3rd Internal

8/3/2019 3rd Internal

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5th

Sem, C, Assignment 3, Srikanth TR, 1MS09EC119

MICROCONTROLLER ASSIGNMENT 3

Question 1:- Explain the registers and pins of an LCD panel and write and 8051 C

program to display the message HELLO on the LCD panel.

LCD pin descriptions:

Pin Symbol I/O Description

1 Vss - Ground

2 Vcc - +5V power supply

3 VEE - power supply to control

4 RS I RS=0 to select code register

RS=1 to select data register5 R/W I R/W=0 for write R/W for read

6 E I/O Enable

7 DB0 I/O The 8-bit data bus








VCC , VSS and VEE :-

While VCC and VSS provide +5V and ground, respectively, VEE is used for controlling LCD

contrast.

RS, register select:-

If RS=0, the instruction command code register is selected, allowing the user to send a

command such as clear display, return home etc.

If RS=1 the data register is selected, allowing the user to send data to be displayed on

the LCD.


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R/W, read/write: R/W input allows the user to write the information to the LCD or read

the information from it.

E, enable:-

The enable pin is used by the LCD to latch information presented to its data pins. When

data is supplied to 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.

D0-D7:-The 8-bit data pins, D0-D7, are used to send information to the LCD or read the

contents of the LCDs internal registers. To display the letters and numbers we send

ASCII codes and also there are instruction command codes we send to these pins while

making RS=1.

We also use RS=0 to check the busy flag bit to see if the LCD is ready to receive

information. The busy flag is D7 and can be read when R/W=1, RS=0.

Registers of LCD panel :-There are main two registers inside the LCD

1.Data register: This register is selected when RS=1. Here LCD stores the messageto be displayed i.e. allowing the user to send data to be displayed on the LCD.

2.Code register: This register is selected when RS=0. Allowing the user to send acommand to LCD. The following table lists the command codes.

Code (hex) Command to LCD instruction register

1 Clear display screen

2 Return home

4 Shift cursor to left

5 Shift display right

6 Shift cursor to right

7 Shift display left


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8 Display off, cursor off

A Display off, cursor on

C Display on, cursor off

E Display on, cursor blinking

F Display on, cursor blinking10 Shift cursor position to left

14 Shift cursor position to right

18 Shift entire display to the left

1C Shift entire display to the right

80 Force cursor to beginning of 1st

line

C0 Force cursor to beginning of 2nd

line

38 2 lines and 5*7 matrix

C-Program:

#include

sfr ldata = 0x90; //P1=LCD data pins

sbit rs = P2^0;

sbit rw = P2^1;

sbit en =P2^2;


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void lcdcmd(unsigned char value)

{

ldata = value; //put the value on the pins

rs = 0; //to select code register

rw =0;en = 1; //strobe the enable pin

MSDelay (1);

en=0;

return;

}

void lcddata (unsigned char value)

{

ldata = value; //put the value on the pins

rs = 1; //to select the data registerrw =0;

en = 1; //strobe the enable pin

MSDelay (1);

en=0;

return;

}

void MSDelay (unsigned int itime){

Unsigned int i, j;

for (i=0; i


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MSDelay (250);

lcdcmd (0x01); // Clear display screen

MSDelay (250);

lcdcmd (0x06); // Shift cursor to right after inserting each //characterMSDelay (250);

lcdcmd (0x86); //line 1, position 6

MSDelay (250);

for (k=0; k


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Q2. Draw the 8051 connection to DAC0808 at port P1 and write

8051 ALP to generate a sine wave


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ALP:

ORG 40

AGAIN: MOV DPTR, #TABLE ;Initialize DPTR to start of table

MOV R2, #COUNT ;Initialize count

BACK: CLR A

MOVC A,@A+DPTR

Angle(Degrees) Sin (angle) Voltage Voltage sent to

Magnitude DAC5V + (Voltage Mag. *

(5V*Sin(angle)) 25.6)

