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Voltage output waveform of a differentiating amplifier


How does this OP-AMP offset voltage measuring circuit work?Does this voltage regulator use “on-off” control, or is it a voltage follower?Signal Processing with Op AmpsCalculate output of Transistor Common Emitter AmplifierNMOS CS Amplifier PSpice Simulation QuestionInconsistency in the calculation of the gain of an emitter followerReduced output of op amp peak detection and hold circuitUnderstanding the necessity of Tail Current Source in Differential Amplifier/Simulation (SPICE)How to adjust gain of differential discrete amplifierVariable gain amplifier output appears on circuit positive and negative supplies






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








2












$begingroup$


I have the following picture of the input and output voltage waveforms from a differentiating amplifier as shown below:



enter image description here



And the schematic is as shown below:



enter image description here



My teacher claimed that these waveforms are correct, but I am starting to feel some doubt.



I calculated the equation of the line for input voltage waveform from time t=0 ms to t = 0.5 ms (with respective voltages of -500 mV and 500 mV) to obtain: y= 2000x - 0.5, and thus the output voltage waveform should be at y=2 mV for t=0 ms to t=0.5 ms, which is not shown in this picture.



Is this something to do with errors, or simply a mistake?



Also, I am aware that when differentiating a triangular waveform the output waveform should be square. So again, is the obtained picture a result of error (eg. from inaccuracies), or "human" mistake?










share|improve this question











$endgroup$











  • $begingroup$
    If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
    $endgroup$
    – jonk
    Apr 24 at 7:13







  • 1




    $begingroup$
    Your opamp is a LOW_PASS_FILTER.
    $endgroup$
    – analogsystemsrf
    Apr 24 at 7:13










  • $begingroup$
    I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
    $endgroup$
    – Michel
    Apr 24 at 7:22






  • 1




    $begingroup$
    This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
    $endgroup$
    – Mattman944
    Apr 24 at 7:35










  • $begingroup$
    If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
    $endgroup$
    – Huisman
    Apr 24 at 7:51


















2












$begingroup$


I have the following picture of the input and output voltage waveforms from a differentiating amplifier as shown below:



enter image description here



And the schematic is as shown below:



enter image description here



My teacher claimed that these waveforms are correct, but I am starting to feel some doubt.



I calculated the equation of the line for input voltage waveform from time t=0 ms to t = 0.5 ms (with respective voltages of -500 mV and 500 mV) to obtain: y= 2000x - 0.5, and thus the output voltage waveform should be at y=2 mV for t=0 ms to t=0.5 ms, which is not shown in this picture.



Is this something to do with errors, or simply a mistake?



Also, I am aware that when differentiating a triangular waveform the output waveform should be square. So again, is the obtained picture a result of error (eg. from inaccuracies), or "human" mistake?










share|improve this question











$endgroup$











  • $begingroup$
    If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
    $endgroup$
    – jonk
    Apr 24 at 7:13







  • 1




    $begingroup$
    Your opamp is a LOW_PASS_FILTER.
    $endgroup$
    – analogsystemsrf
    Apr 24 at 7:13










  • $begingroup$
    I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
    $endgroup$
    – Michel
    Apr 24 at 7:22






  • 1




    $begingroup$
    This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
    $endgroup$
    – Mattman944
    Apr 24 at 7:35










  • $begingroup$
    If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
    $endgroup$
    – Huisman
    Apr 24 at 7:51














2












2








2





$begingroup$


I have the following picture of the input and output voltage waveforms from a differentiating amplifier as shown below:



enter image description here



And the schematic is as shown below:



enter image description here



My teacher claimed that these waveforms are correct, but I am starting to feel some doubt.



I calculated the equation of the line for input voltage waveform from time t=0 ms to t = 0.5 ms (with respective voltages of -500 mV and 500 mV) to obtain: y= 2000x - 0.5, and thus the output voltage waveform should be at y=2 mV for t=0 ms to t=0.5 ms, which is not shown in this picture.



Is this something to do with errors, or simply a mistake?



