UNIT – III MICROPROCESSOR 8085

INTRODUCTION:

Ø The microprocessor is a clock-driven semiconductor device consistingof electronic logic circuits manufactured by using either a large-scaleintegration (LSI) or very-large-scale integration (VLSI) technique.

Ø The microprocessor is capable of performing various computingfunctions and making decisions to change the sequence of programexecution. In large computers, a CPU implemented on one or morecircuit boards performs these computing functions.

Ø The microprocessor is in many ways similar to the CPU, but includes allthe logic circuitry, including the control unit, on one chip.

Ø The microprocessor can be divided into three segments for the sake ofclarity, arithmetic/logic unit (ALU), register array, and control unit.

Arithmetic/Logic Unit: This is the area of the microprocessor where variouscomputing functions are performed on data. The ALU unit performs such arithmeticoperations as addition and subtraction, and such logic operations as AND, OR, andexclusive OR.Register Array: This area of the microprocessor consists of various registersidentified by letters such as B, C, D, E, H, and L. These registers are primarily usedto store data temporarily during the execution of a program and are accessible to theuser through instructions.Control Unit: The control unit provides the necessary timing and control signals toall the operations in the microcomputer. It controls the flow of data between themicroprocessor and memory and peripherals.Memory: Memory stores such binary information as instructions and data, andprovides that information to the microprocessor whenever necessary. To executeprograms, the microprocessor reads instructions and data from memory and

performs the computing operations in its ALU section. Results are either transferredto the output section for display or stored in memory for later use. Read-Onlymemory (ROM) and Read/Write memory (R/WM), popularly known as Random-Access memory (RAM).

1. The ROM is used to store programs that do not need alterations. Themonitor program of a single-board microcomputer is generally storedin the ROM. This program interprets the information entered through akeyboard and provides equivalent binary digits to the microprocessor.Programs stored in the ROM can only be read; they cannot be altered.

2. The Read/Write memory (RIWM) is also known as user memory It isused to store user programs and data. In single-boardmicrocomputers, the monitor program monitors the Hex keys andstores those instructions and data in the R/W memory. Theinformation stored in this memory can be easily read and altered.

I/O (Input/Output): It communicates with the outside world. I/O includes two types of devices: input and output; these I/O devices are also known as peripherals.Ø System Bus: The system bus is a communication path between the

microprocessor and peripherals: it is nothing but a group of wires to carry bits

MICROPROCESSOR ARCHITECTURE AND ITS OPERATIONS:

Architecture of 8085 Microprocessor

Ø Address Bus: The address bus is a group of 16 lines generally identified asA0 to A15. The address bus is unidirectional: bits flow in one direction —fromthe MPU to peripheral devices. The MPU uses the address bus to perform thefirst function: identifying a peripheral or a memory location.

Data Bus:The data bus is a group of eight lines used for data flow. These lines are

bidirectional—data flow in both directions between the MPU and memory andperipheral devices. The MPU uses the data bus to perform the second function:transferring binary information .The eight data lines enable the MPU to manipulate 8-bit data ranging from 00 to FF (28 = 256 numbers). The largest number that canappear on the data bus is 11111111.Control Bus:

The control bus is comprised of various single lines that carry synchronizationsignals, providing timing signals. The MPU generates specific control signals for everyoperation it performs. These signals are used to identify a device type with which theMPU intends to communicate.

Registers:

8085 Bus Structure

The 8085 programmableregister

The 8085 have six general-purpose registers to perform the first operationlisted above; that is, to store 8-bit data during program execution. These registersare identified as B, C, D, E, H, and L. They can be combined as register pairs —BC,DE, and HL—to perform some 16-bit operations.

Ø Accumulator: The accumulator is an 8-bit register that is part of thearithmetic/logic unit (ALU). This register is used to store 8-bit data and toperform arithmetic and logical operations. The result of an operation isstored in the accumulator. The accumulator is also identified asregister A.

Ø Flags:The ALU includes five flip-flops that are set or reset according to the result of

an opera tion. The microprocessor uses them to perform the third operation; namely,testing for data conditions. They are Zero (Z), Carry (CY), Sign (S), Parity (P), andAuxiliary Carry (AC) flags. The most commonly used flags are Sign, Zero, and Carry;the others will be explained as necessary.

The bit position for the flags in flag register is,D7 D6 D5 D4 D3 D2 D1 D0

S Z AC P CY

(1) Sign Flag (S): After execution of any arithmetic and logicaloperation, if D7 of the result is 1, the sign flag is set. Otherwise it isreset. D7 is reserved for indicating the sign; the remaining is themagnitude of number. If D7 is 1, the number will be viewed asnegative number. If D7 is 0, the number will be viewed as positivenumber.

(2) Zero Flag (z): If the result of arithmetic and logical operation iszero, then zero flag is set otherwise it is reset.

(3) Auxiliary Carry Flag (AC): If D3 generates any carry when doingany arithmetic and logical operation, this flag is set. Otherwise it isreset.

(4) Parity Flag (P): If the result of arithmetic and logical operationcontains even number of 1’ s then this flag will be set and if it is oddnumber of 1’ s it will be reset.

(5) Carry Flag (CY): If any arithmetic and logical operation resultsany carry then carry flag is set otherwise it is reset.

Program Counter (PC):ØØ This 16-bit register deals with the fourth operation, sequencing the

execution of instructions. This register is a memory pointer.ØØ The microprocessor uses this register to sequence the execution of

instructions. The function of the program counter is to point to thememory address from which the next byte is to be fetched.

ØØ When a byte (machine code) is being fetched, the program counteris incremented by one to point to the next memory location.

Stack Pointer (Sp):The stack pointer is also a 16-bit register used as a memory pointer; initially,

it will be called the stack pointer register to emphasize that it is a register. It points

to a memory location in R/W memory, called the stack. The beginning of the stack isdefined by loading a 16-bit address in the stack pointer (register).

Temporary Register: It is used to hold the data during the arithmetic and logicaloperations.

Instruction Register: When an instruction is fetched from the memory, it is loadedin the instruction register.

Instruction Decoder: It gets the instruction from the instruction register anddecodes the instruction. It identifies the instruction to be performed.

Serial I/O Control: It has to control signals named SID and SOD for serial datatransmission.

