4
\$\begingroup\$

I have this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The op-amp is being run off 15 and -15V supply rails, and I am using a signal generator to input a sine wave of different frequencies with an amplitude of 2V, then using an oscilloscope to record the output wave. I am using this data to calculate the gain of the filter at different frequencies.

For this project I am required to produce a table of predicted values for the gain of the filter. I produced this table and my maximum gain was about 1.5. In practice, I had a maximum gain of almost 2. My question is, why is the gain higher in practice than in theory? I thought that it could be other impedances in the wires, but I reasoned that that shouldn't affect the gain since the impedance of both the feedback loop and the input would increase equally.

I calculated the gain using the capacitive reactance formula as well as the formulae for resistances in parallel and in series. For example, the expected gain at 2100Hz:

\$R_f = \frac{(2 \pi * 150*10^{-12}*2100)^{-1} * 68000}{68000 + (2 \pi * 150*10^{-12}*2100)^{-1}}\$

\$R_{in} = 33000+((2 \pi * 21000 * 10*10^{-9})^{-1})\$

\$Gain = -\frac{R_f}{R_{in}} \approx -1.5\$

Why is my theoretical result significantly different from the practical result?

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Where did you get 1.5 from? Leaving the capacitors aside, say around 5kHz, the gain is 68/33, assuming also you have a zero input impedance source and a reasonable load impedance. And what do series/parallel resistance formulae have to do with it? \$\endgroup\$
    – user207421
    Jul 9 '15 at 11:40
  • \$\begingroup\$ He may mean impedances in series and parallel : in which case if he linearly added instead of remembering the 90 degree phase shift and taking the hypotenuse ... \$\endgroup\$ Jul 9 '15 at 12:51
  • \$\begingroup\$ Even though you've received an answer, this is a good question and will help others in the future. Would you mind including the specific formula you used, as well as the calculated result that led to the 1.5 gain? \$\endgroup\$
    – Adam Davis
    Jul 9 '15 at 18:49
  • \$\begingroup\$ @AdamDavis of course \$\endgroup\$
    – imulsion
    Jul 9 '15 at 19:01
6
\$\begingroup\$

The transfer function is

$$H(s)=\frac{-sR_1C_2}{1+s(R_1C_1+R_2C_2)+s^2R_1R_2C_1C_2}$$

and the maximum gain is

$$A_{\text{max}}=\frac{R_1C_2}{R_1C_1+R_2C_2}=2.04$$

\$\endgroup\$
3
  • \$\begingroup\$ Absolutely correct and at a frequency of \$\dfrac{1}{2\pi\sqrt{R_1R_2C_1C_2}}\$ \$\endgroup\$
    – Andy aka
    Jul 9 '15 at 12:52
  • \$\begingroup\$ Now there's something I haven't learned yet! I suppose my calculations were wrong, I used capacitive reactance formulae to calculate the resistance of the feedback loop and the input - then used gain = -Rf/Rin. Thanks for the help :) \$\endgroup\$
    – imulsion
    Jul 9 '15 at 18:00
  • \$\begingroup\$ If you replace Rf/Rin by Zf/Zin everything is OK (Z=complex expression). \$\endgroup\$
    – LvW
    Jul 9 '15 at 22:34
3
\$\begingroup\$

I did a quick simulation and maximum gain is 6.007 dB at 2.407kHz: -

enter image description here

6.007 dB is a voltage gain of 1.997 so I guess your mathematical approach is flawed in some respects.

