Microprocessor and Microcontroller (4341101) - Winter 2024 Solution

Question 1(a) [3 marks]

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List Common features of Microcontrollers.

Answer:

Mnemonic: “CRITICS: CPU ROM I/O Timers Interrupts Comm Serial”

Question 1(b) [4 marks]

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Explain the functions of ALU.

Answer:

Diagram:

Mnemonic: “ALFS: Arithmetic Logic Flags Status”

Question 1(c) [7 marks]

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Define: Memory, Operand, Instruction Cycle, Opcode, CU, Machine Cycle, CISC

Answer:

• Memory: Organized array of storage cells with unique addresses
• Operand: Data elements that instructions operate upon
• Instruction Cycle: Fetch-decode-execute sequence for each instruction
• Opcode: Binary code that tells processor what operation to perform

Diagram:

Mnemonic: “MO-ICO-MC: Memory-Operand-Instruction-Control-Operation-Machine-Complex”

Question 1(c OR) [7 marks]

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i) Define: Microprocessor. ii) Compare Von-Neumann and Harvard architecture.

Answer:

i) Microprocessor Definition:

An integrated circuit containing the CPU functionality of a computer,
capable of fetching, decoding, and executing instructions with ALU and
control circuitry on a single chip.

ii) Von-Neumann vs Harvard Architecture:

Diagram:

Mnemonic: “Harvard Has Separate Spaces”

Question 2(a) [3 marks]

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Explain various Registers of 8085 microprocessor.

Answer:

Mnemonic: “AGSF: Accumulator-General-Stack-Flags”

Question 2(b) [4 marks]

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Explain Fetching, Decoding and Execution of Instruction.

Answer:

Diagram:

• Fetching: PC sends address to memory, instruction loaded into IR
• Decoding: Control unit interprets instruction opcode and addressing mode
• Execution: ALU performs arithmetic/logic, data moves between registers/memory

Mnemonic: “FDE: First Get, Then Understand, Finally Do”

Question 2(c) [7 marks]

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Describe block diagram of 8085 microprocessor with the help of neat diagram.

Answer:

Diagram:

• Core Components: ALU and registers form processing core
• Control Path: Instructions flow through register, decoder, control unit
• Data Path: Data moves through buffers to/from external buses
• Timing: Synchronizes all operations via internal clock

Mnemonic: “RAID: Registers-ALU-Instructions-Decoders”

Question 2(a OR) [3 marks]

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Compare Microprocessor & Microcontroller.

Answer:

Mnemonic: “Micro-P Processes, Micro-C Controls”

Question 2(b OR) [4 marks]

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Explain De-multiplexing of Address and Data buses for 8085 Microprocessor.

Answer:

Diagram:

• Multiplexing: AD0-AD7 pins carry address and data at different times
• ALE Signal: Address Latch Enable controls when address is captured
• 8-bit Latch: Holds lower address bits during entire machine cycle
• Timing: Address valid only during high state of ALE pulse

Mnemonic: “ALAD: ALE Latches Address before Data”

Question 2(c OR) [7 marks]

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Describe Pin diagram of 8085 microprocessor with the help of neat diagram.

Answer:

Diagram:

• Address Pins: A8-A15 (8) and multiplexed AD0-AD7 (8) for 16-bit addressing
• Control Pins: Generate timing and control signals for peripheral devices
• Interrupt Pins: Hardware interrupt handling with priority levels
• Clock: X1, X2 for crystal connection, CLK for synchronization
• Power: Vcc (+5V) and Vss (Ground) for power supply

Mnemonic: “ACID-PS: Address-Control-Interrupt-DMA-Power-Serial”

Question 3(a) [3 marks]

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Explain interrupts of 8051 microcontroller.

Answer:

Diagram:

Mnemonic: “ETTES: External-Timer-Timer-External-Serial”

Question 3(b) [4 marks]

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Draw Pin diagram of 8051 microcontroller.

Answer:

Mnemonic: “PORT 0123: Data-General-Address-Special”

Question 3(c) [7 marks]

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Explain Internal RAM Organization of 8051 microcontroller.

