I am relatively new to analyze circuits in the Laplace domain. So I decided to solve some problems as an exercise.

Below presented a problem and my solution to it and I have two questions:

  1. Once I have found Vout/Vin in the laplace domain. What is the actual gain. For example, suppose the input is a sine wave with amplitude 1V and frequency of 1kHz, How do I interpret the answer which is a function of s to an actual gain?

  2. Is my analysis of the transfer function correct? My main concern is that in equation (2) I considered only the resistance of C1 in (Vp - Vout), while the resistance of R2,C2 may also indirectly effect on this potential difference, since Op+ and hence Op- are effected from their resistances. Or, am I overthinking this?

Below The exercise and my solution. Thanks!

enter image description here enter image description here

  • \$\begingroup\$ Just a fun fact: That's a sallen-key topology. \$\endgroup\$ – Harry Svensson May 22 '19 at 6:37
  • \$\begingroup\$ user135172, the midterm in the denominator is not correct. \$\endgroup\$ – LvW May 22 '19 at 10:01

To determine the transfer function of such op-amp-based circuit, it is easy to apply the fast analytical circuits techniques or FACTs. The exercise is quite simple: determine the natural time constants of this circuit when the source (\$V_{in}\$) is reduced to 0 V or replaced by a short circuit in the electrical diagram. To determine time constants, simply "look" at the resistance offered by \$C_1\$ and \$C_2\$ connecting terminals when they are temporarily disconnected from the circuit. This will give you \$\tau_1\$ and \$\tau_2\$. Summing them leads to the first term \$b_1=\tau_1+\tau_2\$. The second high-frequency term \$b_2\$ is obtained by combining \$\tau_2\$ and another term \$\tau_{21}\$. This second term implies that capacitor \$C_2\$ is set in its high-frequency state (a short circuit) while you determine the resistance offered by \$C_1\$ connecting terminals. You finally assemble the terms as follows:


The below sketch shows you the way. Start with \$s=0\$ and open all caps. The gain in this mode is 1: no leading term for the final transfer function. Then proceed by determining the time constants. Once it is done, you have your transfer function without writing a single line of algebra!

enter image description here

You can capture your formulae in a Mathcad sheet and rearrange it to express the final result in a low-entropy format: a quality factor \$Q\$ (or a damping ratio if you like it) and a resonant angular frequency \$\omega_0\$. It is easy to do because the FACTs naturally lead you to a formalized denominator form with \$b_1\$ and \$b_2\$. This is the correct way to express a transfer function.

enter image description here

The FACTs are not only faster than any other methods but they naturally deliver a clear and ordered form fitting the low-entropy format. This format is necessary to let you design your circuit so that you meet some of the desired criteria: quality factor and resonant frequency. Furthermore, for simple circuits like this one, you can determine the transfer function by inspection, without writing a line of algebra. Should you make a mistake, simply solve one of the intermediate sketches without restarting from scratch. A truly powerful skill that I encourage students and EEs to acquire: once you master it, you won't return to classical analysis.

  • \$\begingroup\$ Will try this method now, Thanks! \$\endgroup\$ – user135172 May 23 '19 at 14:05
  • \$\begingroup\$ Good idea! Have a look at this tutorial then cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ – Verbal Kint May 23 '19 at 14:40
  • \$\begingroup\$ Is FACTs related to the Extra element theorem (EET)? I see that your book is linked under the "Further Reading", just wondering if FACTs is the same thing as EET or not. \$\endgroup\$ – Harry Svensson May 23 '19 at 15:09
  • \$\begingroup\$ The FACTs are an extension of the EET and the 2EET. However, you don't need to learn the EET to start applying the FACTs. The best approach is to start with simple circuits and gradually increase complexity. Then, once this is fully understood, learn the EET which is also an extremely valuable tool. \$\endgroup\$ – Verbal Kint May 23 '19 at 15:57

For your analysis please see this website - it implies your derivation is incorrect because your final equation doesn't match their final equation (which I know to be correct): -

enter image description here

Once I have found Vout/Vin in the laplace domain. What is the actual gain. For example, suppose the input is a sine wave with amplitude 1V and frequency of 1kHz, How do I interpret the answer which is a function of s to an actual gain?

The form of the equation shown above is more practical to deal with and it results in these properties: -

  • \$\dfrac{1}{R_1R_2C_1C_2}\$ = \$\omega_n\$ the natural resonant frequency
  • \$2\cdot\zeta\cdot\omega_n\$ = \$\bigg [\dfrac{1}{R_2C_1} +\dfrac{1}{R_1C_1}\bigg ]\$

So, if you understand \$\zeta\$ and \$\omega_n\$ then you should understand my answer. If you don't understand those terms you ought to do a bit of research into Sallen-key filters and 2nd order filters in general (or ask another question).

Try and do some research on the general properties of a 2nd order low pass filter and understand this equation: -

enter image description here

Then, it's just a matter of converting s to jw and calculating out the gain at whatever frequency you want. This website might also provide some help in understanding the above formula and how the bode plot and pole-zero placement fit together.

  • \$\begingroup\$ Thanks for the links and details! \$\endgroup\$ – user135172 May 23 '19 at 14:05

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.