Would reducing the reference voltage of an ADC have any effect on accuracy? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraAVR 1.1V Internal ADC Reference Over-Voltagecan the input voltage to ADC exceed the reference in LPC 2468Implications of INL on the accuracy and resolution of an ADCSTM32F1 Can't I use the internal voltage reference as the ADC reference?Long term effect of over-voltage on clamping diodes and ADC reference voltagewould using the reference voltage of an ADC as a VOLTAGE SOURCE affect my ADC or the stability of the circuit?Voltage divider for Arduino external ADC reference voltage (AREF)STM32 ADC reference voltageADC variable reference voltagedsPIC33E ADC with external voltage reference

How to not starve gigantic beasts

Mistake in years of experience in resume?

"My boss was furious with me and I have been fired" vs. "My boss was furious with me and I was fired"

Implementing 3DES algorithm in Java: is my code secure?

Does Feeblemind produce an ongoing magical effect that can be dispelled?

With indentation set to `0em`, when using a line break, there is still an indentation of a size of a space

Why is an operator the quantum mechanical analogue of an observable?

PIC mathematical operations weird problem

A strange hotel

How to translate "red flag" into Spanish?

AI positioning circles within an arc at equal distances and heights

std::is_constructible on incomplete types

What if Force was not Mass times Acceleration?

What is it called when you ride around on your front wheel?

Could moose/elk survive in the Amazon forest?

Why didn't the Space Shuttle bounce back into space many times as possible so that it loose lot of kinetic energy over there?

Is Diceware more secure than a long passphrase?

How would I use different systems of magic when they are capable of the same effects?

Co-worker works way more than he should

Multiple options vs single option UI

My admission is revoked after accepting the admission offer

Expansion//Explosion and Siren Stormtamer

Why did C use the -> operator instead of reusing the . operator?

What units are pgfphysicalheight and pgfphysicalwidth defined in?



Would reducing the reference voltage of an ADC have any effect on accuracy?



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraAVR 1.1V Internal ADC Reference Over-Voltagecan the input voltage to ADC exceed the reference in LPC 2468Implications of INL on the accuracy and resolution of an ADCSTM32F1 Can't I use the internal voltage reference as the ADC reference?Long term effect of over-voltage on clamping diodes and ADC reference voltagewould using the reference voltage of an ADC as a VOLTAGE SOURCE affect my ADC or the stability of the circuit?Voltage divider for Arduino external ADC reference voltage (AREF)STM32 ADC reference voltageADC variable reference voltagedsPIC33E ADC with external voltage reference



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








5












$begingroup$


Regarding the following information taken from a paper:



enter image description here



The paragraph is telling about what happens when the ADC Vref is reduced.



There is the statement from the above quote:




Note that if you reduce the reference voltage to 0.8V, the LSB would
then represent 100mV, allowing you to measure a smaller range of
voltages (0 to 0.8V) with greater accuracy.




Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?



(I'm asking because if I don't clarify this point, I will be misunderstanding all the rest)










share|improve this question









$endgroup$











  • $begingroup$
    they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
    $endgroup$
    – Neil_UK
    8 hours ago






  • 1




    $begingroup$
    But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
    $endgroup$
    – Elliot Alderson
    7 hours ago










  • $begingroup$
    Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
    $endgroup$
    – sstobbe
    7 hours ago

















5












$begingroup$


Regarding the following information taken from a paper:



enter image description here



The paragraph is telling about what happens when the ADC Vref is reduced.



There is the statement from the above quote:




Note that if you reduce the reference voltage to 0.8V, the LSB would
then represent 100mV, allowing you to measure a smaller range of
voltages (0 to 0.8V) with greater accuracy.




Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?



(I'm asking because if I don't clarify this point, I will be misunderstanding all the rest)










share|improve this question









$endgroup$











  • $begingroup$
    they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
    $endgroup$
    – Neil_UK
    8 hours ago






  • 1




    $begingroup$
    But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
    $endgroup$
    – Elliot Alderson
    7 hours ago










  • $begingroup$
    Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
    $endgroup$
    – sstobbe
    7 hours ago













5












5








5





$begingroup$


Regarding the following information taken from a paper:



enter image description here



The paragraph is telling about what happens when the ADC Vref is reduced.



There is the statement from the above quote:




Note that if you reduce the reference voltage to 0.8V, the LSB would
then represent 100mV, allowing you to measure a smaller range of
voltages (0 to 0.8V) with greater accuracy.




Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?



(I'm asking because if I don't clarify this point, I will be misunderstanding all the rest)










share|improve this question









$endgroup$




Regarding the following information taken from a paper:



enter image description here



The paragraph is telling about what happens when the ADC Vref is reduced.