Also, I am aware that when differentiating a triangular waveform the output waveform should be square. So again, is the obtained picture a result of error (eg. from inaccuracies), or "human" mistake?










share|improve this question











$endgroup$




I have the following picture of the input and output voltage waveforms from a differentiating amplifier as shown below:



enter image description here



And the schematic is as shown below:



enter image description here



My teacher claimed that these waveforms are correct, but I am starting to feel some doubt.



I calculated the equation of the line for input voltage waveform from time t=0 ms to t = 0.5 ms (with respective voltages of -500 mV and 500 mV) to obtain: y= 2000x - 0.5, and thus the output voltage waveform should be at y=2 mV for t=0 ms to t=0.5 ms, which is not shown in this picture.



Is this something to do with errors, or simply a mistake?



Also, I am aware that when differentiating a triangular waveform the output waveform should be square. So again, is the obtained picture a result of error (eg. from inaccuracies), or "human" mistake?







operational-amplifier amplifier differential






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 24 at 7:43









pipe

10.4k42659




10.4k42659










asked Apr 24 at 7:09









MichelMichel

345




345











  • $begingroup$
    If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
    $endgroup$
    – jonk
    Apr 24 at 7:13







  • 1




    $begingroup$
    Your opamp is a LOW_PASS_FILTER.
    $endgroup$
    – analogsystemsrf
    Apr 24 at 7:13










  • $begingroup$
    I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
    $endgroup$
    – Michel
    Apr 24 at 7:22






  • 1




    $begingroup$
    This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
    $endgroup$
    – Mattman944
    Apr 24 at 7:35










  • $begingroup$
    If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
    $endgroup$
    – Huisman
    Apr 24 at 7:51

















  • $begingroup$
    If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
    $endgroup$
    – jonk
    Apr 24 at 7:13







  • 1




    $begingroup$
    Your opamp is a LOW_PASS_FILTER.
    $endgroup$
    – analogsystemsrf
    Apr 24 at 7:13










  • $begingroup$
    I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
    $endgroup$
    – Michel
    Apr 24 at 7:22






  • 1




    $begingroup$
    This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
    $endgroup$
    – Mattman944
    Apr 24 at 7:35










  • $begingroup$
    If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
    $endgroup$
    – Huisman
    Apr 24 at 7:51
















$begingroup$
If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
$endgroup$
– jonk
Apr 24 at 7:13





$begingroup$
If you provide the schematic, someone will point out the details involved. But since you already know that the differential of your triangular waveform is a square wave, imagine that at the output there is some resistance and capacitance. This will tend to "low-pass filter" the square wave, yes? What would that look like?
$endgroup$
– jonk
Apr 24 at 7:13





1




1




$begingroup$
Your opamp is a LOW_PASS_FILTER.
$endgroup$
– analogsystemsrf
Apr 24 at 7:13




$begingroup$
Your opamp is a LOW_PASS_FILTER.
$endgroup$
– analogsystemsrf
Apr 24 at 7:13












$begingroup$
I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
$endgroup$
– Michel
Apr 24 at 7:22




$begingroup$
I got told that the amplifier behaves as a high-pass filter, with a cutoff frequency of 2 kHz. Having said that, is it right to assume that the above waveform is a result of human error? Thank you.
$endgroup$
– Michel
Apr 24 at 7:22




1




1




$begingroup$
This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
$endgroup$
– Mattman944
Apr 24 at 7:35




$begingroup$
This is a high-pass filter. If R1 is small it will approximate a differentiator. How did you calculate the output? You didn't tell us what the values were, but the output looks reasonable to me.
$endgroup$
– Mattman944
Apr 24 at 7:35












$begingroup$
If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
$endgroup$
– Huisman
Apr 24 at 7:51





$begingroup$
If it were a LOW_PASS_FILTER then the ouput would look like more and more to a sine wave with smaller and smaller amplitude (depending on the filtering) .
$endgroup$
– Huisman
Apr 24 at 7:51











2 Answers
2






active

oldest

votes


















6












$begingroup$

Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.



In your case, we have



(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and



(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.



Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).



