Microprocessor & Microcontroller Systems (1333202) - Winter 2023 Solution

Question 1(a) [3 marks]

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Define Microprocessor.

Answer:

A microprocessor is a single-chip CPU that contains all the arithmetic, logic, and control circuitry required to perform the functions of a digital computer’s central processing unit.

Table: Microprocessor Key Features

• Central Processing Unit: Core component that executes instructions
• Integrated Circuit: All functions combined on single silicon chip
• Programmable Device: Can execute different programs based on stored instructions

Mnemonic: “Single Chip CPU = Smart Computer Processor Unit”

Question 1(b) [4 marks]

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Explain Flag register of microprocessor.

Answer:

The Flag register stores status information about the result of arithmetic and logical operations performed by the ALU.

Table: 8085 Flag Register Bits

• Status Indicator: Shows condition of last operation result
• Conditional Instructions: Used for branching and decision making
• 5 Active Flags: Sign, Zero, Auxiliary Carry, Parity, and Carry flags

Mnemonic: “Flags Show Zero, Sign, Parity, Auxiliary, Carry”

Question 1(c) [7 marks]

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Explain format of instruction of microprocessor with example.

Answer:

Microprocessor instructions consist of opcode and operand fields that specify the operation and data locations.

Table: 8085 Instruction Format Types

Diagram:

graph TD
    A[Instruction Format] --> B[Opcode Field]
    A --> C[Operand Field]
    B --> D[Operation Code<br/>Specifies Operation]
    C --> E[Data/Address<br/>Specifies Source/Destination]

• Opcode Field: Defines the operation to be performed (ADD, MOV, JMP)
• Operand Field: Contains data, register, or memory address information
• Variable Length: Instructions can be 1, 2, or 3 bytes long
• Addressing Modes: Different ways to specify operand location

Mnemonic: “Opcode Operations + Operand Objects = Complete Commands”

Question 1(c OR) [7 marks]

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Explain function of ALU, Control Unit and CPU of Microprocessor.

Answer:

The CPU consists of three main functional units that work together to execute instructions.

Table: CPU Components and Functions

Diagram:

graph LR
    A[CPU] --> B[ALU<br/>Arithmetic Logic Unit]
    A --> C[Control Unit]
    A --> D[Register Array]
    B --> E[Math Operations<br/>Logic Operations]
    C --> F[Instruction Control<br/>Signal Generation]

• ALU Functions: Performs all arithmetic calculations and logical operations
• Control Unit Tasks: Manages instruction execution cycle and generates control signals
• CPU Coordination: Integrates ALU and Control Unit for complete processing

Mnemonic: “ALU Adds, Control Commands, CPU Coordinates”

Question 2(a) [3 marks]

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Explain function of ALE signal with diagram.

Answer:

ALE (Address Latch Enable) signal is used to demultiplex the lower-order address and data lines.

Table: ALE Signal Functions

Diagram:

• Active High Signal: ALE goes high during T1 state
• External Latching: Used with 74373 latch to hold address
• System Timing: Provides reference for external devices

Mnemonic: “ALE Always Latches External Addresses”

Question 2(b) [4 marks]

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Compare microprocessor and microcontroller

Answer:

Table: Microprocessor vs Microcontroller Comparison

• Integration Level: Microcontroller has more integrated components
• Application Focus: Microprocessor for computing, microcontroller for control
• System Complexity: Microprocessor needs more external components
• Design Flexibility: Microprocessor offers more expandability

Mnemonic: “Microprocessor = More Power, Microcontroller = More Control”

Question 2(c) [7 marks]

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Draw & explain block diagram of microprocessor.

Answer:

The 8085 microprocessor consists of several functional blocks that work together.

Diagram:

graph TB
    A[Accumulator] --> B[ALU]
    C[Temp Register] --> B
    B --> D[Flag Register]
    E[Instruction Register] --> F[Instruction Decoder]
    F --> G[Timing & Control Unit]
    H[Program Counter] --> I[Address Buffer]
    J[Stack Pointer] --> I
    K[Register Array<br/>B,C,D,E,H,L] 
    I --> L[Address Bus A8-A15]
    M[Address/Data Buffer] --> N[Address/Data Bus AD0-AD7]

Table: Block Functions

• Data Path: Information flows between registers through internal bus
• Control Signals: Generated by timing and control unit
• Bus Interface: Connects to external memory and I/O devices
• Register Operations: Temporary storage for operands and results

Mnemonic: “Blocks Build Better Processing Systems”

Question 2(a OR) [3 marks]

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Explain 16 bits registers of microprocessor.

