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WINTER – 19EXAMINATION
Subject Name: Microcontroller & Application Model Answer Subject Code: 22426
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in themodel answer
scheme.
2) The model answer and the answer written by candidate may vary but the examiner may tryto assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given moreImportance (Not applicable
for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in thefigure. The figures
drawn by candidate and model answer may vary. The examiner may give credit for anyequivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constantvalues may vary
and there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q. Sub Answer Marking

No. Q. N. Scheme

Q.1 Attempt any FIVE of the following: 10M

a) Compare address bus and data bus used in 8051. 2M

Ans: Sr. No. Address Bus Data Bus 1M each
A bus that is used to (Any 2
A bus that is used to transmit data points)
1 specify a physical address
among components
in memory
2 Unidirectional Bidirectional
Helps to transfer memory
3 Helps to send and receive data
address of data and I/O
4 16 bit address bus in 8051 8 bit data bus in 8051
b) Calculate the number of address lines required to access 16 kB ROM. 2M
Ans: 14 address lines required to access 16 KB of ROM as 2M
2 14 = 16KB
c) State features of ADC 0808. 2M

Ans: 1.Easy to interface with all Microprocessors or works Stand alone. 1M each
2.Eight channel 8-bit ADC module. ( Any 2
3.Can measure up to 8 Analog values. points)
4.On chip Clock not available, external Oscillator is needed (Clock).
5.Digital output various from 0 to 255, operating power is 15mW, conversion time
100us.
d) List specifications of 8051 microcontroller. 2M

Ans: 1) 8- bit data bus and 8- bit ALU. 1M each

2) 16- bit address bus – can access maximum 64KB of RAM and ROM. ( Any 2
3) On- chip RAM -128 bytes (Data Memory) points)
4) On- chip ROM – 4 KB (Program Memory)

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5) Four 8-bit bi- directional input/output ports Four 8-bit bi- directional input/ output
ports.
6) Programmable serial ports i.e. One UART (serial port)
7) Two 16- bit timers- Timer 0& Timer 1
8) Works on crystal frequency of 11.0592 MHz
9) Has power down and idle mode in microcontroller when no operation is performed.
10) Six interrupts are available.
e) List any two instructions which makes accumulator zero individually. 2M

Ans: MOV A,#00H 1M each

CLR A
f) Compare data memory and program memory. 2M

Ans: Sr.No. Program Memory Data Memory 1M each

It is used for storing

It is used for storing temporary
the hexadecimal codes of the
1 variable data and intermediate
program to be executed i.e.
results.
instructions.
Program Memory of 8051 is Data Memory of 8051 is 128
2
4kB bytes
g) List SFR in 8051. (any four) 2M
Ans: • ACC and B registers – 8 bit each ½M
• DPTR : [DPH:DPL] – 16 bit combined each
• PC : Program Counter – 16 bits
• Stack pointer SP – 8 bit
• PSW : Program Status Word
• Port Latches
• Serial data buffer, serial control
• Timer Registers (TCON,TMOD,TL0/1,TH0/1)
• Power control
• Interrupt Enable, Interrupt Priority

12-
Q.2 Attempt any THREE of the following: Total
Marks
Compare any three derivatives of 8051 microcontroller on the basis of RAM,ROM,Timer
a) 4M
and Interrupts.

Ans: 1M
89v51 RD2
Features 8051 8052 89c52 8031 8751 each
(Any
RAM 128 256 256 128 128 1k 4
4K 8K 8K 4K (UV- 664KB Points
ROM 0
(mask (EPROM) (Flash) EPROM) (FLASH)(Fla )
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ROM) sh)

TIMER 2 3 3 2 2 3

INTERRUPTS 6 8 8 6 6 8
b) Draw and explain the interfacing of DAC to 8051. 4M
Ans: Diagram: 2M

