Receiving Pulses on INT0 interrupt Pin on the LPC932

"the code is already posted"

You've only posted the bit where YOU think the problem lies.

"The other part of the program are the initialization of ports and the clock source and speed. Nothing else."

Well, post it then.

Stefan

"the code is already posted"

You've only posted the bit where YOU think the problem lies.

"The other part of the program are the initialization of ports and the clock source and speed. Nothing else."

Well, post it then.

Stefan

Here is Stefan, this is the whole program I am testing with out success.
Thank you


#include <Reg932.h>
#include <stdio.h>
#include <intrins.h>



void init(void);
void brkrst_init(void);
void Time_Base(void);

unsigned int Sec_Counter = 0;

void Main(void)
{
P2 = 0x00;
P0 = 0x00;

init(); // configure ports
brkrst_init(); //enable UART break detect
Time_Base();

}

void init(void)
{
P0M1 = 0x00;
P0M2 = 0xCD; // Quasi-bidirectional
P1M1 = 0x00; // Push-pull
P1M2 = 0xFF;
P2M1 = 0x00; // push pull output
P2M2 = 0xFF;
P1M1 |= 0x18; // make INT0 and INT1 input pins
P1M2 &= 0xE7;
ES = 1; // enable UART interrupt
EA = 1;
}

void brkrst_init(void) // This function allows ISP entry through the UART break detect
{
SCON = 0x50; // select the BRG as UART baud rate source, pg. 60
SSTAT = 0x00; //Timer1
BRGR0 = 0xF0; // 9600 BAUD at @ 7.373MHz internal RC oscillator
BRGR1 = 0x02; //timer increments every .0271 microsec.
BRGCON = 0x03; // enable BRG, pg. 59
AUXR1 |= 0xC0; // enable reset on break detect, pg. 102
}

void UART(void) interrupt 4 //Interrupt number for serial port
{
RI = 0; // clear receive interrupt flag
}

void Time_Base(void)
{
ET0 =1;
TR0 = 0;
TMOD |= 0x01;
TAMOD = 0x00;
TH0 = (65536-30720)/256; // 34816
TL0 = (65536-30720)%256; // .2
TF0 = 0;
EA = 1;
TR0 = 1;
}


void timer(void) interrupt 1 using 1
{
Sec_Counter++;
if (Sec_Counter == 20) // if 1 sec have passed
{
Sec_Counter = 0;
P2 = 0xFF;
TR0 = 0;
}
}

This version of the program differs from the earlier one in the timer ISR, which is no longer calling Time_Base(). The ISR needs to clear the overflow flag (TF0) so that the interrupt can re-occur. And since this is a one-shot timer mode, the ISR must also reload the timer registers. I would expect this program to produce one timer interrupt (50 msec after main finishes, assuming the timer values are calcuated correctly).

When posting source code, if you put the text between the "pre" and "/pre" tags, it will preserve indentation on the left.

Try something like:

void T0_Start (void)
    {
    TH0 = (65536-30720)/256; // 34816
    TL0 = (65536-30720)%256; // .2
    TF0 = 0;
    }


void Time_Base(void)
    {
    ET0 =1;
    TR0 = 0;
    TMOD |= 0x01;
    TAMOD = 0x00;

    T0_Start();
    EA = 1;
    TR0 = 1;
    }


void timer(void) interrupt 1 using 1
    {
    T0_Start();  // start timer again

    Sec_Counter++;
    if (Sec_Counter == 20) // if 1 sec have passed
        {
        Sec_Counter = 0;
        P2 = 0xFF;
        TR0 = 0;
        }
    }

Drew Davis I made the changes you posted. It still does not work. I tried it with emulator and it single steps, jumps into the ISR, reloads the timer, but does not clear TF0 flag bit. It increments the sec_counter variable once though. Did you try it yourself. did it work for you. I tried with the actual chip LPC932 revision F and it did not work...I say that because it did not turn on P2 LED's.
What would be the next step.


#include <Reg932.h>
#include <stdio.h>
#include <intrins.h>



void init(void);
void brkrst_init(void);
void Time_Base(void);
void T0_Start(void);

unsigned int Sec_Counter = 0;

void Main(void)
{
P2 = 0x00;
P0 = 0x00;

init(); // configure ports
brkrst_init(); //enable UART break detect
Time_Base();

}

void init(void)
{
P0M1 = 0x00;
P0M2 = 0xCD; // Quasi-bidirectional
P1M1 = 0x00; // Push-pull
P1M2 = 0xFF;
P2M1 = 0x00; // push pull output
P2M2 = 0xFF;
P1M1 |= 0x18; // make INT0 and INT1 input pins
P1M2 &= 0xE7;
ES = 1; // enable UART interrupt
EA = 1;
}

void brkrst_init(void) // This function allows ISP entry through the UART break detect
{
SCON = 0x50; // select the BRG as UART baud rate source, pg. 60
SSTAT = 0x00; //Timer1
BRGR0 = 0xF0; // 9600 BAUD at @ 7.373MHz internal RC oscillator
BRGR1 = 0x02; //timer increments every .0271 microsec.
BRGCON = 0x03; // enable BRG, pg. 59
AUXR1 |= 0xC0; // enable reset on break detect, pg. 102
}

void UART(void) interrupt 4 //Interrupt number for serial port
{
RI = 0; // clear receive interrupt flag
}

void Time_Base(void)
{
ET0 =1;
TR0 = 0;
TMOD |= 0x01;
TAMOD = 0x00;
T0_Start();
EA = 1;
TR0 = 1;
}


void timer(void) interrupt 1 using 1
{
T0_Start(); // start timer again

Sec_Counter++;
if (Sec_Counter == 20) // if 1 sec have passed
{
Sec_Counter = 0;
P2 = 0xFF;
TR0 = 0;
}
}

void T0_Start (void)
{
TH0 = (65536-30720)/256; // 34816
TL0 = (65536-30720)%256; // .2
TF0 = 0;
}

The ISR needs to clear the overflow flag (TF0) so that the interrupt can re-occur
no need, quoting "the bible"
Cleared by hardware as the program vectors to the interrupt routine.

Erik

Erik, TF0 could be cleared by software as well...But can any one try it on his simulator or emulator...please.
Thank you

void Main(void)
{
P2 = 0x00;
P0 = 0x00;

init(); // configure ports
brkrst_init(); //enable UART break detect
Time_Base();

}

What do you think happens after the call to Time_Base()?

Stefan