Question 1(a) [3 marks]

List common features of 8051 microcontroller.

Answer:

Table: Common Features of 8051 Microcontroller

Feature	Description
On-chip Oscillator	Built-in clock generator circuit
Program Memory	4KB internal ROM for code storage
Data Memory	128 bytes internal RAM
I/O Ports	4 bidirectional 8-bit ports (P0-P3)
Timers/Counters	Two 16-bit Timer/Counter units
Serial Port	Full duplex UART communication
Interrupts	5 interrupt sources with priority
SFRs	Special Function Registers for control

Mnemonic: "On Program Data I/O Timers Serial Interrupts SFRs"

Question 1(b) [4 marks]

Define T-State, Machine Cycle, Instruction Cycle and Opcode.

Answer:

Table: Microprocessor Timing Definitions

Term	Definition	Duration
T-State	One clock period of system clock	Basic timing unit
Machine Cycle	Time to complete one memory operation	3-6 T-states
Instruction Cycle	Time to fetch, decode and execute instruction	1-4 Machine cycles
Opcode	Operation code specifying instruction type	1-3 bytes

T-State: Smallest unit of time in microprocessor operation
Machine Cycle: Contains multiple T-states for memory access
Instruction Cycle: Complete instruction execution time
Opcode: Binary code identifying specific instruction

Mnemonic: "Time Machine Instruction Operation"

Question 1(c) [7 marks]

Compare Von-Neumann and Harvard Architecture.

Answer:

Table: Von-Neumann vs Harvard Architecture Comparison

Parameter	Von-Neumann	Harvard
Memory Organization	Single memory for code and data	Separate memory for code and data
Bus Structure	Single bus system	Dual bus system
Speed	Slower due to bus sharing	Faster parallel access
Cost	Lower cost implementation	Higher cost due to dual memory
Flexibility	More flexible memory usage	Less flexible, fixed allocation
Examples	8085, x86 processors	8051, DSP processors

graph TB
    subgraph "Von-Neumann Architecture"
        CPU1[CPU] <--> MEM1[Single Memory<br/>Code + Data]
    end
    
    subgraph "Harvard Architecture"
        CPU2[CPU] <--> PMEM[Program Memory]
        CPU2 <--> DMEM[Data Memory]
    end

Key Differences:

Memory Access: Von-Neumann uses sequential access, Harvard allows simultaneous
Performance: Harvard is faster for embedded applications
Applications: Von-Neumann for general computing, Harvard for real-time systems

Mnemonic: "Von-Single Harvard-Dual"

Question 1(c) OR [7 marks]

Explain Microcomputer System with block diagram.

Answer:

Microcomputer System Components:

graph TB
    subgraph "Microcomputer System"
    direction TB
        CPU[Central Processing Unit<br/>- Control Unit<br/>- ALU<br/>- Registers] 
        MEM[Memory Unit<br/>- RAM<br/>- ROM]
        IO[Input/Output Unit<br/>- Keyboard<br/>- Display<br/>- Ports]
        SYS[System Bus<br/>- Address Bus<br/>- Data Bus<br/>- Control Bus]
        
        CPU <--> SYS
        MEM <--> SYS
        IO <--> SYS
    end

Table: Microcomputer System Components

Component	Function	Examples
CPU	Central processing and control	8085, 8086
Memory	Program and data storage	RAM, ROM, EPROM
I/O Unit	Interface with external world	Keyboard, Display
System Bus	Data transfer pathway	Address, Data, Control

CPU: Executes instructions and controls system operation
Memory: Stores programs and data for processing
I/O: Provides communication with external devices
Bus: Connects all components for data transfer

Mnemonic: "CPU Memory I/O Bus"

Question 2(a) [3 marks]

Draw Bus organization in 8085 Microprocessor.

