{"id":24527,"date":"2020-02-27T11:05:28","date_gmt":"2020-02-27T05:35:28","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=24527"},"modified":"2021-07-02T15:24:09","modified_gmt":"2021-07-02T09:54:09","slug":"2nd-puc-electronics-question-bank-chapter-11","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-question-bank-chapter-11\/","title":{"rendered":"2nd PUC Electronics Question Bank Chapter 11 Microcontrollers"},"content":{"rendered":"

Students can Download 2nd PUC Electronics Chapter 11 Microcontrollers Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Electronics Question Bank Chapter 11 Microcontrollers<\/h2>\n

2nd PUC Electronics Microcontrollers One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is a microcontroller?
\nAnswer:
\nMicrocontroller is a small computer on a single IC consisting of a relatively simple CPU combined with support functions.<\/p>\n

Question 2.
\nWhat is an Accumulator?
\nAnswer:
\nAccumulator is a device which stores a number and which on receipt of another number, adds the two and stores the sum. It can also sense the signal, clear and complement etc.<\/p>\n

Question 3.
\nWhat is a Program Counter?
\nAnswer:
\nProgram counter is a circuit, which holds the address of a byte in memory. It also specifies the address of the next instruction to be fetched and executed.<\/p>\n

\"2nd<\/p>\n

Question 4.
\nWhat is a Stack pointer?
\nAnswer:
\nStack is the area of the memoiy (RAM) that is used in some instructions to store and retrieve data quickly. Stack pointer is used to hold an internal RAM address. When a program jumps to some branch the address of the next instruction is stored in Stack.<\/p>\n

Question 5.
\nWhat is an interrupt circuit?
\nAnswer:
\nIt is an electronic circuit, which breaks in the normal flow of system or routine and flow can be resumed from that point at a later time.<\/p>\n

Question 6.
\nWhat is the need of clock circuit?
\nAnswer:
\nClock circuit generates the internal clock pulses and all the internal operations are synchronised with this pulse.<\/p>\n

Question 7.
\nHow much is the ROM of microcontroller 8051?
\nAnswer:
\nMicrocontroller 8051 has on chip ROM of 4k bytes.<\/p>\n

Question 8.
\nHow much is the RAM of microcontroller 8051?
\nAnswer:
\nMicrocontroller 8051 has 128 bytes of RAM.<\/p>\n

Question 9.
\nHow many serial ports are there in microcontroller 8051?
\nAnswer:
\nMicrocontroller 8051 has one serial port for communication.<\/p>\n

\"2nd<\/p>\n

Question 10.
\nHow many interrupt sources are there in microcontroller 8051?
\nAnswer:
\nMicrocontroller 8051 has six(6) interrupt sources.<\/p>\n

Question 11.
\nWhat is an addressing mode?
\nAnswer:
\nAddressing modes specify the way in which the operands are accessed by the instruction.<\/p>\n

Question 12.
\nWhat is an Operational code (op code)?
\nAnswer:
\nOperational code (op code) is the short form of expressing the instruction. It consists of mnemonic and operands.<\/p>\n

Question 13.
\nWhat is the meaning of MOV A, R0?
\nAnswer:
\nThe content of R0 is moved (copied) to register A (accumulator).<\/p>\n

Question 14.
\nWhat is the meaning of MOV R5, A?
\nAnswer:
\nCopy the contents of register A(Accomulator) to Register 5.<\/p>\n

Question 15.
\nWhat is the purpose of register addressing mode?
\nAnswer:
\nIn register addressing mode data is moved between registers A, DPTR, Carry bit and registers R0-R7 are used as part of opcode mnemonics as source or destination.<\/p>\n

\"2nd<\/p>\n

Question 16.
\nWhat is the use of direct addressing mode?
\nAnswer:
\nIn direct addressing mode 128 bytes of internal RAM and SFRs may be addressed directly using single byte address assigned to each RAM location and each SFR.<\/p>\n

Question 17.
\nWhat is the direct address of working register R5 in register bank 2?
\nAnswer:
\nThe direct address of register R5 in register bank 2 is ID.<\/p>\n

