This is for an automotive project. The hardware is an arduino uno. I am interfacing it with a TI TPIC8101 IC, also a secondary uno is doing the SPI programming and interpreting the TPIC8101 output (irrelevant to my issue).
I have a window as small as .278ms and as large as 2ms. The window is between two interrupts. Basically the first (interrupt 0) is increasing a count by one and the other (interrupt 1) resets the count to 0. The window between the interrupts is not adjustable but is directly related to engine speed (up to 9k rpm). The input to the interrupts are TTL square waves.
I can post a screenshot from an oscilloscope if necessary.
edit: Attached oscilloscope screenshot. Yellow is reset interrupt and blue is the counter.
I am consistently missing the "reset" interrupt.
I have found lots of information about optimizing the speed of arduino code. However, my lack of programming knowledge has left me baffled when trying to apply it.
I'm a second semester electrical/computer engineering student that tinkers between classes and work. I am so close to finishing this but this has completely stalled my progress. Any help on this would be appreciated!
Edit: Updated code 12:07 3/16/14
#define intHold 7
#define waitPin 8
volatile int c = 0;
volatile int r = 0;
void setup()
{
pinMode(intHold, OUTPUT); //intHold output to TPIC8101 on pin 6
digitalWrite(intHold, LOW);
attachInterrupt(0, count, RISING); //24 tooth interrupt on pin 2 calls to function count below
attachInterrupt(1, reSet, RISING); //1 tooth "reset" interrupt on pin 3 calls to function reSet below
Serial.begin(9600);
Serial.println("Setup"); //verify setup completed
}
void loop()
{
//Listen For Knock between teeth four and six
//Holds intHold high between teeth four and six
if(c == 4)
{
digitalWrite(intHold, HIGH);
Serial.println("tooth 4");
}
if(c == 6)
{
digitalWrite(intHold, LOW);
Serial.println("tooth 6");
}
//Listen For Knock Between Teeth 10 and 12
//Holds intHold high between teeth 10 and 12
if(c == 10)
{
digitalWrite(intHold, HIGH);
Serial.println("tooth 10");
}
if(c == 12)
{
digitalWrite(intHold, LOW);
Serial.println("tooth 12");
}
//Verify resets are working
if(r == 1)
{
Serial.println("reset");
r = 0;
}
}
//Increase tooth count by one
void count()
{
c++;
}
//Reset tooth count
void reSet()
{
c = 0;
r = 1;
}
cjdelphi:
I'll take a stab in the dark (I'll wait for mike if I'm wrong lol)
But... I think the voltage is not high enough or is just on the borderline to trigger a rising edge.
If that's correct, you may need to boost the voltage signal.
That's my fault for not setting the V/div to the same for both channels. The yellow channel is on 2v/div. 2.5ish div = 5 volts. So the reset interrupts is right at 5 volts but the counter interrupt is 3.5ish volts. I believe the minimum for a TTL high is 2 volts.
Rather than wasting your time and ours barking up the wrong tree I suggest we all find the tree that actually contains the critter of interest. In other words, do what you should have done in your first post: describe what you want your code to do.
Most likely do to my lack of programming knowledge, your last post is difficult to follow. I'm going to assume when you refer to cjdelphi's post you are talking about the serial.println?
As to what I want the code to do, count the number of pulses at interrupt zero, hold a pin high between teeth (indicated by the count) 4 and 6 and then go low at 6. The same between teeth 10 and 12. Interrupt one needs to reset the count.
Interrupt zero is from a 12 toothed wheel and interrupt one is from a one tooth wheel. Both wheels are turning at the same speed on the same shaft.
The reset interrupt is supposed to set the count so that c==1 correlates to tooth one. Without this reset I have no idea what tooth just caused the interrupt and what the current degree position is.
