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will
Joined: 28 May 2012
Posts: 24
Location: france
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| problem button 16F88 |
 Posted: Sat Feb 23, 2013 10:12 am |
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Hello
I do a small program on a 16F88.
When I press a button it lights a LED LED RED then GREEN (10 seconds each example).
And when I release the button it flashes a blue.
The problem is when I release my button, the program continues its cycle (red and 10 seconds 10 seconds GREEN) before flashing the blue LED.
I want that when I release the button the program goes directly to the LED blinks blue.
if you can help me. An interrupt can be, but I'm not good. You can help me?
Thank you very much.
| Code: |
#include <16F88.h>
#fuses HS,NOWDT,NOPROTECT,NOBROWNOUT,NOLVP
#use delay(clock=20000000)
#use standard_io (B)
#define led_red PIN_A2
#define led_verte PIN_A3
#define led_bleu PIN_A4
#define button PIN_B1
#int_TIMER1
TIMER1_isr()
{
while (1)
{
if (input(PIN_B0))
{
output_toggle(pin_B1);
}
else
{
output_high(PIN_A2);
delay_ms(10000);
output_low(PIN_A2);
output_high(PIN_A4);
delay_ms(10000);
output_low(PIN_A4);
}
}
}
}
void main()
{
setup_timer_1(T1_INTERNAL|T1_DIV_BY_8);
enable_interrupts(INT_TIMER1);
enable_interrupts(GLOBAL);
while(1)
{
}
} |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19537
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| Posted: Sat Feb 23, 2013 10:54 am |
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The reason, is that you are sitting in two ten second delays.
You don't need an interrupt, just break the delays down into smaller bits.
So:
| Code: |
//small part only
int8 count;
else {
output_high(PIN_A2);
for (count=0;count<100;count++) {
if (!input(PIN_B0)) break;
delay_ms(100);
}
output_low(PIN_A2);
for (count=0;count<100;count++) {
if (!input(PIN_B0)) break;
delay_ms(100);
}
}
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If the button releases, it'll terminate within 0.1seconds.
Not as fast as an interrupt response, but as fast as a 'human' needs.
Best Wishes |
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