Question 18.
\nWrite the direct address of the Accumulator.
\nAnswer:
\nThe direct address of accumulator register is OEO.<\/p>\n

Question 19.
\nWrite the mnemonic for the operation \u201cSave content of A in RAM location 45H\u201d.
\nAnswer:
\nThe mnemonic for saving content of register A in location 45H is MOV 45H, A.<\/p>\n

Question 20.
\nWhat is meaning of MOV A, @R1?
\nAnswer:
\nThe meaning of MOV A, @R1 is to copy the contents of RAM location whose address is stored in R1 into accumulator A.<\/p>\n

\"2nd<\/p>\n

Question 21.
\nHow do you write the mnemonic for the operation \u201ccopy the contents of RAM location whose address is stored in R0 into accumulator A\u201d in an indirect addressing mode?
\nAnswer:
\nThe mnemonic for copy the contents of RAM location whose address is stored in R0 into accumulator A is MOV A, @ R0.<\/p>\n

Question 22.
\nWhich sign is used as the mnemonic for immediate data?
\nAnswer:
\nThe mnemonic for immediate data is hash (#).<\/p>\n

Question 23.
\nWhat is meaning of MOV A, #22H?
\nAnswer:
\nThe meaning of MOV A, #22H is to load value 22H into accumulator A.<\/p>\n

Question 24.
\nWrite the mnemonic for the operation \u201cload the decimal value 66 into R3\u201d.
\nAnswer:
\nThe mnemonic for loading decimal value 66 into R3 is MOV R3, #66.<\/p>\n

Question 25.
\nWhat is meaning of MOV DPTR, #1246H?
\nAnswer:
\nThe meaning of MOV DPTR, #1246H is to load 16 bit number (1246H) immediately into register DPTR.<\/p>\n

Question 26.
\nWhat does \u2018jump\u2019 instruction do?
\nAnswer:
\nThe jump instruction permanently changes the contents of program counter either conditionally or unconditionally.<\/p>\n

Question 27.
\nWhat does \u2018call\u2019 instruction do?
\nAnswer:
\nThe call instruction temporarily changes the contents of program counter to allow another part of the program to run either conditionally or unconditionally.<\/p>\n

\"2nd<\/p>\n

Question 28.
\nWhat is the description of the mnemonic JNZ and ACALL?
\nAnswer:
\nJNZ is conditional jump instruction, here contents of PC changes only when the result of operation is non zero. ACALL allows to specify 11- bit address in the instruction and calling subroutine within 2k program memory block, starting from the address of the next instruction is possible.<\/p>\n

Question 29.
\nWhy is ROM called non volatile?
\nAnswer:
\nROM is called non volatile because it does not depend on electrical power to store the numbers.<\/p>\n

Question 30.
\nWhy is RAM called volatile?
\nAnswer:
\nRAM is called volatile because it depends on electrical power to store the numbers, when power is OFF it loses its information.<\/p>\n

Question 31.
\nWhat is a program?
\nAnswer:
\nComputer program is a sequence of operations that the computer is to perform.<\/p>\n

Question 32.
\nExpand EEPROM.
\nAnswer:
\nElectrical Erasable Programmable Read Only Memory.<\/p>\n

Question 33.
\nExpand SRAM.
\nAnswer:
\nStatic Random Access Memory.<\/p>\n

Question 34.
\nExpand RAM.
\nAnswer:
\nRandom Access Memory or Read write memory.<\/p>\n

Question 35.
\nWhat is data address?
\nAnswer:
\nAddress in memory that is used by the CPU to read or write is called data address.<\/p>\n

Question 36.
\nWhat is a machine language?
\nAnswer:
\nMachine language is one in which information is available only in binary form.
\n\"2nd<\/p>\n

Question 37.
\nWhy is assembly language programming called low level programming?
\nAnswer:
\nAssembly language is low level programming because, it is machine dependant.<\/p>\n

Question 38.
\nHow many bits of binary data can a register A hold?
\nAnswer:
\nRegister A can hold 8 bits of binary data.<\/p>\n