There is the statement from the above quote:




Note that if you reduce the reference voltage to 0.8V, the LSB would
then represent 100mV, allowing you to measure a smaller range of
voltages (0 to 0.8V) with greater accuracy.




Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?



(I'm asking because if I don't clarify this point, I will be misunderstanding all the rest)







adc accuracy resolution






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









atmntatmnt

849624




849624











  • $begingroup$
    they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
    $endgroup$
    – Neil_UK
    8 hours ago






  • 1




    $begingroup$
    But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
    $endgroup$
    – Elliot Alderson
    7 hours ago










  • $begingroup$
    Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
    $endgroup$
    – sstobbe
    7 hours ago
















  • $begingroup$
    they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
    $endgroup$
    – Neil_UK
    8 hours ago






  • 1




    $begingroup$
    But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
    $endgroup$
    – Elliot Alderson
    7 hours ago










  • $begingroup$
    Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
    $endgroup$
    – sstobbe
    7 hours ago















$begingroup$
they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
$endgroup$
– Neil_UK
8 hours ago




$begingroup$
they use 'precision' in the next sentence. I'm not sure there's a well defined difference between accuracy and precision, the way there is a well defined difference between accuracy and resolution.
$endgroup$
– Neil_UK
8 hours ago




1




1




$begingroup$
But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
$endgroup$
– atmnt
8 hours ago





$begingroup$
But accuracy and precision are not same thing afaik. Accuracy I guess is deviation of the mean value from true value. But precision is more about the number of decimals a quantity can be represented. 2.123 is more precise than 2.12. This was I thought.
$endgroup$
– atmnt
8 hours ago













$begingroup$
Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
$endgroup$
– Elliot Alderson
7 hours ago




$begingroup$
Your understanding of what "precision" means is not correct. The word precision is usually used to mean repeatability or reproducibility of a measurement. See for example nist.gov/pml/…
$endgroup$
– Elliot Alderson
7 hours ago












$begingroup$
Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
$endgroup$
– sstobbe
7 hours ago




$begingroup$
Perhaps you have not seen the bullseye diagram before: cdn.antarcticglaciers.org/wp-content/uploads/2013/11/…
$endgroup$
– sstobbe
7 hours ago










5 Answers
5






active

oldest

votes


















5












$begingroup$

I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V



More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.



So 1.221 V has greater accuracy (but less precision)



while



1.2300 V has greater precision (but less accuracy).



In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.



Also: instead "precision" engineers more often use "resolution".






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
    $endgroup$
    – Bimpelrekkie
    8 hours ago










  • $begingroup$
    You mean 3-bit instead of 8 bit ?
    $endgroup$
    – atmnt
    8 hours ago











  • $begingroup$
    You're right, 3 bit, 8 values
    $endgroup$
    – Bimpelrekkie
    8 hours ago











  • $begingroup$
    I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
    $endgroup$
    – atmnt
    7 hours ago



















2












$begingroup$


Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?




Oh, good. Someone who knows the difference.



To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.



A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.






share|improve this answer









$endgroup$




















    1












    $begingroup$

    Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.



    I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).



    Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.






    share|improve this answer











    $endgroup$




















      0












      $begingroup$

      I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With $N$ bits, the voltage resolution is $V_ref/2^N$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.



      However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just $V_LSB$. So, if you change the reference voltage then the value of $V_LSB$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of $V_LSB$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).






      share|improve this answer











      $endgroup$












      • $begingroup$
        Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
        $endgroup$
        – atmnt
        7 hours ago











      • $begingroup$
        You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
        $endgroup$
        – Elliot Alderson
        7 hours ago


















      0












      $begingroup$

      With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.



      You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.






      share|improve this answer









      $endgroup$













        Your Answer






        StackExchange.ifUsing("editor", function ()
        return StackExchange.using("schematics", function ()
        StackExchange.schematics.init();
        );
        , "cicuitlab");

        StackExchange.ready(function()
        var channelOptions =
        tags: "".split(" "),
        id: "135"
        ;
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function()
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled)
        StackExchange.using("snippets", function()
        createEditor();
        );

        else
        createEditor();

        );

        function createEditor()
        StackExchange.prepareEditor(
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader:
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        ,
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        );



        );













        draft saved

        draft discarded


















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f435214%2fwould-reducing-the-reference-voltage-of-an-adc-have-any-effect-on-accuracy%23new-answer', 'question_page');

        );

        Post as a guest















        Required, but never shown

























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V



        More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.



