Question 1(a) [3 marks]

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

Feature	Description
Single Chip	Complete CPU on one integrated circuit
Processing Unit	Executes instructions and performs calculations
Control Logic	Manages system operations and data flow

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]

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

Flag	Position	Purpose
S (Sign)	Bit 7	Indicates sign of result (1=negative, 0=positive)
Z (Zero)	Bit 6	Set when result is zero
AC (Auxiliary Carry)	Bit 4	Carry from bit 3 to bit 4
P (Parity)	Bit 2	Even parity flag
CY (Carry)	Bit 0	Carry from MSB

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]

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

Format	Size	Structure	Example
1-Byte	8 bits	Opcode only	MOV A,B
2-Byte	16 bits	Opcode + 8-bit data	MVI A,05H
3-Byte	24 bits	Opcode + 16-bit address	LDA 2000H

Diagram:

graph LR
    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]

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

Component	Primary Function	Key Operations
ALU	Arithmetic & Logic Operations	ADD, SUB, AND, OR, XOR
Control Unit	Instruction Control	Fetch, Decode, Execute
CPU	Overall Processing	Coordinate all operations

Diagram:

graph TD
    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]

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

Function	Description
Address Latching	Captures lower 8-bit address
Demultiplexing	Separates address from data
Timing Control	Provides timing reference

Diagram:

    +--------+     ALE     +--------+
    |  8085  |------------>| Latch  |
    |        |             | 74373  |
    | AD0-7  |<----------->|        |
    +--------+             +--------+
                               |
                           A0-A7 (Address)

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]

Compare microprocessor and microcontroller

Answer:

Table: Microprocessor vs Microcontroller Comparison

Parameter	Microprocessor	Microcontroller
Design	General purpose	Application specific
Memory	External RAM/ROM	Internal RAM/ROM
I/O Ports	External interface	Built-in I/O ports
Timers	External	Built-in timers
Cost	Higher system cost	Lower system cost
Power	Higher consumption	Lower consumption

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]

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

Block	Function
ALU	Arithmetic and logical operations
Register Array	Temporary data storage (B,C,D,E,H,L)
Control Unit	Instruction execution control
Address Buffer	Drive address bus lines

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]

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

Register	Formation	Purpose
PC	Single 16-bit	Program Counter - next instruction address
SP	Single 16-bit	Stack Pointer - top of stack address
HL	H + L registers	Memory pointer - data address

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]

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

Lines	T1 State	T2-T4 States
AD0-AD7	Lower Address A0-A7	Data D0-D7
ALE Signal	High	Low

Diagram:

           8085
    +----------------+
    |                | ALE
    |      AD0-AD7   |---->+
    |                |     |
    +----------------+     |
            |              |
            |         +----v----+
            +-------->| 74373   |
                      | Latch   |
                      +---------+
                           |
                       A0-A7

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]

Draw and explain pin diagram of 8085.

Answer:

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

Diagram:

        8085 Pin Diagram
    +-------------------+
X1  |1               40| Vcc
X2  |2               39| HOLD
RST |3               38| HLDA  
SOD |4               37| CLK
SID |5               36| RESET
TRAP|6               35| READY
RST7|7               34| IO/M*
RST6|8               33| S1
RST5|9               32| RD*
INTR|10              31| WR*
INTA|11              30| ALE
AD0 |12              29| S0
AD1 |13              28| A15
AD2 |14              27| A14
AD3 |15              26| A13
AD4 |16              25| A12
AD5 |17              24| A11
AD6 |18              23| A10
AD7 |19              22| A9
Vss |20              21| A8
    +-------------------+

Table: Pin Groups

Group	Pins	Function
Address/Data	AD0-AD7, A8-A15	Memory addressing and data transfer
Control	ALE, RD, WR, IO/M*	Bus control signals
Interrupts	INTR, RST7-RST5, TRAP	Interrupt handling
Power	Vcc, Vss	Power supply connections

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]

Draw clock and reset circuit of microcontroller

Answer:

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

Diagram:

Clock Circuit:
    +12MHz Crystal
    |
XTAL1 +---||---+ XTAL2
      |       |
     30pF    30pF
      |       |
     GND     GND

Reset Circuit:
     +5V
      |
     10K
      |
RST --+---||---GND
          10µF

Table: Circuit Components

Component	Value	Purpose
Crystal	11.0592 MHz	Clock generation
Capacitors	30pF each	Crystal stabilization
Reset Resistor	10KΩ	Pull-up for reset
Reset Capacitor	10µF	Power-on reset delay

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]

Explain internal RAM of 8051.

Answer:

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

Table: Internal RAM Organization

Address Range	Size	Purpose
00H-1FH	32 bytes	Register Banks (4 banks × 8 registers)
20H-2FH	16 bytes	Bit-addressable area
30H-7FH	80 bytes	General purpose RAM
80H-FFH	128 bytes	Special Function Registers (SFRs)

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]

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

Block	Function
CPU	Instruction execution and control
Memory	4KB ROM + 256B RAM
Timers	Two 16-bit timer/counters
I/O Ports	Four 8-bit bidirectional ports
Serial Port	Full-duplex UART
Interrupts	5-source interrupt system

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]

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

Register	Full Form	Function
DPTR	Data Pointer	Points to external data memory
PC	Program Counter	Points to next instruction address

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]

Explain different timer modes of microcontroller.

Answer:

The 8051 has two timers with four different operating modes.

Table: Timer Modes

Mode	Configuration	Purpose
Mode 0	13-bit timer	Compatible with 8048
Mode 1	16-bit timer	Maximum count capability
Mode 2	8-bit auto-reload	Constant time intervals
Mode 3	Two 8-bit timers	Timer 0 split operation

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]

Explain interrupts of microcontroller.

