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What is the instruction to add the data in register B to the data in accumulator?
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add B
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What the Difference between accumulator and instruction register?
Examples: /360: no accumulator 8080: A 6800: A and B 8086: AX ...
What the Difference between accumulator and register?
The accumulator is a general register that holds a value. It is also a special register that can be used as the target for the result of various arithmetic or logical computations. For instance, if you wanted to add two numbers, you could load the first into the accumulator, add the second to the accumulator, and then store the accumulator where wanted. The instruction register is an internal register that holds the value of the instruction opcode in order for the 8085 to decode and process the instruction. While it is shown on the Intel block diagram for the 8085, it is not directly accessible by the running program - it is for internal use only by the 8085.
What is the purpose of adc instruction according to 8085 microprocessor?
this instruction is used to add the specified register content to that of the accumulator along with the carry flag value. this instruction is used in processes which involve continuous addition.
How does the microprocessor work?
Suppose we give a 8-bit instruction ADD B to the microprocessor then this instruction is not at all understood by microprocessor as it only accepts binary inputs so first of all it stores the instruction in the INSTRUCTION REGISTOR then it decodes this instruction ADD B to its suitable binary code 80H in the INSTRUCTION DECODER.. after converting to 80H then the microprocessor understands that .. yes i have to add the content of the resistor B with that of A(accumulator) and store the result in the accumulator A this is a small example how microprocessor operates facing the instructions
How many number of machine cycles are there in each of the instructions in 8085?
There are following machine cycles of Intel 8085:Input-Output (I/O) Read Machine Cycle:The I/O Read Machine cycle is executed by the microprocessor to read a data from an input device. It consists of 3T states. The IN Instruction uses this machine cycle during the execution. Input - Output (I/O) Write Machine CycleThe I/O write machine cycle is executed by the microprocessor to write a data byte from an output device. It consists of 3T states. The instruction, which sends the data to the output device, comes under this machine cycle. Instruction cycle is defined, as the time required completing the execution of an instruction.An Instruction Cycle will have one to six machine cycles.
If the accumulator has C5H from a previous instruction and the carry flag is set then what will be the contents of the accumulator and carry flag after executing the instruction ADI 94H?
If you add 94H to C5H with the ADI instruction, the result in the accumulator will be 59H and the carry flag will be set. It does not matter what value the carry flag had to start with, because you said ADI, instead of ACI. (For ACI, the result would be 5AH with carry set.)
74 basic instructions set in 8085 microprocessor?
8085 Instruction Set Page 1 8085 INSTRUCTION SET INSTRUCTION DETAILS DATA TRANSFER INSTRUCTIONS Opcode Operand Description Copy from source to destination MOV Rd, Rs This instruction copies the contents of the source M, Rs register into the destination register; the contents of Rd, M the source register are not altered. If one of the operands is a memory location, its location is specified by the contents of the HL registers. Example: MOV B, C or MOV B, M Move immediate 8-bit MVI Rd, data The 8-bit data is stored in the destination register or M, data memory. If the operand is a memory location, its location is specified by the contents of the HL registers. Example: MVI B, 57H or MVI M, 57H Load accumulator LDA 16-bit address The contents of a memory location, specified by a 16-bit address in the operand, are copied to the accumulator. The contents of the source are not altered. Example: LDA 2034H Load accumulator indirect LDAX B/D Reg. pair The contents of the designated register pair point to a memory location. This instruction copies the contents of that memory location into the accumulator. The contents of either the register pair or the memory location are not altered. Example: LDAX B Load register pair immediate LXI Reg. pair, 16-bit data The instruction loads 16-bit data in the register pair designated in the operand. Example: LXI H, 2034H or LXI H, XYZ Load H and L registers direct LHLD 16-bit address The instruction copies the contents of the memory location pointed out by the 16-bit address into register L and copies the contents of the next memory location into register H. The contents of source memory locations are not altered. Example: LHLD 2040H 8085 Instruction Set Page 2 Store accumulator direct STA 16-bit address The contents of the accumulator are copied into the memory location specified by the operand. This is a 3-byte instruction, the second byte specifies the low-order address and the third byte specifies the high-order address. Example: STA 4350H Store accumulator indirect STAX Reg. pair The contents of the accumulator are copied into the memory location specified by the contents of the operand (register pair). The contents of the accumulator are not altered. Example: STAX B Store H and L registers direct SHLD 16-bit address The contents of