Answer:

    +----------+
    |   8085   |
    |   CPU    |
    +----------+
         |
    +----+----+
    |         |
    v         v
+--------+ +--------+
|Address | |  Data  |
|  Bus   | |  Bus   |
|16-bit  | | 8-bit  |
+--------+ +--------+
    |         |
    v         v
+--------+ +--------+
|Memory  | |  I/O   |
|System  | |Devices |
+--------+ +--------+

Table: 8085 Bus Organization

Bus Type	Width	Function
Address Bus	16-bit	Memory addressing (64KB)
Data Bus	8-bit	Data transfer
Control Bus	Multiple	Control signals

Mnemonic: "Address Data Control"

Question 2(b) [4 marks]

List Flags used in 8085 and Explain working of each flags.

Answer:

Table: 8085 Flags Register

Flag	Name	Bit Position	Function
S	Sign	D7	Set if result is negative
Z	Zero	D6	Set if result is zero
AC	Auxiliary Carry	D4	Set if carry from bit 3 to 4
P	Parity	D2	Set if result has even parity
CY	Carry	D0	Set if carry/borrow occurs

D7  D6  D5  D4  D3  D2  D1  D0
+---+---+---+---+---+---+---+---+
| S | Z | - |AC | - | P | - |CY |
+---+---+---+---+---+---+---+---+

Sign Flag: Indicates negative result (MSB = 1)
Zero Flag: Set when arithmetic result is zero
Auxiliary Carry: Used for BCD arithmetic operations
Parity Flag: Checks even number of 1's in result
Carry Flag: Indicates overflow in arithmetic operations

Mnemonic: "Sign Zero Auxiliary Parity Carry"

Question 2(c) [7 marks]

Draw and Explain Block Diagram of 8085.

Answer:

graph TB
    subgraph "8085 Microprocessor"
        ALU[Arithmetic Logic Unit<br/>8-bit operations]
        ACC[Accumulator<br/>Register A]
        REG[General Purpose<br/>Registers B,C,D,E,H,L]
        PC[Program Counter<br/>16-bit]
        SP[Stack Pointer<br/>16-bit]
        FLAGS[Flag Register<br/>5 flags]
        CU[Control Unit<br/>Instruction decode]
        
        ALU <--> ACC
        REG <--> ALU
        PC --> CU
        CU --> ALU
        FLAGS <--> ALU
    end
    
    EXT["External Buses<br/>Address(16), Data(8), Control"]
    8085 <--> EXT

Table: 8085 Block Components

Block	Function	Size
ALU	Arithmetic and logical operations	8-bit
Accumulator	Primary register for operations	8-bit
Registers	Data storage (B,C,D,E,H,L)	8-bit each
Program Counter	Points to next instruction	16-bit
Stack Pointer	Points to stack top	16-bit
Control Unit	Instruction decode and control	-

Data Flow: Instructions fetched via PC, decoded by CU, executed in ALU
Register Operations: Accumulator works with ALU, other registers store data
Address Generation: PC and SP provide 16-bit addresses
Control Signals: CU generates timing and control signals

Mnemonic: "ALU Accumulator Registers Program Stack Control"

Question 2(a) OR [3 marks]

Explain Instruction Fetching, Decoding and Execution Operation in microprocessor.

Answer:

sequenceDiagram
    participant PC as Program Counter
    participant MEM as Memory
    participant IR as Instruction Register
    participant CU as Control Unit
    participant ALU as ALU
    
    PC->>MEM: Send address
    MEM->>IR: Fetch instruction
    IR->>CU: Send opcode
    CU->>ALU: Generate control signals
    ALU->>ALU: Execute operation

Table: Instruction Cycle Phases

Phase	Operation	Duration
Fetch	Get instruction from memory	1 machine cycle
Decode	Interpret instruction opcode	Part of execute
Execute	Perform required operation	1-3 machine cycles

Mnemonic: "Fetch Decode Execute"

Question 2(b) OR [4 marks]

What is Demultiplexing of Lower order Address and Data lines in 8085? Explain using neat sketch.

