MICROCONTROLLER AND EMBEDDED SYSTEMS ...Create project A7/A7.s for the leagcy device LPC2148 w/o startup.s 2. Translate A7.s and build 3. View RO data in memory window 1 at 0x0000

MICROCONTROLLER AND EMBEDDED SYSTEMSLABORATORY

18CL48

ATRIA INSTITUTE OF TECHNOLOGYDEPARTMENT OF COMPUTER SCIENCE & ENGINEERING

Bengaluru - 5600242020

1

Microcontroller And Embedded Systems 18CL48

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Contents

1 A1 Multiply 4

2 A2 Sum 6

3 A3 Factorial 8

4 A4 Array 10

5 A5 Table lookup 12

6 A6 Min/Max 14

7 A7 Sort 16

8 A8 Count 1s 19

9 B1 UART 22

10 B2 DC Motor 26

11 B3 Stepper Motor 30

12 B4 ADC 34

13 B5 DAC 40

14 B6 Keypad 45

15 B7 EINT 49

16 B8 7SD 54

17 BU LCD 60

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18 LPC2148 65

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1 A1 Multiply

Write a program to multiply two 16 bit binary numbers.

Objective

1. Multiply two 16 bit numbers

Design

1. Initialize memory with 16bit data i.e half words

2. Reserve memory space for the results

3. Load half words from memory into registers

4. Store the multiplication result of 2x16 bit words into memory word

5. Algorithm

int main(){

int n, n1, n2;n = n1 * n2;

}

ARM Instructions

1. DCW, DCD directives

2. LDR, LDRH, STR, MUL

Procedure

1. Create project A1/A1.s for the leagcy device LPC2148 w/o startup.s

2. Translate A1.s and build

3. View RO data in memory window 1

4. View RW data in memory window 2 at 0x40000000

5. Trace the program logic step by step (using say F10)

Note: Result is stored in LE format.

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4 3 2 1 1 2 3 40x00 00 01 00 -> 00 01 00 00

; A1.s; Multiplication of 16bit data

AREA AP, CODE, READONLYENTRY

START

LDR R0, =RD

LDRH R1, [R0],#2 ; R1 = mem[R0], R0 = R0 +2, PostincrmentLDRH R2, [R0],#2

MUL R3, R2, R1 ; R3 = R2*R1

LDR R0, =WRSTR R3, [R0] ; mem[R0] = R3

STOP B STOP

AREA RO, DATA, READONLYRD DCW 0x10, 0x10 ; Test with different input values

AREA RW, DATA, READWRITEWR DCD 0x0 ; Result of multiplication here

END

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2 A2 Sum

Write a program to find the sum of first 10 integer numbers.

Objectives

1. Sum of first 10 natural numbers

Design

1. Reserve memory for the result

2. Generate first 10 numbers

3. Compute incremental sum

4. Store the sum

5. Algorithm

int sum = 0;for (int i = 1; i <= 10; i++) {

sum += i;}

ARM Instructions

1. Directives DCD

2. MOV, ADD, CMP, BLE, LDR, STR

Procedure





5. Store the results and verify it to be n(n+1)/2

; A2.s; Sum of first 10 natural numbers

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STARTMOV R0, #0 ; R0=0, accumulates totalMOV R1, #1 ; R1=1, counts from 1 to 10

LOOPADD R0, R0, R1 ; R0 = R0 + R1ADD R1, R1, #1 ; R1 = R1 + 1CMP R1, #10 ; repeat if R1 < 10BLE LOOP


STOP B STOP

AREA RW, DATA, READWRITEWR DCD 0x0 ; Result is stored here

END

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3 A3 Factorial

Write a program to find factorial of a number.

Objectives

1. Fact(n)

Design

1. Input is n

2. Reserve memory for the result n!

3. Computer factorial as n.(n-1). . . 3.2.1

4. Store the results in memory

5. Algorithm

fact = 1;for (i=n; i > 0; i--)

{fact = fact*i;

}

ARM Instructions

1. Directives DCD

2. MOV, MUL, SUB[S], BNE, LDR, STR

Procedure





5. Store the results and verify it to be n.(n-1). . . 3.2.1

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; A3.s; factorial n


STARTMOV R0, #1 ; fact = 1MOV R1, #5 ; n

LOOPMUL R0, R1, R0 ; n*(n-1)(n-2)...3.2.1SUBS R1, R1, #1 ; n = n-1BNE LOOP


STOP B STOP

AREA RO, DATA, READWRITEWR DCD 0x0 ; Result stored here

END

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4 A4 Array

Write a program to add an array of 16 bit numbers and store the 32 bit result in internal RAM

Objective

1. Sum of numbers in an array

Design

1. Load 32bit numbers in RO memory

2. Set the base as begining of array

3. Get the number at the base and advance the base addresss to next number

4. Stop after reading N numbers

5. Algorithm

sum = 0;for (i=0; i < n; i++)

{sum = sum + a[i];

}

ARM Instructions

1. Directives DCD

2. LDR, MOV, ADD, CMP, BNE, LDR, STR

Procedure



3. View RO data in memory window 1 at 0x0000 0000

4. View RW data in memory window 2 at 0x4000 0000


6. Store the results

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; A4.s; Find sum of numbers in array


N EQU 5START

LDR R0, =RDMOV R1, #0 ; i = 0MOV R2, #0 ; sum = 0

LOOPLDRH R3, [R0], #2 ; n = *A++ADD R2, R2, R3 ; sum = sum + nADD R1, R1, #1 ; i = i+1CMP R1, #N ; i < 5, repeatBNE LOOP


STOP B STOP

AREA RO, DATA, READONLYRD DCW 1, 2, 3, 4, 5 ; Array of numbers

AREA RW, DATA, READWRITEWR DCD 0x0 ; Result is tored here

END

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5 A5 Table lookup

Write a program to find the square of a number (1 to 10) using look-up table.