Timing and Control unit: It has control and status signals. It provide control signalto synchronize the components of microprocessor and timing for instruction toperform the operation.

Interrupt Control Unit: It is used to receive an interrupt signal for process theoperation and send an acknowledgement for receiving the interrupt signal.

I/O INTERFACING:

The two methods of data transfer are:• Serial I/O : One bit is transferred using one data line• Parallel I/O : Data can enter (or exit) in groups of eight bits using the

entire data bus.

Thus the I/O devices (Keyboards and displays) can be interfaced using two methodsnamely:

• Peripheral-mapped I/O: The device is identified with an 8-bit address andenabled by I/O – related control signals.

• Memory-mapped I/O: The device is identified with a 16-bit address andenabled by memory related control signals.

Concepts Review for peripheral mapped I/Oa) When an I/O instruction is executed, 8085 places the device address (Port

number) on the de multiplexed low – order and high – order address bus.b) The address is decoded to generate the pulse corresponding to the device.c) The device address pulse is AND ed with the appropriate control signals like

IO/M, RD and WR to assert the I/O device.d) A latch is used for an output port and a tri-state buffer is used for an input

port..e) The address bus can be decoded by using either the absolute or the linear

select decoding reduces the component cost but the I/O device ends up withmultiple addresses.

Memory Mapped I/O :

Here, the input and output devices are assigned and identified by 16-bitaddresses. To transfer data between 8085 and I/C devices, memory – relatedinstructions like LDA, STA etc are used. The control signals MEMR and MEMWshould be connected to I/O devices.

STA 8000H; Address 2050H; stores the contents of A to 8000.

2050 32;2051 00;2052 80;

8085 requires 4 Machine cycles to execute STA; Instruction fetch and decodein M1; read 2051 and 2052 in M2 an M3; In M4 8085 places the entire address(8000H) on the address lines, the contents of the accumulator on data bus andgenerates MEMW

CharacteristicsMemory-Mapped I/O Peripheral I/O

1. Device address 16 Bit 8 Bit2. Control Signals RD / WR (MEMR /

MEMW)IO/M , RD / WR (IOR,IOW)

3. Instructions Memory relatedinstructionsLDA, STA, MOV, ADD,SUB

IN, OUT

4. Data transfer Between any registerand I/O

Between I/O andAccumulator

5.Maximum number of I/O 64K memory sharedbetween I/O and systemmemory.

256 Input; 256 Output

6. Execution 13 T States (LDA, STA)7 T States (MOV)

10 T States

7. Hardware More hardware requiredto decode 16 bit address

Less hardware to decode8 bit address

8. Other features Arithmetic or Logicaloperations can beperformed

Not Possible.

8085 INSTRUCTION SET:

The 8085 instruction set can be classified into the following five functionalheadings.1. DATA TRANSFER INSTRUCTIONS :

Includes the instructions that moves (copies) data between registers orbetween memory locations and registers. In all data transfer operations the contentof source register is not altered. Hence the data transfer is copying operation.2. ARITHMETIC INSTRUCTIONS:

Includes the instructions, which performs the addition, subtraction,increment or decrement operations. The flag conditions are altered after execution ofan instruction in this group.

3. LOGICAL INSTRUCTIONS:The instructions which performs the logical operations like AND, OR,

EXCLUSIVE- OR, complement, compare and rotate instructions are grouped underthis heading. The flag conditions are altered after execution of an instruction in thisgroup.

4. BRANCHING INSTRUCTIONS: The instructions that are used to transfer the program control from one

memory location to another memory location are grouped under this heading.

5. MACHINE CONTROL INSTRUCTIONS:Includes the instructions related to interrupts and the instruction used to halt

program execution.

1. ACI: Add Immediate to Accumulator with Carry.

Description: The 8-bit data (operand) and the Carry flag are added to the contentsof the accumulator, and the result is stored in the accumulator. All flags are modifiedto reflect the result of the addition.

2. ADC: Add Register to Accumulator with Carry

Description: The contents of the operand (register or memory) and the Carry flagare added to the contents of the accumulator and the result is placed in theaccumulator.The contents of the operand are not altered; however, the previous Carry flag isreset.All flags are modified to reflect the result of the addition.

3. ADD: Add Register to Accumulator

Description: The contents of the operand (register or memory) are added to thecontents of the accumulator and the result is stored in the accumulator. If theoperand is a memory location, that is indicated by the 16-bit address in the HLregister. All flags are modified to reflect the result of the addition.

4. ADI: Add Immediate to Accumulator

Description : The 8-bit data (operand) are added to the contents of theaccumulator, and the result is placed in the accumulator. All flags are modified toreflect the result of the addition.

5. ANA: Logical AND with Accumulator

Description: The contents of the accumulator are logically AND ed with thecontents of the operand (register or memory), and the result is placed in theaccumulator. If the operand is a memory location, its address is specified by thecontents of HL registers. Flags S, Z, P are modified to reflect the result of theoperation. CY is reset. In 8085 AC is set.

6. ANI: AND Immediate with Accumulator

Description: The contents of the accumulator are logically AND ed with the 8-bitdata (operand) and the results are placed in the accumulator. Flags S, Z, P aremodified to reflect the results of the operation. CY is reset. In 8085, AC is set.

7. CALL: Unconditional Subroutine Call

Description: The program sequence is transferred to the address specified by theoperand. Before the transfer, the address of the next instruction to CALL (thecontents of the program counter) is pushed on the stack.

8. CMA: Complement Accumulator

Description: The contents of the accumulator are complemented. No flags areaffected.

9. CMC: Complement Carry.

Description: The carry flag is complemented.

10. CMP: Compare with Accumulator.

Description: The contents of the operand (register or memory) are compared withthe contents of the accumulator.

11. CPI: Compare Immediate with Accumulator

Description: The second byte (8-bit data) is compared with the contents of theaccumulator.

12. DAA: Decimal-Adjust Accumulator

Description: The contents of the accumulator are changed from a binary value totwo 4-bit binary-coded decimal (BCD) digits. This is the only instruction that uses theauxiliary flag (internally) to perform the binary-to-BCD conversion. Flags S, Z, AC, P.CY flags are altered to reflect the results of the operation.