Answer:

     8085
+-------------+
|  AD7-AD0    |  Multiplexed
+------+------+  Address/Data
       |
       v
+------+------+
|    ALE      |  Address Latch Enable
+------+------+
       |
       v
+------+------+
|   74LS373   |  Latch IC
+------+------+
   |       |
   v       v
A7-A0    D7-D0
Address   Data

Demultiplexing Process:

ALE Signal: Controls separation of address and data
Latch IC: 74LS373 stores address when ALE is high
Timing: Address appears first, then data on same lines

Table: Demultiplexing Components

Component	Function	Timing
ALE	Address Latch Enable signal	High during T1
74LS373	Octal latch IC	Stores A7-A0
AD7-AD0	Multiplexed lines	Address then Data

Mnemonic: "Address Latch Enable Demultiplexes Lines"

Question 2(c) OR [7 marks]

Draw and Explain Pin Diagram of 8085.

Answer:

        8085 PIN DIAGRAM
    +-------------------+
X1  |1               40| VCC
X2  |2               39| HOLD
RST |3               38| HLDA
SOD |4               37| CLK
SID |5               36| RESET IN
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: 8085 Pin Functions

Pin Group	Function	Count
Address Bus	A8-A15 (Higher order)	8 pins
Address/Data	AD0-AD7 (Multiplexed)	8 pins
Control Signals	ALE, RD, WR, IO/M	4 pins
Interrupts	TRAP, RST7.5, RST6.5, RST5.5, INTR	5 pins
Power	VCC, VSS	2 pins
Clock	X1, X2, CLK	3 pins

Address Lines: 16-bit addressing capability (64KB)
Data Lines: 8-bit data transfer
Control Lines: Memory and I/O operation control
Interrupt Lines: Hardware interrupt handling

Mnemonic: "Address Data Control Interrupt Power Clock"

Question 3(a) [3 marks]

Draw IP SFR of 8051 and Explain function of each bit.

Answer:

IP Register (Interrupt Priority) - Address B8H
MSB                           LSB
+---+---+---+----+---+----+----+----+
| - | - | - | PT2| PS| PT1| PX1| PX0|
+---+---+---+----+---+----+----+----+
 D7  D6  D5  D4  D3  D2  D1  D0

Table: IP Register Bit Functions

Bit	Name	Function
D4	PT2	Timer 2 interrupt priority
D3	PS	Serial port interrupt priority
D2	PT1	Timer 1 interrupt priority
D1	PX1	External interrupt 1 priority
D0	PX0	External interrupt 0 priority

Priority Levels: 1 = High priority, 0 = Low priority
Default: All interrupts have low priority (00H)
Usage: Set bit to 1 for high priority interrupt

Mnemonic: "Timer2 Serial Timer1 External1 External0"

Question 3(b) [4 marks]

Draw and explain Timer/Counter Logic diagram for 8051.

Answer:

graph TB
    subgraph "Timer/Counter Logic"
        OSC[System Clock<br/>÷12] --> MUX1{Timer/Counter<br/>Mode Select}
        T0PIN[T0 Pin<br/>External] --> MUX1
        MUX1 --> TH0[TH0 Register<br/>High Byte]
        TH0 --> TL0[TL0 Register<br/>Low Byte]
        TL0 --> OVF[Overflow<br/>Flag TF0]
        
        TMOD[TMOD Register<br/>Mode Control] --> MUX1
        TCON[TCON Register<br/>Control Bits] --> MUX1
    end

Table: Timer Components

Component	Function	Size
TH0/TL0	Timer 0 high/low byte registers	8-bit each
TMOD	Timer mode register	8-bit
TCON	Timer control register	8-bit
TF0	Timer 0 overflow flag	1-bit

Clock Source: Internal (system clock/12) or External (T0 pin)
Operation: Counts up from loaded value to FFH
Overflow: Sets TF0 flag and generates interrupt
Modes: 4 different timer modes available

Mnemonic: "Timer High-Low Mode Control Flag"

Question 3(c) [7 marks]

Draw and Explain Block Diagram of 8051.