0 0 5 128

30 0.5 7.5 192

60 0.87 9.33 238

90 1 10 255

120 0.87 9.33 238

150 0.5 7.5 192

180 0 5 128

210 -0.5 2.5 64

240 -0.87 0.67 17

270 -1 0 0

300 -0.87 0.67 17

330 -0.5 205 64

360 0 5 128


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MOV P1,A ;Move values from LUT to P1

INC DPTR

DJNZ R2,BACK ;Loop the values

SJMP AGAIN ;Looping sine wave

ORG 300H ;Initializing LUTTABLE: DB 128,192,238,255,192,128,64,17,0,17,64,128

END

Question 3:-Show a simple keyboard interfacing with Port 1 and Port 2 of 8051 and

explain its operation.

Keyboards are organized in a matrix of rows & columns. The cup Accesses both rows &

columns through ports; therefore, with two 8 bit Ports, an 8 8 matrix of keys can be

connected to a microprocessor. When a key is pressed, a row & a column make a

contact; otherwise, there is no connection between rows & columns. In IBM pc

keyboards, a single microcontroller takes care of hardware & software interfacing of

the keyboard. In such system, it is the function of programs stored in the EPROM of the

microcontroller to scan the keys continuously, identify which one has been activated, &

present it to the motherboard.

Fig shows 44 matrix connected to two ports. The rows are connected to anoutput port & the columns are connected to an input port. If no key has been pressed,

reading the input port will yield 1s for all columns since they are

connected to high (Vcc). If all the rows are grounded & a key is pressed, one of the

columns will have 0 since the key pressed provides the path to ground. It is the function

of the microcontroller to scan the keyboard continuously to detect & identify the key

pressed.

Operation:-

To detect the pressed key, the microcontroller grounds all the rows by providing 0 to

the output latch, 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


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detected. However, if one of the column bits has a zero, this means that a key press has

occurred. 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. This should be easy

since the microcontroller knows at any time which row and column are being accessed.


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4. Show a scheme of interfacing ADC0809 to 8051 controller. Write

an ALP to obtain the output from such an interface. Discuss

practical application.

Ans.


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The ADC0809 chip has an 8 bit data output. The 8 analog input channels are

multiplexed and selected according to the table(i) shown using 3 address pins A,B and

C.

Selected Analog

ChannelC B A

IN0 0 0 0

IN1 0 0 1

IN2 0 1 0

IN3 0 1 1

IN4 1 0 0

IN5 1 0 1IN6 1 1 0

IN7 1 1 1

IN0

IN7

D0

D7

EOCOE

Vref(+)Vref(-)

gnd Clk Vcc

SC ALE C B A

(LSB)

ADC0809

Table 1

Fig. 1


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Table 2

Steps for programming the ADC0809:

The following are the steps to get data from an ADC0809

1. Select analog channel by providing bits to A, B and C address according totable(ii).

2. Activate the ALE (address latch enable) pin. It needs an L-H pulse to latch inthe address.

3. Activate SC (start conversion) by an L-H pulse to initiate conversion.4. Monitor EOC (end of conversion) to see whether conversion is finished. H-L

output indicates that the data is converted and is ready to be picked up. If

we do not use EOC, we can read the converted digital data after a brief

time delay. The delay size depends on the speed of the external clock we

connect to CLK pin.

5. Activate OE (output enable) to read data out of the ADC chip. An L-H pulseto the OE pin will bring digital data out of the chip.

Vref (V) Vin (V) Step Size (mV)

Not connected 0 to 5 5 / 256 = 19.534.0 0 to 4 4 / 256 = 15.62

3.0 0 to 3 3 / 256 = 11.71

2.56 0 to 2.56 2.56 / 256 = 10

2.0 0 to 2.0 2 / 256 = 7.81

1.0 0 to 1.0 1 / 256 = 3.90


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In ADC0809 there is no self clock and the clock must be prepared from an

external source to CLK pin. The speed of conversion depends on the frequency

of the clock connected to the CLK pin, it cannot be faster than 100

microseconds.