register L are stored into the memory location specified by the 16-bit address in the operand and the contents of H register are stored into the next memory location by incrementing the operand. The contents of registers HL are not altered. This is a 3-byte instruction, the second byte specifies the low-order address and the third byte specifies the high-order address. Example: SHLD 2470H Exchange H and L with D and E XCHG none The contents of register H are exchanged with the contents of register D, and the contents of register L are exchanged with the contents of register E. Example: XCHG Copy H and L registers to the stack pointer SPHL none The instruction loads the contents of the H and L registers into the stack pointer register, the contents of the H register provide the high-order address and the contents of the L register provide the low-order address. The contents of the H and L registers are not altered. Example: SPHL Exchange H and L with top of stack XTHL none The contents of the L register are exchanged with the stack location pointed out by the contents of the stack pointer register. The contents of the H register are exchanged with the next stack location (SP+1); however, the contents of the stack pointer register are not altered. Example: XTHL 8085 Instruction Set Page 3 Push register pair onto stack PUSH Reg. pair The contents of the register pair designated in the operand are copied onto the stack in the following sequence. The stack pointer register is decremented and the contents of the highorder register (B, D, H, A) are copied into that location. The stack pointer register is decremented again and the contents of the low-order register (C, E, L, flags) are copied to that location. Example: PUSH B or PUSH A Pop off stack to register pair POP Reg. pair The contents of the memory location pointed out by the stack pointer register are copied to the low-order register (C, E, L, status flags) of the operand. The stack pointer is incremented by 1 and the contents of that memory location are copied to the high-order register (B, D, H, A) of the operand. The stack pointer register is again incremented by 1. Example: POP H or POP A Output data from accumulator to a port with 8-bit address OUT 8-bit port address The contents of the accumulator are copied into the I/O port specified by the operand. Example: OUT F8H Input data to accumulator from a port with 8-bit address IN 8-bit port address The contents of the input port designated in the operand are read and loaded into the accumulator. Example: IN 8CH 8085 Instruction Set Page 4 ARITHMETIC INSTRUCTIONS Opcode Operand Description Add register or memory to accumulator ADD R The contents of the operand (register or memory) are M added to the contents of the accumulator and the result is stored in the accumulator. If the operand is a memory location, its location is specified by the contents of the HL registers. All flags are modified to reflect the result of the addition. Example: ADD B or ADD M Add register to accumulator with carry ADC R The contents of the operand (register or memory) and M the Carry flag are added to the contents of the accumulator and the result is stored in the accumulator. If the operand is a memory location, its location is specified by the contents of the HL registers. All flags are modified to reflect the result of the addition. Example: ADC B or ADC M Add immediate to accumulator ADI 8-bit data The 8-bit data (operand) is added to the contents of the accumulator and the result is stored in the accumulator. All flags are modified to reflect the result of the addition. Example: ADI 45H Add immediate to accumulator with carry ACI 8-bit data The 8-bit data (operand) and the Carry flag are added to the contents of the accumulator and the result is stored in the accumulator. All flags are modified to reflect the result of the addition. Example: ACI 45H Add register pair to H and L registers DAD Reg. pair The 16-bit contents of the specified register pair are added to the contents of the HL register and the sum is stored in the HL register. The contents of the source register pair are not altered. If the result is larger than 16 bits, the CY flag is set. No other flags are affected. Example: DAD H 8085 Instruction Set Page 5 Subtract register or memory from accumulator SUB R The contents of the operand (register or memory ) are M subtracted from the contents of the accumulator, and the result is stored in the accumulator. If the operand is a memory location, its location is specified by the contents of the HL registers. All flags are modified to reflect the result of the subtraction. Example: SUB B or SUB M Subtract source and borrow from accumulator SBB R The contents of the operand (register or memory ) and M the Borrow flag are subtracted from the contents of the accumulator and the result is placed in the accumulator. If the operand is a memory location, its location is specified by the contents of the HL registers. All flags are modified to reflect the result of the subtraction. Example: SBB B or SBB M Subtract immediate from accumulator SUI 8-bit data The 8-bit data (operand) is subtracted from the contents of the accumulator and the result is stored in the accumulator. All flags are modified to reflect the result of the subtraction. Example: SUI 45H Subtract immediate from accumulator with borrow SBI 8-bit data The 8-bit data (operand) and the Borrow flag are subtracted from the contents of the accumulator and the result is stored in the accumulator. All flags are modified to reflect the result of the