Answer:

The 8085 has three 16-bit registers formed by combining 8-bit register pairs.

Table: 16-bit Registers

• Program Counter: Automatically increments to next instruction
• Stack Pointer: Points to last pushed data on stack
• HL Pair: Most frequently used for memory addressing

Mnemonic: “PC Points Program, SP Stacks Properly, HL Holds Location”

Question 2(b OR) [4 marks]

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Explain de-multiplexing lower order address and data lines with diagram of microprocessor.

Answer:

The 8085 multiplexes lower 8-bit address with data lines to reduce pin count.

Table: Multiplexed Lines

Diagram:

• Time Division: Same lines carry address then data
• External Latch: 74373 captures address when ALE is high
• Signal Separation: Creates separate address and data buses

Mnemonic: “ALE Always Latches External Address Elegantly”

Question 2(c OR) [7 marks]

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Draw and explain pin diagram of 8085.

Answer:

The 8085 is a 40-pin microprocessor with multiplexed address/data bus.

Diagram:

Table: Pin Groups

• Multiplexed Bus: AD0-AD7 carry both address and data
• Active Low Signals: Signals with * are active low
• Crystal Connections: X1, X2 for clock generation

Mnemonic: “Forty Pins Provide Perfect Processing Power”

Question 3(a) [3 marks]

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Draw clock and reset circuit of microcontroller

Answer:

The 8051 requires external clock and reset circuits for proper operation.

Diagram:

Table: Circuit Components

• Clock Frequency: Commonly 11.0592 MHz for serial communication
• Reset Duration: Must be high for at least 2 machine cycles
• Power-on Reset: Automatic reset when power is applied

Mnemonic: “Crystals Create Clock, Resistors Reset Reliably”

Question 3(b) [4 marks]

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Explain internal RAM of 8051.

Answer:

The 8051 contains 256 bytes of internal RAM organized in different sections.

Table: Internal RAM Organization

Diagram:

graph TD
    A[Internal RAM 256 Bytes] --> B[Bank 0-3<br/>00H-1FH]
    A --> C[Bit Addressable<br/>20H-2FH]  
    A --> D[General Purpose<br/>30H-7FH]
    A --> E[SFRs<br/>80H-FFH]

• Register Banks: Four banks of 8 registers each (R0-R7)
• Bit Addressing: Individual bits can be addressed in 20H-2FH area
• Stack Area: Usually located in general purpose RAM area
• Direct Access: All locations accessible through direct addressing

Mnemonic: “RAM Registers, Bits, General, Special Functions”

Question 3(c) [7 marks]

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Explain block diagram of 8051.

Answer:

The 8051 microcontroller integrates CPU, memory, and I/O on a single chip.

Diagram:

graph TB
    A[CPU Core] --> B[ALU]
    A --> C[Accumulator]
    A --> D[B Register]
    E[Program Memory<br/>4KB ROM] --> F[Program Counter]
    G[Data Memory<br/>256B RAM] --> H[Data Pointer DPTR]
    I[Timer/Counter] --> J[Timer 0/1]
    K[Serial Port] --> L[UART]
    M[Interrupt System] --> N[5 Interrupt Sources]
    O[I/O Ports] --> P[P0, P1, P2, P3]
    Q[Oscillator] --> R[Clock Circuit]

Table: Major Blocks

• Harvard Architecture: Separate program and data memory spaces
• Built-in Peripherals: Timers, serial port, interrupts integrated
• Expandable: External memory and I/O can be added
• Control Applications: Optimized for embedded control tasks

Mnemonic: “Complete Control Chip Contains CPU, Memory, I/O”

Question 3(a OR) [3 marks]

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Explain function of DPTR and PC.

Answer:

DPTR and PC are important 16-bit registers in 8051 for memory addressing.

