How to detect low battery in battery powered F683 circuit

[tekart]

Connect one end of diode to ground, other to pic input and pull up resistor. Just like in schematic on page 8.
They suggesting to connect diode to ground thru mosfet to minimize current consumption. When mosfet is turned on, there is about 0,6V on pic input, and that voltage won't change.
You do not need to use mosfet, you can connect diode to ground permanently, or connect to output pin.

Yeah, I get the hardware side now - but how to implement this on the F683 with no Vref and one A/D? Am I missing something?

[pedja089]

You are missing...
There is Vref on every pic with ADC.
http://ww1.microchip.com/downloads/e...oc/41211d_.pdf Page 63 Bit VCFG(bit 6 of ADCON0)

[tekart]

OK, I see that now. So how do I select between the IO pin and the Vref source in code? No idea how that is done...
Still don't see how the resistor - diod junction will provide me with a reading that represents the changing Vbatt...

[mark_s]

I think the confusion between you and pedja089. The first way you need two external pins from the ADC or comparator.
One of the pins is tied to an external vref source and the other is your analog input. There is no voltage reference in the 683
VCFG bit or ADCON0.6 either connects the ADC vref(input) to VDD internally or connects it to and external (pin) reference you provide. .
The second way only one external pin is needed. (pic12F1822) has an internal reference diode which can be configured to be the vref source for the adc or comparator.

[tekart]

Mark, thank you for clarifying. I thought it was something like that but my inexperience with internal voltage references left me confused. I have ordered some F1822 chips to experiment with.

So would you mind clarifying for me how I would go about using the single pin to take advantage of the internal reference diode. Do I understand correctly that this diode would replace the external diode we have been discussing, so all I would need is a pull-up resistor on the ADC pin? I still don't see how that would give me a reading that would very since I would still just be reading the voltage drop of the reference diode. Forgive me for being dense about this but I still don't get it.

[mark_s]

Your battery positive will be connected thru a 1K resistor to the adc, you may need a divider if v battery is greater than VDD? Then you take a normal acin reading and determine what number between 0 - 1024
equals a low battery. If an < x then low_bat. else..

To start

Fixed voltage reference FVR page 137

ADCIN will take care of the ADCON0 register

FVRCON = %10000001 ' vref enabled 1.024v
ADCON1 = %11110011 'Right just,FRC,+vref connected to FVR

[tekart]

FVRCON = %10000001 ' vref enabled 1.024v
ADCON1 = %11110011 'Right just,FRC,+vref connected to FVR

Mark, thanks for the very helpful configuration settings. I got a test chip working fine using 8-bit reads and left justified results. I am in a similar position to peterdeco1 in that I have a circuit board designed for production and ready to go into volume assembly. Fortunately I already had two resistors connected to the A/D input creating a voltage divider so I am good to go - except for one minor problem. My test code works on the F683 chip just fine but when I try to implement it on my actual design I find that I do not have proper digital I/O on PortA.5 for some reason. My design uses portA.0 for the analog input and portA.2 for PWM out. I am struggling with what configuration error I have that is preventing portA.5 from operating as a digital output. I have been poring over the datasheet for an hour or so now and can't seem to find a configuration that seems to be setting that into something other than digital I/O. Any help would be appreciated.

[gadelhas]

Hi;
The 12f1840 as Internal voltage reference ( Fixed Voltage Reference (FVR) with 1.024V, 2.048V and 4.096V output levels )

[tekart]

Thanks guys! Now I completely understand what I need to do and this solution will work extremely well for me. I have ordered some sample chips and will dig into this later in the week. I appreciate your help and patience.