Answer:

graph TB
    subgraph "8051 Microcontroller"
    direction TB
        CPU[8-bit CPU<br/>ALU + Control]
        ROM[Program Memory<br/>4KB ROM]
        RAM[Data Memory<br/>128B RAM]
        
        T0[Timer 0<br/>16-bit]
        T1[Timer 1<br/>16-bit]
        UART[Serial Port<br/>Full Duplex]
        
        P0[Port 0<br/>8-bit I/O]
        P1[Port 1<br/>8-bit I/O]
        P2[Port 2<br/>8-bit I/O]
        P3[Port 3<br/>8-bit I/O]
        
        INT[Interrupt<br/>System]
        OSC[Oscillator<br/>Clock Gen]
        
        CPU <--> ROM
        CPU <--> RAM
        CPU <--> T0
        CPU <--> T1
        CPU <--> UART
        CPU <--> P0
        CPU <--> P1
        CPU <--> P2
        CPU <--> P3
        CPU <--> INT
        OSC --> CPU
    end

Table: 8051 Block Components

Block	Function	Specification
CPU	Central processing unit	8-bit processor
Program Memory	Code storage	4KB internal ROM
Data Memory	Variable storage	128 bytes RAM
I/O Ports	External interface	4 ports (32 I/O lines)
Timers	Timing operations	2 × 16-bit timers
Serial Port	Communication	Full duplex UART
Interrupts	Event handling	5 interrupt sources

Architecture: Harvard architecture with separate program/data memory
I/O Capability: 32 bidirectional I/O lines
On-chip Features: Timers, serial port, interrupt system
Memory: Von-Neumann for data, Harvard for program

Mnemonic: "CPU Program Data I/O Timer Serial Interrupt"

Question 3(a) OR [3 marks]

Draw PCON SFR of 8051 and Explain function of each bit.

Answer:

PCON Register (Power Control) - Address 87H
MSB                           LSB
+---+---+---+---+---+---+---+---+
|SMOD| - | - | - |GF1|GF0| PD|IDL|
+---+---+---+---+---+---+---+---+
 D7  D6  D5  D4  D3  D2  D1  D0

Table: PCON Register Bit Functions

Bit	Name	Function
D7	SMOD	Serial port mode modifier
D3	GF1	General purpose flag bit 1
D2	GF0	General purpose flag bit 0
D1	PD	Power down mode control
D0	IDL	Idle mode control

SMOD: Doubles serial port baud rate when set
GF1, GF0: User-defined flag bits
PD: Activates power-down mode
IDL: Activates idle mode

Mnemonic: "Serial General Power Idle"

Question 3(b) OR [4 marks]

In 8051 Serial communication Mode 1, For XTAL=11.0592 MHz, find TH1 value needed to have for 9600 and 4800 baud rate.

Answer:

Formula for Mode 1 Baud Rate:

Baud Rate = (2^SMOD/32) × (Timer1 Overflow Rate)
Timer1 Overflow Rate = XTAL/(12 × (256 - TH1))

For 9600 Baud Rate:

9600 = (1/32) × (11059200/(12 × (256 - TH1)))
9600 = 28800/(256 - TH1)
256 - TH1 = 3
TH1 = 253 = FDH

For 4800 Baud Rate:

4800 = (1/32) × (11059200/(12 × (256 - TH1)))
4800 = 28800/(256 - TH1)
256 - TH1 = 6
TH1 = 250 = FAH

Table: TH1 Values for Baud Rates

Baud Rate	TH1 Value (Hex)	TH1 Value (Decimal)
9600	FDH	253
4800	FAH	250

Mnemonic: "Higher Baud Higher TH1"

Question 4(a) [3 marks]

What are the differences in LCALL and LJMP instructions in 8051?

Answer:

Table: LCALL vs LJMP Comparison

Parameter	LCALL	LJMP
Function	Long subroutine call	Long jump
Stack Usage	Pushes return address	No stack operation
Return	RET instruction needed	Direct jump only
Bytes	3 bytes	3 bytes
Address Range	16-bit (64KB)	16-bit (64KB)
PC Action	Saved then loaded	Directly loaded

LCALL: Calls subroutine, saves return address on stack
LJMP: Unconditional jump to specified address
Stack Impact: LCALL uses 2 stack bytes, LJMP uses none
Usage: LCALL for functions, LJMP for program flow control

Mnemonic: "Call Saves Jump Goes"

Question 4(b) [4 marks]

Write 8051 Assembly Language Program to generate square wave on port 1.0 using Timer0.