Question 39.
\nHow many bits of binary data can a register R hold temporarily?
\nAnswer:
\nRegister R can hold 8 bit of binary data temporarily.<\/p>\n

Question 40.
\nHow many bits of address can a register PC hold?
\nAnswer:
\nRegister PC can hold 16- bit of address number.<\/p>\n

Question 41.
\nHow much \u201cPAL register PC\u201d can hold code address?
\nAnswer:
\nPractice Assembly Language (PAL) Register can hold code address from 0000H to FFFFH (64K).<\/p>\n

Question 42.
\nWhat are directives?
\nAnswer:
\nAssembly language instructions like ADD and MOV are statements called directives.<\/p>\n

Question 43.
\nWhat does directive do?
\nAnswer:
\nInstructions tell the computer what to do, while directives give directions to the assembler.<\/p>\n

Question 44.
\nWhat is a PIC microcontroller?
\nAnswer:
\nPIC microcontrollers are a family of microcontroller chips produced by microchip technology.<\/p>\n

\"2nd<\/p>\n

Question 45.
\nExpand PIC.
\nAnswer:
\nPIC is Peripheral Interface Controller.<\/p>\n

Question 46.
\nHow many accumulators are there in PIC microcontroller?
\nAnswer:
\nPIC microcontrollers has one accumulator.<\/p>\n

Question 47.
\nMention the maximum operating frequency of PIC16F877.
\nAnswer:
\nMaximum operating frequency of PIC16F877 is 20MHz.<\/p>\n

Question 48.
\nHow many times data can be written in PIC16F877?
\nAnswer:
\nIn PIC16F877 microcontrollers 8k ROM memory in flash technology upto 105 times chip can be reprogrammed.<\/p>\n

Question 49.
\nWhat is the working voltage range of PIC16F877?
\nAnswer:
\nPIC microcontrollers has working voltage range of 2.0V to 5.5V.<\/p>\n

Question 50.
\nHow much is the EEPROM memory?
\nAnswer:
\nEEPROM memory has 128 bytes.<\/p>\n

\"2nd<\/p>\n

Question 51.
\nWhat is the largest hex value that can be moved into an 8 bit register?
\nAnswer:
\nFFH is the largest value that can be moved into an 8 bit register.<\/p>\n

2nd PUC Electronics Microcontrollers Two Marks Questions and Answers<\/h3>\n

Question 1.
\nMention two differences between microprocessor and microcontroller.
\nAnswer:<\/p>\n\n\n\n\n\n
Microprocessor<\/strong><\/td>\nMicrocontroller<\/strong><\/td>\n<\/tr>\n
Most MPs have very less bit handling instructions<\/td>\nMCs have many bit handling instructions<\/td>\n<\/tr>\n
MPs are intended for general purpose applications<\/td>\nMCs are for a specific application only<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 2.
\nWhat is an addressing mode? Why is it necessary?
\nAnswer:
\nThe CPU can store data may be in registers, memory or in an external source. The ways by which these data source address are specified are called addressing modes. These are necessary because, the data can be accessed using any one of the modes.<\/p>\n

Question 3.
\nMention the different opcodes used in 8051.
\nAnswer:
\nMOV dest, sour; PUSH sour; POP dest; XCH dest, sour; ADD A, Rn etc are some of the opcodes used in 8051.<\/p>\n

Question 4.
\nWhat are the different parts of memory of 8051?
\nAns:
\nThe memory is divided into Internal RAM, Intrenal SFR, External RAM, internal and External ROM.<\/p>\n

Question 5.
\nWhat is the main feature of MOV opcode? Give an example.
\nAnswer:
\nMOV opcode transfers data within the 8051 memory. MOV A, Rl; MOV R0,22H; MOV A, #54H; MOV R2,#40.<\/p>\n

\"2nd<\/p>\n

Question 6.
\nWhat is direct addressing? Explain.
\nAnswer:
\nIn direct addressing mode all 128 bytes of internal RAM and SFRs may be addressed directly using single-byte address assigned to each RAM and SFR registers. MOV R0, 40H save the content of location 40H in R0 register. MOV B, Rl means copy the content of Rl to B.<\/p>\n