Programming ADC0809 in Assembly

ALE BIT P2.4

OE BIT P2.5

SC BIT P2.6

EOC BIT P2.7

ADDR_A BIT 92.0

ADDR_B BIT P2.1

ADDR_C BIT P2.2

MYDATA EQU P1

ORG 0H

MOV MYDATA, #0FFH ;make P1 as input

SETB EOC ;make EOC as input

CLOCK

WR(SC)

RD(OE)

ALE

ADDR

EOC(INTR)

D0-D7

LATCH

ADDRESSLATCH

DATA

Selecting a Channel and Read Timing for ADC0809


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CLR ALE ;clear ALE

CLR SC ;clear WR

CLR OE ;clear RD

BACK: CLR ADDR_C ;C=0

CLR ADDR_B ;B=0SETB ADDR_A ;A=1 (select channel 1)

ACALL DELAY ;make sure address is stable

SETB ALE ;latch address

ACALL DELAY ;delay for fast DS89C4x0 Chip

SETB SE ;start conversion

ACALL DELAY

CLR ALE

CLR SC

HERE: JB EOC, HERE ;wait until doneHERE1:JNB EOC, HERE1 ;wait until done

SETB OE ;enable RD

ACALL DELAY ;wait

MOV A, MYDATA ;read data

CLR OE ;clear RD for next time

ACALL CONVERSION ;hex to ASCII

ACALL DATA_DISPLAY ;display the data

SJMP BACK


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Question 5:- Describe the 8051 connection to stepper motor. A switch is

connected to pin p2.7. Write a c program to monitor the status of SW and

perform the following

If SW=0, the stepper motor moves clockwise.

If SW=1, the stepper motor moves counter-clockwise.

The following steps show the 8051 connection to the stepper motor and its

programming:

1. Use an ohmmeter to measure the resistance of the leads. This should identify

which COM leads are connected to which winding leads.

2. The common wire(s) are connected to the positive side of the motors power

supply. In many motors, +5V is sufficient.

3. The four leads of the stator winding are controlled by four bits of the 8051

port (P1.0 P1.3). However, since the 8051 lacks the sufficient current to drivethe stepper motor windings, we must use a driver such as the ULN2003 to

energize the stator. Instead of the ULN2003, we could have used transistors as

drivers, as shown in the figure below. However, notice that if transistors are used

as drivers, we must also use diodes to take care of inductive current generated

when the coil is turned off. One reason that using the ULN2003 is preferable to

the use of transistors as drivers is that the ULN2003 has an internal diode to take

care of the back EMF.


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Fig. 8051 Connection to Stepper

// A C program to monitor P2.7 (SW) and rotate the stepper motor// clockwise if SW = 0; anticlockwise if SW = 1

#include

sbit SW=P2^7;

void main() {

SW=1;

while(1) {

if(SW==0)

{


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P1=0x66;

MSDelay(100);

P1=0xCC;

MSDelay(100);

P1=0x99;

MSDelay(100);

P1=0x33;

MSDelay(100);

}

else

{

P1=0x66;

MSDelay(100);

P1=0x33;

MSDelay(100);

P1=0x99;

MSDelay(100);

P1=0xCC;

MSDelay(100);

}


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}

}

void MSDelay(unsigned int value)

{

unsigned intx,y;

for(x=0;x


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doing other things. The ability to control the speed of the dc motor using PWM is

one reason that DC motors are preferred over AC motors. AC motor speed is

dictated by the AC frequency of the voltage applied to the motor and frequency is

generally fixed. Therefore we cannot control the speed of AC motor when the

load is increased. We can also change the DC motors direction and torque.

Question 7:- Explain the registers and pins of an LCD panel and write an 8051 ALP

display message MSRIT on the LCD panel.

The Optrex LCD Panel has 14 pins. The function of each pin is given in the table

below.

Pin Symbol I/O Description

1 VSS - Ground

2 VCC - +5V power supply

3 VEE - Power supply to control contrast

1/4 POWER 25% DC

1/2 POWER 50% DC

3/4 POWER 75% DC

FULL POWER 100% DC

Pulse width Modulation Comparision


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4 RS I RS = 0 to control command register

RS = 1 to select data register

5 R/W I R/W = 0 for write, R/W = 1 for read

6 E I/O Enable









VCC, VSS and VEE :-

VCC and VSS provide +5V and ground respectively. VEE is used for controlling LCD

contrast.