Objective

1. Find square(n) from the lookup table a[n]

Design

1. Read n from RO memory

2. Traverse the table for matching ’n’

3. Read out the associated value

4. Store the result in RW memory

5. Algorithm

sum = 0;for (i=0; i < n; i++)

{if (n == a[i]) break;

}

if (i < n) printf("Found");

ARM Instructions

1. Directives DCD

2. LDR, LDMIA, MOV, ADD, CMP, BEQ, BNE, LDR, STR

Procedure






6. Store the results as table[n]

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; A4.s; Find square of a number (1-10) using look up table


N EQU 4START

LDR R0, =RD ; LB R0 = aLDR R1, =RD+40 ; UB R1 = ARRAY a + 5*8MOV R2, #N ; Key n = NMOV R5, #0 ; 0

LOOPLDMIA R0!, {R3, R4} ; Get the tuple {n, n^2}, incr i to next tupleCMP R2, R3 ; Found if n= N, R4 hold the square of nBEQ DONE ; square(n)CMP R0, R1 ; End of the arrayBNE LOOPMOV R4, #-1 ; -1 if not found

DONELDR R5, =WRSTR R4, [R5] ; mem[R5] = R4

STOP B STOP

AREA RO, DATA, READONLYRD DCD 1, 1, 2, 4, 3, 9, 4, 16, 5, 25

AREA RW, DATA, READWRITEWR DCD 0x0 ; Result stored here

END

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6 A6 Min/Max

1. Write a program to find the largest/smallest number in an array of 32 numbers .

Objective

1. Smallest number in the list of numbers

Design

1. Input 32 numbers in RO memory

2. Select first number as smallest

3. Check all other numbers and change smallest if required

4. Store the smallest number in RW memory

5. Algorithm

min = a[0];for (i=0; i < n; i++)

{if (min > a[i]) min = a[i];

}

ARM Instructions

1. Directives DCW, DCD

2. LDR[H], MOV[GT], CMP, BNE, LDR, STR

Procedure






6. Store the results as e.g min(10, 5, 15) is 5

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; A6.s; Find smallest/larget of the number in an array of 32 numbers


N EQU 5

STARTLDR R0, =RD ; location of a[0] i.e aLDR R1, =RD+8 ; location of a[N-1] i.e a + (N-1)*2LDRH R2, [R0] ; min = a[0], i=0

LOOPLDRH R3, [R0, #2]! ; n = A[++i] i.e i = i + 2 (preindexing)CMP R2, R3 ; R2 > R3MOVGT R2, R3 ; min = R3CMP R0, R1 ; i < N, repeatBNE LOOP

LDR R3, =WRSTR R2, [R3] ; mem[R3] = R2, minimum value

STOP B STOP

AREA RO, DATA, READONLYRD DCW 15, 10, 5, 20, 25 ; Array of numbers

AREA RW, DATA, READWRITEWR DCD 0x0 ; Result stored here

END

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7 A7 Sort

Write a program to arrange a series of 32 bit numbers in ascending/descending order.

Objective

1. Sort numbers

Design

1. Input numbers in RO memory

2. Copy them over to RW memory

3. Sort using bubble sort algorithm

4. Results are in RW memory

5. Algorithm

for (int i=0; i < n; i++){

for (int j=0; j < n-1-i; j++){

if (a[j] > a[j+1]){

int t = a[j];a[j] = a[j+1];a[j+1] = t;

}}

}

ARM Instructions

1. Directives DCW

2. LDMIA, STMIA, ADD, SUBS, CMP, BGT, BGE, LDR, STR

Procedure




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6. Sorted results aare in RW memory

; A7.s; Sort numbers (bubble sort)


N EQU 5START

LDR R0, =RDLDR R1, =WRLDMIA R0!, {R2-R6} ; copy numbers from RO areaSTMIA R1!, {R2-R6} ; store in RW area

MOV R1, #0 ; i = 0 LBMOV R7, #N ; i = n UBMOV R8, #N

LOOPILDR R0, =WRMOV R2, #0 ; j = 0 LBSUB R8, #1 ; j = n-1-i

LOOPJ^^ILDM R0, {R5, R6} ; r5,r6 = A[j], A[j+1]CMP R5, R6BLE SKIPMOV R9, R5 ; R6, R5 seq doesn't work for STMSTM R0, {R6, R9} ; A[j], A[j+1] = R6, R5 (swap)

SKIPADD R0, #4 ; j = J+1; Inner LoopADD R2, R2, #1 ; Outer loopCMP R2, R8 ; j < n-1-i, repeatBLT LOOPJ

; Outer LoopADD R1, R1, #1 ; Outer loopCMP R1, R7 ; i < n, repeatBLT LOOPI

STOP B STOP

AREA RO, DATA, READONLYRD DCD 5, 4, 3, 2, 1 ; numbers

AREA RW, DATA, READWRITEWR DCD 0, 0, 0, 0, 0 ; RW <- RD and then sort

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END

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8 A8 Count 1s

Write a program to count the number of ones and zeros in two consecutive memory locations.

Objective

1. Count 0s and 1s

2. In two consecutive locations

Design

1. Write a function to count 1s

2. Invoke function to get the count of two 1s in two locations

3. 1s are counted if there is carry on ROR

4. 0s is derved from 1s count i.e by subtracting 1s count from 64

5. Algorithm

unsigned int bitCount (unsigned int value) {unsigned int count = 0;while (value > 0) { // until all bits are zero

if ((value & 1) == 1) // check lower bitcount++;

value >>= 1; // shift bits, removing lower bit}return count;

}

ARM Instructions

1. Directives DCD

2. LDR, STR, BL, MOV, MOVS, ADDCS, SUBS, BNE, LDMFD, STMFD

3. LDMFD, where FD is Full Descending

Procedure



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5. Store the results 0s in word and 1s in another word

; A8.s; Count 0s and 1 in two consecutive locations


STARTLDR SP, =STACKLDR R2, =RDSTR R2, [SP] ; store R2 before

LDR R0, [R2] ; R0 = numberBL COUNT ; n = count(R0)MOV R3, R0 ; R3 = R0 returned by count()