Question 7.
\nWhat is an assembly language?
\nAnswer:
\nAssembly language fills the gap between low level machine language and high level language. It is written in instruction mnemonics.<\/p>\n

Question 8.
\nWrite ALP to represent \u201cload 15H into R2\u201d
\nAnswer:
\nMOV R2,#15H<\/p>\n

Question 9.
\nWrite the summary of unsigned multipliers
\nAnswer:
\nMultiplication operation use registers A and B as both source and destination addresses for the operation. MUL AB multiplies contents of A and B.<\/p>\n\n\n\n\n
Multiplication<\/strong><\/td>\nOperand 1<\/strong><\/td>\nOperand 2<\/strong><\/td>\nResult<\/strong><\/td>\n<\/tr>\n
Byte x Byte<\/td>\nA<\/td>\nB<\/td>\nA = lower byte
\nB = High byte<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 10.
\nWrite the summary of division of unsigned numbers.
\nAnswer:
\nIn the division of unsigned numbers 8051 supports byte over byte only. DIV AB, after division quotient is in A and remainder in B<\/p>\n\n\n\n\n
Division<\/strong><\/td>\nNumerator<\/strong><\/td>\nDenominator<\/strong><\/td>\nQuotient<\/strong><\/td>\nRemainder<\/strong><\/td>\n<\/tr>\n
Byte\/Byte<\/td>\nA<\/td>\nB<\/td>\nA<\/td>\nB<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"2nd<\/p>\n

Question 11.
\nMention any two core features of PIC microcontroller.
\nAnswer:
\nInstruction set simplicity and programmable timer options.<\/p>\n

Question 12.
\nMention the main features of PIC microcontroller.
\nAnswer:
\nOperating frequency-0-20MHz, Power supply voltage 2.0 to 5.5V, power saving sleep mode, ADC, Watch dog timer, US ART module etc.<\/p>\n

Question 13.
\nMention the power supply and power consumption of PIC16F877.
\nAnswer:
\nPower suppy 2.0V to 5.5V and power consumption is 220mA (2.0V, 4MHz), llmA(2.0V. 32kHz), 50nA standby mode.<\/p>\n

Question 14.
\nBriefly explain Data transfer instruction.
\nAnswer:
\nIn data transfer instructions, data is moved(copied) from the source to destination. Most ofthe instructions in this group does not affect the PSW flags. MOV, PUSH, POP and XCH are the commonly used instructions in this group.<\/p>\n

Question 15.
\nBriefly explain Arithmetic instruction.
\nAnswer:
\nVarious arithmetic operations are carried out in this group of instructions. ADD, SUB, INC, DEC, MUL, DIV are the basic operations used.<\/p>\n

Question 16.
\nBriefly explain Logical instructions.
\nAnswer:
\nLogical operations like AND, OR, XOR, complement, rotate are performed by these instructions. One of the operand is always register A.<\/p>\n

Question 17.
\nBriefly explain programming instructions.
\nAnswer:
\nThese are also called branching instructions. Conditional and unconditional branching instructions are used. JMP, AJMP, LJMP, CALL, ACALL are some examples.<\/p>\n

\"2nd<\/p>\n

Question 18.
\nWhat is the function of register A?
\nAnswer:
\nThe register A is used to store the data temporarily and also to hold the result of arithmetic and logical instructions.<\/p>\n

Question 19.
\nWhat is the function of register R?
\nAnswer:
\nRegister R stores numbers temporarily.<\/p>\n

Question 20.
\nWhat is the function of register PC?
\nAnswer:
\nIt is a 16 bit register. Program instruction bytes are fetched from locations in memory that are addressed by PC. The PC is automatically incremented after every instruction byte is fetched and may also be altered by certain instructions.<\/p>\n

Question 21.
\nBriefly explain about structure of Assembly Language.
\nAnswer:
\nAssembly language is a series of lines of assembly language instructions consisting of mnemonics, one or two operands. The four fields of assembly language are: label; mnemonic; operand; comment.<\/p>\n

