Voltage output waveform of a differentiating amplifier

Question

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

And the schematic is as shown below:

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?

Answer 1

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.

Answer 2

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.