LDR R0, [R2, #4] ; R0 = numnberBL COUNT ; n = count(R0)ADD R3, R3, R0 ; 1s counted to the previous count of 1sMOV R4, #64SUB R4, R4, R3 ; count 0's = 64 - count 1's

LDR R2, =WRSTM R2, {R3, R4} ; mem[2], mem[r2+4] = counts of 0's and 1's

STOP B STOPNOP

; n = count(); this function counts the 1s and returns its count

COUNTSTMFD SP!, {R2, LR} ; save registersMOV R1, #0x0 ; n = 0MOV R2, #32 ; i = 32, number of times to rotate

ONESMOVS R0, R0, ROR #1 ; rotate right R0ADDCS R1, R1, #1 ; n = n+1 if there is carrySUBS R2, R2, #1 ; i = i-1BNE ONES ; i > 32, repeat

MOV R0, R1 ; R0 = n, return valueLDMFD SP!, {R2, PC} ; unsave registers

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AREA RO, DATA, READONLYRD DCD 0x80808080, 0x40404040 ; numbers in consecutive locations

AREA ARW, DATA, READWRITEWR DCD 0x0, 0x0 ;result count of 0,s, count of 1's

DCD 0,0,0,0,0 ; stack area, decreasingSTACK

END

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9 B1 UART

Display "Hello World" message using Internal UART.

Design:

For the transmitter operation, the TX has two special registers called Transmit Holding Register(THR) and Transmit Shift Register (TSR). In order to transmit the data, it is first sent to THRand then moved to TSR.

For the receiver operation, the RX has two special registers called Receiver Buffer Register (RBR)and Receive Shift Register (RSR). When the data is received, it is first stored in the RSR andthen moved to RBR.

The baudrate is configured using UART0 Divisor Latch Registers. Choose the U0DLL andU0DLM such that the baudrate is 9600.

baudrate =PCLK

16 ∗ U0DLL

U0DLL =PCLK

16 ∗ 9600

Where PCLK is 1/4 of the CCLK (60MHz).

The format of the data transmitted or received is configured using the line control register.Configure the line with 1 stop bit, no parity and 8bit word length.

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The Line Status Register is used to check TX ready and RX ready condition. Check the THRis empty (THRE) before writing THR and data ready (DR) before reading RBR.

Hardware

• LPC 2148

• UART 16C550

• USB Cable

• Terminal/PC

Software

• Keil uVision5

• Flash Magic Tool

• Traget options: MicroLIB , HEX file, Memory layout

Procedure

1. Configure the ports P0.0 and P0.1 as Tx and Rx respectively

2. Enable DLAB in LCR to access divisor latch registers

3. Program LCR for 1 stop bit, no parity and word length 8bits

4. Program U0DLL and U0DLM for 9600 baudrate

5. Remove access to DLL by resetting DLAB

6. Transmit data if transmit hold register is empty (THRE)

Simulation

1. Create project B1/B1.c for the leagcy device LPC2148 with startup.s

2. Open output windows View->Serial Window->UART#1

3. Open Peripheral window Pheriperal->UART->UART0

4. Run to main

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//1. Display "Hello World" message using Internal UART.#include <LPC214X.h>

// PINSEL

#define TX (0b01 << 0) //bits(1:0) P0.0 as TX#define RX (0b01 << 2) //bits(3:2) P0.1 as RX

// LCR

#define DLAB (1 << 7) //bit(7)#define WL (0b11 << 0) //bits(1:0)

// LSR

#define THRE (1 << 5) //bit(5)#define DR (1 << 0) //bit(0)

void init(){

PINSEL0 = TX | RX;

U0LCR = DLAB | WL;U0DLM = 0x00;U0DLL = 15000000 / (16*9600);U0LCR = WL; // Clear DLAB

}

void tx(char ch){

while(!(U0LSR & THRE));U0THR = ch; // Transmit the char

}

int main(){

char msg[] = "Hello World\r\n";

init();for (int i=0; msg[i]; i++)

{tx(msg[i]);

}}

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10 B2 DC Motor

Interface and Control a DC Motor.

Design

A DC motor is a device that converts electrical energy (direct current system) into mechanicalenergy. It is of vital importance for the industry today.

The aim of this project is to demonstrate the DC Motor control using LPC2148. Since motorsdraw a significant amount of current, we need a separate IC called Motor Driver IC.

When the system is powered on, the status of the button is read by the MCU. As per theprogram, when the button is not pressed, the motor rotates in forward direction.

Whenever the button is pressed, a change in state at the button pin is detected by the MCUand the motor rotates in reverse directions.

Circuit Diagram

• P0.21 - PWM

• P1.30 - IN1 (driver)

• p1.31 - IN2 (driver)

• Enable 1 - Directly giving 5v

Hardware

• LPC2148

• Motor driver

• Motor

• Powersupply to the driver

Software

• Keil uVision5


• Traget options: MicroLIB , HEX file, Memory layout

Procedure

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1. EN Pin High, driver is enabled

2. Forward (general switch SW25 is ON)

• IN1 is High• IN2 is LOW

Allow motor to run for say 5s

3. Reverse (general switch S25 is OFF)

• IN1 is LOW• IN2 is HIGH

Allow motor to run for say 5sec

4. Repeat step 2 and 3

Simulation


2. Open Peripherals->GPIO slow interface->Port0

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//2. Interface and Control a DC Motor.#include <lpc214x.h>

#define IN1 (1 << 30) // P1.30#define IN2 (1 << 31) // P1.31#define SW1 (1 << 23) // P0.23 Switch

void delay(int n){

for(int i=0; i < n; i++){

for(int j=0; j < 6000; j++); /* At 60Mhz, the loop delays by 1ms */}

}

int main(){

IO1DIR &= ~SW1; //Input port from the switchIO1DIR = IN1 | IN2; //Output ports IN1 and IN2

// Clockwise or anticlockwise based SW1. SW1 is OFF in forward position,// ON if backwards.while (1)

{IO1SET = IO0PIN & SW1 ? IN1 : IN2;IO1CLR = IO0PIN & SW1 ? IN2 : IN1;

delay(1000); // stop after 5s}

}

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11 B3 Stepper Motor

Interface a Stepper motor and rotate it in clockwise and anti-clockwise direction.

