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Fast external interrupt on XC164

Daniel L

I want to setup the fast external interrupt 0 (EXI0)on the Infineon XC16x board (pin P1H.0). My functions look like this:

void EXI0_Int (void) interrupt 24
{
  // Interrupt routine
}

void EXI0_Init (void)
{

    EXICON |= 0x02; // Falling edge EXI0
    CC1_CC8IC = CALC_IC (0, 1, 13, 0);
    // ILVL = 13, Interrupt enable
    PSW_IEN = 1;
}
The interrupt doesn't work. Could anyone help me with the right settings?

Parents Reply Children
  • 0 in reply to Siegfried Seebacher

    You can use the code generated by DAVE (command 0 to 4 as described in the SU user's manual) to get write access to the EXICON register after the EINIT (end of initialization) instruction has been executed:

    
// Unlock procedure
void MAIN_vUnlockProtecReg(void)
{
 unsigned char ubPASSWORD;

//if low protected mode
 if((SCUSLS & 0x1800) == 0x0800)
 {
   ubPASSWORD = SCUSLS & 0x00FF;
   ubPASSWORD = ~ubPASSWORD;
   SCUSLC = 0x8E00 | ubPASSWORD; //command 4
 }

 //if write protected mode
 if((SCUSLS & 0x1800) == 0x1800)
 {
    SCUSLC = 0xAAAA;    //command0
    SCUSLC = 0x5554;    //command1

    ubPASSWORD = SCUSLS & 0x00FF;
    ubPASSWORD = ~ubPASSWORD;

    SCUSLC = 0x9600 | ubPASSWORD; //command 2
    SCUSLC = 0x0800; //command 3
    ubPASSWORD = SCUSLS & 0x00FF;
    ubPASSWORD = ~ubPASSWORD;
    SCUSLC = 0x8E00 | ubPASSWORD; //command 4
 }

}

  • 0 in reply to Daniel L

    Thanks a lot! This was the problem! After I used the unprotected algorithm it works perfectly!