SAADC on nRF52-DK unexpected result

[paul_tanner]

Using SDK_14 and unmodified SAADC example on pca10040.

Initially connected A0 (PO.03) to ground but getting seemingly random values.

<info> app: ADC event number: 586

<info> app: 293

<info> app: 76

<info> app: 24

<info> app: 292

<info> app: 208

Same result on an earlier SDK version.

Possible hardware problem? unilikely.

Any suggestions pls.

[Jørgen Holmefjord]

Hi,

P0.03 is AIN1. If the application is setup to sample AIN0, you need to ground P0.02 to see the desired behavior. Note that the markings in parenthesis on the DK is related to Arduino form factor and not to the nRF pinout. Please have a look at the nRF52832 Pin assignment documentation.

Best regards,
Jørgen

[paul_tanner]

Thx Jørgen

As you may have guessed I did not locate the definition of NRF_SAADC_INPUT_AIN0.  I still have not but amm using PO.02 now.

I'm now getting sensible values in the region of -9  (avg) when grounded and about 813 (avg) when connected to VDD

To convert, I am reading that I need to multiply by 

* GAIN/REFERENCE * 2^RESOLUTION

This doc should probably say 
 * GAIN/(REFERENCE * 2^RESOLUTION)

Given this NRF_DRV_SAADC_DEFAULT_CHANNEL_CONFIG_SE(NRF_SAADC_INPUT_AIN0);

Gain = presumably 6 (NRF_SAADC_GAIN1_6)

Reference presumably about 4v (can't find NRF_SAADC_REFERENCE_INTERNAL)

Resolution = SAADC_CONFIG_RESOLUTION 1 = 10 bit

.. which would put VDD at 1.19v (multimeter tells me it's 2.85)

So I conclude we're almost there but I need to locate the definitions for the other 2 constants.
Thx for pointing me to these.

[Jørgen Holmefjord]

All these defines are defined in the SAADC HAL file (nrf_saadc.h - located in modules\nrfx\hal\nrf_saadc.h in SDK 15.x.0). The defines used in this file is mostly found in the MDK (modules\nrfx\mdk\nrf52_bitfields.h). The MDK defines refer to the registers and register bitfields.

[paul_tanner]

Thx Jørgen,

I found the file ./components/drivers_nrf/hal/nrf_saadc.h

So the reference is .6v and the gain is 1/6

This would translate VDD to 813/6/.6/1024 ie. 0.22v

Do there is still an issue here.

Did I do the math correctly?

[Jørgen Holmefjord]

No, there is an error in your calculations. The output formula is given in Digital output documentation.

To calculate voltage from digital output (single ended input), that would be:

V(P) = RESULT * REFERENCE / (GAIN * 2(RESOLUTION)) or 813 * 0.6 V / (1/6 * 1024) = 2.86 V