Hence, the output is as expected and, therefore, correct.






share|improve this answer











$endgroup$




















    1












    $begingroup$

    This circuit is not a pure differentiator due to the presence of R1.



    Its transfer function (assuming an ideal op amp) is:



    $$
    H(s)=-fracR_2R_1 fracss+frac1R_1C_1
    $$



    What represents a first order high-pass filter with cutoff frequency $f_0=frac12 pi R_1 C_1$



    Another way to interpret this is to break it down into two factors:



    $$
    H(s)=A(s)B(s)
    $$

    where $A(s)$ is the ideal differentiator (with negative gain):
    $$
    A(s)=-R_2C_1s
    $$

    and $B(s)$ is a low-pass filter:
    $$
    B(s)= fracfrac1R_1C_1s+frac1R_1C_1
    $$



    So, in your mental model you can visualize a triangular wave going through an ideal differentiator $A(s)$ and becoming a perfect square wave, which would then go through a low-pass filter $B(s)$ that would round its edges.



    Keep in mind that this analysis assumes an ideal op amp. Even if you remove $R_1$, the limited bandwidth of the op amp will also result in a similar effect where the cutoff frequency will be defined by the gain bandwidth product of the particular op amp you pick.






    share|improve this answer











    $endgroup$












    • $begingroup$
      joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
      $endgroup$
      – LvW
      Apr 24 at 10:23










    • $begingroup$
      @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
      $endgroup$
      – joribama
      yesterday











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.



    In your case, we have



    (1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and



    (2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.



    Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).



    Hence, the output is as expected and, therefore, correct.






    share|improve this answer











    $endgroup$

















      6












      $begingroup$

      Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.



      In your case, we have



      (1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and



      (2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.



      Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).



      Hence, the output is as expected and, therefore, correct.






      share|improve this answer











      $endgroup$















        6












        6








        6





        $begingroup$

        Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.



        In your case, we have



        (1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and



        (2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.



        Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).



        Hence, the output is as expected and, therefore, correct.






        share|improve this answer











        $endgroup$



        Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.



        In your case, we have



        (1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and



        (2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.



        Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).



        Hence, the output is as expected and, therefore, correct.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered Apr 24 at 7:41









        LvWLvW

        15k21330




        15k21330























            1












            $begingroup$

            This circuit is not a pure differentiator due to the presence of R1.



            Its transfer function (assuming an ideal op amp) is:



            $$
            H(s)=-fracR_2R_1 fracss+frac1R_1C_1
            $$



            What represents a first order high-pass filter with cutoff frequency $f_0=frac12 pi R_1 C_1$



            Another way to interpret this is to break it down into two factors:



            $$
            H(s)=A(s)B(s)
            $$

            where $A(s)$ is the ideal differentiator (with negative gain):
            $$
            A(s)=-R_2C_1s
            $$

            and $B(s)$ is a low-pass filter:
            $$
            B(s)= fracfrac1R_1C_1s+frac1R_1C_1
            $$



            So, in your mental model you can visualize a triangular wave going through an ideal differentiator $A(s)$ and becoming a perfect square wave, which would then go through a low-pass filter $B(s)$ that would round its edges.



            Keep in mind that this analysis assumes an ideal op amp. Even if you remove $R_1$, the limited bandwidth of the op amp will also result in a similar effect where the cutoff frequency will be defined by the gain bandwidth product of the particular op amp you pick.






            share|improve this answer











            $endgroup$












            • $begingroup$
              joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
              $endgroup$
              – LvW
              Apr 24 at 10:23










            • $begingroup$
              @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
              $endgroup$
              – joribama
              yesterday















            1












            $begingroup$

            This circuit is not a pure differentiator due to the presence of R1.



            Its transfer function (assuming an ideal op amp) is:



            $$
            H(s)=-fracR_2R_1 fracss+frac1R_1C_1
            $$



            What represents a first order high-pass filter with cutoff frequency $f_0=frac12 pi R_1 C_1$



            Another way to interpret this is to break it down into two factors:



            $$
            H(s)=A(s)B(s)
            $$

            where $A(s)$ is the ideal differentiator (with negative gain):
            $$
            A(s)=-R_2C_1s
            $$

            and $B(s)$ is a low-pass filter:
            $$
            B(s)= fracfrac1R_1C_1s+frac1R_1C_1
            $$



            So, in your mental model you can visualize a triangular wave going through an ideal differentiator $A(s)$ and becoming a perfect square wave, which would then go through a low-pass filter $B(s)$ that would round its edges.



