chardwareinterruptspic
I am trying to program a PIC 16F876A using mikroC. I need the program to run when I push a button to a certain point, stop and wait till the button is pushed again before finishing.
void interrupt() // ISR
{
INTCON.INTF=0; // Clear the interrupt 0 flag
PORTC=~PORTC; // Invert (Toggle) the value at PortD
Delay_ms(1000); // Delay for 1 sec
}
I have read that using a hardware interrupt is the way to do this but can not get mine to work.
Try clearing INTEDG bit of OPTION_REG register. From the datasheet page 30(32):
INTEDG: Interrupt Edge Select bit
1 = Interrupt on rising edge of INT pin
0 = Interrupt on falling edge of INT pin
When PORTD is high, does the LED turn ON or OFF?
Also check out switch contact bounce. Here is a link on it and how to debounce. To check it is the contact bounce causing the thing that is seen as a "problem", add a delay of 500msec or 1 sec after PORTD=0xff; in the interrupt routine. If the port is high, your problem is contact bounce if the above suggested solution doesn't solve your problem -that is clearing INTEDG bit.
You don't need that GIE=1; in the last line of the interrupt block.
I have no experience with Hi-Tech C compiler. Chances are low but try doing INTE=1; before GIE=1; in the main function.
Edit
Try changing void intMain() to void interrupt intMain(). According to
HI-TECH C® for PIC10/12/16 User’s Guide 3.8 INTERRUPT HANDLING IN C (Page 86) Link
Edit After Second Question
According to the same manual mentioned above page(299):
To produce an infinite loop, use for(;;).
You are trying to assign an 'integer' value to a 'character' variable as I quoted below. Define variable i as unsigned char. Compiler should take care of this, but it is worth trying.
...
int i,k;
...
PORTD=i;
...
Firstly, you always need series resistors with the LEDs when driven from a voltage source (e.g. battery, DC supply, etc)
This is because LEDs have a non-linear I-V curve, which looks like a high impedance up to the threshold voltage of the LED then rises very sharply, so it means with a very slight change of voltage the current changes a lot, making it almost impossible to set the current to a stable value in this manner.
By using the correct value series resistors, you ensure the current cannot rise enough to damage the LED.
To calculate the resistor value you need to know the LED forward voltage (Vf) then subtract Vf from supply voltage and divide by desired current, e.g for a 5V supply, a 2V Vf and 15mA:
(5V - 2V) / 0.020A = 200Ω (The standard value of 220Ω will do - if you don't have this then aim for anything between 150Ω and 600Ω for a 20mA to 5mA range)
This is assuming a typical 5mm or 3mm LED of 20mA max operating current.
Although it says "3 x 1.5V Battery in Series" in your schematic, the batteries look like they may be actually connected in parallel. T confirm, the batteries need to be connected end to end like the bottom diagram in this image:
You need decoupling capacitors present between the microcontroller Vdd and ground. I won't go into detail (search on here, there are many good answers on this subject) but they are basically to provide the microcontroller with a low impedance local energy reserve for high frequency current demand, which the power supply cannot respond to quickly enough.
Ideally you should place one (100nF or 1uF ceramic is pretty standard) across the power and ground pins, as near to the pins as possible.
Make sure you are using the internal oscillator if you have no crystal present. Your code does not show the config bit settings, if you have left them out you need to add them to ensure the microcontroller is setup correctly. The CCS manual should tell you how to do this. Also in the config bits, make sure the Watchdog Timer is set to off, otherwise your micro will continually reset (unless you call the WDT clear command regularly)
Make sure that you have the LEDs the right way round.
Make sure you have the MCLR pin tied high, otherwise your PIC will be held in reset (unless you turn MCLR off in the config bits) This is usually done with a resistor to Vdd, value around 10kΩ. The datasheet will have an example of this in the reset section. (thanks to ajs410 for mentioning this)
Best Answer
You do not specify what is meant by "not working", so I'll just point out a few things.
The interrupt service routine posted does not seem to address the program function described. This routine toggles a port, waits a second, then goes back to whatever was running before the interrupt was triggered. Instead you say you want the program to wait for a button press, then do something, then wait for another button press.
If the program is doing nothing but waiting for an input, you don't really need an interrupt, just a polling loop. Interrupts should be used when you need to quickly respond to an event, or do something at a precise time. ISRs should finish quickly; never should one include a one second delay.
Another thing to consider: There is more to using an interrupt than writing an ISR. I'm not familiar with mikroC, but in general you will need to at least set up an interrupt vector, configure the I/O or peripheral to generate an interrupt, and enable interrupts. Have you done all this necessary initialization?
Also, upon entry to an interrupt routine, the CPU context, its status and registers, must be saved, and restored upon return. Typically the compiler is alerted to this by using a keyword, often interrupt, in the function definition. Does the compiler understand that this is an ISR?
Finally, make sure your watchdog doesn't timeout while you are waiting for input.
Oh, and don't forget: you may have to debounce the pushbutton.