SAADC changing value range (signed to unsigned)

RE: SAADC changing value range (signed to unsigned)

Duy — Fri, 26 Jul 2019 12:50:40 GMT

Hello,

Thanks for your clear answer. I am totally clear now.

Should check the datasheet more carefully.

RE: SAADC changing value range (signed to unsigned)

Jared — Fri, 26 Jul 2019 12:38:31 GMT

Hi,

The SAADC output is in the range [0, +2^(resolution)] in single ended mode. If resolution = 12, then the range is [0, 4096]

If the SAADC samples VDD, then the output value is:

RESULT = (VDD) * (GAIN/VDD) * 2(RESOLUTION)

RESULT = VDD * (GAIN/VDD) * 2^(12) = GAIN * 2^12

If GAIN = 1, sampling VDD will output: 0b0001 0000 0000 0000 = 4096

The sample will then be stored with 2's complement applied.

Negative values are due to ground bounce, if you experience this often, then consider to use differential mode instead.

regards

Jared

RE: SAADC changing value range (signed to unsigned)

Duy — Fri, 26 Jul 2019 12:23:33 GMT

awneil

Sometimes I still feel a little confusing when getting the result and reading the datasheet.

RE: SAADC changing value range (signed to unsigned)

Duy — Fri, 26 Jul 2019 12:17:25 GMT

Jared,

Oh I see that I didn't read this part carefully:

RE: SAADC changing value range (signed to unsigned)

awneil — Fri, 26 Jul 2019 11:41:18 GMT

[quote userid="27129" url="~/f/nordic-q-a/50161/saadc-changing-value-range-signed-to-unsigned/200739"]But what is the truly input voltage when I got that small negative result[/quote]

As the post you linked says, the negative results are likely due to slight ground differences, noise, etc.

So the actual input voltage is less than the offset & noise!

RE: SAADC changing value range (signed to unsigned)

Duy — Fri, 26 Jul 2019 11:11:49 GMT

Sorry for making the question note clear enough. Here is my calculation.

For example I got:

+ VDD = 2.4V

+ Reference = VDD/4; Gain = 1/4

+ Input range = (VDD/4)/(1/4) = 2.4V

+ Single-ended ADC

+ ADC resolution = 12 bit (-2048 to 2047?)

So if my ADC is around 0 to 2.4V, the ADC I got will be from 0 to 2047, is it right?

And when the input is 2.4V, the ADC value (I get after calling the convert buffer function) will be = 1111 1000 0000 0000b.

Is this right?

And sometimes I get the result (after calling the convert buffer function): -7, -8, -4...to -1. What is the reason for this result? As I read from this post, this is the offset. But what is the truly input voltage when I got that small negative result?

Thank you and best regards,

Duy

RE: SAADC changing value range (signed to unsigned)

Jared — Fri, 26 Jul 2019 09:03:34 GMT

[quote user="DuyLionTran"]In my case, I can only use 11 bits if the output of my device is 0V -> 1.9V. Is it right? [/quote]

I'm not sure I understand you correctly, the result is stored as an 16 bit int. The number of bits you use is dependent on the chosen resolution. All bits will be set to 1's when you've reached the reference/gain. The reference/gain value should be set to the maximum voltage that you would want to measure.

RE: SAADC changing value range (signed to unsigned)

Duy — Thu, 25 Jul 2019 15:36:42 GMT

Hello Jared,

Thank you for your reply. It is because I am communicating with a device whose output is 0V to 1.9V and I want to use all 12 bits of the SAADC. In my case, I can only use 11 bits if the output of my device is 0V -> 1.9V. Is it right?

Best regards,

Duy

RE: SAADC changing value range (signed to unsigned)

Jared — Thu, 25 Jul 2019 08:34:56 GMT

Hi,

The SAADC stores samples as 16 bits 2's complement, I don't know of any future plans to implement unsigned SAADC atm. Why would you need to store the samples as unsigned SAADC? The SAADC can be used in differential mode if you want to sample something not relative to local ground.

regards

Jared