 Microcontroller generates output which is in digital form but many controlling system 2M
requires analog signal as they don't accept digital data thus making it necessary to use Expla-
DAC which converts digital data into equivalent analog voltage. nation
 In the figure shown, we use 8-bit DAC 0808. This IC converts 8 bit digital data into
equivalent analog current. Hence we require an I to V converter to convert this current
into equivalent voltage.
c) Describe 8051 microcontroller as boolean processor. 4M
Ans:  8051 processor is a CPU that can perform some operation on a data and gives the output. 4M
 The 8051 processor contains a complete Boolean processor for single-bit operations.
 The internal RAM contains 128 addressable bits, and the SFR space supports up to 128
other addressable bits.
 All port lines are bit-addressable, and each can be treated as a separate single-bit port.
 The instructions that access these bits are not only conditional branches but also a
complete set of move, set, clear, complement, OR, and AND instructions.
 The 8051 instruction set is optimized for the one bit operations. The Boolean processor
provides direct support for bit manipulation and testing of individual bit allows the use of
single bit variable to perform logical operations therefore 8051 can be used to solve
Boolean expression. Bits may be set or cleared in a single instruction.
 Eg: CLR C means clear the carry bit
SETB 20h means set the memory bit with bit address 20h.
d) Explain function of following pins of 8051
(i) Pin 31
4M
(ii) Pin 29
(iii) Pin 21-28
Ans: i) Pin 31-EA : It is and active low I/P to 8051 microcontroller. When (EA)= 0, then 1M- EA
8051 microcontroller access from external program memory (ROM) only. When (EA) = 1,
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then it access internal and external program memories (ROMS). 1M-
ii) Pin 29- PSEN : This is an output pin. PSEN stands for “program store enable.” It is active PSEN
low O/P signal. It is used to enable external program memory (ROM). When [PSEN(bar)]=
2M-Pin
0, then external program memory becomes enabled and micro controller read content of
21-28
external memory location. Therefore it is connected to (OE) of external ROM.
iii) Pin 21-28: A8 – A15 : These pins are known as Port 2. It serves as I/O port. Each pin is 1M
bidirectional Input /Output with internal pull – up resistors. Besides the Input /Output, when Port 2
external memory is interfaced, PORT 2 pins act as the higher-order address bus. (A8-A15) & 1M
A8 -
A15

12-
Q.3 Attempt any THREE of the following: Total
Marks

Develop Assembly Language program (ALP) to find the largest number in a block of 10
a) 4M
numbers stored at location 40H onwards in internal RAM.
Ans: (NOTE: Marks to be given for any other correct logic used by students.) 4M
ORG 0000H for
MOV R1, #0AH ; Initialize Byte Counter correc
MOV R0, #40H ; Initialize source pointer R0 to 40H
t
DEC R1 ; decrement counter by one
MOV 60H, @R0 ;Read First Byte progr
UP: INC R0 ; Increment the contents of R0 am
MOV A, @R0 ; Read second number
CJNE A, 60H, DN ;compare the first two numbers, if not equal go to DN
AJMP LARGE ;else go to LARGE
DN: JC LARGE ;check carry
MOV 60H, A ;Store largest number to 60H
LARGE: DJNZ R1, UP ;decrement the counter by one, if count ≠ 0, then go to UP
END
Largest No. is saved in memory 60H. Assume any location to store the result.
OR
MOV R1, #0AH ; initialize the counter
MOV R0, #40H ; initialize the memory pointer
DEC R1 ; decrement counter by one
MOV A,@R0 ; load number in accumulator
MOV B, A ; move that number to register B
UP: INC R0 ; increment the memory pointer
MOV A,@R0 ; read the next number in A
CJNE A, B, DOWN ; compare the first two numbers, if not equal go to DOWN
AJMP NEXT ; else go to NEXT
DOWN: JC NEXT ; if number in A is greater then go to NEXT
MOV B, A ; else move the number in register B
NEXT: DJNZ R1, UP ; decrement the counter by one, if count ≠ 0, then go to UP
INC R0 ; increment the memory pointer
MOV A,B
MOV 50H, A ; store result at memory location 50H(Assume any location)
HERE: SJMP HERE
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b) Sketch the internal memory organization in 8051. 4M