\$\endgroup\$
3
\$\begingroup\$

The expression derived by LvW is correct but there are no steps showing how he got there. Of course, if you compute the gain \$H(s)=-\frac{Z_1(s)}{Z_2(s)}\$ in which \$Z_1(s)=R_1||\frac{1}{sC_1}\$ and \$Z_2(s)=R_2+\frac{1}{sC_2}\$ you'll get there but a) you can make mistakes while developing the expression b) you will need to rearrange the formula to unveil the needed gain in the flat portion. I will show how the fast analytical techniques or FACTs (see this book) will get you there just by inspecting simple schematics, no algebra at all. Look at the below schematic:

enter image description here

You start by determining the transfer function in dc, for \$s=0\$ and you do that by opening all caps. You find \$H_0=0\$ because the first cap blocks the dc and creates a zero at the origin. Then, you reduce the excitation to 0 V and "look" at the resistance offered by the capacitor connection terminals when they are temporarily removed from the circuit. Associating the resistance with the involved capacitor gives you the corresponding time constants. Considering the virtual ground, it is easy to find \$\tau_1=R_1C_1\$ and \$\tau_2=R_2C_2\$. You can sum these two guys to form \$b_1\$ in the denominator. \$b_2\$ is is simply obtained by combining \$\tau_1\$ with a time constant obtained when \$C_1\$ is replaced by a short circuit: \$\tau_{12}=C_2R_2\$: \$b_2=\tau_1\tau_{12}=R_1C_1R_2C_2\$. There is redundancy meaning \$b_2=\tau_2\tau_{21}\$ in which you combine \$\tau_2\$ with a time constant obtained when \$C_2\$ is replaced by a short circuit. Results are identical. The denominator equals:

\$D(s)=1+sb_1+s^2b_2=1+s^2(R_1C_1+R_2C_2)+s^2R_1C_1R_2C_2\$.

From this second-order polynomial form, we can define a quality coefficient \$Q=\frac{\sqrt{b_2}}{b_1}\$ and a resonant frequency \$\omega_0=\frac{1}{\sqrt{b_2}}\$.

The numerator can be derived using the generalized transfer function form obtained by calculating three gains when each capacitor is alternatively set in its high-frequency state (a short circuit) and when both caps are replaced by a short circuit. You can see from the sketches that a gain \$H_2\$ exists only when \$C_2\$ is replaced by a short circuit. The gain in this case is \$H^2=-\frac{R_1}{R_2}\$. The numerator is defined as:

\$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{12}\tau_1\tau_{12}=-sR_1C_2=-\frac{s}{\omega_z}\$ where \$\omega_z=\frac{1}{R_1C_2}\$.

With a proper factorization, you can rework this transfer function in a low-entropy form in which you immediately see the gain you wanted:

\$H(s)=-H_{bp}\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}\$ in which the bandpass gain is simply \$H_{bp}=\frac{R_1C_2}{R_1C_1+R_2C_2}\$. With the given values, that bandpass gain is exactly 1.9988 or 6.015 dB. All calculations appear in the below Mathcad screenshots:

enter image description here

enter image description here

You can apply the FACTs to passive or active circuits. The nice thing is that you individually determine the coefficients of the numerator and the denominator via small sketches. That way, if a deviation exists between the raw formula and what you have derived, it is easy to solve the guilty intermediate step and correct it. With a brute-force analysis, you would have to restart from scratch. An introduction to the FACTs can be found here. I encourage students and engineers to acquire that skill given the ease and speed it provides when analysing transfer functions.

Addition to the original answer following a suggestion from G36:

A question was raised by G36 in the comments section on how to apply the FACTs in case \$C_1\$ was in series with \$R_1\$ rather than in parallel as in the original question? If you try to apply the FACTs and determine the time constants of this new circuit, you will find infinite values for \$\tau_1\$ and \$\tau_2\$ because there is no dc path when both capacitors are removed from the circuit (and \$V_{in}\$ reduced to 0 V). This is an indeterminacy and it must be removed. The simplest trick consists of adding a dummy resistance \$R_3\$ in parallel with \$C_1\$ and update the circuit as below:

enter image description here

The new picture to determine the time constants is shown below and you can see that the dc path created by the addition of \$R_3\$ helps us getting rid of the two indeterminacies:

enter image description here

You can now nicely determine all the time constants including \$R_3\$ and follow the path given in the original reply. If everything goes well, you should be find:

