Input interrupt pin associated with TIMER1 on ATmega168

atmegainterruptstimer

A few days ago I came across the source-code for this ATmega168 based Inductance Meter from Dan's Workshop Blog. Below is the code that I believe sets up an ATmega168 interrupt to measure an external frequency:

// set up the 16 bit timer as an external frequency counter:
TCCR1B |= (1 << CS10)|(1 << CS11)|(1 << CS12); // External clock, rising edge
TIMSK1 |= (1 << TOIE1); // Enable overflow interrupt, 
                        // it will overflow a few times in counting frequency

I'm still studying how to set ATmegas interrupts, so I don't quite understand what that code does.

My questions are:

Does the code above determine which pin will provide the frequency to be measured? How?
Would that same code define the same input interrupt pin on an ATmega328?

From the schematic posted on the site (copied below) I can see that the pin whose frequency is counted/measured is PD5 (pin 11 on the IC). Is that correct?

_{(source: dansworkshop.com)}

Best Answer

Does the code above determine which pin will provide the frequency to be measured? How?

No. Timer 1 is always clocked externally on T1, which is on PD5 on the 'XX8. If timer 0 was used instead then the external pin would be T0 (PD4), and so on. Setting CSn[2:1] to 0b11 merely selects the external input, and CSn0 selects the edge.

Would that same code define the same input interrupt pin on an ATmega328?

The '168 and '328 have the same peripherals and pinout, only the memory differs (hence the "8" at the end, with the varying powers of 2). But no pin directly triggers the interrupt; it is triggered from the timer, which is clocked via PD5.

Related Solutions

Electronic – Using multiple external interrupts in PIC

This is C pseudo-code to explain one idea. It uses and exclusive OR to work out which pins have changed and will call your different handlers within the one RBIE interrupt. Depending on how critical the application is you may want to check how the PIC handles situations such as a port changing while the interrupt is executing to make sure you won't miss any events.

int old_port_b;

void isr_handler()
{
    int new_port_b, changed_pins;
    new_port_b = read_port_b();
    changed_pins = new_port_b ^ old_port_b;
    if (changed_pins | 1)
        rb0_hander();
    if (changed_pins | 2)
        rb1_hander();
        // ... etc
    old_port_b = new_port_b;
}

int main()
{
    old_port_b = read_port_b();
    enable_interrupt();
}

Calculating how many times TIMER1 will overflow

As per page 140 of the datasheet the flag that must be cleared is called \$\mathtt{TOV1}\$ and is the LSB of the register \$\mathtt{TIFR1}\$. That bit should be cleared automatically but that feature might be disabled to use nested interrupts, which seems the case to me. I would have expected an higher value for running, but that's an int and might well be 8 bit wide, so it's overflowing like crazy (that's even signed!).

Try adding this line as the first thing in your overflow ISR:

TIFR1 &= 0x01;

Maybe something like:

TIFR1 &= 1 << TOV1;

will work too, but who knows? You should dive into headers. Try the first solution, that should work.

Best Answer

Related Solutions

Electronic – Using multiple external interrupts in PIC

Calculating how many times TIMER1 will overflow

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