Half gain measured versus configured on signal

Hi marcaon3d ,

After checking the register configuration, I noticed that the data/status register is for channel 5. Channel 5 is for AIN2-AINCOM, and channel 5 is enabled together with channel 7 (AIN4-AINCOM) based on the settings in Configuration register. Could you try isolating it by disabling channel 5 then confirm the value (data register) you are getting for channel 7 so that we can confirm as well the value you obtained, which is 2.839V?

Thanks,
Francis

Hi FMarasigan,

Thank you for the reply.

We actually alternate between reading channel 5 and channel 7 back and forth. I just ended up pulling the register when it was configured on channel 5, that is my bad.

Here is the Status register when reading from channel 7 (AIN4) after disabling channel 5 : 0x00000097

The input is still 2.584V
Gain is configured as x8
And ADC output is 2.84V

Here is the DATA register
Data          : 0x915B1617

This value should convert into 2.856V according to my calculations

Thanks

Hi marcaon3d ,

Apologies for the late response. After checking the value on your DATA register without the appended value for STATUS, which is 0x915B16, it is equivalent to the AIN voltage of 0.084V. Could you share how you got the ADC output of 2.84V? The formula as shown below for computing the output code for bipolar configuration is based on page 33 in the datasheet.

Regards,
Francis

Hi FMarasigan,

Thank you for the reply.

Considering the output code in bipolar, when I solve for the DATA output we got, I do not get an AIN of 0.084V

0x915B16 = 2^23 * [(AIN * 8/2.5)+1]

Solving for AIN will give AIN = 0.354V which matches with our initial problem.

For reference, this is how we compute our result from ADC to Volts :

OutputV = Vref * ((adc_value - adc_half) / adc_half + 1)

For an ADC output of 0x915B16,
we get 2.838V

Please let me know if our conversion from ADC code to voltage makes sense.

Thank you for your help !

Hi marcaon3d ,

Looking at the schematic you've provided, REFIN+ = AVDD, which is connected to +5V. In the computation, it looks like you are using 2.5V.

Using the formula from the datasheet:

(\frac{9526038}{2^{23}}-1)\cdot(\frac{5}{8})\text{ }=\text{ }0.084

Even when using the formula you've provided without the +1 to compute for the equivalent voltage from the code:

5\cdot(\frac{(9526038-8388607)}{8388607}\text{ }=\text{ }0.677

adc_value = 9256038
adc_half = 8388607

From the 0.677 computed value, dividing it to a gain of 8 will result to 0.084V as well.

Regards,
Francis