Table: DPTR and PC Functions

• DPTR Usage: Accessing external RAM and lookup tables
• PC Function: Automatically increments after instruction fetch
• 16-bit Addressing: Both can address 64KB memory space

Mnemonic: “DPTR Data Pointer, PC Program Counter”

Question 3(b OR) [4 marks]

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Explain different timer modes of microcontroller.

Answer:

The 8051 has two timers with four different operating modes.

Table: Timer Modes

• Mode Selection: Controlled by TMOD register bits
• Timer 0/1: Both timers support modes 0, 1, 2
• Mode 3 Special: Only Timer 0 can operate in mode 3
• Applications: Delays, baud rate generation, event counting

Mnemonic: “Modes Make Timers Tremendously Versatile”

Question 3(c OR) [7 marks]

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Explain interrupts of microcontroller.

Answer:

The 8051 has a 5-source interrupt system for handling external events.

Table: 8051 Interrupt Sources

Diagram:

graph TD
    A[Interrupt System] --> B[External INT0]
    A --> C[Timer 0 Overflow]
    A --> D[External INT1] 
    A --> E[Timer 1 Overflow]
    A --> F[Serial Port]
    G[Interrupt Control] --> H[IE Register]
    G --> I[IP Register]

• Interrupt Enable: IE register controls individual interrupt enables
• Priority Control: IP register sets interrupt priorities
• Vector Addresses: Each interrupt has fixed vector location
• Nested Interrupts: Higher priority can interrupt lower priority

Mnemonic: “Five Interrupt Sources Serve System Efficiently”

Question 4(a) [3 marks]

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Explain data transfer instruction with example for 8051.

Answer:

Data transfer instructions move data between registers, memory, and I/O ports.

Table: Data Transfer Instructions

• MOV Variants: Register to register, immediate to register
• External Access: MOVX for external RAM operations
• Code Access: MOVC for reading program memory tables

Mnemonic: “MOV Moves data, MOVX eXternal, MOVC Code”

Question 4(b) [4 marks]

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List and explain different addressing modes of microcontroller.

Answer:

The 8051 supports several addressing modes for flexible data access.

Table: 8051 Addressing Modes

• Immediate Mode: Constant data included in instruction
• Register Mode: Fastest execution using register file
• Direct Mode: Access any internal RAM location
• Indirect Mode: Pointer-based addressing for arrays
• Indexed Mode: Table lookup and array access

Mnemonic: “Immediate, Register, Direct, Indirect, Indexed Addressing”

Question 4(c) [7 marks]

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Write a program to copy block of 8 data starting from location 100h to 200h.

Answer:

Assembly Program:

ORG 0000H           ; Start address
MOV R0,#100H        ; Source address pointer
MOV R1,#200H        ; Destination address pointer  
MOV R2,#08H         ; Counter for 8 bytes

LOOP:
MOV A,@R0           ; Read data from source
MOV @R1,A           ; Write data to destination
INC R0              ; Increment source pointer
INC R1              ; Increment destination pointer
DJNZ R2,LOOP        ; Decrement counter and jump if not zero

END                 ; End of program

Table: Register Usage

• Indirect Addressing: @R0 and @R1 for memory access
• Loop Control: DJNZ instruction decrements and tests
• Block Transfer: Copies 8 consecutive bytes efficiently

Mnemonic: “Read, Write, Increment, Decrement, Jump Loop”

Question 4(a OR) [3 marks]

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Write a program to add two bytes of data and store result in R0 register.

Answer:

Assembly Program:

ORG 0000H           ; Start address
MOV A,#25H          ; Load first byte
ADD A,#35H          ; Add second byte
MOV R0,A            ; Store result in R0
END                 ; End program

Table: Operation Steps

• Addition Result: 25H + 35H = 5AH
• Flag Effects: Carry flag set if result > FFH

Mnemonic: “Move, Add, Move = Simple Addition”

Question 4(b OR) [4 marks]

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Explain indexed addressing mode with example.

Answer:

Indexed addressing uses a base address plus an offset for memory access.