Ans: Daigram: 4M
for
neat
Sketc
h with
label

c) Explain processes of interrupt enabling and disabling in 8051. 4M

Ans: Interrupts are the events that temporarily suspend the main program, pass the control to the 2M
external sources and execute their task. It then passes the control to the main program where it forma
had left off.8051 has 5 interrupt signals, i.e. INT0, TF0, INT1, TF1, RI/TI. Each interrupt can t
be enabled or disabled by setting bits of the IE register and the whole interrupt system can be
disabled by clearing the EA bit of the same register.
IE (Interrupt Enable) Register:
This register is responsible for enabling and disabling the interrupt. EA bit is set to 1 for
enabling interrupts and set to 0 for disabling the interrupts. Its bit sequence and their meanings
are shown in the following figure.
EA _ _ ES ET1 EX1 ET0 EX0

2M
functi
on of
each
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bit

Explain following instructions of 8051.

d) (i) ADDC 4M
(ii) L CALL
Ans: (i) ADDC: The ADDC instruction adds a byte value and the value of the carry flag to 2M
the accumulator. The results of the addition are stored back in the accumulator. each
Several of the flag registers are affected. instru
ADDC
ction
Function: Add with Carry
Syntax: ADDC A, source byte
Flags affected: OV,AC,CY
Description: ADDC simultaneously adds the byte variable indicated, the carry flag and the
Accumulator contents, leaving the result in the Accumulator (A = A + byte +CY). The carry and
auxiliary-carry or bit flags are set, respectively. If CY = 1 prior to this instruction, CY is also
added to A.
Addressing modes supported for ADDC instruction :
 Immediate: ADDC A,#data
 Register: ADDC A, Rn
 Direct: ADDC A, address
 Register Indirect: ADDC A, @Ri
(ii) LCALL
Function: Long call, Transfers control to a subroutine
Syntax: LCALL 16 bit addr
Flags affected : None
No. of bytes used: 3 byte( 1 byte is opcode and other two bytes are the 16 bit address of the
target subroutine)
Description: This instruction is used to transfers control to a subroutine To reach the target
address in the 64 Kbytes maximum ROM space of the 8051, LCALL instruction is used. For
calling a subroutine, the PC register (which has the address of the instruction after the LCALL)
is pushed onto the stack, and the stack pointer (SP) is incremented by 2. Then the program
counter is loaded with the new address and control is transferred to the subroutine.

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Q.4 Attempt any THREE of the following : 12
Marks

a) Draw the format of TCON register of 8051 and describe the function of each bit of it. 4M
Ans: TCON: TIMER/COUNTER CONTROL REGISTER.BIT ADDRESSABLE 2M
forma
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0 t
TF1 TCON. 7 Timer 1 overflows flag. Set by hardware when the Timer/Counter 1
Overflows. Cleared by hardware as processor vectors to the interrupt
Service routine. 2M
TR1 TCON. 6 Timer 1 run control bit. Set/cleared by software to turn Timer/Counter1 Functi
ON/OFF.
on of
TF0 TCON. 5 Timer 0 overflow flag. Set by hardware when the Timer/Counter 0
Overflows. Cleared by hardware as processor vectors to the service routine. each
TR0 TCON. 4 Timer 0 run control bit. Set/cleared by software to turn Timer/Counter 0 bit
ON/OFF.
IE1 TCON. 3 External Interrupt 1 edge flag. Set by hardware when External Interrupt edge is
detected. Cleared by hardware when interrupt is processed.
IT1 TCON. 2 Interrupt 1 type control bit. Set/cleared by software to specify
falling edge/low level triggered External Interrupt.
IE0 TCON. 1 External Interrupt 0 edge flag. Set by hardware when External
Interrupt edge detected. Cleared by hardware when interrupt is processed.
IT0 TCON. 0 Interrupt 0 type control bit. Set/cleared by software to
Specify falling edge/low level triggered External Interrupt

