In the subtraction of two numbers, 8085 imposes the restriction that Accumulator will have one of the operands from which the other operand specified by one of the following will be subtracted.—Contents of an 8-bit register;—Contents of memory location pointed by HL pair;—Eight-bit immediate data.In 8085 Instruction, SUB is a mnemonic that stands for ‘SUBtract contents of R from Accumulator’. Here R stands for any of the following registers, or memory location M pointed by HL pair.R = A, B, C, D, E, H, L, or MMnemonics, Operand Opcode (in HEX)BytesSUB A971SUB B901SUB C911SUB D921SUB E931SUB H941SUB L951SUB M961In this instruction ... Read More

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 More

In the ALS-SDA-85M kit, we have 74138 EPROM of minimum size 16K x 8 and size of RAM IS 2K x 8, there are empty sockets for the EPROM to get expanded. The selection of these four chips is done by the integrated circuit 74138. Every time the selection of 74138 is done when IO/M* turns to 0, the election of 27138 is done when A15A14 turns to 0. 27138 have the lowest address range likeLowest address: 00 00000000000000 is 0000H.Highest address: 00 11111111111111 is 3FFFH.Whereas in 74138 the address ranges are as follows -Lowest address: 11 xxx 00000000000Highest Address: 11 xxx 11111111111.The ... Read More

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 ... Read More

Let’s take an Integrated Circuit decoder. We take the popular 3 to 8 decoder Integrated Circuit 74138. The Integrated Circuit is of 16 pins.We have three input pins which are actively in high state and are classified as I2, I1 and I0. The outputs are actively in low state and are eight in number and are classified as O7*, O6*, …, O0*. A power supply of +5 V DC is needed by the chip and is Grounded.                                                ... Read More

Suppose we have the chip select circuit as shown in the below figure, then what is the address range for the RAM?Fig. Chip selection that results in multiple address range.A15 and A14 pins are used for the selection of Random Access Memory, and pins ranging from A10 to A0 are used for the selection of location in RAM. But the value of the pins A13, A12 and A11 have non-dependent value. They play no role for selection of chip or location on RANDOM ACCESS MEMORY. Hence they are termed as don't cares and called x for more signification.We have eight ... Read More

Let us assume that we have a RAM chip of 2K x 8, It signifies that, we have 2 x 1024 = 2048 memory locations having information of 8 bits at each location. For selecting one of 2K = 211 memory locations in chip, there should be exactly input pins of 11 address lines. For bidirectional data transfer scheme, there should be eight pin involved in this process. The processor action involved in it is signified by the input pins RD* and WR*. Two pins are specified for grounding process of the pin and connecting of Direct Current of +5 ... Read More

The master of microcomputer system is the microprocessor since all the operations of a computer are controlled by the microprocessor, the control unit often called as (CU) is found in the microprocessor. In the microcomputer system apart from processor there are several chips of RAM, CHIPS of EPROM and other Input Output port chips, timer 8253 is also present. But surprisingly at any moment of time the remains with only one of the chip. By activation of the chip selection the processor, the selection of the chip of the processor is done. Hence except the processor all chips must have ... Read More

Running the program in Single-Step Mode:The single step mode is very useful for finding our program which has not yielded the desired outcomes after being executed in a single burst. We use it for program debugging. Similarly, like the operation to be executed singly ‘S’ is typed at the ‘>’ prompt. We can notice clearly that the change of address occurs after S. The given system clearly prompts which is stated below:The address of starting is: xxxx - yy/where the memory address is xxxx and the content of the located memory is yy which allows for responding the user ... Read More

In a single operation, we can run the entire program of 8085. We run the entire program in this mode so that we can get the accurate results. If the process is unsuccessful a single-stepping by the entire program is attempted.We type ‘G’ at the prompt ‘>’. By noticing the absence of change of address after G. The prompting of the system are as follows:The address of starting is: xxxx - yy/where the memory address is xxxx and the content of the memory location is yy. Which allows the user for responding with the desired address of starting. If ... Read More