Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Code for UART with interrupt for LPC2138

Status
Not open for further replies.

embedtronics

Newbie level 4
Joined
Mar 24, 2013
Messages
5
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,313
Hello everyone,

I need a code for LPC2138 to run UART with interrupt. I have pasted my code below. I am switching on an LED inside the interrupt if '1' is received. But the led constantly stays on and that too its dim with a voltage of 3.3V across it. Please help!!


Code C - [expand]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
#include<lpc21xx.h>
__irq void UART0_ISR(void);
int main()
{
    PINSEL0 = 0x05;
    U0LCR = 0x83;
    U0DLL = 124;
    U0LCR = 0x03;
    U0IER = 0x01;
    IODIR0 = 0x00000010;
    IOCLR0 = 0x00000010;
    VICVectAddr0 = (unsigned int)UART0_ISR;
    VICVectCntl0 =  0x20|6;
    VICIntEnable = 0x00000040;
    while(1)
    {
    }
}
__irq void UART0_ISR(void)
{
    if((U0IIR & 0x04)==0x04)
    {
        if(U0RBR == '1')
            IOSET0 = 0x00000010;
        else
        IOCLR0 = 0x00000010;
    }
    VICVectAddr = 0;
}

 
Last edited by a moderator:

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top