13. DAD: Add Register Pair to H and L Registers

Description: The 16-bit contents of the specified register pair are added to thecontents of the HL register and the sum is saved in the HL register. The contents ofthe source register pair are not altered. If the result is larger than 16 bits the CYflag is set. No other flags are affected.

14. DCR: Decrement Source by 1

Description: The contents of the designated register/memory are decremented by 1and the results are stored in the same place. If the operand is a memory location, itis specified by the contents of the HL register pair. Flags S. Z, P, AC are modified toreflect the result of the operation. CY is not modified.

15. DCX: Decrement Register Pair by 1

Description: The contents of the specified register pair are decremented by 1. Thisinstruction views the contents of the two registers as a 16-bit number. No flags areaffected.

16. DI: Disable Interrupts

Description: The Interrupt Enable flip-flop is reset and all the interrupts except theTRAP (8085) are disabled. No flags are affected.

17. El: Enable Interrupts

Description: The Interrupt Enable flip-flop is set and all interrupts are enabled. Noflags are affected.

18. HLT: Halt and Enter Wait State

Description: The MPU finishes executing the current instruction and halts anyfurther execution. The MPU enters the Halt Acknowledge machine cycle and Waitstates are inserted in every clock period. The address and the data bus are placed inthe high impedance state. The contents of the registers are unaffected during theHLT state. An interrupt or reset is necessary to exit from the Halt state. No flags areaffected.

19. IN: Input Data to Accumulator from a Port with 8-bit Address

Description: The contents of the input port designated in the operand are read andloaded into the accumulator. No flags are affected.

20. INR: Increment Contents of Register/Memory by 1

Description: The contents of the designated register/memory are incremented by Iand the results are stored in the same place. If the operand is a memory location, itis specified by the contents of HL register pair. Flags S. Z, P, AC are modified toreflect the result of the operation. CY is not modified.

21. INX: Increment Register Pair by 1

Description: The contents of the specified register pair are incremented by 1. The instruction views the contents of the two registers as a 16-bit number. No flags areaffected.

22. JMP: Jump Unconditionally.Description: The program sequence is transferred to the memory location specifiedby the 16-bit address. This is a 3-byte instruction; the second byte specified the low-order byte and the third byte specifies the high-order byte. No flags are affected.

Jump Conditionally:Description:Jump on carryJump on No carryJump on PositiveJump on minusJump on Parity Even

Jump on parity OddJump on ZeroJump on No Zero

23. LDA: Load accumulator Direct

Description: The contents of a memory location, specified by a 16-bit address inthe operand, are copied to the accumulator. The contents of the source are notaltered. This is a 3-byte instruction; the second byte specifies the low-order addressand the third byte specifies the high-order address. No flags are affected.

24. LDAX: Load Accumulator Indirect.

Description: The contents of the designated register pair point to a memorylocation. This instruction copies the contents of that memory location into theaccumulator. The contents of either the register pair or the memory location are notaltered. No flags are affected.

25. LD: Load H and L Registers Direct

Description: The instruction copies the contents of the memory location pointed outby the 16-bit address in register L and copies the contents of the next memorylocation in register H. The contents of source memory locations are not altered. Noflags are affected.

26. Load Register Pair Immediate

Description: The instruction loads 16-bit data in the register pair designated in theoperand. This is a 3-byte instruction; the second byte specifies the low-order byteand third byte specifies the high-order byte.

27. MOV: Move – Copy from Source to Destination.

Description: This instruction copies the contents of the source register into thedestination register; the contents of the source register are not altered. If one of theoperand is a memory location, it is specified by the contents of HL registers. No flagsare affected.

28. MVI: Move Immediate 8-Bit

Description: The 8-bit data is stored in the destination register or memory. If theoperand is a memory location, it is specified by the contents of HL registers. No flagsare affected.

29. NOP: No Operation

Description: No operation is performed. The instruction is fetched and decoded;how ever, no operation is executed. No flags are affected.

30. ORA: Logically OR with Accumulator

Description: The contents of the accumulator are logically OR with the contents ofthe operand (register or memory), and the results are placed in the accumulator. If

the operand is a memory location, its address is specified by the contents of HLregisters. Flags Z, S, P are modified to reflect the results of the operation. AC and CYare reset.31. ORI: Logically OR Immediate

Description: The contents of the accumulator are logically OR with the 8-bit data inthe operand and the results are placed in the accumulator. Flags S, Z, P are modifiedto reflect the results of the operation. CY and AC are reset.

32. OUT: Output Data from Accumulator to a Port with 8-Bit Address

Description: The contents of the accumulator are copied into the output portspecified by the operand. Flags No flags are affected.

33. PCHL: Load Program Counter with HL Contents

Description: The contents of registers H and L are copied into the program counter.The contents of H are placed as a high-order byte and of L as a low-order byte. Noflags are affected.

34. POP: Pop off Stack to Register Pair

Description: The contents of the memory location pointed out by the stack pointerregister are copied to the low-order register (such as C, E, L, and flags) of theoperand. The stack pointer is incremented by 1 and the contents of that memorylocation are copied to the high-order register (B, D, H, A) of the operand. The stackpointer register is again incremented by 1. No flags are modified.

35. PUSH: Push Register Pair onto Stack

Description: The contents of the register pair designated in the operand are copiedinto the stack in the following sequence. The stack pointer register is decrementedand the contents of the high-order register (B, D, H, A) are copied into that location.The stack pointer register is decremented again and the contents of the low-orderregister (C, E, L, flags) are copied to that location. No flags are modified.

36. RAL: Rotate Accumulator Left through Carry

Description: Each binary bit of the accumulator is rotated left by one positionthrough the Carry flag. Bit D is placed in the bit in the Carry flag and the Carry flag isplaced in the least significant position D. Flags CY is modified to bit D S, Z, AC, P arenot affected.

37. RAR: Rotate Accumulator Right through CarryDescription: Each binary bit of the accumulator is rotated right by one positionthrough the Carry flag. Bit D is placed in the Carry flag and the bit in the Carry flag isplaced in the most significant position, D. Flags CY is modified according to bit D S,Z, P, AC are not affected.

38. RLC: Rotate Accumulator LeftDescription: Each binary bit of the accumulator is rotated left by one position. Bit Dis placed in the position of D as well as in the Carry flag. Flags CY is modifiedaccording to bit D S, Z, P, AC are not affected.

