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newguy
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| interrupt on change |
 Posted: Wed Jun 23, 2004 11:33 am |
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A question for the masters here regarding the RB interrupt.
I have a really simple routine that I'm testing, and it's not behaving like I thought that it should.
I'm playing with a 4x4 keypad hooked up to port B, the lower 4 bits are outputs (low), and the upper 4 bits are inputs. Pullups are enabled. Pressing a key will connect a lower port B bit to an upper port B bit and pull down one of the upper 4 bits, and trigger the RB interrupt on change.
My interrupt routine is as follows:
| Code: | #int_RB
void RB_isr(void) {
key_pressed = TRUE;
} |
My main routine simply checks the status of key_pressed, and if true, prints "Key pressed" on an lcd module I've hooked up to it.
What has me scratching my head is that if I press and release a button, "Key pressed" gets displayed. However, if I press and hold the button, nothing happens. As soon as I release the button, then "Key pressed" appears.
What am I missing? Shouldn't a button press instantly cause the interrupt routine to run, and shouldn't it run again once the button is released, since there are two changes happening?
How do I get the thing to recognize the initial press, and also recognize the subsequent release? What I'm ultimately after is an interrupt driven routine that recognizes momentary button presses, but will also support press-and-hold actions as well. |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Wed Jun 23, 2004 1:30 pm |
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You will get an RB interrupt if the logic level on one of the pins changes
from the previous state it was in, based on the last time Port B was
read (or written to, according to the data sheet).
In other words, if you want to detect a falling edge, you need to read
the port when the inputs are at a high level. This will load a logic '1'
into a latch. Then when you press a key and put a '0' on the pin,
the logic will now detect this "mismatch" condition and cause an
interrupt.
So I think you need to do several things:
Read Port B initially, and ensure that the value read is all logic 1's
on the RB4-RB7 pins before you proceed. (I assume you have
Port B pullups enabled).
Manually clear the RBIF flag before enabling interrupts.
Also clear RB interrupts by reading Port B within your ISR.
This post contains a demo program to show how to use Port B.
http://www.ccsinfo.com/forum/viewtopic.php?t=15126
Please don't take this as an example of how to do an interrupt driven
keypad. The example is only intended to "get you going" on INT_RB
interrupts by showing some aspects of how to use it. |
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PICCER
Joined: 10 Oct 2005
Posts: 1
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| Posted: Mon Oct 10, 2005 9:16 am |
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Hello
I'm wondering what I should add the PCM PROGRAMMER?s code that it regonize right button in PORTB what I have pressed. Buttons are connected to RB4,RB5,RB6. So can someone put me some sample code?
| Code: |
#include "18F4520.h"
#device ICD=true
#fuses HS, NOWDT, NOPROTECT, BROWNOUT, NOLVP
#use Delay(Clock=20000000)
#zero_ram
#use RS232(DEBUGGER)
#byte port_b = 0xF81
void int_rb_isr(void);
char got_interrupt;
char interrupt_count;
void main(void)
{
set_tris_b(0x30);
port_b = 0;
port_b_pullups(TRUE);
delay_us(10);
interrupt_count = 0;
got_interrupt = FALSE;
enable_interrupts(INT_RB);
enable_interrupts(GLOBAL);
while(1)
{
disable_interrupts(INT_RB);
if(got_interrupt == TRUE)
{
printf("\%d\n\r", interrupt_count);
got_interrupt = FALSE;
}
enable_interrupts(INT_RB);
}
}
/
#int_rb
void int_rb_isr(void)
{
static char i;
i = port_b;
interrupt_count++;
got_interrupt = TRUE;
}
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Cheers
John |
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newguy
Joined: 24 Jun 2004
Posts: 1907
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