nrf52 Battery Measurement 4*AA batteries

[Bilal]

Hello,

I need to measure battery on my nrf52 beacon. My input voltage is 4*AA batteries(each 1.2V so in total 4.8V). The schematic provided by the hardware developer is as follows

Now I read few posts about battery measurements and looks like I need to add some additional hardware? Kindly tell me whether its possible to measure voltage using existing hardware, if so please describe, the calculations or at least point me to the right direction.

Thank You

Regards

Bilal

[awneil]

Bilal said:

I read few posts about battery measurements and looks like I need to add some additional hardware

Yes - that should be clear from those posts!

Surely, this is something that your hardware developer should be doing? It is a hardware question!

[Bilal]

I think it involves both, hardware and software. As far as I know , there must be some kind of voltage regulator used to bring down voltage to 3V. Can we use that to do our battery measurements?

[awneil]

Of course - the whole thing involves both!

But it should be the hardware designer's job to feed an appropriate signal to a pin on the chip.

You need a voltage divider - not a regulator.

[cbd]

I agree with awneil.

I see on your circuit diagram that you are running your nRF52 at 3.3V.

Although you can run the nRF52 on more than 3.3V, you can only allow an input to be 0.3V above that.

If you wish to detect the battery voltage directly then you will need a divider of sorts to step down the voltage, but these tend to impose their own problems due to leakage currents.

These are all points that your hardware engineer should take into account.

So you may look at measuring a voltage that has a maximum of 2.4V say and then internally calculate that as 4.8V

Alternatively you may wish to just measure the voltage on the MCU supply line using one of your Analogue Inputs (AIN#), but that will limit your maximum value to 3.3V.

[Bilal]

Thank You @cbd and @awneil.  So i just checked, apparently we have a voltage divider having R1 = 470k, R2 = 110k and capacitor = 1nf, we also feed the output to AIN7. So do we need to just start the adc pin and start sampling with configuration NRF_ADC_CONFIG_REF_VBG   ?

[awneil]

You haven't said what specific chip you are using.

Whatever it is, study the ADC chapter in its Product Specification - that will tell you the options for gain, reference, etc - and you choose whatever is appropriate to the signal you're measuring.

You may also want to look at the ADC example in the SDK ...

[Bilal]

nrf52832

[awneil]

nRF52832 Product Specification, then.

[Bilal]

So I took reference voltage , vdd/4 (my vdd is 3.3V), gain = 1/5 and resolution 12 bit, here is my initialization for it

err_code = nrf_drv_saadc_init(nullptr,saadc_callback);
APP_ERROR_CHECK(err_code);
nrf_saadc_channel_config_t channel_config = NRF_DRV_SAADC_DEFAULT_CHANNEL_CONFIG_SE(NRF_SAADC_INPUT_AIN7);
channel_config.gain = NRF_SAADC_GAIN1_5;
channel_config.reference = NRF_SAADC_REFERENCE_VDD4;

err_code = nrf_drv_saadc_channel_init(0, &channel_config);
APP_ERROR_CHECK(err_code);
nrf_saadc_resolution_set(NRF_SAADC_RESOLUTION_12BIT);
err_code = nrf_drv_saadc_buffer_convert(m_buffer, 1);

while I convert it as follows:

constexpr double REF_VOLTAGE_IN_VOLTS = 0.825; // Maximum Internal Reference Voltage
constexpr double VOLTAGE_DIVIDER_IN_VOLTS = 0.2; //Internal voltage divider
constexpr double ADC_RESOLUTION_12BIT = 4095;
#define RESULT_IN_VOLTS(ADC_VALUE) (ADC_VALUE*REF_VOLTAGE_IN_VOLTS)/(VOLTAGE_DIVIDER_IN_VOLTS*ADC_RESOLUTION_12BIT)

The measurement from my voltage divider is feeded to AIN7 but it measures 0.9V on the battery while my input is 5.2V and 0.7V when my input 4.5V. Can you suggest what am doing wrong in here

[Einar Thorsrud]

Hi,

Can you try using the internal reference instead of VDD? I would expect that to be more stable, even though you must have a regulator in your power supply circuit since the battery power is too high.

Then there is also the point that depending on your circuitry and battery etc the voltage will probably drop a bit when you do ADC sampling compared to when you measure on the battery when the nRF does not draw much current (if that is what you are ding). Therefor you should have a capacitor in the circuit as well, as described here.

Depending on what I have mentioned above, your result might not be that far from expected.

The voltage out of the voltage divider in your case is 5.2 V / (470+110)*110 = 0.986 V.

Then you put that into

RESULT = [V(P) – V(N) ] * GAIN/REFERENCE * 2^(RESOLUTION - m)

So for single ended it is:

RESULT = V * GAIN/REFERENCE * 2^RESOLUTION

And get something like:

RESULT = 0.986 * 1 / 5 * 2^12 = 808.

Is that roughly the sample value you get?

Then you swap around so that you solve for V you get back the same result:

V = RESULT / (GAIN/REFERENCE * 2^RESOLUTION)

V = 808/(1 / 5 * 2^12) = 0.986

Then you have to remember that this is the voltage you measure over R2 in the voltage divider, so you have to account for that as well, and find that the voltage you measure is 5.2 V in this example.

[Bilal]

Yes, my measurements are close to what you said, now we need to map our voltage divider measurement to input voltage?