Answer:

ORG 0000H           ; Start address
LJMP MAIN           ; Jump to main program

ORG 0030H           ; Main program start
MAIN:
    MOV TMOD, #01H  ; Timer0 mode1 (16-bit)
    MOV TH0, #HIGH(-50000)  ; Load high byte
    MOV TL0, #LOW(-50000)   ; Load low byte
    SETB TR0        ; Start Timer0

LOOP:
    JNB TF0, LOOP   ; Wait for overflow
    CLR TF0         ; Clear overflow flag
    CPL P1.0        ; Toggle P1.0
    MOV TH0, #HIGH(-50000)  ; Reload timer
    MOV TL0, #LOW(-50000)   ; Reload timer
    SJMP LOOP       ; Repeat

END

Program Explanation:

Timer Setup: Mode 1 (16-bit timer)
Count Value: -50000 for specific delay
Square Wave: Toggle P1.0 on each overflow
Continuous: Loop maintains square wave

Mnemonic: "Mode Load Start Wait Toggle Reload"

Question 4(c) [7 marks]

Explain any three Logical and any four Data Transfer Instruction of 8051 with example.

Answer:

Table: Logical Instructions

Instruction	Function	Example	Result
ANL	Logical AND	ANL A, #0FH	A = A AND 0FH
ORL	Logical OR	ORL A, #F0H	A = A OR F0H
XRL	Logical XOR	XRL A, #FFH	A = A XOR FFH

Table: Data Transfer Instructions

Instruction	Function	Example	Operation
MOV	Move data	MOV A, #50H	Load 50H into A
MOVX	Move external	MOVX A, @DPTR	Load from external memory
PUSH	Push to stack	PUSH ACC	Push accumulator to stack
POP	Pop from stack	POP ACC	Pop from stack to accumulator

Detailed Examples:

; Logical Instructions
ANL A, #0FH     ; Mask upper nibble
ORL P1, #80H    ; Set bit 7 of Port1
XRL A, #FFH     ; Complement accumulator

; Data Transfer Instructions
MOV R0, #30H    ; Load immediate data
MOVX @DPTR, A   ; Store to external memory
PUSH B          ; Save B register
POP PSW         ; Restore status word

Mnemonic: "AND OR XOR Move External Push Pop"

Question 4(a) OR [3 marks]

Explain Instructions: (i) RRC A (ii) POP (iii) CLR PSW.7

Answer:

Table: Instruction Explanations

Instruction	Function	Operation	Example
RRC A	Rotate right through carry	A→C, C→A(MSB)	A=85H,C=0 → A=42H,C=1
POP	Pop from stack	SP→Register, SP-1	POP ACC
CLR PSW.7	Clear bit 7 of PSW	PSW.7 = 0	Clear CY flag

RRC A Operation:
Before: A = [D7 D6 D5 D4 D3 D2 D1 D0] C = [C]
After:  A = [C  D7 D6 D5 D4 D3 D2 D1] C = [D0]

RRC A: Rotates accumulator right through carry flag
POP: Removes top stack element into specified register
CLR PSW.7: Clears carry flag (bit 7 of Program Status Word)

Mnemonic: "Rotate Pop Clear"

Question 4(b) OR [4 marks]

Write 8051 Assembly Language Program to Divide data stored in location 30H by data stored in location 31H and store remainder in 40h and quotient in 41h memory location.

Answer:

ORG 0000H           ; Program start
LJMP MAIN

ORG 0030H
MAIN:
    MOV A, 30H      ; Load dividend
    MOV B, 31H      ; Load divisor
    DIV AB          ; Divide A by B
    MOV 41H, A      ; Store quotient
    MOV 40H, B      ; Store remainder
    SJMP $          ; Stop here

END

Program Steps:

Load Data: Move dividend and divisor to A and B
Division: Use DIV AB instruction
Store Results: Quotient in A, remainder in B
Save: Store results in specified memory locations

Table: DIV AB Instruction

Before	After
A = Dividend	A = Quotient
B = Divisor	B = Remainder

Mnemonic: "Load Divide Store"

Question 4(c) OR [7 marks]

List Addressing Modes of 8051 Microcontroller and Explain each with Example.