39. RRC: Rotate Accumulator Right

Description: Each binary bit of the accumulator is rotated right by one position. BitD is placed in the position of D as well as in the Carry flag. Flags CY is modifiedaccording to bit Do. S. Z, P, AC are not affected.

40. RET: Return from Subroutine Unconditionally.

Description: The program sequence is transferred from the subroutine to th4ecalling program. The two bytes from the top of the stack are copied into theprogram, counter and the program execution begins at the new address. Theinstruction is equivalent to POP Program Counter. No flags are affected.

41. RIM: Read Interrupt Mask.

Description: This is a multipurpose instruction used to read the status of interrupts7.5, 6.5, 5.5 and to read serial data input bit. No flags are affected.

42. RST: Restart.

Description: The RST instructions are equivalent to 1-byte call instruction to one ofthe eight memory locations on page 0. The instructions are generally used inconjunction with interrupts and inserted using external hardware. However, thesecan be used as software instructions in a program to transfer program execution toone of the eight locations. No flags are affected.

43. SBB: Subtract Source and Borrow from Accumulator.

Description: The contents of the operand (register or memory) and the Borrow flagare subtracted from the contents of the accumulator and the results are placed in theaccumulator. The contents of the operand are not altered; however, the previousBorrow flag is reset. All flags are altered to reflect the result of the subtraction.

44. SBI: Subtract Immediate with Borrow.

Description: The 8-bit data (operand) and the borrow are subtracted from thecontents of the accumulator, and the results are placed in the accumulator. All flagsare altered and the results are placed in the accumulator.

45. SHLD: Store H and L Register Direct.

Description: The contents of register L are stored in the memory location specifiedby the 16 bit address in the operand, and the contents of H register are stored in thenext memory location by incrementing the operand. The contents of registers HL arenot altered. This is a 3-byte instruction; the second byte specifies the low-orderaddress and the third byte specifies the high order address. No flags are affected.

46. SIM: Set Interrupt Mask.

Description: This is a multipurpose instruction and used to implement the 8085interrupts (RST 7.5,6.5 and 5.5) and serial data output.

47. SPHL: Copy H and L Registers to the Stack Pointer.

Description: The instruction loads the contents of the H and L registers into thestack 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 theH and L registers are not altered. No flags are affected.

48. STA: Store Accumulator Direct.

Description: The contents of the accumulator are copied to a memory locationspecified by the operand. This is a 3-byte instruction; the second byte specifies thelow order address and the third byte specifies the high order address. No flags areaffected.

49. STAX: Store Accumulator Indirect.

Description: The contents of the accumulator are copied into the memory locationspecified by the contents of the operand (register pair). The contents of theaccumulator are not altered. No flags are affected.

50. STC: Set Carry.

Description: The carry flag is set to 1. No other flags are affected.

51. SUB: Subtract Register or Memory from Accumulator.

Description: The contents of the register or the memory location specified by theoperand are subtracted from the contents of the accumulator, and the results areplaced in the accumulator. The contents of the source are not altered. All flags areaffected to reflect the result of the subtraction.

52. SUI: Subtract Immediate from Accumulator.

Description: The 8-bit data (the operand) are subtracted from the contents of theaccumulator and the results are placed in the accumulator. All flag are modified toreflect the results of the subtraction.

53. XCHG: Exchange H and L with D and E.

Description: The contents of register H are exchanged with the contents of registerD and the contents of register L are exchanged with the contents of register E. Noflags are affected.

54. XRA: Exclusive OR with Accumulator.

Description: The contents of the operand (register or memory) are Exclusive ORwith the contents of the accumulator, and the results are placed in the accumulator.The contents of the operand are not altered. Z,S,P are altered to reflect the results ofthe operation. CY and AC are reset.

55. XRI: EXCLUSIVE OR Immediate with Accumulator.

Description: The 8 bit data (operand) are exclusive ORed with the contents of theaccumulator, and the result are placed in the accumulator. Z,S,P are altered toreflect the results of the operation. CY and AC are reset.

56. XTHL: Exchange H and L with Top of Stack.Description: The contents of the L register are exchanged with the stack locationpointed out by the contents of the stack pointer register. The contents of the Hregister are exchanged with the next stack location (SP+1); however, the contentsof the stack pointer register are not altered. No flags are affected.

ADDRESSING MODES:Every instruction of a program has to operate on a data. The method of

specifying the data to be operated by the instruction is called Addressing. The 8085has the following 5 different types of addressing.

1. Immediate Addressing2. Direct Addressing3. Register Addressing4. Register Indirect Addressing5. Implied Addressing

Immediate Addressing:In immediate addressing mode, the data is specified in the instruction itself.

The data will be a part of the program instruction. Ex: MVI B, 3EH - Move the data 3EH given in the instruction to B register.

Direct Addressing:In direct addressing mode, the address of the data is specified in the

instruction. The data will be in memory. In this addressing mode, the programinstructions and data can be stored in different memory.

Ex: LDA 1050H - Load the data available in memory location 1050H in toaccumulator..

Register Addressing:In register addressing mode, the instruction specifies the name of the register

in which the data is available.Ex: MOV A, B - Move the content of B register to A register.

Register Indirect Addressing:In register indirect addressing mode, the instruction specifies the name of the

register in which the address of the data is available. Here the data will be inmemory and the address will be in the register pair.

Ex; MOV A, M - The memory data addressed by H L pair is moved to Aregister.

Implied Addressing:In implied addressing mode, the instruction itself specifies the data to be

operated.Ex: CMA - Complement the content of accumulator.

ASSEMBLY LANGUAGE PROGRAMMING:ASSEMBLER:

An ASSEMBLER is a program, which is used to translate assembly languageprogram to correct binary code for each instruction.Types of assembler:

1. One pass assembler:

• It is an assembler in which the source codes are processed only once.• Very fast.• Backward reference only used.• It issues an error message if it encounters a label or variable that is

defined at a later end of a program. So it cannot have forwardreferences.

2. Two pass assembler:• The source codes are processed two times.• In the first pass it assigns addresses to all the labels and attach values

to all the variables used in the program.• In the second pass it converts the source code into machine code.

Advantages of assembler:• Translates mnemonics into binary code with speed and accuracy.• Allows the programmer to use variables in the program.• It is easier to alter the program and reassemble.• It identifies the syntax error.• It can reserve memory locations for data or result.• It provides list file for documentation.

