Code for UART with interrupt for LPC2138

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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]
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#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;
}
 
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