Answer:

Table: 8051 Addressing Modes

Mode	Description	Example	Explanation
Immediate	Data in instruction	MOV A, #50H	Load 50H into A
Register	Use register	MOV A, R0	Move R0 content to A
Direct	Memory address specified	MOV A, 30H	Load from address 30H
Indirect	Address in register	MOV A, @R0	Load from address in R0
Indexed	Base + offset	MOVC A, @A+DPTR	A = content of (A+DPTR)
Relative	PC + offset	SJMP HERE	Jump relative to PC
Bit	Bit address	SETB P1.0	Set bit 0 of Port 1

Detailed Examples:

; Immediate Addressing
MOV A, #25H         ; Load immediate value 25H

; Register Addressing  
MOV A, R7           ; Move register R7 to A

; Direct Addressing
MOV A, 40H          ; Load from memory location 40H

; Indirect Addressing
MOV R0, #50H        ; R0 points to address 50H
MOV A, @R0          ; Load from address pointed by R0

; Indexed Addressing
MOV DPTR, #TABLE    ; Point to lookup table
MOVC A, @A+DPTR     ; Load from table[A]

; Relative Addressing
SJMP NEXT           ; Jump to label NEXT

; Bit Addressing
SETB P2.5           ; Set bit 5 of Port 2

Mnemonic: "Immediate Register Direct Indirect Indexed Relative Bit"

Question 5(a) [3 marks]

Draw Interfacing of Relay with 8051 microcontroller.

Answer:

    8051 Port Pin
         |
         |
    +----+----+
    |  2.2K   |  Resistor
    +----+----+
         |
    +----+----+  Base
    |  NPN    |  Transistor
    |   BC547 |  
    +----+----+
         |     Collector
         |
    +----+----+
    |  Relay  |  12V Relay
    |  Coil   |
    +----+----+
         |
       +VCC (12V)
    
    Relay Contacts
    +----+----+
    | NO | NC |  Load Connection
    +----+----+

Table: Interface Components

Component	Function	Value
Transistor	Current amplifier	BC547 NPN
Resistor	Base current limiter	2.2KΩ
Relay	Electromagnetic switch	12V DC
Diode	Back EMF protection	1N4007

Operation: Port pin HIGH → Transistor ON → Relay energized
Protection: Diode prevents back EMF damage
Isolation: Relay provides electrical isolation

Mnemonic: "Transistor Resistor Relay Diode"

Question 5(b) [4 marks]

Interface 7 Segment display with 8051 microcontroller and write a program to print "1" on it.

Answer:

    8051 Port 1
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

    7-Segment Display
     aaaa
    f    b
    f    b
     gggg
    e    c
    e    c
     dddd  dp

Program to Display "1":

ORG 0000H
LJMP MAIN

ORG 0030H
MAIN:
    MOV P1, #06H    ; Display "1" (segments b,c ON)
    SJMP $          ; Stop here

; Pattern for "1": 00000110 = 06H
; Only segments b and c are ON

END

Table: 7-Segment Display Components

Component	Function	Value
Current Limiting Resistor	Protect LED segments	330Ω
Port Connection	Digital output control	Port 1
Display Pattern	Segment control	Binary pattern

Mnemonic: "Current Limit Segment Pattern"

Question 5(c) [7 marks]

Interface DAC 0808 with 8051 microcontroller and write a program to generate Square wave.

Answer:

    8051           DAC0808
Port 2     +---------------+
P2.0 ----->| D0        IOUT|---[10KΩ]---+---> Vout
P2.1 ----->| D1            |            |
P2.2 ----->| D2        IREF|---[10KΩ]---+
P2.3 ----->| D3            |            |
P2.4 ----->| D4         VCC|---(+5V)    |
P2.5 ----->| D5         VEE|---(-5V)    |
P2.6 ----->| D6            |            |
P2.7 ----->| D7         GND|---(GND)    |
           +---------------+            |
                                        |
                                 Op-Amp Buffer
                                      |
                                   Output

Program to Generate Square Wave:

ORG 0000H
LJMP MAIN

ORG 0030H
MAIN:
    MOV A, #00H     ; Minimum value (0V)
    MOV P2, A       ; Output to DAC
    CALL DELAY      ; Wait period
    