2nd PUC Electronics Microcontrollers Three Marks Questions and Answers<\/h3>\n

Question 1.
\nMention the reason to write computer instructions in assembly language.
\nAnswer:
\na. To speed computer operation. Programs written in assembly language can be stored compactly, and less time is spent fetching the code. High level languages are converted to code by utility programs named compilers. Because of general nature of high level languages, the compilers often produce excess or over head-code.<\/p>\n

b. To reduce the size of the program. Assembly language requires no extra overhead code because the assembly language programmer is aware of extra needs of the program for any given situation.<\/p>\n

c. To write programs for special situations. Often, particularly when dealing with machine control, no standard programs exist. Robot arms and antilock brakes, for instance, have no standard drivers. It is generally more efficient to write nonstandard driver programs in assembly code, also, when speed of response is critical, assembly-coded programs execute rapidly because of the exact fit of program code to task requirements.<\/p>\n

d. To make economical. Small computer systems such as those that are embedded inside other machines, are often produced in large numbers. Reducing code size also reduces the cost of associated ROM chips.<\/p>\n

e. To better understand how computer operates. In order to fully understand what is going on “under the hood” of the CPU, you should learn to program the CPU in assembly language.<\/p>\n

\"2nd<\/p>\n

Question 2.
\nMention the different programs required ALPs converted into code memory bit charges,
\nAnswer:
\nThe different programs required ALPs converted into code memory bit charges are:
\na. An operating system program, which controls the operation of the personal computer used for the entire programming process. DOS is the operating system program used by many PCs. Windows is another.<\/p>\n

b. A word processing, often called a text editor program. Programs in assembly language mnemonics are written using the text editor and stored on a disk as files that, normally, end in the extension asm.The asm files are intended for the use of another program called the assembler programs. Any text editor that can produce an ASCII (text) file is suitable for writing assembly language, asm files.<\/p>\n

c. An assembler program, which takes the. asm assembly language program file and converts it to a machine code, obj file. The assembler converts the ASCII mnemonic text file into an. obj file that contains machine code instructions to the CPU, in binary form.<\/p>\n

d. A testing program, which lets you run and test your program under controlled conditions. Testing your program is the most important step of the programming process. To be able to test program one must be able to execute each instruction and see the results. Utility programs that allow the user to test programs are called debuggers or simulators.<\/p>\n

Question 3.
\nBriefly explain the different bits of binary memories of different registers.
\nAnswer:
\nFigure shows the different bits of binary memories of different registers.
\n\"2nd<\/p>\n

Question 4.
\nWhat are the different fields present in ALP (PAL – Practice Assembly language)?
\nAnswer:
\nPAL contains three internal registers:
\nRegister A: Performs all operations in the CPU Register B: Stores numbers temporarily
\nRegister PC: Holds the address of the next instruction to be executed in code memory. Register A and R can hold 8 bits of binary data and register PC can hold a 16-bit address number. PAL is called an 8-bit computer because the working registers A and R can hold a 1-byte number. PAL register PC limits the number of code byte addresses to 64k because it can hold code address from address 0000H to address FFFFH.<\/p>\n

Question 5.
\nWrite ALP to represent load 15H into R2
\nAnswer:
\nMOV R2, #15h<\/p>\n

Question 6.
\nWrite the description of ALP ADDA, R2
\nAnswer:
\nThe content of register R2 is added with the content of accumulator and the the sum is stored in the accumulator.<\/p>\n

\"2nd<\/p>\n

Question 7.
\nMention the different types of addressing modes.
\nAnswer:
\nThere are four addressing modes in 8051<\/p>\n

    \n
  • Immediate addressing mode<\/li>\n
  • Register addressing mode<\/li>\n
  • Direct addressing mode<\/li>\n
  • Indirect addressing mode<\/li>\n<\/ul>\n

    Question 8.
    \nBriefly explain data transfer instructions.
    \nAnswer:
    \nThese instructions move the content of one register to another one. Data can also be transferred to stack with the help of PUSH and POP instructions.
    \n\"2nd<\/p>\n