Answer:

Diagram:

• Register Banks: Four banks of 8 registers (R0-R7) selectable by PSW
• Bit-addressable: 16 bytes (128 bits) individually addressable as bits
• General Purpose: User variables and stack space
• SFRs: Control and status registers at higher addresses

Mnemonic: “RBBS: Registers Bits Buffer Special”

Question 3(a OR) [3 marks]

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List SFRs with their addresses.

Answer:

Mnemonic: “PDPT-SP: Ports-Data-Program-Timers-Serial-Prioritized”

Question 3(b OR) [4 marks]

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Explain Timers/Counters logic diagram of 8051 microcontroller.

Answer:

Timer/Counter Diagram:

• Timer: Uses internal clock, counts machine cycles
• Counter: Counts external events on T0/T1 pins
• Control Bits: Set in TMOD and TCON registers
• Modes: Different timer configurations (13/16/8-bit)

Mnemonic: “TCG: Timer-Counter-Gate”

Question 3(c OR) [7 marks]

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Explain block diagram of 8051 microcontroller.

Answer:

Diagram:

• Harvard Architecture: Separate program and data memory
• CISC Design: Rich instruction set (over 100 instructions)
• In-built Peripherals: No need for external components
• Single-chip Solution: Complete system on one chip

Mnemonic: “CAPITALS: CPU Architecture Ports I/O Timer ALU LS-Interface Serial”

Question 4(a) [3 marks]

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Write an 8051 Assembly Language Program to add two bytes of data and store result in R4 register.

Answer:

      MOV  A, #25H    ; Load first value (25H) into accumulator
      MOV  R3, #18H   ; Load second value (18H) into R3
      ADD  A, R3      ; Add R3 to accumulator
      MOV  R4, A      ; Store result in R4 register

Key Steps:

• Load first operand into Accumulator
• Load second operand into register R3
• Perform addition using ADD instruction
• Store result from Accumulator to R4

Mnemonic: “LLAS: Load-Load-Add-Store”

Question 4(b) [4 marks]

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Write an 8051 Assembly Language Program to OR the contents of Port-1 and Port-2 then put the result in external RAM location 0200H.

Answer:

      MOV  A, P1      ; Load contents of Port-1 into accumulator
      ORL  A, P2      ; OR Port-2 contents with accumulator
      MOV  DPTR, #0200H ; Load DPTR with external RAM address
      MOVX @DPTR, A    ; Store result in external RAM location 0200H

Key Steps:

• Read Port-1 into Accumulator
• Perform OR operation with Port-2
• Set up Data Pointer (DPTR) for external RAM
• Write result to external memory

Mnemonic: “PORT: Port-OR-Register-Transfer”

Question 4(c) [7 marks]

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List Addressing Modes of 8051 Microcontroller and explain them with at least one example.

Answer:

Examples:

• Immediate: MOV A, #55H (Load A with 55H)
• Register: ADD A, R3 (Add R3 to A)
• Direct: MOV 40H, A (Store A at address 40H)
• Indirect: MOV @R0, #5 (Store 5 at address in R0)
• Indexed: MOVC A, @A+DPTR (Read code memory)
• Bit: CLR C (Clear carry flag)
• Relative: JZ LOOP (Jump if A is zero)

Mnemonic: “I’M DIRBI: Immediate Register Direct Bit Indexed”

Question 4(a OR) [3 marks]

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Explain following instructions: (i) DJNZ (ii) POP (iii) CJNE.

Answer:

Examples and Explanation:

• DJNZ R7, LOOP: Decrements R7 and jumps to LOOP if R7≠0

  • Used for loop counters and delays

• POP 30H: Copies data from stack to address 30H

  • Increments SP after data retrieval

• CJNE A, #25H, NOTEQUAL: Compares A with 25H, jumps if not equal

  • Also sets Carry flag if A < data

Mnemonic: “DPC: Decrement-Pop-Compare”

Question 4(b OR) [4 marks]

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For 8051 Microcontroller with a crystal frequency of 12 MHz, generate a delay of 4ms.