    MOV A, #0FFH    ; Maximum value (approx 5V)
    MOV P2, A       ; Output to DAC
    CALL DELAY      ; Wait period
    
    SJMP MAIN       ; Repeat for square wave

DELAY:
    MOV R0, #200    ; Delay counter
LOOP1:
    MOV R1, #250    ; Inner loop counter
LOOP2:
    DJNZ R1, LOOP2  ; Inner delay loop
    DJNZ R0, LOOP1  ; Outer delay loop
    RET

END

Table: DAC Interface Specifications

Parameter	Value	Function
Resolution	8-bit	256 output levels
Reference Voltage	5V	Full scale output
Output Range	0-5V	Analog voltage range
Interface Type	Parallel	8-bit data bus

Square Wave Generation:

Low Level: 00H produces approximately 0V output
High Level: FFH produces approximately 5V output
Frequency: Determined by delay routine duration
Output: Clean analog square wave at DAC output

Mnemonic: "Digital Analog Convert Square"

Question 5(a) OR [3 marks]

Interface of Push button Switch with 8051 microcontroller.

Answer:

         +5V
          |
          |
     +----+----+
     |   10K   |  Pull-up Resistor
     +----+----+
          |
          +-------- P1.0 (8051 Port Pin)
          |
     +----+----+
     |   SW    |  Push Button Switch
     +----+----+
          |
         GND

Table: Push Button Interface Components

Component	Value	Function
Pull-up Resistor	10KΩ	Ensures logic HIGH when switch open
Push Button	SPST Momentary	User input device
Logic Levels	HIGH=1, LOW=0	Switch open=1, pressed=0

Sample Program:

CHECK_SWITCH:
    JB P1.0, SW_RELEASED    ; Jump if switch not pressed
    ; Switch pressed code here
    CALL SWITCH_PRESSED
    SJMP CHECK_SWITCH

SW_RELEASED:
    ; Switch not pressed code here
    SJMP CHECK_SWITCH

SWITCH_PRESSED:
    ; Action when switch is pressed
    RET

Operation:

Switch Open: Pull-up resistor makes pin HIGH (logic 1)
Switch Pressed: Pin connected to GND, becomes LOW (logic 0)
Debouncing: May require software debouncing for reliable operation

Mnemonic: "Pull-up Switch Ground"

Question 5(b) OR [4 marks]

Interface DC Motor with 8051 microcontroller.

Answer:

    8051 Port Pin (P1.0)
         |
    +----+----+
    |   1K    |  Base Resistor
    +----+----+
         |
    +----+----+  Base
    |  NPN    |  Power Transistor
    | TIP122  |  (Darlington)
    +----+----+
         |     Collector
         |
    +----+----+
    |   DC    |  12V DC Motor
    |  Motor  |
    +----+----+
         |
       +VCC (12V)
    
    +----+----+  Freewheeling Diode
    | 1N4007  |  (Across Motor)
    +----+----+

Motor Control Program:

MOTOR_ON:
    SETB P1.0       ; Turn motor ON
    RET

MOTOR_OFF:
    CLR P1.0        ; Turn motor OFF
    RET

MOTOR_SPEED_CONTROL:
    ; PWM for speed control
    SETB P1.0       ; Motor ON
    CALL DELAY_ON   ; ON time duration
    CLR P1.0        ; Motor OFF  
    CALL DELAY_OFF  ; OFF time duration
    RET

DELAY_ON:
    MOV R0, #100    ; ON time delay
    DJNZ R0, $
    RET

DELAY_OFF:
    MOV R0, #50     ; OFF time delay  
    DJNZ R0, $
    RET

Table: DC Motor Interface Components

Component	Function	Specification
Power Transistor	Current amplification	TIP122 (Darlington pair)
Base Resistor	Current limiting	1KΩ
Freewheeling Diode	Back EMF protection	1N4007
DC Motor	Load device	12V DC Motor

Operation Principle:

Motor ON: Port pin HIGH → Transistor saturated → Motor runs
Motor OFF: Port pin LOW → Transistor cut-off → Motor stops
Speed Control: PWM technique varies average power to motor
Protection: Diode protects transistor from back EMF

Mnemonic: "Transistor Resistor Diode Motor"

Question 5(c) OR [7 marks]

Interface LCD with 8051 microcontroller and write a program to display "Hello".

