Help with computing correct SAADC output value

[davidb]

I am using an Nrf52840 and sdk V14.2.0.

My question is how to compute the battery voltage based on my hardware.

The hardware uses AIN0 to measure.

The hardware has a resistor divider: GPIO output pin >---806K---X---->2M------>GND.

AIN0 is connected at point X above. The GPIO pin is configured as an output and set to a logic 1.

How would I scale the reading taken this way to represent a supply voltage of 3V?

With 3V as VDD AIN0 would see 2V I believe.

I use default config so gain is 1/6.

Battery has no protection diode in series.

Thanks.

[davidb]

Thanks.

We are using a 3V and saadc provides 10-bit conversions. 

with 10-bit readings and 3V then (2.14V/3V) = .713. -- .713 * 1024 = 730.

730 represents the value when VDD is 3V if I understand correctly.

To display 3V to the user as the measured value how would I scale 730 to result in a 3V value?

[Jørgen Holmefjord]

davidb said:

with 10-bit readings and 3V then (2.14V/3V) = .713. -- .713 * 1024 = 730.

No, this is not correct if you are using internal 0.6V reference and 1/6 Gain, then the input range will be 3.6V and your input will be related to this, not the supply voltage of 3V.

Correct calculation will be:

(2.14V/3.6V)*1023 = 0.59*1023 = 609.

davidb said:

To display 3V to the user as the measured value how would I scale 730 to result in a 3V value?

Not sure I understand what you ask about here. With your current voltage divider and SAADC configurations, a digital output of 609 will correspond to the 3V supply voltage.

[davidb]

I guess I need to know use the digital output of 609 in a printf: printf("Voltage is %d", digital_output);

if digital_output is 609 then the printed voltage will be 609 and not 3.

[Jørgen Holmefjord]

If you want to calculate the input voltage from the digital output, you just need to divide the sample (609) with the resolution (1023) and multiply this with the input range (3.6V), i.e. 609/1023*3.6 = 2.14V. To get the voltage before your voltage resistor, you should multiply by ((806K+2M)/2M)=1.403. You can use the same calculations with any digital output to get the voltage, not just 609 as I used in this example.