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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?

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  • \$\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
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    \$\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
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    \$\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
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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.

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

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

$$ H(s)=-\frac{R_2}{R_1} \frac{s}{s+\frac{1}{R_1C_1}} $$

What represents a first order high-pass filter with cutoff frequency \$f_0=\frac{1}{2 \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: $$ B(s)= \frac{\frac{1}{R_1C_1}}{s+\frac{1}{R_1C_1}} $$

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.

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  • \$\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 Apr 25 at 21:40

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