SAADC differential input range?

[Yi-Wei]

Hi:

In the datasheet, there have a  description

Input range = (+- 0.6 V or +-VDD/4)/Gain

for a example, if Gain = 2, reference = 0.6V,

input range = 0.6 / 2 = 0.3V = 300mV

so, this range is +/- 150mV or +/- 300mV?

[Einar Thorsrud]

Hi,

The input range of the ADC core is nominally ± (0.6 V / Gain) if you use the internal reference or ± ((VDD / 4) / Gain) if you use VDD as reference. So, with internal reference and Gain = 2, the input range is from -300 mV to +300 mV.

[Yi-Wei]

so, in the single ended input with a same calculate

gain = 1/6

Input range = ± (0.6 V)/(1/6) = 3.6 V

range is -3.6V to 3.6V?

but, I had tried a SAADC with nrf52832, only 0~3.6V

I use those setting

    {                                                  \
        .resistor_p = NRF_SAADC_RESISTOR_DISABLED,     \
        .resistor_n = NRF_SAADC_RESISTOR_DISABLED,     \
        .gain       = NRF_SAADC_GAIN1_6,               \
        .reference  = NRF_SAADC_REFERENCE_INTERNAL,    \
        .acq_time   = NRF_SAADC_ACQTIME_10US,          \
        .mode       = NRF_SAADC_MODE_SINGLE_ENDED,     \
        .burst      = NRF_SAADC_BURST_DISABLED,        \
        .pin_p      = (nrf_saadc_input_t)(PIN_P),      \
        .pin_n      = NRF_SAADC_INPUT_DISABLED         \
    }
    
    
    // and
    
    nrf_saadc_resolution_set(NRF_SAADC_RESOLUTION_12BIT);

and use this formula to calculate my input

float vf = p_event->data.done.p_buffer[0] * 3.6f / 4096.0f; // 12bit resolution

and I get a correct result, also it is means range only 0~3.6V