# How to predict the magnitude of output voltage in a practical differentiator?

I designed the following differentiator circuit in LT Spice

and the output waveform was as following:

Now my question is how do I predict the magnitude of output theoretically? The best I have been able to do so far is look at Wikipedia's page on Differentiator and see the frequency response to figure out gain. However, the calculations do not seem to match.

The first pole $$\f_1=\dfrac 1{2\pi R_1C}\$$ comes out to be $$\\approx 16kHz\$$, so some crude calculations suggest that at $$\f=1kHz(\approx 1.6kHz)\$$, $$\ V_{out} \approx \dfrac{V_{in}}{10}\$$.

but this is not the case.

So what am I doing wrong? and is there a better way to solve this?

• Your circuit offers a transfer function made of $R_2$ divided by the series connection of $R_1$ and $C_1$. Compute the magnitude of $H(s)=-\frac{R_2}{R_1+\frac{1}{sC_1}}$ at 1 kHz (please use a small k for kilo) and it should do well. Commented May 23, 2021 at 8:36
• @VerbalKint sorry for using 'K' my bad (I've corrected it now). However, I don't know what $H(s)$ is or what $s$ means in $\dfrac 1 {sC_1}$. if you could explain, that would be helpful. Thanks! Commented May 23, 2021 at 8:45
• Ooops, then things become more complicated and won't be easily answered in a comment window : | $s$ refers to the Laplace transform. Do you know how to derive the transfer function of a simple op-amp-based inverter with two resistors? Is this a hobby or a school exercise? Commented May 23, 2021 at 8:52
• @VerbalKint This is a school exercise but we haven't been taught Laplace transformations .Also, all I know is transfer function is the ratio of output voltage to input voltage. So I am guessing there is another way to solve this but I cannot find it. Commented May 23, 2021 at 9:05