Assembler directives:• They are the instructions to the assembler regarding the program

being assembled.• Also called as pseudo instructions or pseudo op-codes.• They will give information’ s like start and end of program, values of

variables used in the program, storage locations of output and inputdata etc.

• Assemblers areORG origin of a programEND End of programEQU EquateDB Define ByteDW Define WordDS Define Storage

SUBROTINE:• It is a group of instructions written separately from the main program

to perform a function that occurs repeatedly in the main program.• It is called in the main program by using CALL addr16 instruction. The

addr16 is the starting address of subroutine.• It should be terminated by RET instruction.

Advantage of subroutine:• Modular programming: The various tasks in a program can be

developed as separate modules and called in the main program.• Reduction in the amount of work and program development time.• Reduces memory requirement for program storage.

DELAY ROUTINE:It is a subroutine used for maintaining the timings of various operations in

microprocessor.List: List is a linked data structure used in programming techniques. The linked datastructure will have a number of components linked in a particular fashion. Eachcomponent will consists of a string data and a pointer to next component. Types oflist are,

• Linear linked lists• Linked list with multiple pointers• Circular inked list & Tress

Array: it is a series of data of the same type stored in successive memory locations.EACH value in the array is referred to as an element of the array.Flow chart: It is a graphical representation of the operation flow of the program. Itis a graphical form of algorithm.

Symbol Operation

Race track shape boxTo indicate the start or end of the program

ParallelogramTo represent input or output operation

Rectangular box To represent simple operations other than I/Ooperations

Rectangular box withdouble lines on verticalsides To represent a subroutine or procedure

Diamond shaped box

To represent the decision point

Small circle It is used as a connector to show theconnections between various part of flowchartwithin a page. Identical numbers are entered

inside the boxes that represent the sameconnecting point.

Five sided box It is used as a off-page connector to show theconnections between various sections offlowchart in a different page. Identical

numbers are entered inside the boxes thatrepresent the same connecting point.

LineArrow

The lines are drawn between boxes anddiamonds to indicate the program flow and

the arrow is placed on the lines to indicate thedirection of flow.

ASSEMBLY LANGUAGE PROGRAMMING DEVELOPMENT TOOLS:1. Editor: It is a program which when run on a microcomputer system, allows

the user to type and modify the assembly language program statements. Themain use of editor is to help the user to construct the assembly languageprogram in the right format and save as a file.

2. Assembler: An ASSEMBLER is a program, which is used to translateassembly language program to correct binary code for each instruction.

3. Linker: A program used to join together several object files into one largeobject file.

4. Locator: A program used to assign specific addresses to the object codes tobe loaded into memory.

5. Debugger: It is a software used to locate and troubleshoot the errors in aprogram.

6. Simulator: A program, which can be run on the development system tosimulate the operations of the newly designed system. Some of theoperations that can be simulated are given below.

• Execute a program and display the result.• Single step execution of a program.• Break-point execution of a program.• Display the content of register/memory.

7. Emulator: A system that can be used to test the hardware and software of anewly developed microprocessor based system.

Counter:Ø A counter is designed simply by loading an appropriate number into one of

the registers and using the INR (Increment by one) or the DCR (Decrementby one) instructions.

Ø A loop is established to update the count and each count is checked todetermine whether it has reached the final number; if not, the loop isrepeated.

Ø The flowchart illustrates the following steps.Ø This counter has one major drawback; the counting is performed at such high

speed that only the last count can be observed. To observe counting theremust be an appropriate time delay between counts.

No

YesTime delay:Ø The procedure used to design a specific delay is similar too that used to set

up a counter.Ø A register is loaded with a number, depending on the time delay required and

then the register is decremented until it reaches zero by setting up a loopwith a conditional jump instruction.

Ø The loop causes the delay, depending upon the clock period of the system.Time delay using one register:Ø The flow chart shows a time delay loop.Ø A count is loaded in a register, and the loop is executed until the count

reaches zero.Ø The set of instructions necessary to set up the loop is shown below.

Initialize

Display

Update

Is thisfinalcount

No

Yes

Label Opcode Operand Comments T-states

MVI C,FFH ;Load register C 7

Loop: DCR C ;Decrement C 4

JNZ Loop ;Jump back to 10/7;decrement C

• The last column shows the T-states (clock period ) required by the 8085microprocessor to execute each instruction. The instruction MVI requiresseven clock periods. An 8085 based micro computer with 2 MHz clockfrequency will execute the instruction MVI in 3.5µs as follows.

Clock frequency of the system f=2 MHzClock period T=1/f=1/2x10-6 =0.5µsTime to execute MVI = 7T states * 0.5 =3.5µs.

• However if the clock frequency of the system is 1MHZ , the microprocessorwill require 7µs to execute the same instruction. To calculate the time delay ina loop , we must account for the T-states required for each instruction and forthe number of times the instructions are execute in the loop.

• In above figure , register C is loaded with the count FFH (22510) by theinstruction MVI, which is executed once and takes seven T-states. The nexttwo instructions DCR and JNZ, form a loop with a total of 14(4+10) T-states.The loop is repeated 255 times until register C=0.

The time delay in the loop TL with 2 MHZ clock frequency iscalculated as

TL=(T x Loop T-States x N10). Where TL= Time delay in the loop

T = System clock period N10 = Equivalent decimal number of the hexadecimal count

loaded in the delay register.TL = (0.5x10-6x14x225) =1785µs = 1.8ms.

In most applications, this approximate calculation of the time delay is consideredreasonably accurate. However, to calculate the time delay more accurately, we need

Load delay Register

Decrement Register

IsRegister=0

to adjust for the execution of the JNZ instruction and add the execution time of theinitial instruction.

• The T-states for JNZ instruction are shown as 10/7. This can be interpretedas follows: The 8085 micro processor requires ten T-states to execute aconditional jump instruction when it jumps or changes the sequence of theprogram and seven T-states when the program falls through the loop (goesto the instruction following the JNZ). In the above figure the loop isexecuted 255 times; in the last cycle, the JNZ instruction will be executed inseven T-states . This difference can be accounted for in the delay calculationby subtraction the execution time of three states. There fore , the adjustedloop delay is

TLA= TL-(3 T states x clock period) =1785.0µs - 1.5µs =1783.5µs.