    Question 9.
    \nBriefly explain arithmetic instructions.
    \nAnswer:
    \nThese instructions perform several basic arithmetic operations. After execution, the result is stored in the first operand. 8 bit addition, subtraction, multiplication, increment-decrement instructions can be performed in 8051.<\/p>\n

    Example:
    \nADDA, Rn
    \nADDA, #8 bit data
    \nADDC A, Direct
    \nADDC A, Rn
    \nADDC A, @Ri
    \nADDC A, #8 bit data
    \nSUBB A, Rn
    \nSUBB A, Direct
    \nSUBB A, @Ri
    \nSUBB A, #8 bit data
    \nINC A
    \nINC Rn
    \nINC Direct
    \nINC @Ri
    \nDEC A
    \nDEC Rn
    \nDEC Direct
    \nDEC @Ri
    \nINC DPTR
    \nMUL A B
    \nDIVA B
    \nDA A<\/p>\n

    Question 10.
    \nBriefly explain logical instructions.
    \nAnswer:
    \nThese instructions perform logical operations between two register contents on a bit by bit\u2019 basis. After execution, the result is stored in the first operand.
    \nExample:
    \nANL A, Rn
    \nANL A, Direct
    \nANL A, @Ri
    \nANL A, #8 bit data
    \nANL Direct, A
    \nANL Direct, #8 bit data
    \nORLA, Rn
    \nQRL A, Direct
    \nORLA, @Ri
    \nORL A, #8 bit data
    \nORL Direct, A
    \nORL Direct, #8 bit data
    \nXRL A, Rn
    \nXRL A, Direct
    \nXRL A, @Ri
    \nXRL A, #8 bit data
    \nXRL Direct, A
    \nXRL Direct, #8 bit data
    \nCLR A
    \nCPLA
    \nRLA
    \nRLC A
    \nRR A
    \nRRC A
    \nSWAP A<\/p>\n

    \"2nd<\/p>\n

    Question 11.
    \nBriefly explain programming instructions.
    \nAnswer:
    \nAssembly language programming consists of writing programs using instr uction mnemonics that are specified by the manufacturer. An instruction consists of a label, an instruction mnemonic, operands and operational comments. The instruction mnemonic specifies the action to be taken by the CPU and the operands specify the addresses of data used in the action. Labels are the names given to the address numbers in program code memory.<\/p>\n

    Question 12.
    \nMention three key features,
    \ni. Max operating frequency
    \nii. Flash programming memory
    \niii. No. of pins of 16F877.
    \nAnswer:
    \ni. Operating speed:<\/p>\n

      \n
    • DC – 20 MHz oscillator\/clock input<\/li>\n
    • DC – 200 ns instruction cycle<\/li>\n<\/ul>\n

      ii. Flash memory size:<\/p>\n

        \n
      • 100,000 write Flash endurance<\/li>\n
      • 1,000,000 write EEPROM endurance<\/li>\n
      • Flash\/Data EEPROM retention: > 40 years<\/li>\n<\/ul>\n

        iii. No. of pins of 16F877: 28 pins<\/p>\n

        Question 13.
        \nMention the value of ROM, RAM and EEPROM of PIC16F877.
        \nAnswer:<\/p>\n

          \n
        • 100,000 write Flash endurance<\/li>\n
        • 1,000,000 write EEPROM endurance<\/li>\n
        • Flash\/Data EEPROM retention: > 40 years<\/li>\n<\/ul>\n