subtracion. Example: SBI 45H Increment register or memory by 1 INR R The contents of the designated register or memory) are M incremented by 1 and the result is stored in the same place. If the operand is a memory location, its location is specified by the contents of the HL registers. Example: INR B or INR M Increment register pair by 1 INX R The contents of the designated register pair are incremented by 1 and the result is stored in the same place. Example: INX H 8085 Instruction Set Page 6 Decrement register or memory by 1 DCR R The contents of the designated register or memory are M decremented by 1 and the result is stored in the same place. If the operand is a memory location, its location is specified by the contents of the HL registers. Example: DCR B or DCR M Decrement register pair by 1 DCX R The contents of the designated register pair are decremented by 1 and the result is stored in the same place. Example: DCX H Decimal adjust accumulator DAA none The contents of the accumulator are changed from a binary value to two 4-bit binary coded decimal (BCD) digits. This is the only instruction that uses the auxiliary flag to perform the binary to BCD conversion, and the conversion procedure is described below. S, Z, AC, P, CY flags are altered to reflect the results of the operation. If the value of the low-order 4-bits in the accumulator is greater than 9 or if AC flag is set, the instruction adds 6 to the low-order four bits. If the value of the high-order 4-bits in the accumulator is greater than 9 or if the Carry flag is set, the instruction adds 6 to the high-order four bits. Example: DAA 8085 Instruction Set Page 7 BRANCHING INSTRUCTIONS Opcode Operand Description Jump unconditionally JMP 16-bit address The program sequence is transferred to the memory location specified by the 16-bit address given in the operand. Example: JMP 2034H or JMP XYZ Jump conditionally Operand: 16-bit address The program sequence is transferred to the memory location specified by the 16-bit address given in the operand based on the specified flag of the PSW as described below. Example: JZ 2034H or JZ XYZ Opcode Description Flag Status JC Jump on Carry CY = 1 JNC Jump on no Carry CY = 0 JP Jump on positive S = 0 JM Jump on minus S = 1 JZ Jump on zero Z = 1 JNZ Jump on no zero Z = 0 JPE Jump on parity even P = 1 JPO Jump on parity odd P = 0 8085 Instruction Set Page 8 Unconditional subroutine call CALL 16-bit address The program sequence is transferred to the memory location specified by the 16-bit address given in the operand. Before the transfer, the address of the next instruction after CALL (the contents of the program counter) is pushed onto the stack. Example: CALL 2034H or CALL XYZ Call conditionally Operand: 16-bit address The program sequence is transferred to the memory location specified by the 16-bit address given in the operand based on the specified flag of the PSW as described below. Before the transfer, the address of the next instruction after the call (the contents of the program counter) is pushed onto the stack. Example: CZ 2034H or CZ XYZ Opcode Description Flag Status CC Call on Carry CY = 1 CNC Call on no Carry CY = 0 CP Call on positive S = 0 CM Call on minus S = 1 CZ Call on zero Z = 1 CNZ Call on no zero Z = 0 CPE Call on parity even P = 1 CPO Call on parity odd P = 0 8085 Instruction Set Page 9 Return from subroutine unconditionally RET none The program sequence is transferred from the subroutine to the calling program. The two bytes from the top of the stack are copied into the program counter, and program execution begins at the new address. Example: RET Return from subroutine conditionally Operand: none The program sequence is transferred from the subroutine to the calling program based on the specified flag of the PSW as described below. The two bytes from the top of the stack are copied into the program counter, and program execution begins at the new address. Example: RZ Opcode Description Flag Status RC Return on Carry CY = 1 RNC Return on no Carry CY = 0 RP Return on positive S = 0 RM Return on minus S = 1 RZ Return on zero Z = 1 RNZ Return on no zero Z = 0 RPE Return on parity even P = 1 RPO Return on parity odd P = 0 8085 Instruction Set Page 10 Load program counter with HL contents PCHL none The contents of registers H and L are copied into the program counter. The contents of H are placed as the high-order byte and the contents of L as the low-order byte. Example: PCHL Restart RST 0-7 The RST instruction is equivalent to a 1-byte call instruction to one of eight memory locations depending upon the number. The instructions are generally used in conjunction with interrupts and inserted using external hardware. However these can be used as software instructions in a program to transfer program execution to one of the eight locations. The addresses are: Instruction Restart Address RST 0 0000H RST 1 0008H RST 2 0010H RST 3 0018H RST 4 0020H RST 5 0028H RST 6 0030H RST 7 0038H The 8085 has four additional interrupts and these interrupts generate RST instructions internally and thus do not require any external hardware. These instructions and their Restart addresses are: Interrupt Restart Address TRAP 0024H RST 5.5 002CH RST 6.5 0034H RST 7.5 003CH 8085 Instruction Set Page 11 LOGICAL INSTRUCTIONS Opcode Operand Description Compare register or memory with accumulator CMP R The contents of the operand (register or memory) are M compared with the contents of the accumulator. Both contents are preserved . The result of the comparison is shown by setting the flags of the