IR detector

[peterdeco1]

Hi geckogrotto. Try something like this.
START:
LET BROKEN = 0
LOW PORTB.1

Main:
IF PORTB.0 = 1 then ACTIVATE 'BEAM BROKEN
GOTO MAIN 'BEAM NOT BROKEN

ACTIVATE:
IF PORTB.0 = 1 THEN ACTIVATE 'WAIT FOR BEAM TO RETURN
LET BROKEN = BROKEN + 1
IF BROKEN >= 10 THEN LIGHTUP 'BEAM NEEDS TO BE BROKEN 10 TIMES
GOTO MAIN

LIGHTUP:
HIGH PORTB.1
GOTO LIGHTUP

[geckogrotto]

Thanks for your suggestion.

Can anyone tell me how to turn on a timer then turn it off and get the amount of time that was timed?

[Archangel]

Hi, yes you can reduce the voltage to any set level you choose.
Using Ohm's Law
V. out = V in X (r2 / r1 + r2)

where r1 is the resistor closest to the power in
and r2 is the resistor closest to ground

Example v out = v in or 10 volts X r2, say 500 ohms . . . 5000
divided by r2 + r1, so if r1= 500 ohms you would get 1000 ohms,
so 5000 / 1000 = 5 volts. Simple huh? The problem here is if you lose the ground
foe even an instant then your input sees full supply voltage, also a voltage divider uses power all the time, in this example .01 amp.
OK the attachment didn't work so basicly two resistors in series, one end attached to + and the otherend to minus and you pull power from the connection in the center.

[geckogrotto]

Hi, yes you can reduce the voltage to any set level you choose.
Using Ohm's Law
V. out = V in X (r2 / r1 + r2)

where r1 is the resistor closest to the power in
and r2 is the resistor closest to ground

Example v out = v in or 10 volts X r2, say 500 ohms . . . 5000
divided by r2 + r1, so if r1= 500 ohms you would get 1000 ohms,
so 5000 / 1000 = 5 volts. Simple huh? The problem here is if you lose the ground
foe even an instant then your input sees full supply voltage, also a voltage divider uses power all the time, in this example .01 amp.
OK the attachment didn't work so basicly two resistors in series, one end attached to + and the otherend to minus and you pull power from the connection in the center.

Ok I don't think these actually exist but lets say they do.

If I had exactly 9V in and used
R2 950
R1 760
That would give me exactly 5V out?

About the connection what your saying is resistor to + and the other resistor connected to that and then to the negative and sandwiched in between the 2 would be my output of 5v?

Also if you want shoot me over the attachment to [email protected]

Thanks

[Archangel]

That is correct as would 400/500 (notice it adds up to 900) or any variant of above say 1200 /1500 12k/15k what changes as you go upscale is the circuit impeadance (the ability to load the circuit without significant loss of voltage).
If you only want to drive a PIC input pin and not light leds or run motors then use the larger (higher resistance) resistors, that way if you mess up your code and cause an input to become an output, set to low, you will not create a big short.

[geckogrotto]

Thanks again Joe.

Now I just need the most basic code I can find to start / stop the timer and get the results and I will be able to finish the rest.

I searched around the forums but I'm not sure how to implement the code with my chip.

There are internal timers right that you can turn on or off or am I thinking this whole thing wrong and I need to figure out how fast the chip is running then make my own count loop?

[mister_e]

one way in my mind now... Use HPWM on your led emitter, place the IR receiver on RA.4/T0CKI. Use Timer0 to count, Timer1 or Pause to do your sample time. The read the TMR0 register.

Another wich i don't really like but i saw many times... Led emitter directly to Battery via resistor, IR detector to RA.4/T0CKI. Jump in a interupt routine on a TMR0 overflow. Easy neat... may kill the IR Led soon.. battery too.

Do a search for T0CKI, you should find some tips to begin.