        So 1.221 V has greater accuracy (but less precision)



        while



        1.2300 V has greater precision (but less accuracy).



        In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.



        Also: instead "precision" engineers more often use "resolution".






        share|improve this answer











        $endgroup$








        • 1




          $begingroup$
          I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
          $endgroup$
          – Bimpelrekkie
          8 hours ago










        • $begingroup$
          You mean 3-bit instead of 8 bit ?
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          You're right, 3 bit, 8 values
          $endgroup$
          – Bimpelrekkie
          8 hours ago











        • $begingroup$
          I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
          $endgroup$
          – atmnt
          7 hours ago
















        5












        $begingroup$

        I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V



        More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.



        So 1.221 V has greater accuracy (but less precision)



        while



        1.2300 V has greater precision (but less accuracy).



        In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.



        Also: instead "precision" engineers more often use "resolution".






        share|improve this answer











        $endgroup$








        • 1




          $begingroup$
          I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
          $endgroup$
          – Bimpelrekkie
          8 hours ago










        • $begingroup$
          You mean 3-bit instead of 8 bit ?
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          You're right, 3 bit, 8 values
          $endgroup$
          – Bimpelrekkie
          8 hours ago











        • $begingroup$
          I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
          $endgroup$
          – atmnt
          7 hours ago














        5












        5








        5





        $begingroup$

        I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V



        More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.



        So 1.221 V has greater accuracy (but less precision)



        while



        1.2300 V has greater precision (but less accuracy).



        In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.



        Also: instead "precision" engineers more often use "resolution".






        share|improve this answer











        $endgroup$



        I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V



        More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.



        So 1.221 V has greater accuracy (but less precision)



        while



        1.2300 V has greater precision (but less accuracy).



        In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.



        Also: instead "precision" engineers more often use "resolution".







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 8 hours ago

























        answered 8 hours ago









        BimpelrekkieBimpelrekkie

        51.9k246115




        51.9k246115







        • 1




          $begingroup$
          I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
          $endgroup$
          – Bimpelrekkie
          8 hours ago










        • $begingroup$
          You mean 3-bit instead of 8 bit ?
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          You're right, 3 bit, 8 values
          $endgroup$
          – Bimpelrekkie
          8 hours ago











        • $begingroup$
          I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
          $endgroup$
          – atmnt
          7 hours ago













        • 1




          $begingroup$
          I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
          $endgroup$
          – Bimpelrekkie
          8 hours ago










        • $begingroup$
          You mean 3-bit instead of 8 bit ?
          $endgroup$
          – atmnt
          8 hours ago











        • $begingroup$
          You're right, 3 bit, 8 values
          $endgroup$
          – Bimpelrekkie
          8 hours ago











        • $begingroup$
          I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
          $endgroup$
          – atmnt
          7 hours ago








        1




        1




        $begingroup$
        I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
        $endgroup$
        – atmnt
        8 hours ago





        $begingroup$
        I also thought similar but then saying "with greater accuracy" in that context is wrong correct? Better resolution causes better precision but not better accuracy.
        $endgroup$
        – atmnt
        8 hours ago













        $begingroup$
        It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
        $endgroup$
        – Bimpelrekkie
        8 hours ago




        $begingroup$
        It is the other way round, precision stays the same due to the ADC being 8 bit. Accuracy increases as the error between reported value and real value becomes smaller when Vref is decreased.
        $endgroup$
        – Bimpelrekkie
        8 hours ago












        $begingroup$
        You mean 3-bit instead of 8 bit ?
        $endgroup$
        – atmnt
        8 hours ago





        $begingroup$
        You mean 3-bit instead of 8 bit ?
        $endgroup$
        – atmnt
        8 hours ago













        $begingroup$
        You're right, 3 bit, 8 values
        $endgroup$
        – Bimpelrekkie
        8 hours ago





        $begingroup$
        You're right, 3 bit, 8 values
        $endgroup$
        – Bimpelrekkie
        8 hours ago













        $begingroup$
        I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
        $endgroup$
        – atmnt
        7 hours ago





        $begingroup$
        I see. Just one more thing. If the Vref would remain the same but the ADC resolution was increased lets say to 6-bit from 3-bit; for that case can we say the precision gets better but accuracy remains the same? Or they both get better? I hope you have comment on this because this is very critical to clarify.
        $endgroup$
        – atmnt
        7 hours ago














        2












        $begingroup$


        Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?




        Oh, good. Someone who knows the difference.



        To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.



        A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.






        share|improve this answer









        $endgroup$

















          2












          $begingroup$


          Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?




