Active LPF or Integrator w/feedback Resistor

[Joshy]

Summary: Why would you choose a RC with a non-inverting op-amp instead of an integrator with feedback resistor for an active LPF?



I was at an interview not long ago and one of the classic questions popped up: Draw me an active LPF with gain. I've seen it plenty in my book (Sedra and Smith) and as well as feeling very comfortable with doing the hand-calculations myself, and so this is typically an acceptable answer. I just picked arbitrary values.

$$H(s) = \left ( -\frac{R_5}{R_4}\right ) \left ( \frac{1}{1+sC_2 R_5} \right )$$

Spoiler

I'll even scan it from Sedra and Smith this is from the 6th edition.

The interviewer strongly insisted that this was wrong and that everybody makes this mistake because the above is an integrator (with a feedback resistor). I was aware of it being very similar to an integrator, but I was a little bit shocked and have never had my previous answer rejected before. Maybe I've been wrong all this time? I'm okay with that, but I wanted to look into this a little bit more. I'm finding bits and pieces here that a lot more people than I was expecting insist on this too (even on these forums here); furthermore: I am noticing that a lot schematics in my other textbooks they seem to prefer using the filter she proposed instead.

$$H(s) = \left ( \frac{R_3 + R_2}{R_3} \right ) \left ( \frac{1}{1 + sCR_1} \right ) $$
I ran an ac simulation on the two. The results do not overlap perfectly, but I felt like they were close enough differences might be due to the non-ideal op-amp with a slight voltage off-set (one is inverting and the other is non-inverting). I wanted to try the transient simulation, but my computer just wasn't happy with it for tonight.

If I had to choose between the two, then I feel like I would personally prefer the one with less components. I feel like that this would be tempting for other engineers too, and so why would they choose otherwise as I have seen in some of their schematics? I haven't stumbled upon an answer I feel comfortable with. Thoughts? What am I doing wrong or possibly overlooking? Is it just as the old thread said, that I will only notice possible issues with real life application?

Thanks!

 

[Baluncore]

Thoughts? What am I doing wrong or possibly overlooking?

The two circuits will have the same response, except that one is inverting. That will show as a 180° phase shift in the AC analysis, and as an inverted signal in the transient analysis.
The interviewer may not understand the subject themselves, and so they may not accept that there are multiple correct solutions to the standard questions.

 

[Averagesupernova]

Draw me an active LPF with gain.

That seems like a loaded question/problem to me. As far as I'm concerned, the schematic that you drew is more correct than the one your interviewer drew. To me, active filter implies the actual filter elements in part of a feedback path.
-
One of the first things taught is that a resistor and capacitor in the form of an L network is an integrator when drawn the way your interviewer drew it. Then has the audacity to claim the opposite? Wow...

 

[LvW]

To me, active filter implies the actual filter elements in part of a feedback path.

Yes - I share this view. The second circuit (non-inverting) is a PASSIVE RC-lowpass with an additinal gain stage.

 

[DaveE]

There may be performance differences related to non-ideal op-amp behavior, such as limited GBW, et. al. Otherwise, they do the same thing. If you are operating in a realm where the difference matters, then there are about a million things to consider.

edit: For example, for a 1MHz GBW op-amp, the second circuit will have an additional pole at 100KHz that the first avoids.

 

[DaveE]

BTW, all op-amp integrators are LPFs, because the amp has limited gain. There's always an inverted pole at low frequency, although we always (safely) ignore it.

 

[Joshy]

Interesting feedback so far. I really appreciate it.

 

[LvW]

Yes - I share this view. The second circuit (non-inverting) is a PASSIVE RC-lowpass with an additinal gain stage.

One additional point just came into my mind:
In case you use a Current-Feedback Amplifier (CFA) instead of a classical Voltage-Opamp, you cannot use the inverting circuit - unless you place an additional resistor in series to C2 (but this will degrade the intergating function additionally). This requirement is due to the fact, that each CFA needs a resistor in the feedback loop (minimum values are given in the data sheets)

 

[onatirec]

My understanding is that integrators are necessarily a type of low pass filter.

Looking at a steady state frequency response in the Laplace domain (s->jw, where w is frequency), introducing an additional integration term (1/s = 1/(jw)) in the transfer function necessarily drives down the response magnitude at higher frequencies. Inversely, taking the derivative (multiplying by s=jw) amplifies the response at high frequencies.