Battery monitoring - ever tried AN1072?

[flotulopex]

Hi,

As some other threads around the forum about battery monitoring, I want to setup a battery monitoring system based on MICROCHIP's AN1072 "Measuring VDD Using the 0.6V Reference".

In short, it says:
"To measure VDD, VDD should be selected as the reference to the ADC via VCFG, and the 0.6V reference selected as the input using the channel select bits, CHS<3:0>. A measurement of the 0.6V input is taken with the ADC, and the result represents 0.6 Volts as a percentage of VDD. As VDD increases, the resulting number will decrease and vice versa. This yields a direct “1/x” relationship between V DD and the produced digital value as seen in Equation 1. In short, given a specific VDD, the digital value is always the same. Working backwards, if the digital value is known, VDD may be calculated."



It tells you what you have to do and even how. Now guess: I can't make it work

I have a circuit using a 16F690 (great! same one as used in the AN!). For the test, the circuit is connected to my variable power-supply so I can simulate a low battery condition (i.e. from 6VDC, I go down to 5,5VDC*).

According to the AN1072, the ADCON0 register has been set to %10110101.

Now when I vary the voltage on the power-supply, I get always the same ADC reading (no value change at all).

What am I doing wrong?

NB*: Yes, I'm using a LDO voltage regulator so I could use another way to check the battery's condition (i.e. comparator or resistor-divider)...but I want to have a go with the AN1072

[flotulopex]

That's the ADCON0 register:

[richard]

I might have the wrong end of the stick here , but if you have a ldo reg suppling vdd won't you always read the regulated voltage. to read the battery voltage that way requires the battery be connected to vdd directly.

[flotulopex]

Sorry Richard, I don't get what you mean.

[Charlie]

You need to make sure you measure on the input side of the LDO, not the output side. It's job it to keep the voltage constant. You will need a resistor divider to reduce the battery sense voltage to within the range of the PIC. Post the schematic of how you connected the PIC.

[towlerg]

The point is that the AN1072 method measures Vdd as supplied to the 16F690 - it is purely internal. If you want to measure some other voltage, like the battery side of the LDO you need to use a potential divider (for example) and an ADC inputs that accepts an external input.

George

[flotulopex]

So I'll have to go for this one?

[flotulopex]

Thank you George.

I did unfortunately not understand the AN the way you explain.

Let's go for a divider then

[longpole001]

better off using higher value resistors to reduce the over all current draw of the voltage divider , but if you have too higher values the current will not be enough for accruate reading but after you do testing you will find the avg voltage range with the resistors used and you can apply that offset to the reading.

also use the inbuild voltage ref for the ad readings so that when you transfer between various input voltage on differnat pics , the same code and voltage tables can be used for either 10 bit or 12 bit ad cpus

[towlerg]

the same code and voltage tables....

Just in case anyone should be put of by mention of tables, the math is real simple - VDD = 6144/Ad_Result (0.6 * 1024 = 614.4 - 10bit - result in 0.1V).

It will come as no surprise to here that Daryl posted this info.

George

[flotulopex]

This is how I made it for now

Vbatt is 6VDC (4x AAA batteries) and VREG is 5VDC (hence, VREG is the reference voltage).

To measure the batteries, I draw 10mA; this seems to be okay to me - if not okay, please suggest another current I should have.

[Ioannis]

Roger, as I stated earlier, when the FET is OFF you supply +6 to the PIC pin. This is dangerous!

Ioannis

[flotulopex]

Wow! I missed that!

What's the solution then? Shall I go with P-channel MOSFET?

[longpole001]

also consider that the voltage drop from a 6v incoming source to your LDR output at 5v /pic is going to give you much margin (.5v at best ) before your looking to have to replace the batteries, no cos they are that low just cos your voltage usage margin is small

consider running the app at 3.3v , take a closer look at thge items that require the 5v externally to the PIC to look at using devices at 3.3v as this will give a larger usage margin for the batteries and device

just a thought

[flotulopex]

After my first tests, I changed my circuit from 3,3VDC to 5VDC hence accepting the "margin" reduction.

The project I'm working on now is a cooking timer and I have a piezo-sounder that sounds best (= loudest) at pseudo* 10VAC (*I use a full bridge inverter for higher volume).

BTW, I use a LP2950ACZ5.0 and it's dropout voltage is around 0,2V @ 10mA. 0,8V leave me far enough accuracy to determine when it is time to change the batteries

[flotulopex]

I still would like to know how to "read" the 0,6V (VP6) channel. This voltage reference is provided by the PIC itself.

If I set the ADSCON0 register to (i.e.) %10110100 (right just., VDD, 0,6V Reference, ADC is ON), what is the PICBASIC command to read that channel? If it is ADCIN, then ADCIN what?



I tried with:

ADCIN 13, myvar

because I see channels from 0 to 11 in the DS, maybe "0,6 V ref" would be the 13th...and PBP accepts the compilation but the ADCIN result shows always 0

[Ioannis]

ADCIN x, myvar

where x is the AD channel where you connected the voltage divider we were talking before.

If it is on pin 18, channel 1 then x=1 and so on.

Ioannis

[flotulopex]

I'm sorry, but I still don't understand

Looking at the ADC bloc diagram I've attached in post#1, I understand there is a "0,6V Reference" channel that I have to select to get an ADCIN result of 122 @ 5VDC or 186 @ 3,3VDC.

This channel doesn't seem to be linked to any pin of the PIC - this is confusing me at most.

Does this "0,6V Reference" channel really exist?

----

Hum...just found I have to ENABLE the 0,6Voltage Reference in VRCON register, bit 4 "VP6EN". I'll have a try in a moment...