          Oh, good. Someone who knows the difference.



          To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.



          A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.






          share|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$


            Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?




            Oh, good. Someone who knows the difference.



            To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.



            A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.






            share|improve this answer









            $endgroup$




            Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?




            Oh, good. Someone who knows the difference.



            To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.



            A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 7 hours ago









            TimWescottTimWescott

            7,1791416




            7,1791416





















                1












                $begingroup$

                Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.



                I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).



                Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.






                share|improve this answer











                $endgroup$

















                  1












                  $begingroup$

                  Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.



                  I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).



                  Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.






                  share|improve this answer











                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.



                    I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).



                    Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.






                    share|improve this answer











                    $endgroup$



                    Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.



                    I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).



                    Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 7 hours ago

























                    answered 7 hours ago









                    nsayernsayer

                    523924




                    523924





















                        0












                        $begingroup$

                        I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With $N$ bits, the voltage resolution is $V_ref/2^N$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.



                        However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just $V_LSB$. So, if you change the reference voltage then the value of $V_LSB$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of $V_LSB$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).






                        share|improve this answer











                        $endgroup$












                        • $begingroup$
                          Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                          $endgroup$
                          – atmnt
                          7 hours ago











                        • $begingroup$
                          You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                          $endgroup$
                          – Elliot Alderson
                          7 hours ago















                        0












                        $begingroup$

                        I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With $N$ bits, the voltage resolution is $V_ref/2^N$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.



                        However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just $V_LSB$. So, if you change the reference voltage then the value of $V_LSB$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of $V_LSB$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).






                        share|improve this answer











                        $endgroup$












                        • $begingroup$
                          Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                          $endgroup$
                          – atmnt
                          7 hours ago











                        • $begingroup$
                          You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                          $endgroup$
                          – Elliot Alderson
                          7 hours ago













                        0












                        0








                        0





                        $begingroup$

                        I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With $N$ bits, the voltage resolution is $V_ref/2^N$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.



                        However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just $V_LSB$. So, if you change the reference voltage then the value of $V_LSB$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of $V_LSB$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).






                        share|improve this answer











                        $endgroup$



                        I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With $N$ bits, the voltage resolution is $V_ref/2^N$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.



                        However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just $V_LSB$. So, if you change the reference voltage then the value of $V_LSB$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of $V_LSB$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).







                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited 7 hours ago

























                        answered 7 hours ago









                        Elliot AldersonElliot Alderson

                        8,23521022




                        8,23521022











                        • $begingroup$
                          Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                          $endgroup$
                          – atmnt
                          7 hours ago











                        • $begingroup$
                          You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                          $endgroup$
                          – Elliot Alderson
                          7 hours ago
















                        • $begingroup$
                          Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                          $endgroup$
                          – atmnt
                          7 hours ago











                        • $begingroup$
                          You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                          $endgroup$
                          – Elliot Alderson
                          7 hours ago















                        $begingroup$
                        Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                        $endgroup$
                        – atmnt
                        7 hours ago





                        $begingroup$
                        Your "precision is directly a function of the converter's reference voltage" does not match with the other answer which says the precision stays the same. There is still some confusion.
                        $endgroup$
                        – atmnt
                        7 hours ago













                        $begingroup$
                        You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                        $endgroup$
                        – Elliot Alderson
                        7 hours ago




                        $begingroup$
                        You're right...but we shouldn't be using the word precision at all. We should be talking about the resolution of the converter. I will edit.
                        $endgroup$
                        – Elliot Alderson
                        7 hours ago











                        0












                        $begingroup$

                        With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.



                        You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.






                        share|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.



                          You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.






                          share|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.



                            You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.






                            share|improve this answer









                            $endgroup$



                            With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.



                            You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 5 hours ago









                            analogsystemsrfanalogsystemsrf

                            16.4k2823




                            16.4k2823



























                                draft saved

                                draft discarded
















































                                Thanks for contributing an answer to Electrical Engineering Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid


                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.

                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f435214%2fwould-reducing-the-reference-voltage-of-an-adc-have-any-effect-on-accuracy%23new-answer', 'question_page');

                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                六本木駅

                                Integral that is continuous and looks like it converges to a geometric seriesTesting if a geometric series converges by taking limit to infinitySummation of arithmetic-geometric series of higher orderGeometric series with polynomial exponentHow to Recognize a Geometric SeriesShowing an integral equality with series over the integersDiscontinuity of a series of continuous functionsReasons why a Series ConvergesSum of infinite geometric series with two terms in summationUsing geometric series for computing IntegralsLimit of geometric series sum when $r = 1$

                                Joseph Lister