Question about EmonTX firmware. BATTERY_VOLTAGE_PIN

Getting Started
1

Hi,

I am trying to understand this line of code in EmonTX firmware.

int battery_voltage=analogRead(battery_voltage_pin) * 0.681322727;                // 6.6V battery = 3.3V input = 1024 ADC

I don’t understand where “0.681322727” cames from.

I understand that the battery pack is connected at the input pin of MCP1702 IC; so ATMEGA is powered by 3.3V source.

EmonTX has a voltage divider (470k x 2) and an input to A7 to read the battery voltage.

So, as I understand, … battery_voltage wil be: analog_read(A7) * 3.3 / 1023 * 2 = analog_read * 0.0064516129

And 0.0064516129 * 100 to generate a integer: 0.64516129

But where does 0.681322727 come from ???

I am use that 0.681322727 will be the correct one, but I wish understand where it cames from.

Regards,

2

I don’t know the answer to your question, so hopefully someone who does will chime in. Your maths seems sound to me. With regards:

I think you’re unlikely to get a sensible reading with a source impedance that high. Have a look at the scope trace in the second last post in this thread:

https://openenergymonitor.org/emon/node/12440

Maybe the larger than expected multiplier is an attempt to compensate for that, but I’d hope not, as the reading you get with such a high source impedance may well depend on what you last read with the ADC. A better solution might be to call analogRead() several times in quick succession.

3

Those resistor values are indeed correct according to the emonTx schematic, and the reason for choosing values as high as that is clear: battery drain. But as you say, it’s unlikely to properly charge the S&H capacitor in the ADC. I can’t provide any enlightenment regarding the derivation of the constant, it may even have been obtained empirically.

4

Hi again both of you.

the 470k values are taken from emontx v3.4 schematic, but my question is about the firmware. I don’t know if this “getting started” category is the rigth place to ask this question. Where does 0.681322727 come from ???

Thanks.

5

The short answer is: we don’t know. @glyn.hudson might be able to tell you.

6

Tbh I can’t remember! I believe it was a result of calculating the value as you have done above then multiplied by a calibration factor after empirical measurement with a calibrated multimeter. This was many years ago, I would have to revisit. Let me know if you think the calibration could be improved.

7

I think @dBC’s suggestion is the one to follow up: Set the calibration to the correct number as calculated, then do several measurements in succession until the value stabilises. Then compare that with your meter reading. Expect up to 2% uncertainty due to the resistor tolerances, plus some uncertainty due to tolerances in the 3.3 V regulators - another 1% if your emonTx is a.c. powered, or up to 1.75% (worst case) if it is powered from the USB connector or programmer. Then of course there’s the accuracy of your meter to consider.

8

Hi,

After reading all of you I think the solution is to have a little cap as jeelab sugest here: http://jeelabs.org/2013/05/16/measuring-the-battery-without-draining-it/

And about firmware, I think ATMEGA is powered by 3.3V, so 1023 anaog value is 3.3V. As voltage divider is a x2 factor, so I think that it will have to be:

int battery_voltage = analogRead(battery_voltage_pin) / 1024 * 3.3 * 2 * 100; //1023 or 1024 (there is always discussion about this). (x100 is to save 2 decimales in the int battery_voltage value

int battery_voltage = analogRead(battery_voltage_pin) * 0.64453125; //1023 or 1024 (there is always discussion about this)

Max value for the battery is 6.6V.

As @glyn.hudson says, the diference between 0.681322727 and 0.64453125 is about 5.7% so may be he check several times with a multimeter. But 5.7% is big.

Conclusion: I understand that 0.681322727 is an empirical value. I will add a 0.1uF cap.

THANKS A LOT.

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