 200BHCalling the ... Read More

In 8085 Instruction set, RP is a mnemonic, which stands for “Return if Positive”. This instruction is used to return to the main program, only if S flag value is 0. If S flag value is 1, program flow continues in the subroutine sequentially. It is a 1-Byte instruction.Mnemonics, OperandOpcode(in HEX)BytesRNZF01Let us consider the following sample code for a better explanation –AddressHex CodesMnemonicComment200031LXI SP, 5000HSP ← 5000H.Initializing the SP200100Low order Byte of the address200250High order Byte of the address200321LXI H, 4050HHL ← 4050H, Initializing the HL register pair200450Low order Byte of the address200540High order Byte of the address2006CDCALL 200BHCalling the ... Read More