RS, register select:-

The RS pin is used to select between different registers of the LCD. If RS = 0, the

instruction command code register is selected, allowing the user to send a


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command such as clear display, cursor at home, etc. to the LCD. If RS = 1 the data

register is selected, allowing the user to send data to be displayed on the LCD.

R/W, Read/Write:-

R/W input allows the user to write information to the LCD or read information

from it. R/W=1 when reading; R/W=0 when writing.

E, Enable:-

The enable pin is used by the LCD to latch information presented to its data pins.

When data is supplied to 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.

D0-D7:-

The 8-bit data pins, D0-D7, are used to send information to the LCD or read the

contents of the LCDs internal registers.

To display letters and numbers, we send ASCII codes for the letters A-Z, a-z, and

numbers 0-9 to these pins while making RS=1.

There also instruction command codes that can be sent to the LCD to clear the

display or force the cursor to home position or blink the cursor. The table below

lists the instruction command codes.


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RS = 0 is used to check the busy flag bit to find out if the LCD is ready to receive

information. The busy flag is D7 and can be read when R/W = 1 and RS = 0. When

D7 = 1, the LCD is busy taking care of internal operations and will not accept anynew information. When D7 = 0, the LCD is ready to receive new information.

Code

(Hex)

Command to LCD instruction register

1 Clear display screen

2 Return home

4 Decrement cursor (shift cursor to left)

6 Increment cursor (shift cursor to right)

5 Shift display right

7 Shift display left

8 Display off, cursor off

A Display off, cursor on

C Display on, cursor off

E Display on, cursor blinking

F Display on, cursor blinking

10 Shift cursor position to left

14 Shift cursor position to right

18 Shift the entire display to the left


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1C Shift the entire display to the right

80 Force cursor to beginning of 1st

line

C0 Force cursor to beginning of 2nd

line

38 2 lines and 5X7 matrix

The LCD has data and command (instruction) registers. The data register is used

to store the data to be displayed on the LCD. The command (instruction) register

is used to store the instructions to the LCD.

Program to display MSRIT on the LCD Panel

ORG 0000H

MOV DPTR, #MYCOM

C1: CLR A

MOVC A,@A+DPTR

ACALL COMNWRT ; call command subroutine

ACALL DELAY ; give LCD some time

JZ SEND_DAT

INC DPTR

SJMP C1

SEND_DAT: MOV DPTR, #MYDATA


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D1: CLR A

MOVC A,@A+DPTR

ACALL DATAWRT ; call command subroutine


INC DPTR

JZ AGAIN

SJMP D1

AGAIN: SJMP AGAIN ; stay here

COMNWRT: MOV P1, A ; SEND COMND to P1

CLR P2.0 ; RS=0 for command

CLR P2.1 ; R/W=0 for write

SETB P2.2 ; E=1 for high pulse


CLR P2.2 ; E=0 for H-to-L

RET

DATAWRT: MOV P1, A ; SEND DATA to P1

SETB P2.0 ; RS=1 for data

CLR P2.1 ; R/W=0 for write

SETB P2.2 ; E=1 for high pulse


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CLR P2.2 ; E=0 for H-to-L pulse

RET

DELAY: MOV R3, #250 ; long delay for fast CPUs

HERE2: MOV R4, #255

HERE: DJNZ R4, HERE

DJNZ R3, HERE2

RET

ORG 0300H ; lookup table

MYCOM: DB 38H,0EH,01H,06H,84H,00H ; commands and null

MYDATA: DB MSRIT, 00H ; data and null

END

8.DRAW 8051 CONNECTION T0 DAC0808 AT PORT P1 ANDWRITE A C PROGRAM TO GENERATE SINE WAVE.

Ans.