Design:

Stepper motor is brushless DC motor, which can be rotated in small angles, these angles arecalled steps. We can rotate stepper motor step by step by giving digital pulses to its pins. Speedof the motor can be controlled by changing frequency of digital pulses.

There are two types of stepper motors available based on the type of stator winding: UNIPOLARand BIPOLAR. Here we are using UNIPOLAR stepper motor which is the most commonly usedstepper motor. To rotate the stepper motor we need to energise the coils of the stepper motorin a sequence. Based on the rotational operation they are classified into two modes:

1. Full Step Mode: (4-Step Sequence)

• One-Phase On Stepping (WAVE STEPPING)• Two-Phase On Stepping

2. Half Step Mode (8-Step Sequence)

Here we will use FULL STEP: ONE PHASE ON or WAVE STEPPING mode to rotate theStepper Motor.

In this method we will energise one coil (one pin of LPC2148) at a time. The sequence ofenergising the coils A, B, C and D is shown below.

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Step Coil A Coil B Coil C Coil DP0.16 P0.17 P0.18 P0.19

1 H L L L2 L H L L3 L L H L4 L L L H

Hardware

• LPC2148

• ULN2803

• STEPPER MOTOR (28BYJ-48)?

Software

• Keil uVision5


Procedure

• Connect 4 output ports p0.16-19 to the stepper motor

• Set the IOPIN0 (PORT pins P0.16-19) to HIGH for clockwise operation

• Set the IOPIN0 (PORT pins P0.19-16) to LOW for anticlockwise operation

• Set the delay 10ms (Full speed), 20ms (Gets slow), 30ms (Gets slower)

Simulation



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//3. Interface a Stepper motor and rotate it in clockwise and anti-clockwise direction.#include <lpc214x.h>

#define STEPS 556 //550 gives one complete rotation. May choose fewer steps#define DELAY 10 //10 ms, increased delay cause slowdown in speed

void delay(int n){



}

int main(){

//PINSEL1 = 0x00000000; Setting PORT0 pinsIO0DIR |= 1 << 16 | 1 << 17 | 1 << 18 | 1 << 19; //Output ports P0.16-19

while(1){

for (int i=0; i < STEPS; i++){

for(int n=16; n <= 19; n++){

IOPIN0 = 1 << n; // Set the P0.n value HIGHdelay(DELAY);

}}

for (int i=0; i < STEPS; i++){

for(int n=19; n>=16; n--){

IOPIN0 = 1 << n;delay(DELAY);

}}

}}

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12 B4 ADC

Determine Digital output for a given Analog input using Internal ADC of ARM controller.

Design

ADC is used by most of the sensors for e.g. temperature sensor, force sensor, IR sensor, poten-tiometer, and many more. All these sensors send the data to the microcontroller in analogueform, which then converts it into digital values.

ADCs in LPC2148 use Successive Approximation technique to convert analog signal into digitalform.

The ADC reference voltage is measured across GND to VREF, meaning it can do the conversionwithin this range. Usually, the VREF is connected to VDD. As LPC2148 works on 3.3 volts,this will be the ADC reference voltage.

Now the resolution of ADC = 3.3/(210) = 3.3/1024 =0.003222 = 3.2mV

ADCR A/D Control register, used for configuring the ADC

31:28 27 26:24 23:22 21 20 19:17 16 15:8 7:0Reserved EDGE START Reserved PDN Reserved CLKS BURST CLCKDIV SEL

1. Bit 7:0 - SEL : Channel SelectThese bits are used to select a particular channel for ADC conversion. One bit is allottedfor each channel. Setting the Bit-0 will make the ADC to sample AD0[0] for conversion.Similary setting bit-7 will do the conversion for AD0[7].

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2. Bit 15:8 - CLCKDIV : Clock DivisorThe APB clock (PCLKADC0) is divided by (this value plus one) to produce the clock forthe A/D converter, which should be less than or equal to 13 MHz.

1. Bit 21 - PDN : Power Down ModeSetting this bit brings ADC out of power down mode and makes it operational. Clearingthis bit will power down the ADC.

2. Bit 24:26 - STARTWhen the BURST bit is 0, these bits control whether and when an A/D conversion isstarted:000 - Conversion Stopped001- Start Conversion Now

ADGDR

31 30 26:24 23:16 15:6 5:0DONE OVERRUN CHN Reserved RESULT Reserved

1. Bit 15:6 - RESULTThis field contains the 10bit A/D conversion value for the selected channel in ADCR.SELThe value from this register should be read only after the conversion is complete ie DONEbit is set.

2. Bit 26:24 - CHN : ChannelThese bits contain the channel number for which the A/D conversion is done and theconverted value is available in RESULT bits(e.g. 000 identifies channel 0, 011 channel3. . . ).

3. Bit 30 - OVERRUNThis bit is set during the BURST mode where the previous conversion data is overwrittenby the new A/D conversion value.

4. Bit 31 - DONEThis bit is set to 1 when an A/D conversion completes. It is cleared when this register isread and when the ADCR is written. If the ADCR is written while a conversion is still inprogress, this bit is set and a new conversion is started.