\$H(s)=-\frac{sC_2(R_1+R_3)+s^2R_1R_3C_1C_2}{1+s(R_3C_1+R_2C_2)+s^2R_3R_2C_1C_2}\$

Now factor \$R_3\$ in the numerator and the denominator then simplify:

\$H(s)=-\frac{sC_2(\frac{R_1}{R_3}+1)+s^2R_1C_1C_2}{\frac{1}{R_3}+s(C_1+\frac{R_2C_2}{R_3})+s^2R_2C_1C_2}\$

Now, when \$R_3\$ approaches infinity, this expression becomes:

\$H(s)=-H_{LF}\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ with \$H_{LF}=\frac{C_2}{C_1}\$, \$\omega_z=\frac{1}{R_1C_1}\$ and \$\omega_p=\frac{1}{R_2C_2}\$.

The below Mathcad pictures confirm these calculations. First, \$R_3=100\;k\Omega\$:

enter image description here

It clearly confirms the gain equal to 0 in dc:

enter image description here

When \$R_3\$ is now pushed to infinity, the dynamic response turns into the response of a zero and a pole while the zero at the origin has disappeared:

enter image description here

Which clearly isn't true as if you dc-bias this circuit (without \$R_3\$), you should get 0 V in the output. Considering the absence of dc path in this case, I am not sure if this circuit has any physical sense or application as such?

\$\endgroup\$
10
  • \$\begingroup\$ But what if C1 will also be in series? We will have a pole and zero. Pole at fp = 0.16/(R2C2) and Zero at Fz = 0.16/(R1C1). Is Ho still equal to 0? If so, H2 is also zero and H12 = -R1/R2 ? \$\endgroup\$
    – G36
    Jul 16 '17 at 15:43
  • \$\begingroup\$ Hello G36, it is good to consider the FACTs as a tool to study this filter. If \$C_1\$ is now in series with \$R_1\$, you introduce another zero at the origin. \$H_0\$ is indeed 0 (all caps are open in dc), \$H^2\$ is also 0 and when both caps are shorted, \$H^{12}=-\frac{R_1}{R_2}\$. \$D(s)\$ is also affected by this change and you must rework the natural time constants. \$\endgroup\$ Jul 16 '17 at 17:27
  • 1
    \$\begingroup\$ I think both time constants in this case are infinite as there is no dc path at all. \$H^2\$ when \$C_2\$ is a short equals \$-A_{OL}\$ and not 0 as I wrongly said. I am not sure a circuit like that has a physical meaning actually? If you compute \$H(s)=-\frac{Z_1(s)}{Z_2(s)}\$ it gives \$H(s)=-\frac{C_2}{C_1}\frac{1+sR_1C_1}{1+sR_2C_2}\$ and it does not predict a dc gain of 0. \$\endgroup\$ Jul 16 '17 at 21:43
  • 1
    \$\begingroup\$ @G36, in case you want to apply the FACTs to a configuration in which \$C_1\$ is now in series with \$R_1\$, you can see that there is no dc path when trying to compute the natural time constants \$\tau_1\$ and \$\tau_2\$. The trick is to add a dummy resistor \$R_3\$ in parallel with \$C_1\$ which provides a dc path. Once all parameters are derived, bring \$R_3\$ to infinity. You then have a perfect match: \$\tau_1=R_3C_1\;\tau_2=C_2R_2\;\tau_{12}=C_2R_2\;H_0=0\;H^1=0\;H^2=-\frac{R_1+R_3}{R_2}\$ and \$H^{12}=-\frac{R_1}{R_2}\$. Do not hesitate to simulate these sketches if a doubt exists. \$\endgroup\$ Jul 17 '17 at 7:28
  • 1
    \$\begingroup\$ @G36, following your question, I have updated my original answer with the circuit of yours and derived the new transfer function with and without \$R_3\$. Let me know if it now makes sense. \$\endgroup\$ Jul 17 '17 at 10:26

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.