Table: Indexed Addressing Details

Example:

MOV DPTR,#1000H     ; Base address
MOV A,#05H          ; Index value
MOVC A,@A+DPTR      ; Read from address 1005H

• Table Access: Ideal for lookup tables and arrays
• Program Memory: MOVC reads from code memory only
• Dynamic Indexing: Index can change during execution

Mnemonic: “Base + Index = Dynamic Access”

Question 4(c OR) [7 marks]

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Explain stack operation of microcontroller, PUSH and POP instruction.

Answer:

The stack is a LIFO memory structure used for temporary data storage.

Table: Stack Operations

Diagram:

Example Program:

MOV SP,#30H         ; Initialize stack pointer
PUSH ACC            ; Save accumulator
PUSH B              ; Save B register
POP B               ; Restore B register
POP ACC             ; Restore accumulator

• LIFO Structure: Last In, First Out data organization
• SP Auto-increment: Stack pointer automatically adjusts
• Subroutine Calls: Stack saves return addresses
• Register Preservation: Save/restore register contents

Mnemonic: “PUSH Puts Up, Stack Holds, POP Pulls Out”

Question 5(a) [3 marks]

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Explain branching instruction with example.

Answer:

Branching instructions alter program flow based on conditions or unconditionally.

Table: Branching Instructions

Example:

MOV A,#00H          ; Load zero
JZ ZERO_FOUND       ; Jump if A is zero
LJMP CONTINUE       ; Jump to continue
ZERO_FOUND:
    MOV R0,#01H     ; Set flag
CONTINUE:
    NOP             ; Continue execution

• Program Control: Changes execution sequence
• Conditional Jumps: Based on flag register status
• Address Range: Can jump to any program memory location

Mnemonic: “Jump Changes Control Flow”

Question 5(b) [4 marks]

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Interface 8 leds with microcontroller and write a program to turn on and off.

Answer:

Circuit Diagram:

Program:

ORG 0000H
MAIN:
    MOV P1,#0FFH        ; Turn ON all LEDs
    CALL DELAY          ; Wait
    MOV P1,#00H         ; Turn OFF all LEDs  
    CALL DELAY          ; Wait
    SJMP MAIN           ; Repeat

DELAY:
    MOV R0,#0FFH        ; Outer loop counter
LOOP1:
    MOV R1,#0FFH        ; Inner loop counter  
LOOP2:
    DJNZ R1,LOOP2       ; Inner delay loop
    DJNZ R0,LOOP1       ; Outer delay loop
    RET                 ; Return
END

Table: Components

• Current Limiting: Resistors protect LEDs from overcurrent
• Port Configuration: P1 used as output port for LED control
• Delay Routine: Creates visible ON/OFF timing

Mnemonic: “Port Controls LEDs with Resistance and Delay”

Question 5(c) [7 marks]

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

Answer:

Circuit Connections:

Program:

ORG 0000H
    CALL LCD_INIT       ; Initialize LCD
    CALL DISPLAY_MSG    ; Display message
    SJMP $              ; Stop here

LCD_INIT:
    MOV P2,#38H         ; Function set: 8-bit, 2-line
    CALL COMMAND
    MOV P2,#0EH         ; Display ON, Cursor ON
    CALL COMMAND  
    MOV P2,#01H         ; Clear display
    CALL COMMAND
    MOV P2,#06H         ; Entry mode set
    CALL COMMAND
    RET

DISPLAY_MSG:
    MOV DPTR,#MESSAGE   ; Point to message
NEXT_CHAR:
    CLR A
    MOVC A,@A+DPTR      ; Read character
    JZ DONE             ; If zero, end of string
    CALL SEND_CHAR      ; Send character to LCD
    INC DPTR            ; Next character
    SJMP NEXT_CHAR
DONE:
    RET

COMMAND:
    CLR P1.0            ; RS = 0 for command
    SETB P1.1           ; EN = 1
    CLR P1.1            ; EN = 0 (pulse)
    CALL DELAY
    RET

SEND_CHAR:
    MOV P2,A            ; Put character on data lines
    SETB P1.0           ; RS = 1 for data
    SETB P1.1           ; EN = 1
    CLR P1.1            ; EN = 0 (pulse)
    CALL DELAY
    RET

DELAY:
    MOV R0,#50          ; Delay routine
DELAY_LOOP:
    MOV R1,#255
DELAY_INNER:
    DJNZ R1,DELAY_INNER
    DJNZ R0,DELAY_LOOP
    RET

MESSAGE:
    DB "WELCOME",0       ; Message string with null terminator
END

Table: LCD Interface Pins

• 4-bit Mode: Uses only upper 4 data lines to save pins
• Control Signals: RS selects command/data, EN provides timing pulse
• Character Display: Each character sent as ASCII code
• Initialization: Required command sequence for proper operation

Mnemonic: “LCD Displays Characters with Commands and Data”

Question 5(a OR) [3 marks]

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Explain logical instruction with example.

