How to detect low battery in battery powered F683 circuit

[tekart]

I do a lot of battery powered designs with li-po (3.7v) batteries - mostly using 8 pin chips like the F683. I am trying to figure out if there is a way to use an AD input - or comparator configuration to determine if the battery that is powering the chip is running low. A threshold of 3.2 V would be good. A resistor divider connected to the A/D input will not work because it will always read the resistor ratio voltage as a constant value as the battery voltage changes. I have been fooling around with a zener diode/resistor combination with no luck. I don't have any experience using the comparators inside PICs and wonder if there is some way to configure a comparator to detect a low battery threshold. Does anyone have an elegant solution for detecting a low battery condition? In my current design I only have AN0 available for battery sensing on the F683.

I really need a very low part count/low cost solution! No external comparators or circuits, is this practical?

[mark_s]

Hello

This might give you some ideas

http://ww1.microchip.com/downloads/en/DeviceDoc/41215C.pdf

[tekart]

Yes, I saw that but the best circuit (on p.8) uses an external comparator. Not sure if that can be implemented with the INTERNAL one?

[pedja089]

Why not? It will work.
But you can only use R3 and D1 connected to ADC pin.
Put Vdd as reference, take sample voltage, and, for larger Vdd ADC result will be lower, and for lower Vdd result will bi higher.

[tekart]

I don't see how this is any different from putting a resistor divider on the ADC input. The problem is that the divided voltage will track Vdd and the A/D reading will always represent the resistor divider ratio (or resistor/diode drop ratio) and thus will maintain a consistent reading despite changing Vdd. Am I missing something here?

[pedja089]

Hey, where I was mention voltage divider???

But you can only use R3 and D1 connected to ADC pin.
That isn't voltage divider.
Biased D1 will produce almost constant voltage on ADC input pin. By keeping constant voltage on ADC input, and changing reference voltage(Vdd)

for larger Vdd ADC result will be lower, and for lower Vdd result will bi higher.
At least try to read response when you ask for help. That is all from me...

[tekart]

Sorry pedja089,
I am having trouble understanding your very terse sentence. I would appreciate it if you could elaborate what you mean by this


Put Vdd as reference, take sample voltage, and, for larger Vdd ADC result will be lower, and for lower Vdd result will bi higher.
As I understand it, my A/D reading will always be defined by the voltage drop of the diode forward voltage drop and so will not change. Remember I only have the one A/D input to work with in my very simple circuit.

[mark_s]

Remember I only have the one A/D input to work with in my very simple circuit.

In that case you need a pic12f1822 or similar with an internal vref. The internal vref will work with the adc or comparator module.

[pedja089]

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.

[tekart]

In that case you need a pic12f1822 or similar with an internal vref. The internal vref will work with the adc or comparator module.

OK, now that is helpful. I was not aware of an 8-pin chip with all those features and at a good price too! Now I'll just have to learn how to configure the internal Vref to get the results I need. Another slog through data sheets that can take hours!

[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/en/devicedoc/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

[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.

[Dick Ivers]

Hi Tekart,
Do you have PBP3 ? I believe you will need it to use the 12F1822 or any xxF1xxx part.

[tekart]

Hi Tekart,
Do you have PBP3 ? I believe you will need it to use the 12F1822 or any xxF1xxx part.
Of course! Been using PIC Basic for over 14 years! Early adopter that came over from Basic Stamps. - thanks for checking though :)

[peterdeco1]

Maybe this will help. A while back we needed a low battery warning for a circuit that was already in production so it had to be done in reverse. The PIC was powered by a 6V battery with a diode to drop it to 5.4V. A jumper from an ADC pin was attached to a 3.3V regulator already on another section of the circuit board. Now as the battery voltage drops, the ADC result will rise, since the 255 reference will always be the battery voltage. The lower the battery voltage the higher the 3.3V ADC reading becomes.

[pedja089]

That is what I was suggested, just to to use diode instead of regulator...

[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.

[mark_s]

These "enhanced" pics don't use the GPIO like the old ones. Use porta.5 to read
and lata.5 to write. See page 13 and page 121 of the data sheet


Add: See Hecklers code and configs
http://www.picbasic.co.uk/forum/showthread.php?t=18537&highlight=pic12f1822

[tekart]

Wow. Thanks Mark - I missed that. I just HATE all these whiz bang "features" that force us mere mortal coders to spend *days* learning a new chip! Grrrr. <end rant>
SO to implement a pin as output only I use:
GreenLED var lata.1 ' high = light power LED
and also set:
' config
TRISa = %011001 ' set I/O directions (0 = output, 1 = input)
LATa = %011001

- still not getting it to work with above settings. Do I have bit polarity right for LATA?

[mark_s]

I think you would declare your variable as

GreenLED var PORTA.1

[tekart]

I think you would declare your variable as
GreenLED var PORTA.1
Been testing that and the compiler seems to accept either
GreenLED var PORTA.1
or
GreenLED var LATA.1

[mark_s]

At this point without posting your complete code it's hard to guess what's wrong.

[tekart]

Got all the OUTPUTS working fine - no need for LATA configurations apparently. Struggling with some kind of hardware bug - so I'm basically good. No idea how or why I would configure LATA at this point as it all seems to work fine without any direct reference to it.

[flotulopex]

Maybe this application note (http://ww1.microchip.com/downloads/en/AppNotes/01072A.pdf) from MICROCHIP can help ;-)

[peterdeco1]

I did this a long time ago with a 6V circuit. The pic was running from the battery (with a series diode = 5.4V) but there was a 3.3V regulator elsewhere on the circuit board. I ran a jumper wire from an ADC pin to the 3.3V reference voltage. As the battery voltage begins to drop, the ADC reading will begin to rise. When the ADC reading reached the desired point, the SOUND command alerted you.

[peterdeco1]

I just realized I posted the above last October.