          2nd PUC Electronics Microcontrollers Five Marks Questions and Answers<\/h3>\n

          Question 1.
          \nCompare a microprocessor with a microcontroller
          \nAnswer:<\/p>\n\n\n\n\n\n\n\n
          Microprocessor<\/strong><\/td>\nMicrocontroller<\/strong><\/td>\n<\/tr>\n
          \u03bcP have many operational codes for moving data and performing operations, but it has only few bit-handling instructions.<\/td>\n\u03bcC have one or two operational codes for moving data from and performing operations, but it has many bit- handling instructions.<\/td>\n<\/tr>\n
          \u03bcP is concerned with rapid movement of code and data from external addresses to the chip.<\/td>\n\u03bcC is concerned with rapid movement of bits within the chip.<\/td>\n<\/tr>\n
          \u03bcP need external digital parts (Hardware) to perform its tasks as computer.<\/td>\n\u03bcC can function as computer without addition of external digital parts.<\/td>\n<\/tr>\n
          \u03bcP are used in general purpose computing devices, supports wide range of applications with general hardware.<\/td>\n\u03bcC are used in embedded systems and supports only dedicated applications with customized hardware.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Question 2.
          \nMention the features of 8051.
          \nAnswer:
          \n1. Eight-bit CPU with registers A (Accumulator) and B.
          \n2. Sixteen-bit program counter (PC) and data pointer(DPTR)
          \n3. Eight-bit program status word (PSW)
          \n4. Eight-bit stack pointer(SP)
          \n5. Internal ROM or EPROM (8751) of 4k bytes
          \n6. Internal RAM of 128 bytes :<\/p>\n

            \n
          • Four register banks, each containing eight registers<\/li>\n
          • Sixteen bytes, which may be addressed at the bit level<\/li>\n
          • Eight bytes of general purpose registers<\/li>\n<\/ul>\n

            7. Four 8-bit I\/O ports (P0-P3)
            \n8. Two 16-bit timer\/counter TO and T1
            \n9. Full duplex serial data receiver\/ transmitter : SBUF
            \n10. Control register: TCON, TMOD,SCON,PCON,IP and IE
            \n11. Two external and two internal interrupt sources
            \n12. Oscillator and clock circuits<\/p>\n

            \"2nd<\/p>\n

            Question 3.
            \nDraw the pin diagram of 8051
            \nAnswer:
            \n\"2nd<\/p>\n

            Question 4.
            \nBriefly explain the core features of pic microcontroller(PIC 16F887)
            \nAnswer:
            \n\"2nd<\/p>\n

            Question 5.
            \nBriefly explain the steps used in creating a program.
            \nAnswer:
            \nThe steps to create a program are:<\/p>\n

              \n
            • Program editor is used to type in the program. The editor must be able to produce ASCII file for many assembler and will have the extension \u201c.asm\u201d or \u201csrc\u201d.<\/li>\n
            • Assembler on receiving the file in \u201c.asm\u201d form converts the instructions into steps to creat a program machine code. Assembler will produce an object file \u201c.obj\u201d and list file \u201c.1st\u201d.<\/li>\n
            • Linker program links one or more object files and produce an absolute object file \u201c.abs\u201d.<\/li>\n
            • The \u201c.abs\u201d file is fed to object to hex converter(OH) and create an extension \u201c.hex\u201d and is ready to bum ROM.<\/li>\n<\/ul>\n

              Question 6.
              \nBriefly explain signed and unsigned subprogram.
              \nAnswer:
              \nSubtraction of unsigned numbers: In 8051 SUBB is used to subtract two numbers by using unsigned representation by setting the bit CY=0 or CY=1. Whereas, in signed representation the sum or difference should not be greater than or less than -128 or 127.<\/p>\n

              Question 7.
              \nExplain the core features of PIC16F887 microcontroller.
              \nAnswer:
              \n24\/35 I\/O pins with individual direction control:<\/p>\n

                \n
              • High current source\/sink for direct LED drive<\/li>\n
              • Interrupt-on-Change pin<\/li>\n
              • Individually programmable weak pull-ups<\/li>\n
              • Ultra Low-Power Wake-up (ULPWU)<\/li>\n<\/ul>\n

                Analog Comparator module with:<\/p>\n

                  \n
                • Two analog comparators<\/li>\n
                • Programmable on-chip voltage reference (CVREF) module (% of VDD)<\/li>\n
                • Fixed voltage reference (0.6V)<\/li>\n
                • Comparator inputs and outputs externally accessible<\/li>\n
                • SR Latch mode<\/li>\n
                • External Timer 1 Gate (count enable)<\/li>\n<\/ul>\n

                  A\/D Converter:<\/p>\n

                    \n
                  • 10-bit resolution and 11\/14 channels<\/li>\n
                  • Timer O: 8-bit timer\/counter with 8-bit programmable prescaler<\/li>\n<\/ul>\n