PSW as follows: if (A) < (reg/mem): carry flag is set if (A) = (reg/mem): zero flag is set if (A) > (reg/mem): carry and zero flags are reset Example: CMP B or CMP M Compare immediate with accumulator CPI 8-bit data The second byte (8-bit data) is compared with the contents of the accumulator. The values being compared remain unchanged. The result of the comparison is shown by setting the flags of the PSW as follows: if (A) < data: carry flag is set if (A) = data: zero flag is set if (A) > data: carry and zero flags are reset Example: CPI 89H Logical AND register or memory with accumulator ANA R The contents of the accumulator are logically ANDed with M the contents of the operand (register or memory), and the result is placed in the accumulator. If the operand is a memory location, its address is specified by the contents of HL registers. S, Z, P are modified to reflect the result of the operation. CY is reset. AC is set. Example: ANA B or ANA M Logical AND immediate with accumulator ANI 8-bit data The contents of the accumulator are logically ANDed with the 8-bit data (operand) and the result is placed in the accumulator. S, Z, P are modified to reflect the result of the operation. CY is reset. AC is set. Example: ANI 86H 8085 Instruction Set Page 12 Exclusive OR register or memory with accumulator XRA R The contents of the accumulator are Exclusive ORed with M the contents of the operand (register or memory), and the result is placed in the accumulator. If the operand is a memory location, its address is specified by the contents of HL registers. S, Z, P are modified to reflect the result of the operation. CY and AC are reset. Example: XRA B or XRA M Exclusive OR immediate with accumulator XRI 8-bit data The contents of the accumulator are Exclusive ORed with the 8-bit data (operand) and the result is placed in the accumulator. S, Z, P are modified to reflect the result of the operation. CY and AC are reset. Example: XRI 86H Logical OR register or memory with accumulaotr ORA R The contents of the accumulator are logically ORed with M the contents of the operand (register or memory), and the result is placed in the accumulator. If the operand is a memory location, its address is specified by the contents of HL registers. S, Z, P are modified to reflect the result of the operation. CY and AC are reset. Example: ORA B or ORA M Logical OR immediate with accumulator ORI 8-bit data The contents of the accumulator are logically ORed with the 8-bit data (operand) and the result is placed in the accumulator. S, Z, P are modified to reflect the result of the operation. CY and AC are reset. Example: ORI 86H Rotate accumulator left RLC none Each binary bit of the accumulator is rotated left by one position. Bit D7 is placed in the position of D0 as well as in the Carry flag. CY is modified according to bit D7. S, Z, P, AC are not affected. Example: RLC Rotate accumulator right RRC none Each binary bit of the accumulator is rotated right by one position. Bit D0 is placed in the position of D7 as well as in the Carry flag. CY is modified according to bit D0. S, Z, P, AC are not affected. Example: RRC 8085 Instruction Set Page 13 Rotate accumulator left through carry RAL none Each binary bit of the accumulator is rotated left by one position through the Carry flag. Bit D7 is placed in the Carry flag, and the Carry flag is placed in the least significant position D0. CY is modified according to bit D7. S, Z, P, AC are not affected. Example: RAL Rotate accumulator right through carry RAR none Each binary bit of the accumulator is rotated right by one position through the Carry flag. Bit D0 is placed in the Carry flag, and the Carry flag is placed in the most significant position D7. CY is modified according to bit D0. S, Z, P, AC are not affected. Example: RAR Complement accumulator CMA none The contents of the accumulator are complemented. No flags are affected. Example: CMA Complement carry CMC none The Carry flag is complemented. No other flags are affected. Example: CMC Set Carry STC none The Carry flag is set to 1. No other flags are affected. Example: STC 8085 Instruction Set Page 14 CONTROL INSTRUCTIONS Opcode Operand Description No operation NOP none No operation is performed. The instruction is fetched and decoded. However no operation is executed. Example: NOP Halt and enter wait state HLT none The CPU finishes executing the current instruction and halts any further execution. An interrupt or reset is necessary to exit from the halt state. Example: HLT Disable interrupts DI none The interrupt enable flip-flop is reset and all the interrupts except the TRAP are disabled. No flags are affected. Example: DI Enable interrupts EI none The interrupt enable flip-flop is set and all interrupts are enabled. No flags are affected. After a system reset or the acknowledgement of an interrupt, the interrupt enable flipflop is reset, thus disabling the interrupts. This instruction is necessary to reenable the interrupts (except TRAP). Example: EI 8085 Instruction Set Page 15 Read interrupt mask RIM none This is a multipurpose instruction used to read the status of interrupts 7.5, 6.5, 5.5 and read serial data input bit. The instruction loads eight bits in the accumulator with the following interpretations. Example: RIM Set interrupt mask SIM none This is a multipurpose instruction and used to implement the 8085 interrupts 7.5, 6.5, 5.5, and serial data output. The instruction interprets the accumulator contents as follows. Example: SIM
What is the function of immedidate instructions in the Intel 8085?