Hardware

• LPC2148

• ADC (builtin)

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• LCD (2 x 16 Character LCD Display) P0.10-13

Software

• Keil uVision5


Procedure

1. Configure the port P0.30 to function as AD0.3 (bit 2*n-32) PINSEL1 register.

2. Enable the CLock to ADC module.

3. Deselect all the channels and Power on the internal ADC module by setting ADCR.PDNbit.

4. Select the Particular channel for A/D conversion by setting the corresponding bits inADCR.SEL

5. Set the ADCR.START bit for starting the A/D conversion for selected channel.

6. Wait for the conversion to complete, ADGR.DONE bit will be set once conversion is over.

7. Read the 10-bit A/D value from ADGR.RESULT.

8. Use it for further processing or just display on LCD.

Simulation



3. Open Peripherals->A/D converter-> A/D converter 0

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//4. Determine Digital output for a given Analog input using Internal ADC of ARM controller.#include <lpc214x.h>#include <stdio.h>#include <string.h>

extern void initLcd(void);extern void lcds(char s[]);extern void delay(int n);

//PINSEL1#define ADC3 (0b01 << 28) // 29:28 to AD0.3 (fn=01). P0.30 hint 2n-32

// AD0CR// | 31:28 | 27 | 26:24 | 23:22 | 21 | 20 | 19:17 | 16 | 15:8 | 7:0 |// | Reserved | EDGE | START | Reserved | PDN | Reserved | CLKS | BURST | CLCKDIV | SEL |

#define CHNSEL (1 << 3) // Channel 7:0 i.e 7 6 5 3 2 1 0#define CLKDIV (3 << 8) // Divide the clock by n = 3MHz (<= 13 MHz)#define PDN (1 << 21) // PDN is 1 i.e operational#define BURST (1 << 16)#define START (000 << 24) // Start conversion

// AD0DR3// | 31 | 30 | 26:24 | 23:16 | 15:6 | 5:0 |// | DONE | OVERRUN | CHN | Reserved | RESULT | Reserved |

#define DONE (1 << 31) // set bit 31

void initAdc(){

PINSEL1 = ADC3;AD0CR = PDN | CLKDIV | CHNSEL | BURST | START; //0x00210308

}

int readAdc(){

AD0CR |= START;while (!(AD0DR3 & DONE)); /* Wait till conversion DONE */

return ((AD0DR3 >> 6) & 0x3FF); // 10bit ADC}

int main(void){

int rv;float voltage;char s[10];

initAdc();

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initLcd();lcds("ADC...");while(1)

{rv = readAdc();voltage = (rv/1023.0)*3.3; // Convert ADC value to equivalent voltagesprintf(s, "%.2f", voltage);lcds(s);delay(1000);

}}

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13 B5 DAC

Interface a DAC and generate Triangular and Square waveforms.

Design

Digital to Analog Converter (DAC) are mostly used to generate analog signals (e.g. sine wave,triangular wave etc.) from digital values.

LPC2148 has 10 bit DAC and offers changeable analog o/p. LPC2148 has Analog output pin(AOUT) on chip, where we can get digital value in the form of Analog output voltage.

The Analog voltage on AOUT pin is calculated as ((VALUE/1024) * VREF). Hence, we canchange voltage 0 to 3.3V by changing 10-bit digital value field in DACR (DAC Register).

DACR

Reserved Bias Value Reserved31:17 16 15:6 5:0

1. Bit 5:0 - RESERVED

2. Bits 15:6 - VALUE This field contains the 10-bit digital value that is to be converted in toAnalog voltage. We can get Analog output voltage on AOUT pin and it is calculated withthe formula (VALUE/1024) * VREF.

3. Bit 16 - BIAS 0 = Maximum settling time of 1ţsec and maximum current is 700µA1 = Settling time of 2.5ţsec and maximum current is 350Note that, the settling times are valid for a capacitance loada on the AOUT pin notexceeding 100 pF. A load impedance value greater than that value will cause settling timelonger than the specified time.

4. Bit 31:17 - RESERVED

Hardware

• LPC2148

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• CRO

Software

• Keil uVision5


Procedure

1. First, configure P0.25/AOUT pin as DAC output using PINSEL Register.

2. Then set settling time using BIAS bit in DACR Register.

3. Now write 10-bit value (which we want to convert into analog form) in VALUE field ofDACR Register.

Simulation


2. Open Peripherals->D/A converter

3. Setup LA to monitor amplitude of AOUT

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//5. Interface a DAC and generate Triangular and Square waveforms.#include <LPC214x.h>

//PINSEL1#define AOUT (0b10 << 18) // P0.25 as AOUT (fn=10) is selected using PINSEL1 19:18. P0.25 => 2*25+1 - 32 = 19

// DACR// | Reserved | Bias | Value | Reserved |// | 31:17 | 16 | 15:6 | 5:0 |

#define BIAS (1 << 16) // settling time#define VALUE(d) ( d << 6) // 10 bit data

void delay(int n){



}

void triangle(){

int d;

d = 0;while (d != 1023)

{DACR = VALUE(d) & 0x0001FFC0;d++;//delay(1);

}

while (d != 0){

DACR = VALUE(d) & 0x0001FFC0;d--;//delay(1);

}}

void square(){

int d;

// Square wave

d = 1023;DACR = VALUE(d) & 0x0001FFC0;

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delay(1);

d = 0;DACR = VALUE(d) & 0x0001FFC0;delay(1);

}

int main(){

PINSEL1 = AOUT;

while(1){

for (int i=0; i <5; i++)//triangle();

for (int i=0; i <5; i++)square();

}}

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14 B6 Keypad

Interface a 4x4 keyboard and display the key code on an LCD.

Design:

At the lowest level, keyboards are organized in a matrix of rows and columns. The CPU accessesboth rows and columns through ports. When a key is pressed, a row and a column make acontact. otherwise, there is no connection between rows and columns. So this is the logic we aregoing to use.

The status of each keys can be determined by a process called Scanning. The column pins (Col1- Col4) are connected to the input pins and all the row pins are connected to the output pins ofthe microcontroller.

Now we can read the status of each switch through scanning.

A logic LOW is given to Row1 and others (Row2 - Row-4) HIGH. Now each Column is scanned.If any switch belongs to 1st row is pressed corresponding column will pulled down (logic LOW)and we can detect the pressed key. This process is repeated for all rows.