Answer:

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

Table: 8051 Interrupt Sources

Interrupt	Vector Address	Priority	Trigger
Reset	0000H	Highest	Power-on/External
External 0	0003H	High	INT0 pin
Timer 0	000BH	Medium	Timer 0 overflow
External 1	0013H	Medium	INT1 pin
Timer 1	001BH	Low	Timer 1 overflow
Serial	0023H	Lowest	Serial communication

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]

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

Instruction	Example	Function
MOV	MOV A,#55H	Move immediate data to accumulator
MOVX	MOVX A,@DPTR	Move external RAM to accumulator
MOVC	MOVC A,@A+PC	Move code memory to accumulator

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]

List and explain different addressing modes of microcontroller.

Answer:

The 8051 supports several addressing modes for flexible data access.

Table: 8051 Addressing Modes

Mode	Example	Description
Immediate	MOV A,#55H	Data specified in instruction
Register	MOV A,R0	Use register contents
Direct	MOV A,30H	Direct memory address
Indirect	MOV A,@R0	Address stored in register
Indexed	MOVC A,@A+DPTR	Base address plus offset

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]

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

Register	Purpose
R0	Source address pointer (100H)
R1	Destination address pointer (200H)
R2	Loop counter (8 bytes)
A	Temporary data storage

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]

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

Step	Instruction	Result
1	MOV A,#25H	A = 25H
2	ADD A,#35H	A = 5AH
3	MOV R0,A	R0 = 5AH

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]

Explain indexed addressing mode with example.

Answer:

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

Table: Indexed Addressing Details

Component	Description	Example
Base Address	DPTR or PC register	DPTR = 1000H
Index	Accumulator contents	A = 05H
Effective Address	Base + Index	1000H + 05H = 1005H

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]

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

Operation	Instruction	Function
PUSH	PUSH 30H	Store data on stack
POP	POP 30H	Retrieve data from stack
Stack Pointer	SP register	Points to top of stack

Diagram:

Stack Operation:
    
Before PUSH:     After PUSH 30H:    After POP 30H:
SP → 07H            SP → 08H           SP → 07H
     06H                 08H: 30H           06H
     05H                 07H: old           05H
     
Stack grows upward in memory

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]

Explain branching instruction with example.

Answer:

Branching instructions alter program flow based on conditions or unconditionally.

Table: Branching Instructions

Type	Instruction	Example
Unconditional	LJMP address	LJMP 2000H
Conditional	JZ address	JZ ZERO_LABEL
Call/Return	LCALL address	LCALL SUBROUTINE

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]

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

Answer:

Circuit Diagram:

8051        LEDs
P1.0 ----[330Ω]----LED1----+5V
P1.1 ----[330Ω]----LED2----+5V  
P1.2 ----[330Ω]----LED3----+5V
P1.3 ----[330Ω]----LED4----+5V
P1.4 ----[330Ω]----LED5----+5V
P1.5 ----[330Ω]----LED6----+5V
P1.6 ----[330Ω]----LED7----+5V
P1.7 ----[330Ω]----LED8----+5V

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

Component	Value	Purpose
Resistor	330Ω	Current limiting
LEDs	8 pieces	Visual indicators
Port	P1	8-bit output port

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]

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

Answer:

Circuit Connections:

8051        16x2 LCD
P2.0 --------> D4
P2.1 --------> D5  
P2.2 --------> D6
P2.3 --------> D7
P1.0 --------> RS (Register Select)
P1.1 --------> EN (Enable)
GND  --------> R/W (Write mode)

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

8051 Pin	LCD Pin	Function
P2.0-P2.3	D4-D7	4-bit data lines
P1.0	RS	Register select (0=command, 1=data)
P1.1	EN	Enable pulse
GND	R/W	Read/Write (tied to ground for write)

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]

Explain logical instruction with example.

Answer:

Logical instructions perform bitwise operations on data.

Table: Logical Instructions

Instruction	Example	Function
ANL	ANL A,#0FH	Bitwise AND operation
ORL	ORL A,#F0H	Bitwise OR operation
XRL	XRL A,#FFH	Bitwise XOR operation

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]

Interface 7 segment with microcontroller.

Answer:

Circuit Diagram:

8051          7-Segment Display
P1.0 ----[330Ω]----a
P1.1 ----[330Ω]----b  
P1.2 ----[330Ω]----c
P1.3 ----[330Ω]----d
P1.4 ----[330Ω]----e
P1.5 ----[330Ω]----f
P1.6 ----[330Ω]----g
P1.7 ----[330Ω]----dp (decimal point)

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

Digit	Hex Code	Binary	Segments Lit
0	3FH	00111111	a,b,c,d,e,f
1	06H	00000110	b,c
2	5BH	01011011	a,b,g,e,d

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]

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

Answer:

Circuit Diagram:

LM35 Temperature Sensor Interface:

+5V ----+---- LM35 ----+---- ADC0804 ----+---- 8051
        |     (Vout)   |     (Vin)       |     (P1)
       GND             |                 |
                      GND                |
                                         |
Relay Control:                           |
8051 P3.0 ----[ULN2003]---- Relay -------+
                                         |
                                    Load (Heater/Fan)

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

Component	Function
LM35	Temperature sensor (10mV/°C)
ADC0804	Analog to digital converter
8051	Main controller
Relay	Switch high power loads
Display	Show current temperature

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

Mnemonic: "Sense, Convert, Compare, Control Temperature Automatically"