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In DAC 808 the digital inputs are converted to current (Lout) and by connecting a

resistor to the Lout pin we convert the result to voltage. Lout is a function of digital

inputs and reference current as below.

The Lrefcurrent is generally 2mA.The maximum output current, (if all inputs to

DAC are high) is 1.99mA (from the above figure).

Generating Sine Wave:

Full scale output of DAC is achieved when all data inputs of DAC are high.

Therefore, to achieve full-scale output we use the following equation.

Vout=5V*(1 + sin).


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Vout for DAC for various angles is calculated as below at increment of 30 degrees.

Angle @

(degrees)

sin@ Vout(voltage

mag)

5V*(1 + sin)

Values sent to

DAC(decimal)

(Voltage mag * 25.6)

0 0 5 128

30 0.5 7.5 192

60 0.866 9.33 238

90 1.0 10 255

120 0.866 9.33 238

150 0.5 7.5 192

180 0 5 128

210 -0.5 2.5 64240 -0.866 0.669 17

270 -1.0 0 0

300 -0.866 0.699 17

330 -0.5 2.5 64

360 0 5 128

C program to generate sine wave:

#include

sfr DACDATA =P1;

void main()

{

Unsigned char table[12]={128,192,238,255,238,192,128,64,17,0,17,64};

Unsigned char x;

while(1)

{

for(x=0;x


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Question9:- How can a microcontroller be used to control automatically the speed

of a dc motor. Explain the concept clearly.

The speed of motor depends on three factors:

a)load

b) Current

c) Voltage

For a fixed load we can maintain a steady speed using a method called pulse

width modulation.

By changing (modulating) the width if the pulse applied to the DC motor we can

increase or decrease the amount power provided to the motor, thereby

increasing o decreasing the motor speed. Although the voltage has a fixed

amplitude, it has a variable duty cycle. That means the wider the pulse, the higher

the speed.PWM is so widely used in DC motor control that some microcontrollers

come with PWM circuitry embedded in the chip. In such microcontrollers all we

have to do is to load the proper registers with values of high and low portions of

the desired pulse, and rest is taken care by the microcontroller. This allows the

microcontroller to do other things. For microcontroller without PWM circuitry, wemust create the various duty cycle pulses using software which prevents the

microcontroller from doing other things.

Pulse width modulation comparison:

.25 power 25% DC

.50 power 50% DC

.75 power 75% DC

Full power 100% DC


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EX: C program to control speed of DC motor based switches connected to p2.6

and p2.7 as per table below

SW2(p2.7) SW(p2.6) Comments

0 0 DC motor moves slowly(25% duty cycle)

0 1 DC motor moves moderately(50% duty cycle)

1 0 DC motor moves fast(75% duty cycle)

1 1 DC motor moves very fast(100% duty cycle)

#include

sbit mtr=P1^0;

void msdelay(unsigned int value);

void main()

{

Unsigned char x;

P2=0xFF;

z=P2;

z=z&0x03;

mtr =0;

while (1)

{

switch (z)


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{

case(0):

{

mtr=1;

msdelay(25);

mtr=0;

msdelay(75);

break;

}

case(1):

{

mtr=1;

msdelay(50);

mtr=0;

msdelay(50);

break;

}

case(2):

{

mtr=1;


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msdelay(75);

mtr=0;

msdelay(25);

break;

}

default:

mtr=1;

}

}

}

Q10. Show a scheme of interfacing an 8-bit ADC to 8051

controller. Write the code required in C to obtain the output

from such an interface. Discuss practical application.

Ans.


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Code:

#include

sbit RD = P2^5;

sbit WR= P2^6;

sbit INTR = P2^7;

sfr MYDATA = P1;void main()

{

unsigned char value;

MYDATA = 0xFF; /* make P1 as input */

INTR = 1; /* make INTR input */


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RD = 1; /* set RD high */

WR = 1; /* set WR high */

while(1)

{

WR = 0; /*send WR pulse */WR = 1;

while(INTR==1); /* wait for EOC*/

RD = 0; /*send RD pulse

value = MYDATA; // read value

ConvertAndDisplay(value); //function to convert and display the

output

RD = 1;

}

}

Practical Applications:

Analog-to-digital converters are among the most widely used devices for data

acquisition. Digital computers use binary (discrete) values, but in the physical

world everything is analog (continuous). Temperature, pressure (wind or liquid),

humidity, and velocity are a few examples of physical quantities that we deal with

every day. A physical quantity is converted to electrical (voltage, current) signalsusing a device called a transducer. Transducers are also referred to as sensors.

