# Voltage output waveform of a differentiating amplifier

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?

• 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?
– jonk
Apr 24, 2019 at 7:13
• Your opamp is a LOW_PASS_FILTER. Apr 24, 2019 at 7:13
• 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. Apr 24, 2019 at 7:22
• 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. Apr 24, 2019 at 7:35
• 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) . Apr 24, 2019 at 7:51

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.

(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.

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.

• joribama....are you sure? A lowpass filter? What will happen for s=0? Lowpass?
– LvW
Apr 24, 2019 at 10:23
• @LvW - I wasn't very clear. I've updated my answer breaking down my thought process Apr 25, 2019 at 21:40