Answer:

Logical instructions perform bitwise operations on data.

Table: Logical Instructions

Example:

MOV A,#55H          ; A = 01010101B
ANL A,#0FH          ; A = 00000101B (mask upper bits)
ORL A,#F0H          ; A = 11110101B (set upper bits)
XRL A,#FFH          ; A = 00001010B (complement all bits)

• Bit Manipulation: Used for setting, clearing, and testing bits
• Masking Operations: ANL clears unwanted bits
• Flag Effects: Updates parity flag based on result

Mnemonic: “AND Masks, OR Sets, XOR Toggles”

Question 5(b OR) [4 marks]

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Interface 7 segment with microcontroller.

Answer:

Circuit Diagram:

Program to Display 0-9:

ORG 0000H
    MOV DPTR,#DIGIT_TABLE   ; Point to lookup table
    MOV R0,#0               ; Start with digit 0

MAIN_LOOP:
    MOV A,R0                ; Get current digit
    MOVC A,@A+DPTR          ; Get 7-segment code
    MOV P1,A                ; Display on 7-segment
    CALL DELAY              ; Wait 1 second
    INC R0                  ; Next digit
    CJNE R0,#10,MAIN_LOOP   ; Check if reached 10
    MOV R0,#0               ; Reset to 0
    SJMP MAIN_LOOP          ; Repeat

DIGIT_TABLE:
    DB 3FH, 06H, 5BH, 4FH, 66H    ; 0,1,2,3,4
    DB 6DH, 7DH, 07H, 7FH, 6FH    ; 5,6,7,8,9
END

Table: 7-Segment Codes

• Common Cathode: Segments light when port pin is high
• Current Limiting: Resistors prevent segment damage
• Lookup Table: Efficient storage of segment patterns

Mnemonic: “Seven Segments Show Digits Clearly”

Question 5(c OR) [7 marks]

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Interface LM 35 with microcontroller and explain block diagram of temperature controller.

Answer:

Circuit Diagram:

Temperature Controller Block Diagram:

graph LR
    A[LM35 Sensor] --> B[ADC0804]
    B --> C[8051 Controller]
    C --> D[Display Unit]
    C --> E[Relay Driver]
    E --> F[Heater/Cooler]
    F --> G[Controlled Environment]
    G --> A

Control Program:

ORG 0000H
MAIN:
    CALL READ_TEMP      ; Read temperature from ADC
    CALL DISPLAY_TEMP   ; Show temperature on display
    CALL TEMP_CONTROL   ; Control heating/cooling
    CALL DELAY          ; Wait before next reading
    SJMP MAIN

READ_TEMP:
    CLR P2.0            ; Start ADC conversion
    SETB P2.0           ; Pulse to start
    JNB P2.1,$          ; Wait for conversion complete
    MOV A,P1            ; Read temperature data
    RET

TEMP_CONTROL:
    CJNE A,#30,CHECK_HIGH   ; Compare with setpoint (30°C)
CHECK_HIGH:
    JC TEMP_LOW             ; If A < 30, temperature is low
    SETB P3.0               ; Turn ON cooling (fan)
    CLR P3.1                ; Turn OFF heating
    RET
TEMP_LOW:
    CLR P3.0                ; Turn OFF cooling
    SETB P3.1               ; Turn ON heating
    RET
END

Table: System Components

• Temperature Sensing: LM35 provides 10mV per degree Celsius
• ADC Conversion: Converts analog voltage to digital value
• Control Logic: Compares with setpoint and controls relays
• Feedback System: Continuous monitoring and adjustment
• Safety Features: Over-temperature protection possible

Mnemonic: “Sense, Convert, Compare, Control Temperature Automatically”