[mbed] What exactly does the return value of AnalogIn::read_u16() means?

[Hieu]

According to this post, and this GitHub source file, in the nRF51 chip's mbed implementation:

• The ADC reference voltage is 1/3 power supply
• read_u16() seems to return NRF_ADC->RESULT
• NRF_ADC->RESULT maximum possible value would be 0x3FF, when ADC input equals to ADC reference voltage.

From an experiment I just ran, I observed the following

• At ADC input of ~897mV, read_u16() returns ~0x011a, or 282 decimal.
• At ADC input of ~520mV, read_u16() returns ~0x00b9, or 185 decimal.
• Leaving the ADC input pin floating, read_u16() returns ~0x00af, or 175 decimal.

Using linear approximation with the first two data points, at read_u16() value of 0x03FF, the ADC input should be >4V.

If using only one of those data points with the assumption that at 0V input, read_u16() returns 0, then 0x03FF still means a voltage >3V.

These numbers certainly cannot be 1/3 power supply! What could be going on here? Am I misunderstanding something?

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Just in case this fact is the reason to the problem: To conduct my experiment, I use a (1M - 220k) voltage divider to supply 1/6 of VDD or 5V to the analog input.

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On another note, according to mbed's Analog In reference and the AnalogIn.h header file, read_u16() returns "input voltage, represented as an unsigned short in the range [0x0, 0xFFFF]". This description sucks. But at the same time it doesn't seem to be what nRF51's implementation of the function is returning. Is there any comment on this?

[Anders Strand]

Hello.

From analogin_init in the github document you linked:

(ADC_CONFIG_INPSEL_AnalogInputOneThirdPrescaling << ADC_CONFIG_INPSEL_Pos) |
(ADC_CONFIG_REFSEL_SupplyOneThirdPrescaling << ADC_CONFIG_REFSEL_Pos) |

This prescales both the input and reference by 1/3. The reference is VDD (3v). I find this to match your numbers.

The maximum value is 0x3FF because the nRF51 ADC support maximum 10 bit resolution.

Update 1:

Let me give you an example.

• You have 3v VDD, this is scaled to 1V.
• The input voltage is 500 mV = 0.5 V. This is scaled to about 167 mV.
• The adc will give you (0.167/1) * 1023 = ~170.

With your numbers:

• We have about 3V VDD. We apply the scaling in the equation (*0.33)
• 897 mV input gives (897 * 0.33/3000 * 0.33) *1023 = ~305. Your result of 282 is given because VDD is not exactly 3v, but a bit higher.
• 520 mV input gives (520 * 0.33/3000 * 0.33) *1023 = ~177. Here your result is slightly higher. This can either be explained by a lower VDD (unlikely), or uncertainty in the measurement of the 520 mV. It can also be other factors in the external circuit, which might be worth investigating.
• Wee see that the scaling of 1/3 can be removed from the equation. This means that the result you get from the adc is the input, divided by VDD, multiplied by 1023.

-Anders

[Anders Strand]

I have updated my answer, with some examples. I agree that from the two data points, your numbers do not add up 100%. Can you do some more tests where you measure VDD, ADC input, and note dwn adc output?