QN908x: getting simple 16bit ADC result

[arpad_toth]

How to format the ADC result value to a simple unsigned 16bit preferably without float operation on QN908x?

Datasheet says:

The ADC result is 23-bit signed fractional data, with the MSB (bit 22) as the sign, while the output data to MCU is 32-bit.

ADC_OUTPUT[31:9]: ADC_DATA[22:0]

ADC_OUTPUT[8:0]: 0x00

I'm interested in 0-65535 range.

  rawADCval = ADC_GetConversionResult(ADC);

[FelipeGarcia]

Hi Arpad,

As you mentioned it's not possible to change the format in the ADC. The user manual gives you the following formats:

However, if you check the user guide in chapter 30.2 it says the ADC is a 16-bit sigma-delta ADC.

Which means the ADC resolution is already 16 bits.

QN908x user manual

Best regards,

Felipe.

[arpad_toth]

Hi Felipe,

I doesn't work

 adcConfigStruct.dataFormat = kADC_DataFormat0WithoutIdx;

s16 result = ADC_GetConversionResult(ADC) >> 16u

gives -63..116 range for GND..3V input, so I loose too much resolution.

and it's not even linear  1.6V gives 310.

Please provide an unsigned float free example how to use the ADC.

Thank you,

[FelipeGarcia · Accepted solution]

Hi Arpad,

Sorry for the late reply.

If you want to have only positive numbers please check the configuration below:

static void ADC_Configuration(void)
{
    adc_config_t adcConfigStruct;
    adc_sd_config_t adcSdConfigStruct;

    /**
     * Initial ADC to default configuration.
     */
    ADC_GetDefaultConfig(&adcConfigStruct);
    adcConfigStruct.channelEnable = (1U << DEMO_ADC_CHANNEL);
    adcConfigStruct.channelConfig = (DEMO_ADC_CFG_IDX << DEMO_ADC_CHANNEL);
    adcConfigStruct.triggerSource = DEMO_ADC_TRIGGER;
    adcConfigStruct.convMode = kADC_ConvModeSingle;
    adcConfigStruct.dataFormat = kADC_DataFormat0WithoutIdx;
    adcConfigStruct.refSource = kADC_RefSourceExtWithoutDriver;
    ADC_Init(DEMO_ADC_BASE, &adcConfigStruct);

    /* Initial ADC Sigma Delta(SD) configuration */
    ADC_GetSdDefaultConfig(&adcSdConfigStruct);
    adcSdConfigStruct.vinnSelect = kADC_VinnSelectAvss;
    adcSdConfigStruct.refGain = kADC_RefGain1;

    ADC_SetSdConfig(DEMO_ADC_BASE, DEMO_ADC_CFG_IDX, &adcSdConfigStruct);

    /* Bandgap voltage */
    g_AdcBandgap = ADC_GetBandgapCalibrationResult(DEMO_ADC_BASE, DEMO_ADC_CFG_IDX);

    /* Calibration VINN value */
    g_AdcVinn = ADC_GetVinnCalibrationResult(DEMO_ADC_BASE, &adcConfigStruct);

    /* Enable ADC */
    ADC_Enable(DEMO_ADC_BASE, true);
}‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

Here I'm using a 1.5V external reference source and I am able to get values from 0 to 3232767 by shifting result_16bits = (g_AdcConvResult >> 16); 

Hope this helps!

Have a great day,
Felipe

[arpad_toth]

Hi Felipe,

Yes it is a 16bit ADC, but the output is represented as 23-bit signed fractional data, so for 0V input I would get weird big signed numbers 0x123456.

Do you have an example C code how to convert it back to unsigned 16bit?

Thanks,

Arpad

[FelipeGarcia]

Hi Arpad,

Unfortunately you cannot get a 16 bits unsigned value cause this ADC is a differential sigma-delta ADC. This means that for single-ended usage, the analog input is connected to the positive end, while the negative end input is connected to an internal generated voltage, VINN.

However, you can approximate to that behavior if you shift the g_AdcConvResult register as follows:

result_16bits = (g_AdcConvResult >> 16); 

By doing this you will be able to get a value from -32768 to 32767.

Finally, you should change the VINN value to 0 if you want only positive numbers. This is possible if you configure the ADC as follows:

Best regards,

Felipe.