Hardware

• LPC2148

• LCD

Software

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• Keil uVision5


Procedure

1. Set port directions. Input ports P0.16-19, Output ports P0.20-23

2. Configure LCD for 4-bit mode and turn on using the commands 0x28 and 0x0f

3. Scan columns for each row set to LOW, sequentially.

4. keypad[row][col] is pressed if col output is LOW

5. Display the key on LCD

Simulation


2. Add scatter file for LCD. Gto Options and press Linker. Select scatter scripting file.

3. Open Peripherals->GPIO slow interface

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//6. Interface a 4x4 keyboard and display the key code on an LCD.

#include <lpc21xx.h>

extern void initLcd(void);extern void lcds(char s[]);extern void lcd(int d, int c);extern void delay(int n);

#define COL1 (1 << 16)#define COL2 (1 << 17)#define COL3 (1 << 18)#define COL4 (1 << 19)

#define ROW1 (1 << 20)#define ROW2 (1 << 21)#define ROW3 (1 << 22)#define ROW4 (1 << 23)

char keypad[] = // keypad[20:23, 16:19]{'0', '4', '8', 'C','1', '5', '9', 'D','2', '6', 'A', 'E','3', '7', 'B', 'F'

};

char keypadscan(void) //Keypad Scan{

int key;

IOSET1 = COL1 | COL2 | COL3 | COL4; //Set the cols to '1'IOSET1 &= ~(ROW1 | ROW2 | ROW3 | ROW4);

key = 0;for (int col=16; col <= 19; col++)

{IOCLR1 = (1 << col); //Set this col to LOW while other cols at '1'

for (int row=20; row <= 23; row++){

if(!(IOPIN1 & (1 << row))) //Scan for key press{

while(!(IOPIN1 & (1 << row)));return keypad[key];

}key++;

}IOSET1 = (1 << col); //Set this col to HIGH again, before trying next column

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}return('X');

}

void initKeypad(){

IO1DIR = COL1 | COL2 | COL3 | COL4; // Output columns}

int main(void){

char c;char s[2] = {'x', '\0'};

initKeypad();initLcd();lcds("Keypad...");

while(1){

c = keypadscan(); //Obtaining values from keypadif (c == 'X') continue;s[0] = c;lcds(s);

}}

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15 B7 EINT

Demonstrate the use of an external interrupt to toggle an LED On/Off.

Design

Interrupt caused by an external source such as external switch, sensor or monitoring device.These interrupt are special events that require immediate attention.

When an IRQ arrives to the CPU, it stops executing the current code and start executing theISR. After the ISR execution has finished the CPU gets back to where it had stopped. Interruptsin LPC214x are handled by Vectored Interrupt Controller (VIC).

VIC has 16 VIRQ slots, Slot-0 to Slot-15. Any IRQ configured interrupts can be assigned to anyslot. Priorities are in the order of slot number. Slot-0 has highest priority than slot-15. VICregisters are:

Hardware

1. LPC2148

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2. Switch

3. LED

Software

• Keil uVision5


Procedure

1. Configure Pin Function P0.15 as EINT2, received from switch

2. Configure P1.16 as GPIO connected to LED

3. Write the ISR(), don’t forget to clear EINT2 flag at the end

4. Select the interrupt, slot and enable it for EINT2

5. Choose Signal mode and Polarity

6. Clear EINT Flag

Simulation



3. Set break point in ISR

4. Set P0.15 to trigger the interrupt

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//7. Demonstrate the use of an external interrupt to toggle an LED On/Off.#include <LPC214x.h>#include <stdio.h>

#define EINT2 (0b10 << 30) //Configure P0.15 to receive EINT2 (0b10)#define IRQ 16 //Interrupt vector number#define ENVEC (1 << 5) //Enabale interrupt vector#define MODE 0b100 // Edge sensitive EINT2 (EINT2.1.0)#define DI 0b100 // EINT2

void delay(int n){



}

void togleLed(int n){

for (int i=0; i < n; i++){

IO1SET = 0x00010000; // Turn ONdelay(100);IO1CLR = 0x00010000; // Turn OFFdelay(100);

}}

void initLed(){

IO1DIR = 1 << 16; // Select P1.16delay(10);IO1CLR = 1 << 16; // Clr P1.16

}

void isr() __irq{

togleLed(10);EXTINT |= DI; // Clear interrupt flag by writing 0b100 i.e EINT2VICVectAddr=0; // Finished ISR, dummy write

}

void initIsr(){

PINSEL0 = EINT2; //0x80000000; warnings 16 << 30

EXTMODE = MODE; // Edge sensitive mode on EINT2EXTPOLAR = 0; // Falling edge sensitive

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VICIntSelect = 0 << IRQ; // default is 0 anywayVICVectAddr0 = (unsigned long)isr; //Address of isr()VICVectCntl0 = ENVEC | IRQ; //Assign INTVEC to slot0 and enable itVICIntEnable = 1 << IRQ; // Enable ENT2 IRQ

}

int main(){

initIsr();initLed();

while(1){}

}

Before you start building the program in KEIL goto the Target options and in theLinker tab check: "Use Memory Layout from the Target Dialog".

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16 B8 7SD

Display the Hex digits 0 to F on a 7-segment LED interface, with an appropriate delay in between.

Design

7 segment displays are among the simplest display units to display the numbers and characters.It is generally used to display numbers and has brighter illumination and simpler constructionthan dot matrix display. And because of brighter illumination, the output can be viewed fromlarger distance than LCD.

It consists of 8 LEDs, each LED used to illuminate one segment of unit and the 8th LED usedto illuminate DOT in 7 segment display. 8th LED is used when two or more 7-segment modulesare used, for example to display (0.1).

The 7-segment display has connections for 8 LEDs. We set logic HIGH to illuminate the partic-ular and set LOW to turn OFF LED.

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Below table shows the HEX values and corresponding digit according to LPC2148 pins for com-mon cathode configuration.

