Following on from this thread I've been trying to get my head around converting the result from analogRead to a voltage. Let's suppose: int value = analogRead (A0); And imagine we get the value of 1000. Now (ignoring the fact that integers don't have decimal places) is the voltage (assuming we have a 5V reference): 1000.0 / 1024 * 5.0 = 4.8828125 or: 1000.0 / 1023 * 5.0 = 4.8875855 In other words, divide by 1023 or 1024? I should point out that dividing by 1024 immediately gives a "wrong" answer for measuring 5V: 1023.0 / 1024 * 5.0 = 4.9951171 This is because the maximum reading is 1023. Browsing the web: http://arduino.cc/en/Tutorial/ReadAnalogVoltage To scale the numbers between 0.0 and 5.0, divide 5.0 by 1023.0 and multiply that by sensorValue One vote for 1023. http://forum.arduino.cc/index.php?topic=200447.0 Multiple schools of thought, including: mrburnette: You divide by 1024 if you are a mathematician or if you truly understand ADC successive approximation. http://blog.codeblack.nl/post/NetDuino-Getting-Started-with-ADC.aspx Hedging his bets: The number 0 represents 0V and 1023 represents Aref. The voltage level at any ADC port can be determined as follows: float analogValue = (float)digitalValue * (3.3f/ 1023f) In theory, you should divide by 1024, because there are that many steps. [image] ADC value divide by 1024 or 1023 - Netduino 2 (and Netduino 1) ADC value divide by 1024 or 1023 - posted in Netduino 2 (and Netduino 1): Hi all, I'm using an ADC port to measure voltage (of a solar charged battery). I use the standard 3.3V Aref and the default 0-1023 range. With a voltage divider (5k6 + 30k)... Different opinions: The ADC returns a value from 0 to 1023 not 0 to 1024. 0 to 1023 is 1024 different values because zero is a value too. ... That is why you divide by 1023 and not 1024. Not exactly. Consider the nominal "width" of the ADC reading: it is 1024. That is strictly related to Vref (typically 3.3V). Now we must slice that segment in many smaller parts: in total are 1024 "tiles" (if you like numbered from 0 to 1023). http://forum.arduino.cc/index.php?topic=252276.0 Why ADC/1024 is correct, and ADC/1023 is just plain wrong! It's common to need to scale a microcontroller ADC result to give a value in some other scale. One example is to scale the ADC to read in actual volts like read from 0.00v -> 5.00v. People sometimes use ADC *500 /1023 as this gives a reading of... The correct scaling math; *x/n or *x/1024 correctly scales all the output data in size, but gives an average rounding error of 0.5 ADC counts (always rounding down). So, how come we get a "wrong" result for a value of 1023? It seems that the ADC can return 1024 "slots" of values (from 0 to 1023) where each slot represents 1/1024 of the reference voltage. So for a 5V reference voltage the slot width is: 5 / 1024 = 0.0048828125V Thus a result of 0 could be: 0V to 0.00488V A result of 1 could be: 0.00488V to 0.009766V ... A result of 1023 could be: 4.99511V to 5V So really the "wrong" result for 5V is more that the voltage is not necessarily 5V, it could be 4.995. In addition to that, the hardware rounds down the result, so you could compensate by adding in an average error of 0.5. 1023.5 / 1024 * 5.0 = 4.99756V So this is reasonably close to 5V. Similarly a reading of 0 should probably be treated as possibly halfway between 0 and 1, thus it could on average be: 0.5 / 1024 * 5.0 = 0.002441V Oh yes, and the Atmega328 datasheet mentions in the section "23.7 ADC Conversion Result" that the figure is 1024.

Ps991: But is it not valid to assume that 4.995 to 4.999...V is still part of the 1024th bin? Exactly. 1023 (the 1024th bin) represents 4.995 to 5V.

but...but...then you can measure 0 to 4.9999...V, its just you can only represent then in 1024 0.0048828125V increments, you can still map 4.999...V to 0 to 1023

I can but the processor can't because it rounds down.

Then we can both agree that the processor is stupid and obsolete and that we must introduce organic brains into microprocessors! Finally somebody that agrees with me on this!

If you are going to argue with me again, explain what the datasheet meant about "minus one LSB."

0x3FF represents the selected reference voltage minus one LSB. This means that 0x3ff represents 5V (reference voltage) minus 1 LSB (5 / 1024)V, thus 5 - 0.005V = 4.995V It can represent the voltage as 0 to 4.995V, but it can measure/map 0 to 4.999...V

I assume they mean that the last bin is the "trash bin". 1023 represents a voltage range that extends to Vcc+0.5

Maybe this will help. The converter does two things... Compare two voltages. The comparison answers one question ... is the sample-and-hold voltage (which is the analog input voltage if everything works as desired) less than the current comparison voltage. Change the current comparison voltage in discrete steps. Each step is Vref (5V) / 1024. We know #1 and #2 are true because that's how successive approximation converters work. We know the step size is correct because the company that made the converter told us the step size. The last step is over-sized (reaches Vcc+0.5) because the converter can only make a less-than comparison. The first step does not include a precise zero because the converter can only make a less-than comparison.

You can avoid the conundrum entirely if you just work directly with the result of analogRead. Generally, I'd expect any thresholds relating to making decisions in a sketch to be #defined or declared const, so there's no reason not to express them in the 0-1023 range. Comments can relate them back to voltage if desired. Given that, I don't see much (if any) benefit in expending CPU cycles converting to voltage.

After reviewing the replies and datasheets (AVR and SAM), for the AVR, I will now use 1024 as the divisor. AVR: Referring to the datasheet for the Atmel 328P, page 252 24.7 ADC Conversion Result: Here it states that the result is ADC = (Vin * 1024) / Vref and that 0x000 represents analog ground and 0x3FF (1023) represents the selected reference voltage minus one LSB. Therefore, the ADC binary range of 0-1023 represents an input voltage range of 0-4.9951171875 volts Converting a 1023 reading to volts, I agree here that using (1023 * 5) / 1024 = 4.9951171875 volts is the accurate method to use. SAM: Now, in the datasheet for the SAM3X (Due), page 1329 44.1 (ADC) mentions: A digital error correction circuit based on the multi-bit redundant signed digit (RSD) algorithm is employed in order to reduce INL and DNL errors. Here, WIKIPEDIA: Successive approximation ADC under Algorithm, it mentions: The objective is to approximately digitize x to an accuracy of 1/2n. Does this mean that here, for 10-bit ADC, one should use 1023.5 and for 12-bit ADC (Arduino default) one should use 4095.5 as the divisor? (it appears so to me).

analogRead - divide by 1023 or 1024? - #65 by system - Programming

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