Now the total delay must take into account the execution time of the instructionsoutside the loop. In the above example we have only one instruction (MVI C) outsidethe loop. Therefore, the total delay is

Total delay = Time to execute instruction + Time to execute theloop instructions.

TD =To+ TLA

=(7x0.5µs) + 1783.5µs =1787µs. ≅1.8ms.• The difference between the loop delay TL and these calculations is only 2µs

and can be ignored in most instances.• The time delay can be varied by changing the count FFH; however to

increase the time delay beyond 1.8ms in a 2MHZ micro computer system, aregister pair or a loop within a loop technique should be used.

Time delay using a register pair:

The time delay can be considerably increased by setting a loop and using aregister pair with a 16-bit number (maximum FFFFH). The 16 bit number isdecremented by using the instruction DCX.Label Opcode Operand Comments T-States

LXI B,2384H ;Load BC with 16 bit count 10Loop: DCX B ;Decrement (BC) by one 6

MOV A,C ;Place contents of C in A 4ORA B ;OR (B) with (C)

to set Zero flagJNZ Loop ;if result =

0,jump back to Loop 10/7Time delay:

• The time delay in the loop is calculated as in the previous example. The loopincludes four instruction: DCX, MOV, ORA, and JNZ, and takes 24 clockperiods for execution. The loop is repeated of 2384H times which is convertedto decimal as

2384H= 2x (16)3 + 3x(16)2 + 8x(16)1 + 4(16)0 = 909210

If the clock period of the system =0.5µs, the delay in the loop TTL= (0.5 x 24 x 909210)

≈109 ms (without adjusting for the last cycle)Total delay Td = 109 ms + T0:

≈ 109 ms (The instruction LXI adds only 5µs.

Time delay using a loop within a loop technique: A time delay similar to that of a register pair can also be achieved by usingtwo loops; one-loop insides the other loop is shown in below figure. For example,register C is used in the inner loop (LOOP 1) and register B is used for the outer loop(LOOP 2). The following instructions can be used to implement the flow chart isshown in figure.

MVI B, 38H 7TLoop2: MVI C,FFH 7TLoop1: DCR C 4T

JNZ Loop1 10/7TDCR B 4TJNZ Loop2 10/7T

Delay calculations:The delay in Loop 1 is TL1=1783.5µs. We are replace loop1 by

TL1. Now we can calculate these delay in Loop2 as if it is one loop; this loop isexecuted 56 times because of the count (38H) in register B:TL2=56(Tl1+21 T states x 0.5µs) = 56(1783.5µs+10.5µs) = 100.46ms.

Flow chart for time delay with two loops.MVI

C, FF

Loop 2

Loop1

No DCRB

JNZLoop2

Yes

No

Load Register B

Load register C

DecrementRegister C

Is registerC=0

Decrement registerB

Is registerB=0

TL, Delayin Loop

• The total delay should include the execution time of the first instruction (MVIB,&T); however, the delay outsides these loops insignificant. The time delaycan be increased considerably by using register pairs in the above example.

TIMING DIAGRAM OF 8085 INSTRUCTIONS:

• The 8085 instructions consist of one to five machine cycles.• Actually the execution of an instruction is the execution of the machine cycles

of that instruction in the predefined order. Therefore, from the knowledge ofthe timing diagrams of each machine cycle of an instruction, the timingdiagram of that instruction can be obtained.

• The timing diagram of an instruction ate obtained by drawing the timingdiagrams of the machine cycles of that instruction, one by one in the order ofexecution.

MICROPROCESSOR APPLICATIONS:

1.TEMPERATURE CONTROL SYSTEM:The microprocessor based temperature control system can be used for

automatic control of the temperature of a body. A simplified block diagram of 8085microprocessor based temperature control system is,

8085 Microprocessor based Temperature Control System

• The system consist of 8085 microprocessor as CPU, EPROM & RAMmemory for program & data storage, INTEL 8279 for keyboard anddisplay interface, ADC, DAC, INTEL 8255 for I/O ports, Amplifiers,Signal conditioning circuit, Temperature sensor and Supply controlcircuit.

• The EPROM memory is provided for storing system program and RAMmemory for temporary data storage & stack operation.

• Using INTEL 8279, a keyboard and six numbers of 7-segment LEDs areinterfaced to the system. The system has been designed to accept thedesired temperature and various control commands through keyboard.

• The 7-segment display has been provided to display the temperatureof the body at any time instant.

• The temperature of the body is measured using a temperature sensor.The different types of temperature sensors that can be used fortemperature measurement are Thermo-couples, Thermistors, PN-junctions, IC sensors like AD590, etc.

• The sensors will convert the input temperature to proportional analogvoltage or current. The output signal of the sensor will be a weaksignal and so it has to be amplified using high input impedance op-amp.

• The analog signal is scaled to suitable level by the signal conditioningcircuit.

• The microprocessor can process only digital signals and so theprocessor cannot read the analog signal from signal conditioning circuitdirectly. The system has an analog-to-digital converter (ADC) toconvert the analog signal to proportional digital data.

• In this system the ADC is interfaced to 8085 processor through port-Aof 8255. The 8085 processor send signal to ADC to start conversionand at the end of conversion it read the digital data from the port-A of8255.

• The 8085 processor calculate the actual temperature using the inputdata and display it on the 7-segment LED.

• Also, the processor compare the desired temperature with actualtemperature (The operator can enter the desired temperature throughkeyboard) and calculate the error (the difference between actualtemperature and desired temperature).

• The error is used to compute a digital control signal, which is converted toanalog control signal by DAC. The DAC is interfaced to the system throughport-B of 8255.

• The analog control signal produced by DAC is used to control the powersupply of the heating element of the body.

• The digital control signal can be computed by the 8085 processor usingdifferent digital control algorithms (P/PL’ PID/FUZZY logic control algorithms).s

• The control circuit for power supply can be either thyristor - based circuit orrelay. In case of thyristor control circuits the firing angle can be varied by thecontrol signal to control the power input to the heater.

• In case of relay the control signal can switch ON/OFF the re to control thepower input to the heater.