                    Enhanced Timer 1:<\/p>\n

                      \n
                    • 16-bit timer\/counter with prescaler<\/li>\n
                    • External Gate Input mode<\/li>\n
                    • Dedicated low-power 32 kHz oscillator<\/li>\n<\/ul>\n

                      Timer2: 8-bit timer\/counter with 8-bit period register, prescaler and postscaler<\/p>\n

                      Enhanced Capture, Compare, PWM+ module:<\/p>\n

                        \n
                      • 16-bit Capture, max. resolution 12.5 ns<\/li>\n
                      • Compare, max. resolution 200 ns<\/li>\n
                      • 10-bit PWM with 1, 2 or 4 output channels, programmable \u201cdead time\u201d, max. frequency 20 kHz<\/li>\n
                      • PWM output steering control<\/li>\n<\/ul>\n

                        Capture, Compare, PWM module:<\/p>\n

                          \n
                        • 16-bit Capture, max. resolution 12.5 ns<\/li>\n
                        • 16-bit Compare, max. resolution 200 ns<\/li>\n
                        • 10-bit PWM, max. frequency 20 kHz<\/li>\n<\/ul>\n

                          Enhanced USART module:<\/p>\n

                            \n
                          • Supports RS-485, RS-232, and LIN 2.0<\/li>\n
                          • Auto-Baud Detect<\/li>\n
                          • Auto-Wake-Up on Start bit<\/li>\n
                          • In-Circuit Serial Programming TM (ICSPTM) via two pins<\/li>\n
                          • Master Synchronous Serial Port (MSSP) module – supporting 3-wire SPI (all 4 modes) and 12C\u2122 Master and Slave Modes with I2C address mask<\/li>\n<\/ul>\n

                            \"2nd<\/p>\n

                            Question 8.
                            \nBriefly explain different addressing modes in the 8051.
                            \nAnswer:
                            \nThere are four addressing modes in 8051.<\/p>\n

                              \n
                            • Immediate addressing mode: the data source is available immediately as a part of instruction itself. Example: MOV A, #8bit data<\/li>\n
                            • Register addressing mode: the source and destination register names are parts of the opcode in instruction. Example: MOV A, R0<\/li>\n
                            • Direct addressing mode: the source and\/or destination may be the internal RAM locations. Example: MOV 90h, #50h<\/li>\n
                            • Indirect addressing mode: the source or destination address may be indicated the content of index register. Example: MOV @Rl,#40h<\/li>\n<\/ul>\n

                              Problems with Solutions<\/span><\/p>\n

                              Question 1.
                              \nWrite the instructions to move value 34H into register A and value 3FH into register B, then add them together.
                              \nAnswer:
                              \nMOV A, #34h
                              \nMOV 0F0, #3Fh
                              \nADD A, 0F0
                              \nLCALL 0003h<\/p>\n

                              Question 2.
                              \nWrite the instructions to add the values 16H and CDH, place the result in register R2.
                              \nAnswer:
                              \nMOV A, #16h
                              \nMOV R2, #CDh
                              \nADD A, R2
                              \nMOV R2,A
                              \nLCALL 0003 h<\/p>\n

                              \"2nd<\/p>\n

                              Question 3.
                              \nAdd 25H and 34H and put the result in register A.
                              \nAnswer:
                              \nMOV A, #25h
                              \nADDA, #34h
                              \nLCALL 0003<\/p>\n

                              Question 4.
                              \nWrite a program to add two 8-bit numbers and store it in R6. The numbers are 01EH and 01CH.
                              \nAnswer:
                              \nMOV A, #lEh
                              \nADD A, #lCh
                              \nMOV R6, A
                              \nLCALL 0003h<\/p>\n

                              Question 5.
                              \nSubtract 21H from 30H and write the program and solve.
                              \nAnswer:
                              \nMOV A, #30h
                              \nSUBB A, #21h
                              \nLCALL 0003h<\/p>\n","protected":false},"excerpt":{"rendered":"

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