girish09
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I am using ARM LPC 2294, i have interfaced LCD and keypad for keypad interfacing i have used interrupt method (because my schematic is like that so i have to write code like that only).......whenever i power on the circuit welcome message is displaying on LCD properly but when press the key of keypad it is showing nothing on LCD but after that if i RESET the micro-controller then it is again showing welcome message properly but not responding to interrupt properly.......what should i supposed to do???? .........I am sure my hardware is ok ...Is any term i am missing to activate interrupt?? plz help me......
HTML:
void initInterrupt()
{
PCB_PINSEL1 = PCB_PINSEL1 | 0x00000001; /// ///////////EINT0 = Pin P0.16////////
EXTMODE = 0x00000000; // when EXTMODE= 0 means it is level sensitive n if EXTMODE=1 then EINT0 is edge sensitive
EXTPOLAR = 0x0000000; //polarity register level or edge
VICIntSelect = (0<<14); // VICIntSelect register pin no 14 (EINT0)=0 to set IRQ register if 1 then FIQ will set
VICVectCntl0 = 0x0000002E ;
VICVectAddr0 = (unsigned)EXTINTVectoredIRQ;
VICIntEnable = VICIntEnable | 0x00004000; //Enable EINT0
}