Sensors for temperature, velocity, pressure, light, and many other natural

quantities produce an output that is voltage (or current). Therefore, we need an

analog-to-digital converter to translate the analog signals to digital numbers so

that the microcontroller can read and process them. An ADC has n-bit resolution

where n can be 8, 10, 12, 16 or even 24 bits. The higher-resolution ADC provides a

smaller step size, where step size is the smallest change that can be discerned by

an ADC.


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Question 11:- Describe the 8051 connection to stepper motor. A switch is

connected to pin p2.7. Write a c program to monitor the status of SW and

perform the following

If SW=0, the stepper motor moves clockwise.

If SW=1, the stepper motor moves counter-clockwise.

The following steps show the 8051 connection to the stepper motor and its

programming:

1. Use an ohmmeter to measure the resistance of the leads. This should identify

which COM leads are connected to which winding leads.

2. The common wire(s) are connected to the positive side of themotors power

supply. In many motors, +5V is sufficient.

3. The four leads of the stator winding are controlled by four bits of the 8051

port (P1.0 P1.3). However, since the 8051 lacks the sufficient current to drivethe stepper motor windings, we must use a driver such as the ULN2003 to

energize the stator. Instead of the ULN2003, we could have used transistors as

drivers, as shown in the figure below. However, notice that if transistors are used

as drivers, we must also use diodes to take care of inductive current generated

when the coil is turned off. One reason that using the ULN2003 is preferable to

the use of transistors as drivers is that the ULN2003 has an internal diode to take

care of the back EMF.


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Fig. 8051 Connection to Stepper

ORG 0H ; starting address

MAIN: SETB P2.7 ; make an input

MOV A, #66H ; starting phase value

MOV P1, A ; send value to port

TURN: JNB P2.7, CW ; check switch results

RR A ; rotate right

ACALL DELAY ; call delay

MOV P1, A ; write value to port

SJMP TURN ; repeat

CW: RL A ; rotate left

ACALL DELAY ; call delay

MOV P1, A ; write value to port

SJMP TURN ; repeat

DELAY: MOV R2, #100

H1: MOV R3, #255

H2: DJNZ R3, H2

DJNZ R2, H1

RET

END


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Q12. Show a simple keyboard interface with port P1 and P2 of

8051 and explain its operation.

Ans: Keyboards are organized in a matrix of rows & columns. The cup Accesses

both rows & columns through ports; therefore, with two 8 bit Ports, an 8 8

matrix of keys can be connected to a microprocessor. When a key is pressed, a

row & a column make a contact; otherwise, there is no connection between rows

& columns. In IBM pc keyboards, a single microcontroller takes care of hardware

& software interfacing of the keyboard. In such system, it is the function of

programs stored in the EPROM of the microcontroller to scan the keys

continuously, identify which one has been activated, & present it to the

motherboard.

Fig shows 44 matrix connected to two ports. The rows are connected to

an output port & the columns are connected to an input port. If no key has been

pressed, reading the input port will yield 1s for all columns since they are

connected to high (Vcc). If all the rows are grounded & a key is pressed, one of

the columns will have 0 since the key pressed provides the path to ground. It isthe function of the microcontroller to scan the keyboard continuously to detect &

identify the key pressed.


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Operation:

To detect the pressed key, the microcontroller grounds all the rows by providing 0

to the output latch, 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 column bits has a zero, this means that a

key press has occurred. 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


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of the row in which the key has been pressed, the next task is to find out which

column the pressed key belongs to. This should be easy since the microcontroller

knows at any time which row and column are being accessed.

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