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Articles by George John
Page 42 of 79
Restart instructions (RSTn) in 8085 Microprocessor
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 MoreHLT instruction in 8085
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 MorePin description of 6800
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 MoreProgramming examples of 6800
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 MoreALE pin in 8085 Microprocessor
Intel 8085 is an 8-bit microprocessor which has 16 address line for 16-bit address of a memory location. 8 higher order address bits are transferred through 8 bit lines out of this 16 address line while remaining lower order 8 bits of the address are sent through another 8 lines multiplexed with the 8-bit data lines. ALE (Address Enable Latch) is the control signal which is nothing but a positive going pulse generated when a new operation is started by microprocessor. So when pulse goes high means ALE=1, it makes address bus enable and when ALE=0, means low pulse makes ...
Read MoreProgrammer's view of 8085 Microprocessor
Intel 8085 receives 8-bit information on AD7-0 from memory or in-port which resides inside the microprocessor via“register”. A register is a group of flip-flops, where each flip-flop can store a bit of information. To store 8 bitsof information, the size of a register in 8085 has to be 8 bitsThe advantages of a register over a memory location is the contents of a register can be accessed much faster by the microprocessor, compared with the contents of a memory location.However, the disadvantages of a register over a memory location are as follows.If there are too many registers, they occupy a lot ...
Read MoreProgram memory structure of Intel 8051
We have already discussed that the 8051 microcontroller has the internal program memory. In this chip, there is EA pin. It indicates the External Access. So by using this pin, we can check whether the internal program memory is used or not. So when there is a low signal in this pin, the internal 4K bytes of program memory is not used, but in this situation, it can access only the ROM.When we are trying to access the external data memory, then the read RD or write WR will be the output from 8051. So for reading the data from ...
Read More2's complement fractions
As an example, the value of 1 001, if the interpretation is that it is a 2's complement fraction will be as follows -It is 1.001 assuming the binary point after the MS bit. As the MS bit is 1, it is a negative number. Then the remaining bits do not specify the magnitude directly. The 2's complement of 1 001 is 0110+ 1 = 0 111. This is a positive fraction with the value 1 × 2−1 +1 × 2−2 +1 × 2−3 = 0.5 + 0.25 + 0.125 = 0.875 decimal. Thus, -0.875 is the value of 1001.If ...
Read MoreSign Magnitude notation
The sign-magnitude binary format is the simplest conceptual format. In this method of representing signed numbers, the most significant digit (MSD) takes on extra meaning.If the MSD is a 0, we can evaluate the number just as we would any normal unsigned integer. And also we shall treat the number as a positive one.If the MSD is a 1, this indicates that the number is negative.The other bits indicate the magnitude (absolute value) of the number. Some of the signed decimal numbers and their equivalent in SM notation follows assuming a word size of 4 bits.Signed decimalsign-magnitude +6 0110 ...
Read MoreBinary to BCD conversion in 8051
In this problem, we will see how to convert an 8-bit binary number to its BCD equivalent. The binary number is stored at location 20H. After converting, the results will be stored at 30H and 31H. The 30H will hold the MS portion, and 31H will hold the LS portion. So let us assume the data is D5H. The program converts the binary value of D5H to BCD value 213D.AddressValue...20HD521H...ProgramMOVR1, #20H;Takethe address 20H into R1 MOVA, @R1;Takethe data into Acc MOVB, #0AH;LoadB with AH = 10D DIVAB ;DivideA with B MOVR5, B;Storethe remainder MOVB, #0AH;LoadB with AH = 10D DIVAB ;DivideA ...
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