2.TRAFFIC LIGHT CONTROL SYSTEM:

The traffic lights placed at the road crossings can be automatically switchedON/OFF in the desired sequence using the microprocessor system. The system canalso have a manual control option, so that during heavy traffic (or during traffic jam)the duration of ON/OFF time can be varied by the operator.

A typical traffic light control system (demonstration type) is,

8085 Microprocessor based Traffic Light Control demonstration System

• The system has been developed using 8085 as CPU. The system hasEPROM memory for system program storage and RAM memory forstack operation. For manual control a keyboard have been provided. Itwill be helpful for the operator if the direction of traffic flow isdisplayed during manual control. Hence 7 segment LEDs are interfaced

to display the direction of traffic flow both during manual andautomatic mode.

• The primary function of the microprocessor in the system is to switchON/OFF the Red/Yellow/Green lights in the specified sequence. In thedemonstration system of fig shown, Red/Yellow/Green LEDs areprovided instead of lights (lamps). The LEDs are interfaced to thesystem through buffer (74LS245) and ports of 8255.

Ø In the practical implementation scheme the lights can be turned ON/ OFFusing driver transistors and relays.

Ø In practical implementation the output of buffer (74LS245) can be connectedto the driver transistor.

Ø A relay placed at the collector of the transistor can be used to switch ON/ OFFthe light as shown in fig.

Ø A reverse biased diode is connected across relay coil to prevent relaychattering (for free-wheeling action).

The microprocessor sends HIGH through a port line to switch ON the light andLOW to switch OFF the light. A switching schedule (or sequence) can be developed asshown in table. In this switching sequence it is assumed that the traffic is allowedonly in one direction at a time.

Switching Circuit for Traffic light

Ø The processor can output the codes for switching the lights forschedule-I and then waits. After a specified time delay the processoroutput the codes for schedule-I and so on.

Ø For each schedule the processor can wait for a specified time. Afterschedule-XH, the processor can again return to schedule-I. Onobserving the schedules we can conclude that three different delayroutines are sufficient for implementing the twelve switchingschedules.

3.STEPPER MOTOR CONTROL SYSTEM:

Ø The stepper motors are popularly used in computer peripherals,plotters, robots and machine tools for precise incremental rotation.

Ø In stepper motor, the stator windings are excited by electrical pulsesand for each pulse the motor shaft advances by one angular step.(Since digital pulses can drive the stepper motor, it is also calleddigital motor).

Ø The step size in the motor is determined by the number of poles in therotor and the number of pairs of stator windings (one pair of statorwinding is called one phase). The stator windings are also calledcontrol windings.

Ø The motor is controlled by switching ON/OFF the control winding. Thepopular stepper motor used for demonstration in laboratories has astep size of 1.8° (i.e, 200 steps per revolution).

Ø The basic step size of the motor is called full-step. By altering theswitching sequence, the motor can be made to run with incrementalmotion of half the full-step value.

8085 Microprocessor Based Stepper Motor Control System

Ø A two phase or four winding stepper motor control system is shown inabove figure. The system consists of 8085 microprocessor as CPU,EPROM and RAM memory for program & data storage and for stack.

Ø Using INTEL 8279, a keyboard and six number of 7-segment LED displayhave been interfaced in the system. Through the keyboard the operator canissue commands to control the system. The LED displays have beenprovided to display messages to the operator.

Ø The windings of stepper motor are connected to the collector ofDarlington pair transistors. The transistors are switched ON/OFF by themicroprocessor through the ports of 8255 and buffer (74LS245).

Ø A freewheeling diode is connected across each winding for fast switching.The flowchart for the operational flow of the stepper motor control systemis shown.

Ø The processor has to output a switching sequence and wait for 1 to 5 msecbefore sending next switching sequence.

Flow Chart for Stepper Motor Control Program

Unit-I

Part A1. Distinguish I/O mapped I/O and memory mapped I/O.2. Explain the execution of the instruction CMA M in 8085.3. What is the function performed by SIM instruction?4. What is meant by processor cycle?5. Explain the different types of flags in 8085.6. What are the two compare instructions available in 8085.7. When the READY signal of 8085 processor is sampled by the processor?8. What is DAD and what are the flags, affected by this instruction?

9. List the Software and Hardware interrupts of 8085?10. How to calculate the vector address of software interrupts and calculate for all.

Part B1. Explain the 8085 architecture in detail2. Draw the pin configuration of 8085 processor & explain the signals3. Draw the Timing diagram for the following machine cycles.

(i) Opcode fetch (ii) Memory Read (iii) Memory Write(iv) I/O Read (v) I/O write

4. (i) Explain the classification of 8085 Instruction set with some examples. (ii) Write about the addressing modes of 8085.

5. (i) How do you Interface a memory with 8085? (ii) Explain the memory mapped I/O Interface.6. (i) Draw the Timing diagram of STA 8080 instruction and explain each cycle. (ii) Draw the Timing diagram of IN 10 and MVI A, 70H instruction & explaineach cycle.7. How will you interface a peripheral device with 8085 processor? Explain.

8. (i) Discuss the interrupt system of 8085. (ii) Draw the timing diagram for memory read operation performed by 8085 (Use associated signals). 12. Write a program to count continuously in hexadecimal from FFH to OOH in a

system with a 5 µsec clock period. Use register C to set up a one milliseconddelay between each count and display the number at one of the output ports.

13. (i)List all the control signals of the Timing and control unit. Explain the use of eachof these signals.(ii)Write an ALP to evaluate the expression C=A2+B2.

14. (i)Compare the similarities and differences between PUSH /POP and CALL/RETinstructions.(ii)What are the features of FIFO and LIFO memory structure?

(iii)Draw and explain the timing diagram for the execution of the instruction LDA2080 H.

15. What is the memory mapped I/O and I/O mapped I/O. Explain.

16. (i)Explain with the help of suitable diagram how the INTR pin can be used tointerrupt the 8085 and how it to the signal.(ii)Bring out the differences between memory mapped I/O and I/O mapped I/O.

17. (i)Explain the logic instructions of 8085 microprocessor with examples.(ii)Write an ALP using 8085 instruction set to add two n-byte numbers stored atmemory locations starting and respectively. Store the result at memorylocation starting from . Draw the flow chart.