b) Describe serial communication in 8051. Explain the use of SCON register. 4M

Ans: 8051 micro controller communicate with another peripheral device through RXD and TXD pin 2M
of port3.controller have four mode of serial communication. mode
1. Serial Data Mode-0 (Baud Rate Fixed) descri
In this mode, the serial port works like a shift register and the data transmission works ption
synchronously with a clock frequency of fosc /12. Serial data is received and transmitted through in
RXD. 8 bits are transmitted/ received at a time. Pin TXD outputs the shift clock pulses of
short
frequency fosc /12, which is connected to the external circuitry for synchronization. The shift
frequency or baud rate is always 1/12 of the oscillator frequency. (½
2. Serial Data Mode-1 (standard UART mode)(baud rate is variable)
mark
In mode-1, the serial port functions as a standard Universal Asynchronous Receiver Transmitter
for
(UART) mode. 10 bits are transmitted through TXD or received through RXD. The 10 bits
each
consist of one start bit (which is usually '0'), 8 data bits (LSB is sent first/received first), and a
mode)
stop bit (which is usually '1'). Once received, the stop bit goes into RB8 in the special function
register SCON. The baud rate is variable &
3. Serial Data Mode-2 Multiprocessor (baud rate is fixed) 2M
In this mode 11 bits are transmitted through TXD or received through RXD. The various bits are forma
as follows: a start bit (usually '0'), 8 data bits (LSB first), a programmable 9 th (TB8 or RB8)bit t with
and a stop bit (usually '1'). While transmitting, the 9 th data bit (TB8 in SCON) can be assigned functi
the value '0' or '1'. For example, if the information of parity is to be transmitted, the parity bit (P) on
in PSW could be moved into TB8.On reception of the data, the 9 th bit goes into RB8 in 'SCON',
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while the stop bit is ignored. The baud rate is programmable to either 1/32 or 1/64 of the
oscillator frequency.
f baud = (2 SMOD /64) fosc
4. Serial Data Mode-3 - Multi processor mode(Variable baud rate)
In this mode 11 bits are transmitted through TXD or received through RXD.The various bits are:
a start bit (usually '0'), 8 data bits (LSB first), a programmable 9 th bit and a stop bit (usually '1').
Mode-3 is same as mode-2, except the fact that the baud rate in mode-3 is variable (i.e., just as in
mode-1).
f baud = (2SMOD /32) * ( fosc/ 12 (256-TH1))
SM0 SM1 SM2 REN TB8 RB8 TI RI

SM0 SCON.7 Serial port mode specifier

SM1 SCON.6 Serial port mode specifier.
SM0 SM1
0 0 Serial Mode 0
0 1 Serial Mode 1, 8-bit data, 1 stop bit, 1 start bit
1 0 Serial Mode 2
1 1 Serial Mode 3
SM2 SCON.5 Used for multiprocessor communication
REN SCON.4 Set/ cleared by software to enable/ disable reception.
TB8 SCON.3 – the 9th bit that will be transmitted in mode 2/3 set/clear by software.
RB8 SCON.2– in mode 2/3 it is the 9th bit that was received .
TI SCON.1 Transmit interrupt flag. Set by hardware at the beginning of the
stop Bit in mode 1.
RI SCON.0 Receive interrupt flag. Set by hardware halfway through the
stop bit time in mode 1.
c) Draw interfacing of 16 × 2 LCD with 8051 and state the function of EN and RS of LCD 4M

Ans: Diagram: 2M
for
diagra
m

2Mar
ks for
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functi
on of
two
pins(
1Mar
k each
pin
functi
on)

RS: RS is the register select pin. We need to set it to 1, if we are sending some data to be
displayed on LCD. And we will set it to 0 if we are sending some command instructions during
the initializing sequence like clear the screen etc.
EN: The enable pin is used by the LCD to latch information presented to its data pins. When
data is supplied to the data pins, a high-to-low pulse must be applied to this pin in order for the
LCD to latch in the data present at the data pins. This pulse must be a minimum of 450ns wide.
d) Explain the use of following assembler directives.
(i) EQU 4M
(ii) ORG
Ans: 2
(i) EQU: Equate Marks
It is used to define constant without occupying a memory location. for
each
Syntax: Label EQU Numeric value
directi
By means of this directive, a numeric value is replaced by a symbol. ve
For e.g. MAXIMUM EQU 99 After this directive every appearance of the label MAXIMUM in
the program, the assembler will interpret as number 99
(MAXIMUM=99).