8 7 6 5 4 3 2 (P0.8:P0.2);x=a b c d e f g0 0 0 0 0 0 0 11 1 0 0 1 1 1 12 0 0 1 0 0 1 03 0 0 0 0 1 1 0 1 = off4 1 0 0 1 1 0 05 0 1 0 0 1 0 0 0 = on6 0 1 0 0 0 0 07 0 0 0 1 1 1 18 0 0 0 0 0 0 09 0 0 0 0 1 0 0A 0 0 0 1 0 0 0B 1 1 0 0 0 0 0C 0 1 1 0 0 0 1D 1 0 0 0 0 1 0E 0 1 1 0 0 0 0F 0 1 1 1 0 0 0

hex value = 0b11 « 9 | x « 2

Hardware

1. LPC2148

2. SSD

Software

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• Keil uVision5


Procedure

1. Generate HEX digits 0-F

2. Set port pins to the digit pattern

3. Dely 1s

4. Clear port pins

Simulation



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//8. Display the Hex digits 0 to F on a 7-segment LED interface, with an//appropriate delay in between.

#include <lpc214x.h>

void delay(int n){



}

// using (P0.2, P0.3, P0.4, P0.5, P0.6, P0.7, P0.8, and P0.9, P).10)int a[]=

{/* 1 = off 0 = on *//* 8765432 ports *//*x abcdefg display*//*0*/ 0b01 << 9 | 0b0000001 << 2,/*1*/ 0b01 << 9 | 0b1001111 << 2,/*2*/ 0b01 << 9 | 0b0010010 << 2,/*3*/ 0b01 << 9 | 0b0000110 << 2,/*4*/ 0b01 << 9 | 0b1001100 << 2,/*5*/ 0b01 << 9 | 0b0100100 << 2,/*6*/ 0b01 << 9 | 0b0100000 << 2,/*7*/ 0b01 << 9 | 0b0001111 << 2,/*8*/ 0b01 << 9 | 0b0000000 << 2,/*9*/ 0b01 << 9 | 0b0000100 << 2,/*A*/ 0b01 << 9 | 0b0001000 << 2,/*B*/ 0b01 << 9 | 0b1100000 << 2,/*C*/ 0b01 << 9 | 0b0110001 << 2,/*D*/ 0b01 << 9 | 0b1000010 << 2,/*E*/ 0b01 << 9 | 0b0110000 << 2,/*F*/ 0b01 << 9 | 0b0111000 << 2

};

int main(){

IO0DIR = 0b01 << 9 | 1<<8 | 1<<7 | 1<<6 | 1<<5 | 1<<4 | 1<<3 | 1<<2;

while(1){

for(int i=0; i < 16; i++){

IO0CLR = 0x00000FFF;IO0SET = a[i];delay(1000);

}

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}}

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17 BU LCD

LCD display is common to ADC and Keypad experiments. The code is included into theseprojects.

P0.13 DB7 Data bit7P0.12 DB6 :P0.11 DB5P0.10 DB4P0.09 DB3P0.08 DB2P0.07 DB1 :P0.06 DB0 Data bit0P0.28 E EN Enable. Falling edge triggerP0.29 R/W RW Read Write mode; 0-write to LCD, 1-read from LCDP0.22 RS RS Register select; 0-command register, 1-data register

LCD commands

No. Hex Command to LCD instruction Register1 01 Clear display screen

11 0F Display on, cursor blinking16 80 Force cursor to beginning ( 1st line)16 86 Force cursor to 6th position of first row17 C0 Force cursor to beginning ( 2nd line)19 38 2 lines and 57 matrix (8-bit mode)

Hardware

1. LPC2148

2. LCD

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Software

• Keil uVision5


Procedure

1. Initialize LCD for 4-bit mode, display on, cursor to home and blinking

2. Write IOPIN0 (c | 0xFFFFFF0F) « 6; IOCLR0=RS if c is a command

3. Delay 100ms before characters

Simulation

1. Create project BU/BU.c for the leagcy device LPC2148 with startup.s

2. Include main code for simulation

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// LCD#include <lpc214x.h>#define W4 10 //13:10#define W8 6 //13:6#define RS (1 << 22) //P0.22 -> Register Select. 0:Command, 1: Data#define RW (1 << 29) //P0.29 -> Read/Write, R/W=0: Write & R/W=1: Read#define EN (1 << 28) //Enable. Falling edge trigger#define DATA (0xff << 6) //Data P0.06:13 -> DB0:DB7

void delay(int n){



}

void wd(int x){

IOCLR0 = RW;if (x==0) IOCLR0 = RS;IOSET0 = EN;delay(100);IOCLR0 = EN;

}

void lcd(char d, int x){

int n;

n = d | 0xFFFFFF0F;IOPIN0 = n<<6;wd(x);

d = d << 4;n = d | 0xFFFFFF0F;^^IIOPIN0 = n<<6;wd(x);

}

void lcds(char s[]){

lcd(0x01,0); // clear the display before printingfor (int i=0; s[i]; i++)

{lcd(s[i],1);

}}

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void initLcd(void){

IODIR0 = EN | RW | RS | DATA ; // output ports for control

lcd(0x28,0); // 4-bit modelcd(0x0F,0); // Display on, cursor blinkinglcd(0x80,0); // Cursor to begininglcd(0x01,0); // Clear display

}

int main(){

int n = 0;char s[10];

initLcd();lcds("Hello...");delay(1000);while(1)

{sprintf(s, "%d", n);n = n%10;lcds(s);delay(10);n++;

}}

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18 LPC2148

18.1 ARM IDE

[

]]

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18.2 ARM development board

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18.3 ARM Microcontroller

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18.4 GPIO

Port number ’n’ cooresponding to,

• PINSEL0[2n+1:2n] if n <= 15

• PINSEL1[2n-32+1:2n-32] if n > 15

Lab Port DIR PINSEL0 00 01 10 11UART P0.0 -> 1:0 TX

P0.1 3:2 RXDC P1.30 ->

P1.31 ->P1.23 <-

SM P0.16 ->P0.17 ->P0.18 ->P0.19 ->

ADC P0.30 29:28 AD0.3DAC P0.25 19:18 AOUTKPAD P0.20 ->

P0.21 ->P0.22 ->P0.23 ->P0.16 <-P0.17 <-P0.18 <-P0.19 <-

EINT P0.15 31:30 EIN2P1.16} ->

SSD P0.2 ->P0.3 ->P0.4 ->P0.5 ->P0.6 ->P0.7 ->P0.8 ->P0.9 ->

LCD P0.22 ->P0.28 ->P0.29 ->P0.6 ->:P0.13 ->

SW p0.23 <-

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18.5 PINSEL

1. PINSEL0

• Configure ports P0.0 to P0.30• Bits 2n+1:2n, where n is the port number

2. PINSEL1

• Configure ports beyond P0.30• Bits 2n+1-32:2n-32, where n is the port number

3. GPIO registers

• IOxPIN read the logic value on a I/O pin• IOxSET set the logic value on a I/O pin to HIGH• IOxCLR set the logic value on a I/O pin to LOW• IOxDIR select the input/output function (0 is output)

