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

Question

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? 
 
 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. 
 
 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 · Accepted Answer

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

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

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