(ii) ORG:-ORG stands for Origin

Syntax: ORG Address
The ORG directive is used to indicate the beginning of the address.The origin directive tells the
assembler where to load instructions and data into memory. It changes the program counter to
the value specified by the expression in the operand field. The number thatcomes after ORG can
be either in hex or in decimal. If the number is notfollowed by H, it is decimal and the assembler
will convert it to hex.

e) State the alternate pin functions of port 3 of 8051. 4M

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Ans: 4
Marks
Pin Name Alternate Function
for 8
P3.0 RXD Serial input line pins(
½
P3.1 TXD Serial output line mark
for
P3.2 ̅̅̅̅̅̅̅ External interrupt 0 each
pin
P3.3 ̅̅̅̅̅̅̅ External interrupt 1
functi
P3.4 T0 Timer0 external input on)

P3.5 T1 Timer1 external input

P3.6 ̅̅̅̅̅ External data memory

write strobe
P3.7 ̅̅̅̅̅ External data memory
read strobe

Q.5 Attempt any TWO of the following 12

Total
Marks
(a) Explain with sketch the interfacing of 4 ×4 matrix keypad with 8051 microcontroller. 6M

Ans: sketch
-3M

Expla
Interfacing keypad nation
Fig. shows how to interface the 4 X 4 matrix keypad to two ports in microcontroller. The rows – 3M
are connected to an output port and the columns are connected to an input port.
To detect a pressed key, the microcontroller grounds all rows by providing 0 to the output
latch, and then it reads the columns. If the data read from the columns is D3-D0=1111, no key
has been pressed and the process continues until a key press is detected. However, if one of the

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column bits has a zero, this means that a key press has occurred. For example, if D3-D0=1101,
this means that a key in the D1 column has been pressed.
After a key press is detected, the microcontroller will go through the process of identifying the
key. Starting with the top row, the microcontroller grounds it by providing a low to row D0
only; then it reads the columns.
If the data read is all 1s, no key in that row is activated and the process is moved to the next row.
It grounds the next row, reads the columns, and checks for any zero. This process continues until
the row is identified. After identification of the row in which the key has been pressed, the next
task is to find out which column the pressed key belongs to.
Differentiate between
(b) (i) Harvard and Von-neuman architecture 6M
(ii) Microprocessor and Microcontroller
Ans: Von
i) Harvard Architecture and Von-neuman architecture Nuem
ann
Harva
rd 3
M
(any
three
points
)

ii) Microprocessor and Microcontroller

Micro
proces
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sor ,
Micro
contro
ller –
3M
(any
three
points
)

Develop an ALP to generate square wave of 3 KHz using 8051 microcontroller on port pin
(c) 6M
P2.3 (Assume Xtalfreqn=12 MHz)

Ans: Crystal frequency= 12 MHz

Count
I/P clock = (12*10^6 )/12= 1 MHz
calcul
Tin = 1μ sec
ation
For 3 kHz square wave
– 2M,
Fout = 3 KHz Tout = 1/ (3X 103) = 0.3msec =333 μ sec
Corre
So TON = TOFF = 333/2 = 166.5 μ sec
ct
N = TON / Tin = 166.5 μ sec /1 μ sec = 166.5 167
progr
65535 – 167+1 = (65369)10 = (FB71)16
am –
Program:-
4M
MOV TMOD, # 01H ; Set timer 0 in Mode 1, i.e., 16 bit timer

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L2: MOV TL0, # 71 H ; Load TL register with LSB of count
MOV TH0, # 0FB H ; load TH register with MSB of count
SETB TR0 ; start timer 0
L1: JNB TF0, L1 ; poll till timer roll over
CLR TR0 ; stop timer 0
CPL P2.3 ; complement port 2.3 line to get high or low
CLR TF0 ; clear timer flag 0
SJMP L2 ; re-load timer with count as mode 1 is not auto reload
12Tot
Q.6 Attempt any TWO of the following: al
Marks
Draw interfacing of stepper motor with 8051 and write an ALP to rotate it in clockwise
(a) 6M
direction.