Answer:

; Crystal frequency = 12 MHz
; Machine cycle = 1 μs
; Required delay = 4 ms = 4000 machine cycles

      MOV  R7, #16    ; Outer loop counter (16 x 250 = 4000)
DELAY1: MOV  R6, #250  ; Inner loop counter 
DELAY2: NOP            ; 1 machine cycle
        NOP            ; 1 machine cycle
        DJNZ R6, DELAY2 ; 2 machine cycles (250 x 4 = 1000 cycles)
        DJNZ R7, DELAY1 ; 16 x 250 = 4000 cycles total
        RET            ; Return from subroutine

Calculation:

• Each inner loop: 4 cycles × 250 iterations = 1000
• Outer loop: 16 iterations × 1000 cycles = 16,000 cycles
• At 12MHz, 1 machine cycle = 1μs
• Total delay = 4ms (4000 cycles)

Mnemonic: “LNDD: Load-NOP-Decrement-Decrement”

Question 4(c OR) [7 marks]

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Explain any seven Logical instructions with example for 8051 Microcontroller.

Answer:

Examples with Explanation:

1. ANL A, #0FH: Masks high nibble (A = A AND 0FH)

   • Before: A = 95H, After: A = 05H

2. ORL 20H, A: Sets bits in memory (20H = 20H OR A)

   • Before: 20H = 55H, A = 0AH, After: 20H = 5FH

3. XRL A, #55H: Toggles specific bits (A = A XOR 55H)

   • Before: A = AAH, After: A = FFH

4. CLR C: Clears carry flag (C = 0)

   • Used before subtraction operations

5. CPL A: Inverts all bits (A = NOT A)

   • Before: A = 55H, After: A = AAH

6. RL A: Rotates accumulator left one bit

   • Before: A = 85H (10000101), After: A = 0BH (00001011)

7. RR A: Rotates accumulator right one bit

   • Before: A = 85H (10000101), After: A = C2H (11000010)

Mnemonic: “A-OX-CCR: AND OR XOR Clear Complement Rotate”

Question 5(a) [3 marks]

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List Applications of microcontroller in various fields.

Answer:

Mnemonic: “I-MACS: Industrial-Medical-Automotive-Consumer-Security”

Question 5(b) [4 marks]

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Interface Push button Switch and LED with 8051 microcontroller.

Answer:

Circuit Diagram:

Program:

AGAIN:  JB    P1.0, LED_OFF  ; If P1.0=1 (not pressed), LED off
        SETB  P1.7           ; If P1.0=0 (pressed), LED on
        SJMP  AGAIN          ; Repeat
LED_OFF:CLR   P1.7           ; Turn LED off
        SJMP  AGAIN          ; Repeat

Mnemonic: “PLIC: Push-LED-Input-Control”

Question 5(c) [7 marks]

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Interface LCD with microcontroller and write a program to display “HELLO”.

Answer:

Circuit Diagram:

Program:

ORG 0000H               ; Start address
    
; Initialize LCD
    MOV A, #38H         ; 8-bit, 2 lines, 5x7 dots
    ACALL COMMAND       ; Send command
    MOV A, #0EH         ; Display ON, cursor ON
    ACALL COMMAND       ; Send command
    MOV A, #01H         ; Clear display
    ACALL COMMAND       ; Send command
    MOV A, #06H         ; Increment cursor
    ACALL COMMAND       ; Send command
    MOV A, #80H         ; First line, first position
    ACALL COMMAND       ; Send command
    
; Display "HELLO"
    MOV A, #'H'         ; Load 'H'
    ACALL DISPLAY       ; Display it
    MOV A, #'E'         ; Load 'E'
    ACALL DISPLAY       ; Display it
    MOV A, #'L'         ; Load 'L'
    ACALL DISPLAY       ; Display it
    MOV A, #'L'         ; Load 'L'
    ACALL DISPLAY       ; Display it
    MOV A, #'O'         ; Load 'O'
    ACALL DISPLAY       ; Display it
    
    SJMP $              ; Stay here
    
; Command subroutine
COMMAND:
    MOV P2, A           ; Put command on data bus
    CLR P3.0            ; RS=0 for command
    CLR P3.1            ; R/W=0 for write
    SETB P3.2           ; E=1
    ACALL DELAY         ; Wait
    CLR P3.2            ; E=0
    RET                 ; Return
    
