1
\$\begingroup\$

A practical differentiator has two cut-off frequency but can no longer perform differentiaiton after first cut-off. After first cut-off there is a 20db per decade gain reduction but why does this affect the mathematical function itself. The DC equation still remains $$ V_{out}=-R_fC_f \frac{dV_{in}}{dt}. $$ In a low-pass filter the signal just attenuates and does not change mathematically, is something different here. Is it just that it is still differentiating but for some reason my output is distorted or is there a mathematical explanation behind this. I am using a standard practical differentiator as shown below and I am sure the values I have are correct.

enter image description here

enter image description here

\$\endgroup\$
  • \$\begingroup\$ What is the purpose of Cf? You can design a stable (practical, non-ideal) diff. circuit even for Cf=0. \$\endgroup\$ – LvW Sep 2 at 8:26
2
\$\begingroup\$

Since you are talking about a non-ideal case, then you can no longer use the ideal transfer function. What you have shown has this transfer function:

$$\begin{align} Z_f&=\frac{1}{\frac{1}{R_f}+sC_f} \\ Z_1&=R_1+\frac{1}{sC_1} \\ H(s)&=\frac{Z_f}{Z_1} \\ H(s)&=\frac{1}{\left(R_1+\frac{1}{sC_1}\right)\left(R_f+\frac{1}{sC_f}\right)} \\ H(s)&=\frac{\frac{s}{R_1C_f}}{s^2+\frac{R_fC_f+R_1C_1}{R_fR_1C_fC_1}s+\frac{1}{R_fR_1C_fC_1}} \end{align}$$

So you can see that it's a bandpass now, which matches the frequency plot in your second picture. Even if you have this transfer function, you cannot talk about a corner frequency for a differentiator, since, if there is a pole, it's no longer a differentiator, it's a highpass, at least. And things get a bit more complicated due to the internal pole of the opamp, parasitic indictances of the elements, themselves, but for a general (and, usually, whithin the necesssary bandwidth) case, this will do, and you can call it a differentiator.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ hey, thanks for answer. Also, I wanted to ask that after cutoff, where can I read more about what happens to output or is it just complete nonsense once cutoff is reached. \$\endgroup\$ – vikas sajanani Sep 2 at 6:59
  • 1
    \$\begingroup\$ @vikassajanani I can't see what's wrong and why it won't format it. At any rate, it's of the form s/(s^2+s+1). I've modifed my answer to include that you can no longer talk about it being a differentiator once a pole is reached. \$\endgroup\$ – a concerned citizen Sep 2 at 7:01
0
\$\begingroup\$

Removing Cf should flatten the curve when Z matches Z(Cin)=Rin until f=GBW/G of OA , OpAmpBut BUT with Z(Cf)=Rf at the same f =GBW/G you have a 2ndorder LPF

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.