6602: Rising edge capturing hardware interrupt

Question

I have this signal in input:
I have to do something on rising edge as soon as possible.
My idea is to receive an hw interrupt from the board, and to write an isr soubroutine.
I read about callback function so i write some code.
But when i call:
    DAQmxErrChk ( DAQmxCfgChangeDetectionTiming ( udpTaskHandle,
                                                  "Dev1/Ctr0",
                                                  "Dev1/Ctr0",
                                                  DAQmx_Val_ContSamps,
                                                  1));
I have an error for the ctr0, it seems i can use onli the port0.
Now my question.... is there on port0 a software thread that read continuosly the value of the pins?
Or is there an hardware interrupt anyway ?
I think that a software thread has a big latency for me, at this point don't need the 6602 board,
i could read  the signal on LPT standard port.
Thank you for the answer.
There are 10 kinds of people. Those who understand binary notation, and those who do not.
Solved!
Go to Solution.

Answer 1

Sorry if it's passed a long time, i tryed to do as you suggested, but if i set  DAQmx_Val_HWTimedSinglePoint how can i have a continuos acquisition?
that's my code, it seems that thecallback function is called only once.
#include <stdio.h>
#include <NIDAQmx.h>
#define DAQmxErrChk(functionCall) if( DAQmxFailed(error=(functionCall)) ) goto Error; else
int32 CVICALLBACK TriggerDetectionCallback(TaskHandle taskHandle, int32 signalID, void *callbackData);
void Cleanup (void);
static TaskHandle  taskHandle=0;
int main(void)
    int         error=0;
    int32       read;
    uInt32      data[1000];
    char        errBuff[2048]={'\0'};
    // DAQmx Configure Code
    DAQmxErrChk (DAQmxCreateTask("",&taskHandle));
    DAQmxErrChk (DAQmxCreateCICountEdgesChan(taskHandle,
                                             "Dev1/ctr1",
                                             DAQmx_Val_Rising,
                                             0,
                                             DAQmx_Val_CountUp));
    DAQmxErrChk (DAQmxCfgSampClkTiming(taskHandle,
                                       "/Dev1/PFI35",
                                       1000.0,
                                       DAQmx_Val_Rising,
                                       DAQmx_Val_HWTimedSinglePoint ,
                                       1000));
    DAQmxErrChk (DAQmxRegisterSignalEvent(taskHandle,
                                          DAQmx_Val_SampleClock ,
                                          0,
                                          TriggerDetectionCallback,
                                          NULL)
    // DAQmx Start Code
    DAQmxErrChk (DAQmxStartTask(taskHandle));
    printf("Continuously reading. Press Ctrl+C to interrupt\n");
    while( 1 ) {
Error:
    puts("");
    if( DAQmxFailed(error) )
        DAQmxGetExtendedErrorInfo(errBuff,2048);
    if( taskHandle!=0 ) {
        // DAQmx Stop Code
        DAQmxStopTask(taskHandle);
        DAQmxClearTask(taskHandle);
    if( DAQmxFailed(error) )
        printf("DAQmx Error: %s\n",errBuff);
    printf("End of program, press Enter key to quit\n");
    getchar();
    return 0;
int32 CVICALLBACK TriggerDetectionCallback(TaskHandle taskHandle, int32 signalID, void *callbackData)
    int32   error=0;
    uInt32  data[200]={0};
    int32   numRead;
    uInt32  i=0;
    char    errBuff[2048]={'\0'};
    static int count =0;
    count++;
    printf("\n\n%d", count);
    if( taskHandle ) {
        // DAQmx Read Code
        DAQmxErrChk (DAQmxReadCounterU32(taskHandle,1000,10.0,data,1000,&numRead,NULL));
        printf("\n%d",data);
    return 0;
Error:
    if( DAQmxFailed(error) )
        DAQmxGetExtendedErrorInfo(errBuff,2048);
        Cleanup();
        printf("DAQmx Error: %s\n",errBuff);
    return 0;
void Cleanup (void)
    if( taskHandle!=0 )
        // DAQmx Stop Code
        DAQmxStopTask(taskHandle);
        DAQmxClearTask(taskHandle);
        taskHandle = 0;
Message Edited by blacksocket on 30-03-2010 03:56 PM
There are 10 kinds of people. Those who understand binary notation, and those who do not.