Answer:

    8051        16x2 LCD Display
Port 2      +-------------------+
P2.0 ------>| D4          VCC   |---(+5V)
P2.1 ------>| D5          VDD   |---(GND)
P2.2 ------>| D6          V0    |---(Contrast Adj)
P2.3 ------>| D7          RS    |<--- P3.0
            |             EN    |<--- P3.1
            |             R/W   |---(GND)
            +-------------------+

Complete LCD Interface Program:

ORG 0000H
LJMP MAIN

ORG 0030H
MAIN:
    CALL LCD_INIT       ; Initialize LCD
    MOV DPTR, #MESSAGE  ; Point to message string
    CALL DISPLAY_STRING ; Display the message
    SJMP $              ; Stop execution

LCD_INIT:
    CALL DELAY_15MS     ; Wait 15ms after power on
    MOV A, #38H         ; Function set: 8-bit mode, 2 lines, 5x7 matrix
    CALL COMMAND_WRITE
    MOV A, #0EH         ; Display on, cursor on, blink off
    CALL COMMAND_WRITE
    MOV A, #01H         ; Clear display
    CALL COMMAND_WRITE
    MOV A, #06H         ; Entry mode: increment cursor, no shift
    CALL COMMAND_WRITE
    RET

COMMAND_WRITE:
    MOV P2, A           ; Send command to data lines (D4-D7)
    CLR P3.0            ; RS = 0 for command
    SETB P3.1           ; Enable pulse high
    CALL DELAY_1MS
    CLR P3.1            ; Enable pulse low
    CALL DELAY_1MS
    RET

DATA_WRITE:
    MOV P2, A           ; Send data to data lines (D4-D7)
    SETB P3.0           ; RS = 1 for data
    SETB P3.1           ; Enable pulse high
    CALL DELAY_1MS
    CLR P3.1            ; Enable pulse low
    CALL DELAY_1MS
    RET

DISPLAY_STRING:
    CLR A
    MOVC A, @A+DPTR     ; Get character from string
    JZ STRING_END       ; If zero, end of string
    CALL DATA_WRITE     ; Display character
    INC DPTR            ; Point to next character
    SJMP DISPLAY_STRING ; Continue until end

STRING_END:
    RET

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

DELAY_1MS:
    MOV R0, #4          ; Outer loop counter
DEL1: 
    MOV R1, #250        ; Inner loop counter  
DEL2: 
    DJNZ R1, DEL2       ; Inner delay loop
    DJNZ R0, DEL1       ; Outer delay loop
    RET

DELAY_15MS:
    MOV R2, #15         ; 15ms delay counter
DEL15: 
    CALL DELAY_1MS      ; Call 1ms delay
    DJNZ R2, DEL15      ; Repeat 15 times
    RET

END

Table: LCD Control Signals

Signal	Pin	Function
RS	P3.0	Register Select (0=Command, 1=Data)
EN	P3.1	Enable pulse for data latch
R/W	GND	Read/Write (tied to GND for write only)
D4-D7	P2.0-P2.3	4-bit data bus (upper nibble)

Table: Important LCD Commands

Command	Hex Code	Function
Function Set	38H	8-bit mode, 2 lines, 5x7 matrix
Display Control	0EH	Display ON, cursor ON, blink OFF
Clear Display	01H	Clear entire display
Entry Mode	06H	Increment cursor, no display shift

LCD Display Process:

Initialization: Configure LCD parameters and clear display
Command Mode: Send commands with RS=0
Data Mode: Send characters with RS=1
Enable Pulse: Latch data/command with EN signal
String Display: Loop through message characters until null terminator

Character Display Steps:

Set RS=1 for data mode
Put character code on data bus
Generate enable pulse (HIGH to LOW)
Wait for LCD to process (1ms delay)
Repeat for next character

Mnemonic: "Initialize Command Data Enable Display"