Ans: Diagram: 3M

Progr
am-
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3M

( Other programs with similar logic can be given marks)

(b) Describe with sketches the procedure to troubleshoot the traffic light controller. 6M

Ans: Considerations of Traffic Signal Any

1) Traffic light may have sensors integrated to provide real time traffic information other
2) Based on the traffic information provided by the sensor, the duration of the green/Red correc
LED light for each direction may vary so that the traffic for both the directions are t
roughly balanced. Time left for the green light should be displayed troubl
3) When the traffic light for one signal is green, then the traffic for the other directions e
should be red (with duration displayed in red) shooti
4) The red light will be switched to yellow when the timer value is 5 sec before switching ng
to red. proce
dure
may
be
given
marks

5) Violations happen when user expectancy is not met. A user like pedestrian does not
expect to stand for more than a minute or two at a signal, when this user expectancy is
not met, the pedestrian tries to venture out and violate the signal
6) The smooth movement of conflicting vehicles is determined by the availability of gaps
in traffic. This is true for both pedestrians and vehicular traffic. Understanding of gaps is
important for justifying the type of traffic control device, including a traffic signal.
Points to consider for determining signal timings
1) The signal operational parameters are reviewed and updated (if needed) on a regular
basis to maximize the ability of the traffic control signal to satisfy current traffic
demands

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2) Geometry of the intersection is the next step in the signal timing process. Determining
the lane use (which traffic mode), dedicated vs. shared lane, type of roads interacting
(arterial with arterial etc.), type of road infrastructure (ramps, one way streets, etc.) will
impact the timing.
3) Basic signal timing parameters comes next. Pedestrian walk times, flashing don’t walk,
yellow time, all red clearance interval, detector gap times all need to calculated or
established.
4) Identify bottlenecks, review conditions, conduct warrant analysis and use engineering
judgment in determining traffic signal installation
5) Determine AM and PM peak hour traffic volumes.
6) Condition diagram which includes roadway geometrics, parking, driveways, sidewalks,
signing, pavement markings, development of intersection quadrants, and any other
features pertinent to the study peak hour delay study
7) A conflict analysis
8) Capacity analysis of the intersection for current and future years using growth

(c) Draw and explain Internal port structure of Port 0 and Port 1 of 8051 microcontroller. 6M

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Ans:

Port-0 can be configured as a normal bidirectional I/O port or it can be used for address/data
Port 0
interfacing for accessing external memory. When control is '1', the port is used for address/data
– 3M
interfacing. When the control is '0', the port can be used as a normal bidirectional I/O port.
Port 1
Let us assume that control is '0'. When the port is used as an input port, '1' is written to the latch. – 3M
In this situation both the output MOSFETs are 'off'. Hence the output pin floats. This high
impedance pin can be pulled up or low by an external source. When the port is used as an output
port, a '1' written to the latch again turns 'off' both the output MOSFETs and causes the output
pin to float. An external pull-up is required to output a '1'. But when '0' is written to the latch, the
pin is pulled down by the lower MOSFET. Hence the output becomes zero.

When the control is '1', address/data bus controls the output driver MOSFETs. If the
address/data bus (internal) is '0', the upper MOSFET is 'off' and the lower MOSFET is 'on'. The
output becomes '0'. If the address/data bus is '1', the upper transistor is 'on' and the lower
transistor is 'off'. Hence the output is '1'. Hence for normal address/data interfacing (for external
memory access) no pull-up resistors are required.
Port-0 latch is written to with 1's when used for external memory access.

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Port-1 does not have any alternate function i.e. it is dedicated solely for I/O interfacing. When
used as output port, the pin is pulled up or down through internal pull-up. To use port-1 as input
port, '1' has to be written to the latch. In this input mode when '1' is written to the pin by the
external device then it read fine. But when '0' is written to the pin by the external device then the
external source must sink current due to internal pull-up. If the external device is not able to sink
the current the pin voltage may rise, leading to a possible wrong reading.

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