            Keep in mind that this analysis assumes an ideal op amp. Even if you remove $R_1$, the limited bandwidth of the op amp will also result in a similar effect where the cutoff frequency will be defined by the gain bandwidth product of the particular op amp you pick.






            share|improve this answer











            $endgroup$












            • $begingroup$
              joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
              $endgroup$
              – LvW
              Apr 24 at 10:23










            • $begingroup$
              @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
              $endgroup$
              – joribama
              yesterday













            1












            1








            1





            $begingroup$

            This circuit is not a pure differentiator due to the presence of R1.



            Its transfer function (assuming an ideal op amp) is:



            $$
            H(s)=-fracR_2R_1 fracss+frac1R_1C_1
            $$



            What represents a first order high-pass filter with cutoff frequency $f_0=frac12 pi R_1 C_1$



            Another way to interpret this is to break it down into two factors:



            $$
            H(s)=A(s)B(s)
            $$

            where $A(s)$ is the ideal differentiator (with negative gain):
            $$
            A(s)=-R_2C_1s
            $$

            and $B(s)$ is a low-pass filter:
            $$
            B(s)= fracfrac1R_1C_1s+frac1R_1C_1
            $$



            So, in your mental model you can visualize a triangular wave going through an ideal differentiator $A(s)$ and becoming a perfect square wave, which would then go through a low-pass filter $B(s)$ that would round its edges.



            Keep in mind that this analysis assumes an ideal op amp. Even if you remove $R_1$, the limited bandwidth of the op amp will also result in a similar effect where the cutoff frequency will be defined by the gain bandwidth product of the particular op amp you pick.






            share|improve this answer











            $endgroup$



            This circuit is not a pure differentiator due to the presence of R1.



            Its transfer function (assuming an ideal op amp) is:



            $$
            H(s)=-fracR_2R_1 fracss+frac1R_1C_1
            $$



            What represents a first order high-pass filter with cutoff frequency $f_0=frac12 pi R_1 C_1$



            Another way to interpret this is to break it down into two factors:



            $$
            H(s)=A(s)B(s)
            $$

            where $A(s)$ is the ideal differentiator (with negative gain):
            $$
            A(s)=-R_2C_1s
            $$

            and $B(s)$ is a low-pass filter:
            $$
            B(s)= fracfrac1R_1C_1s+frac1R_1C_1
            $$



            So, in your mental model you can visualize a triangular wave going through an ideal differentiator $A(s)$ and becoming a perfect square wave, which would then go through a low-pass filter $B(s)$ that would round its edges.



            Keep in mind that this analysis assumes an ideal op amp. Even if you remove $R_1$, the limited bandwidth of the op amp will also result in a similar effect where the cutoff frequency will be defined by the gain bandwidth product of the particular op amp you pick.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered Apr 24 at 7:53









            joribamajoribama

            3615




            3615











            • $begingroup$
              joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
              $endgroup$
              – LvW
              Apr 24 at 10:23










            • $begingroup$
              @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
              $endgroup$
              – joribama
              yesterday
















            • $begingroup$
              joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
              $endgroup$
              – LvW
              Apr 24 at 10:23










            • $begingroup$
              @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
              $endgroup$
              – joribama
              yesterday















            $begingroup$
            joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
            $endgroup$
            – LvW
            Apr 24 at 10:23




            $begingroup$
            joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
            $endgroup$
            – LvW
            Apr 24 at 10:23












            $begingroup$
            @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
            $endgroup$
            – joribama
            yesterday




            $begingroup$
            @LvW - I wasn't very clear. I've updated my answer breaking down my thought process
            $endgroup$
            – joribama
            yesterday

















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