The immediate instructions in the 8085 contain the operand in the instruction, as the second byte, rather than in a register. For instance, if you want to add 3 to the accumulator, you can use ADI 3. Without the immediate instructions you would need two instructions, LHLD {address-of-a-3}, ADD M, but you might also need to save and restore HL.
Assembly language pgm to find sum of n numbers?
C100 LXI H C200 21 ; Load the HL register pair immediately C101 00 ; C102 C2 ; C103 MOV B M 46 ; Move the memory content (Total number of data) in to B register C104 MVI C 00 0E ; Initialize the C register with 00 H C 105 00 ; C106 SUB A 97 ; Clear the ac cumulator C107 INX H 23 ; Increment the HL register pair (to get the next input data) C108 MOV A M 7E ; Move the memory content in to accumulator C109 DCR B 05 ; Decrement the B register c o n tent C10A INX H 23 ; Increment the HL register pair C10B ADD M 86 ; Add the memory content to accumulator C10C JNC C110 D2 ; Jump if carry = 0, to C110 H C10D 10 ; C10E C1 ; C10F INR C 0C ; Increment the C register content C110 DCR B 05 ; Decrement the B register content C111 JNZ C10A C2 ; Jump if no zero to C10A H C112 0A ; C113 C1 ; C114 STA C300 32 ; Store the accumulator content (sum) at C300 H C115 00 ; C116 C3 ; C117 MOV A C 79 ; Move the C register content to accumulator C118 STA C301 32 ; Store the accumulator content ( carry ) at C300 H C119 01 ; C11A C3 ; C11B HL T 76 ; Halt the exection
What is xlat in 8086?
XLAT instruction converts the contents of AL register into a number stored in a memory table,this instruction perform the direct table look up technique often used to convert one code to another.An XLAT instruction 1st add the contents of this AL to BX to form a memory address within a data segment .It then copies the contents of this address into AL.This is the only instruction that adds an 8 bit number to 16-bit number.
Difference between direct addressing mode and indirect addressing mode in detail?
Direct addressing mode means the operand address is contained in the instruction. In the 8085, an example is LDA 1234H, which loads the accumulator with the contents of memory location 1234H. In the 8086/8088, an example is MOV AL,[1234H], which accomplishes nearly the same thing.Indirect addressing mode means the operand address is contained in a register. In the 8085, an example is LDAX B, which loads the accumulator with the contents of the memory location specified in register BC. In the 8086/8088, an example is MOV AL,[BX], which accomplishes nearly the same thing.
How do you draw timing diagram of ADD M instruction?
I'm assuming that you mean "Add the contents of memory to an accumulator and place the sum back in the accumulator". First, the address of the memory location needs to be placed on the address bus and the control lines to memory need to be driven in such a way as to tell the memory that a write operation has been requested (usually enabling OE and synchronizing with a clock). Next (one or more clock cycles later depending on the type of memory being used), the data is returned from the memory. It must then be routed to the ALU. The value in the accumulator must also be moved to the ALU and can be performed in either the first or second CPU cycle, depending on the internal architecture of the CPU. It is probably safest to do this in the first cycle. Next the ALU must perform the addition instruction (third CPU cycle). Finally, in the fourth cycle, the result can be moved from the ALU back to the accumulator.