Unit-II

Part A1. Distinguish microprocessor and micro controller.2. List any two applications of micro controller.3. Write and ALP for 8051 to implement a BCD, counter and store the count in

memory location starting from .4. What is the significance of Interrupt Priority control register in 8051

microcontrollers?5. Write an ALP for time delay using a register pair available in 8085.6. What is the job of the TMOD register?7. What is the function of DPTR register?8. Name the interrupts of 8051 micro controllers.

9. What is the importance of special function registers available in 8051 microcontrollers?

10.What is the voltage level used in RS 232?

Part B1. Explain briefly the architecture or 8051.2. (i)List the various special function registers in 8051 and explain its usage.

(ii)With neat diagram explain ports 1 pin configuration.3. (i)Explain the significance of SFRs in 8051 micro controller.

(ii)Explain how to interface external memory devices with 8051 micro controller.4. (i)Write an ALP in 8051 to sort the numbers stored in an array.

(ii)Compare the features of 8 bit and 16 bit micro controller.5. Explain with neat diagram of port pin and circuits that connect the 8051 micro

controller to the outside world.6. With neat sketch, describe the hardware features of 8051 microcontroller.7. (i)Draw the bit pattern of program status word of 8051 and explain the

significance of each bit with examples.(ii)List the special function registers of 8051 micro controller and explain theirfunctions.

8. (i)Explain the various modes available for timer in 8051 microcontroller.(ii) Discuss the interrupt structure of 8051 microcontroller.

9. Draw the pin configuration of 8051 and explain the signals. 10. (i)Explain the hardware circuits used in 8051 microcontrollers.

(ii)Explain the operation of the following hardware circuits in 8051. (i) Timers (ii) Serial data I/O (iii) Interrupts (iv) External memory

11. Explain the architecture of 16 bit microcontroller in detail (80196). 12. With necessary hardware and software details explain how to interface

LCDs with 8051 micro controller. 13. (a) Explain with neat diagram of port pin & circuits that connect the 8051 microcontroller to the outside world. (b) With neat sketch, describe the hardware features of 8051 microcontroller.

Unit-III

Part A

1. What are the segment registers in8086?2. List the additional features of 486 processor in comparison with 386 processor.3. List the merits of Memory segmentation.4. Name the external hardware synchronization instruction 8086 processor.5. What is the job of the TMOD register?6. What is segment override prefix? Give an example.7. What is the function of TEST pin in 8086 processor?8. How does the 8086 processor access a word at on odd address?9. What are the differences between 8085 and 8086?10. Briefly explain the interrupts in 8086.

Part B 1. With neat diagram explain the architecture of 8086 processor.

2. (i)Explain the instruction set of 8086 with examples.(ii)Briefly explain the advanced design features in Intel Pentium processor whencompared with 486 processor.

3. (i)Explain with examples the addressing modes of 8086 processor.(ii)Explain the interrupt structure or 8086 processor.

4. Explain the minimum and maximum mode of operation of o8086 microprocessors5. (i)Explain any three addressing modes of 8086.6. (ii)Discuss the interrupt mode configuration of 8086 processor.7. (i)Design an 8086 based system in minimum modes to intsesrfac3e 64 KB

EPROM and 64KB RAM with starting address 00000H and 80000H respectively.

(ii) Draw the pin configurations of 8086 and explain the signals8. Discuss the features and architecture of Pentium processor with necessary

diagrams. 9. Explain the architecture of 80386 (32-bit) 10. Explain the architecture of 80486 (32-bit) 11. Explain the working of 8086 in maximum mode systems. 12. Explain the working of 8086 in minimum mode systems. 13. (i) Explain the Instruction set of 8086. (ii) Write any 5 features of 8086.

Unit-IV

Part A

1. What is 8259? What are its functions?2. What is handshake port?3. How a keyboard matrix is formed in keyboard interface using 8279?4. What is resolution and conversion time in ADC?5. Write a program segment that will carry out the following binary operation using

8086 instructions. W< X+Y+24-Z.6. Specify the bit of a control word of the 8255, which differentiates between I/O

mode and the BSR mode.7. What is cycle stealing in DMA?8. List the functions performed by 8279.

9. What are the modes of data transmission of 8251?10. List the signals used by transmitter and receiver unit of 8251.

Part B

1. List the major components of the 8279 keyboard/ display interface and explaintheir functions, with neat diagram.

2. Design an interfacing circuit to read data from an A/D converter using 8255 inmemory mapped I/O form.

3. (i)Explain the important features of programmable DMA controller. (ii)Show how the 8255 PPI can be interfaced to the 8085 processor based system

with 2 K bytes of EPROM.4. What are the different modes of serial communication? Draw the internal block

diagram of 8251 and explain how the CPU uses it for serial communication.5. (i)What are the different modes of DMA transfer?

(ii)With a neat sketch show how a DMA controller is connected to the 8085processor. Outline the sequence of operation needed in performing a DMAtransfer.

6. (i)Interface 8251 A to 8086 in I/O mapped I/O mode and memory mapped I/Omode.(ii)Discuss the features of programmable DMA controller with functional blockdiagram.

7. Explain how to interface:a. ADC andb. DAC with 8051 micro controller.

Unit-V

Part A

1. Mention the hardware requirement to interface an LCD using 8255.2. What is a smart scale?3. Draw the circuit using MOSFET to drive solenoids or motor windings.4. Name any two alphanumerical display devices.5. What Is The Voltage Levels Used In RS232?6. What is the shaft encoder?7. List the advantages of robotic control.8. What do you mean by real time digital signal processing?9. Write short notes on stepper motor.

10. List the applications of stepper motor.

Part-B

1. With neat block diagram explain the operation of 8085 based stepper motorcontrol system.

2. With neat diagram explain the operation of 8085 based industrial temperaturecontrol system.

3. With necessary hardware and software details explain how to interface LCDs with8051 microcontroller.

4. (i)Explain with diagram how to interface stepper motor with a microprocessor.(ii)How optical motor shaft encoders are used in motor control applications.

5. (i)With a block diagram and flow chart, explain the operation of microcomputerbased smart scale.(ii)Show how will interface a stepper motor to the microprocessor.

6. (i)Develop a hardware and necessary software algorithm to run a stepper motor inthe forward direction through five revolutions in 10 seconds. Assume necessaryspecification for the stepper motor.(ii)With block diagram explain briefly the microprocessor based scale.