; Display subroutine
DISPLAY:
    MOV P2, A           ; Put data on data bus
    SETB P3.0           ; RS=1 for data
    CLR P3.1            ; R/W=0 for write
    SETB P3.2           ; E=1
    ACALL DELAY         ; Wait
    CLR P3.2            ; E=0
    RET                 ; Return
    
; Delay subroutine
DELAY:
    MOV R7, #50         ; Load counter
DELAY_LOOP:
    DJNZ R7, DELAY_LOOP ; Loop until R7=0
    RET                 ; Return
    
END                     ; End of program

• Initialization: Configure LCD for 8-bit, 2 lines, cursor options
• Data Transfer: Commands sent with RS=0, characters with RS=1
• Characters: ASCII values sent one by one to display text
• Timing: Delay routine ensures proper signal timing

Mnemonic: “DICE: Data-Instruction-Control-Enable”

Question 5(a OR) [3 marks]

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Draw Interfacing of LM35 with 8051 microcontroller.

Answer:

Circuit Diagram:

Key Points:

• LM35 produces analog voltage proportional to temperature
• ADC0804 converts analog voltage to digital value
• 8051 controls ADC and reads temperature data
• Resolution: 10mV/°C → ~0.2°C resolution with 8-bit ADC

Mnemonic: “TAC: Temperature-Analog-Convert”

Question 5(b OR) [4 marks]

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Interface Stepper motor with 8051 microcontroller.

Answer:

Circuit Diagram:

Program:

      ORG 0000H

; Stepper Motor Sequence for Clockwise Rotation
SEQ:  DB 00001000B  ; Step 1
      DB 00001100B  ; Step 2
      DB 00000100B  ; Step 3
      DB 00000110B  ; Step 4
      DB 00000010B  ; Step 5
      DB 00000011B  ; Step 6
      DB 00000001B  ; Step 7
      DB 00001001B  ; Step 8

MAIN: MOV R0, #00H  ; Initialize sequence pointer
      
STEP: MOV A, R0     ; Get current sequence number
      ANL A, #07H   ; Keep within 0-7 range (8 steps)
      MOV DPTR, #SEQ ; Point to sequence table
      MOVC A, @A+DPTR ; Get sequence pattern
      MOV P1, A     ; Output to stepper motor
      
      ACALL DELAY   ; Wait between steps
      
      INC R0        ; Next sequence
      SJMP STEP     ; Repeat
      
DELAY: MOV R6, #250 ; Delay loop
LOOP:  MOV R7, #250
LOOP2: DJNZ R7, LOOP2
       DJNZ R6, LOOP
       RET
       
      END

Key Points:

• Stepper motor requires specific sequence of pulses for rotation
• ULN2003 provides current amplification for motor coils
• 8-step sequence provides smoother rotation
• Delay between steps controls rotation speed

Mnemonic: “PDCS: Port-Driver-Current-Sequence”

Question 5(c OR) [7 marks]

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Interface ADC0804 with 8051 microcontroller.

Answer:

Circuit Diagram:

Program:

ORG 0000H

START:  CLR P1.0      ; CS = 0 (Chip Select active)
        CLR P1.1      ; RD = 0 
        CLR P1.2      ; WR = 0
        SETB P1.2     ; WR = 1 (Start conversion)
        
WAIT:   JB P1.3, WAIT ; Wait for conversion (INTR = 0)
        
        CLR P1.1      ; RD = 0 (Read data)
        MOV A, P0     ; Read converted data into A
        MOV 30H, A    ; Store result in RAM
        
        SETB P1.0     ; CS = 1 (Chip deselect)
        
PROCESS:
        ; Process the data (e.g., display, compare, etc.)
        ; ...
        
        ACALL DELAY   ; Wait before next conversion
        SJMP START    ; Repeat
        
DELAY:  MOV R7, #200  ; Delay routine
DLOOP:  DJNZ R7, DLOOP
        RET
        
END

ADC0804 Features:

• 8-bit resolution (256 steps)
• 0-5V input range
• Single-channel operation
• ~100μs conversion time
• Interface protocol:
  1. Activate CS, pulse WR to start conversion
  2. Wait for INTR to go low (conversion complete)
  3. Activate RD to read data
  4. Deactivate CS when done

Mnemonic: “CRIW: Control-Read-Interrupt-Write”