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18.6 UART Registers

1. U0DLM, U0DLMDivisor latch registers DLM and DLL are the upper and lower bytes of the word register.

• U0DLM = 0• U0DLL = pclk

16∗baudrate

2. U0LCR Line control register

|------+-----+--------+-----------+-------------|| 7 | 6:4 | 3 | 2 | 1:0 ||------+-----+--------+-----------+-------------|| DLAB | | Parity | Stop bits | Word length |

U0LCR(DLAB=1«7, 1 stop bit=0, no parity=0, 8bit WL=0b11)

3. U0LSR Line status register

|-----+------+-----+----|| 7:6 | 5 | 4:1 | 0 ||-----+------+-----+----|| | THRE | | DR |

U0LSR(1«5)

4. U0THR Transmit Holding RegisterU0THR = data

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18.7 ADC

1. AD0CR A/D Control register

|----------+------+-------+----------+-----+----------+-------+-------+---------+-----|| 31:28 | 27 | 26:24 | 23:22 | 21 | 20 | 19:17 | 16 | 15:8 | 7:0 ||----------+------+-------+----------+-----+----------+-------+-------+---------+-----|| Reserved | EDGE | START | Reserved | PDN | Reserved | CLKS | BURST | CLCKDIV | SEL |

• SEL = 100, select 3rd channerl• CLKDIV = 3MHz• PDN = 1, power down mode 1 is operations• BURST = 1, ADC does repeated conversions at the rate selected by the CLKS field

for the analog inputs selected by SEL field• START = 000 start conversion

2. AD0DR3

|------+---------+-------+----------+--------+----------|| 31 | 30 | 26:24 | 23:16 | 15:6 | 5:0 ||------+---------+-------+----------+--------+----------|| DONE | OVERRUN | CHN | Reserved | RESULT | Reserved |

• DONE = 1, data is available• CHN = 000, ADC channel (should be 011?)• RESULT is ADC data

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18.8 DAC

1. DACR D/A control register

|----------+------+-------+----------|| 31:17 | 16 | 15:6 | 5:0 ||----------+------+-------+----------|| Reserved | Bias | Value | Reserved |

• BIAS=1, setling time is 2.5 µS• Value = DA value

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18.9 EINT

1. VICIntSelect

|--------+----+----+-----+-------+----+-----+---|| Bit | 31 | 30 | ... | 16 | 15 | ... | 0 ||--------+----+----+-----+-------+----+-----+---|| Symbol | | | | EINT2 | | | |

• EINT2=0 IRQ (default)

2. VICIntEnable

|--------+----+----+-----+-------+----+-----+---|| Bit | 31 | 30 | ... | 16 | 15 | ... | 0 ||--------+----+----+-----+-------+----+-----+---|| Symbol | | | | EINT2 | | | |

• EINT2=1, Enable interrupt• eg. EINT2 has 1 « 16

3. VICVectCntl0-15Vector Control registers 0-15. Each of these registers controls one of the 16 vectored IRQslots. Slot 0 has the highest priority and slot 15 the lowest.

|----------+------------------+------------|| 31:6 | 5 | 4:0 ||----------+------------------+------------|| Reserved | IRQ slot enabled | IRQ number |

• IRQ slot enabled = 1 e.g 1«5• IRQ number=16

4. VICVectAddrx

|--------------------|| 31:0 ||--------------------|| IRQ vector address |

• IRQ vector address = isr()

5. EXTINT Register

|-----+-------+-------+-------+-------|| 8:4 | 3 | 2 | 1 | 0 ||-----+-------+-------+-------+-------|| | EINT3 | EINT2 | EINT1 | EINT0 |

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Clear EINTx after the ISR.

• EINT1=1, clear EINT1 e.g EXTINT=2

6. EXTMODE External Interrupt Mode register

|----------+------------|| 7:4 | 3:0 ||----------+------------|| Reserved | EXTMODE0-3 |

• EXTMODE1 = 0 e.g EXTMODE = 4 for edge sensitive

7. EXTPOLAR External Interrupt Polarity register

|----------+-------------|| 7:4 | 3:0 ||----------+-------------|| Reserved | EXTPOLAR0-3 |

• EINT1 = 0 e.g EXTPOLAR=0 falling edge sensitive

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18.10 LCD commands

No. Hex Command to LCD instruction Register1 01 Clear display screen2 02 Return home3 04 Decrement cursor (shift cursor to left)4 06 Increment cursor (shift cursor to right)5 05 Shift display right6 07 Shift display left7 08 Display off, cursor off8 0A Display off, cursor on9 0C Display on, cursor off

10 0E Display on, cursor blinking11 0F Display on, cursor blinking12 10 Shift cursor position to left13 14 Shift cursor position to right14 18 Shift the entire display to the left15 1C Shift the entire display to the right16 80 Force cursor to beginning ( 1st line)17 C0 Force cursor to beginning ( 2nd line)18 28 2 lines and 57 matrix (4-bit mode)19 38 2 lines and 57 matrix (8-bit mode)

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Documents

MICROCONTROLLER AND EMBEDDED SYSTEMS ...Create project A7/A7.s for the leagcy device LPC2148 w/o startup.s 2. Translate A7.s and build 3. View RO data in memory window 1 at 0x0000