SAADC - nRF52832 - questions

RE: SAADC - nRF52832 - questions

Jørgen Holmefjord — Thu, 08 Feb 2018 12:43:35 GMT

You should try to avoid sampling while the radio is active, to avoid the increased noise levels.

RE: SAADC - nRF52832 - questions

Arepa — Wed, 07 Feb 2018 18:26:11 GMT

Hi Jorgen, Thanks for the recommendations :) the root of my error was the moment when I was reading, it looks the ADC is reading perfectly, I compared the conversion value=165 using 8bit resolution with values obtained using Oscilloscope and it matches with the voltage drop when my device wakes up and starts a connection.

Is it a good practice measure the battery voltage when starting a connection? or should be it before when the device is still with a low consumption?

Thanks

RE: SAADC - nRF52832 - questions

Jørgen Holmefjord — Tue, 06 Feb 2018 15:35:39 GMT

Your sample might be affected by noise and errors. As described in this post, you should expect a deviation of a few LSBs. Have you tried increasing the resolution, enabling oversampling, and doing offset calibration? Your calculations are correct, but the readings could be the source of error.

RE: SAADC - nRF52832 - questions

Arepa — Tue, 06 Feb 2018 03:05:04 GMT

Hi Jorgen, I am using the equations on the documentation resolution 8bits, gain 1/6 and reference internal 0.6V, the input range is from 0V - 3.6V, but, for example if my converter result it 128, using the equation you posted above my voltage is 1.8V, but using a tester and oscilloscope I got 2V.

For example I use the following equation:

Input range = (0.6 V)/(1/6) = 3.6 V

Voltage = (RESULT * (3.6)) / (2^(RESOLUTION - m))

both equation give me the same result, and my calculated voltage value is different to the voltage measured using tester and oscilloscope for +-200mV

RE: SAADC - nRF52832 - questions

Jørgen Holmefjord — Thu, 25 Jan 2018 17:08:38 GMT

Hi,

The digital output is calculated using the formula in the documentation. I'm not sure I understand what you expect to get out of your formula.
With an input voltage of 1.5V and your configuration, you should get a result of ~1706 (1.5 * (1/6)/0.6V * 2^12). If you want to calculate the corresponding voltage, you the formula will be V(P) = RESULT / (GAIN/REFERENCE * 2^(RESOLUTION - m)).
I don't see any better solution than the one you propose.
Unfortunately, we do not have any data on this. The pin low output are only specified to be between VSS and 0.3*VDD. This should be easy to measure on your device though.

Best regards,

Jørgen