writing bits to PORTA on p16f84

I know this is true of some PICs, maybe all. It catches just about everyone. :-)

By default, the port A pins are set to be inputs to the analog to digital converters, not digital I/O. Looks at the documentation on the ADCON register(s) to see how to do that.

The 16F84 doesn't have an ADC.

Leon

Ah, good point.

I'm wondering why RA0 is reading 0 instead of 1. It looks like if the pin is set as an input, it should always read as a 1. I'm wondering if the simulator defaults to assuming the pin is being pulled low by something.

When set as an input, a read on an I/O line on PICs always gives the actual (external) logic state on the pin, and not at all whatever you write in the output register.

If this were a real PIC and RA0 was not connected to anything, would it read 1 when RA0 is set as an input?

I don't use gpsim, but the standard MPLAB SIM. MPLAB SIM seems to assume that input pins are tied low unless told otherwise by the Pin Stimulator. I tried setting RA0 as an output and setting the latch, then setting RA0 as an input and RA0 turned off at that point.

I don't know if gpsim has some way for you to force the value of RA0 to what you want when it's configured as an input. MPLAB SIM refuses to do that, but then it has the Pin Stimulator.

^^^^^^^^^^^^^

here you set it to be an input. i.e. you can only intelligbly(sp?)

there is no problem. you cannot turn an input on and off (high and low) from *within* the chip, you need to do it from outside.

try this:

1. bridge RA0 and RA1.
2. replace lines [1] and [2] above with this: bsf TRISA, 1 ; RA0 is now input bcf TRISA, 0 ; RA1 is now output
3. replace line [3] with bsf PORTA, 1 ; this will turn on RA1, which will ; make RA0 high, which you can then read

hth

goose, hand

Maybe I'm overthinking this now. What happens if you set RA0 as an output bit. When you change RA0, does reading RA0 reflect the change?

This works in MPSIM.

It would read an undefined value. Might be 1 or 0, who knows?

Hi goose,

Thanks for the information. I thought that might have been the cause but I was uncertain, as I'm still somewhat new to PIC software development

I don't have breadboard components at the moment, and the simulator I am using (gpsim) is still under development and its features are somewhat limited. As per your suggestion, is bridging RA0 and RA1 possible via some instructions on the PIC? Or were you thinking about doing this via a simulator?

Thanks

-v

Hello Gary,

gpsim allows me to turn bits on and off on various registers, when I set certain bits on TRISA to 1 (output), gpsim _will not_ let me turn the appropriate bit on in PORTA register (I can turn on the other bits though, when the appropriate TRISA bits are set to 0)

Is this what is happening to you in MPLAB SIM?

-v

Right. If TRISA,0 is set to 1 then MPSIM won't let me set PORTA,0. If TRISA,0 is 0, then MPSIM allows me to set the value of PORTA,0.

So, why not just clear TRISA,0 so you can set the value or PORTA,0? You should still be able to read PORTA,0. You just have to remember to change your code once you get real hardware.

From the diagram in the datasheet, if the bit is set as an input, the output from the latch would be 1. If nothing is driving the input pin high or low, I would think that is the value that would be read. What would be a reason for the value to be 0? Doesn't something need to pull the signal low to make it read 0?

Hi Gary,

This is precisely what I have done for now. As well as put in comments to remind myself later to set the bits. I suppose I wasn't so confuzzled after all... Now if I can get some kind of stimulus system working to simulate voltage on the pins I'll be gtg :)

Thanks

-v

The TRIS latch, yes. But it would put the ouput driver (the P/N pair) in Hi-Z, exactly as is stated in the data sheet:

"Setting a TRISA bit (=1) will make the corresponding PORTA pin an input, i.e., put the corresponding output driver in a hi-impedance mode."

The upper transistor drawn in the diagram is a "P" type. It won't act as a weak pull-up if the TRIS bit is 1, as you seem to mean. Port B pins have weak pull-ups, see PORTB description to see the difference.

No reason for it to be 1 either. See above. In input mode, the pin is simply left "floating". A floating pin is a no-no in the "real world". It could read anything, and possibly even oscillate in some cases, with the side effect of consuming more current. That's why it's important that you connect unused inputs either to ground or Vcc, depending on the circuit.

In a simulator, they obviously have to choose a fixed state in this case, since they probably won't simulate a real circuit with all its complexity. That's why you have to use those simulators with a bit of caution.

Ah. Thanks for clearing that up. I'm a hardware dunce.