When I put the following narrowband active bandpass circuit on a breadboard, using a MCP601 single-supply CMOS op-amp, it just oscillates at about 339 kHz with a Vpp of about 1.36 V at a supply of 5 V. When I apply a signal generator (sine wave) to the input, sweeping between 100 and 900 kHz, it doesn't seem to do much at the output other than mix with the oscillation sine wave. How do I solve the oscillation problem and get a working bandpass filter? The objective is to have an active bandpass filter with a center somewhere in the low to mid kilohertz range. Right now I don't care about precision of the center frequency nor precision of the bandwidth. Tuning that can come later once I get a better grasp on the oscillation issue.

MCP601 active bandpass attempt

I realize I can design a passive bandpass and amplify the output of that and call it a day, but I'd prefer first to understand what I'm doing wrong with this active design before I give up on it.

Side note: I don't have a working LTSpice MCP601 model but when I simulate with other single-supply CMOS op-amp models that come with LTSpice, I get good results (i.e, it acts as a bandpass in kHz range without oscillating).

UPDATE: I made the mistake of using R6 to model the next stage of the circuitry on my breadboard, which was in actuality a higher impedance than 5k ohms. Once I removed that circuitry and replaced it with an actual 5k resistor, I had better luck. Perhaps the lower-Z load is what solves the oscillation problem? Although I'm not entirely sure why. So if that's the case, I think the question is still valid in terms of understanding how to properly size the output load on this active filter to avoid oscillation (where next stage is high-Z)?

UPDATE #2: Just swapping various resistors out for R6, the best seemed to be R6=10k (which, notably, is the same as R3). I say best because it didn't oscillate at 10k and there was least amount of voltage attenuation.

Further updates:

  • I have discovered this circuit is called a Multiple Feedback bandpass filter. Reference: Linear Circuit Design Handbook, 2008, Analog Devices, Ch. 8, Analog Filters, pp. 8.75-8.76 and p. 8.94. There are design equations on p. 8.94.

Closing thoughts:

  • There is unanimous consensus that the breadboard is one of the primary problems.
  • The accepted answer mentions the TI Filter designer. I found this tool to be helpful in looking at the design of this kind of filter. It shows minimum op-amp specs needed in order to achieve results for the desired filter response.
  • The TI Filter designer shows that the MCP601 op-amp has a GBW specification that is not even close to the minimum needed for the specifications I provided in the comments section. The accepted answer mentions the possibility of cascading lower Q filters to achieve results but I think another reasonable conclusion is that a multiple feedback filter is not the right approach for the Q at the center frequency that I need. The accepted answer mentions more realistic filter types, such as ceramic filters or crystal filters.
  • Despite the hunt n' peck approach being not ideal, it's still notable that I was able to solve the oscillation problem with a low enough resistor-to-ground (R6) on the output. However, as stated above, even though it's not oscillating anymore, clearly I will still have problems with the filter. Rather than hunt n' peck, it's better to design with equations and for that I found equations for multiple feedback filters in the Analog Devices Linear Circuit Design Handbook, cited above.
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    \$\begingroup\$ What type of breadboard are you referring to? The common plug-in ones have terrible parasitics. It's not obvious where coupling could cause a problem though. That 50 ohms value of R2 looks rather low - the opamp will need lots of GBW and low noise to function.reasonably \$\endgroup\$ Commented Sep 1, 2021 at 22:57
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    \$\begingroup\$ 10uF decoupling alongside that 0.1uF (C3) is probably a good idea. \$\endgroup\$
    – user16324
    Commented Sep 1, 2021 at 23:04
  • \$\begingroup\$ Typical hobbyist breadboard (so yeah, inexpensive type). Looks to be about 64 rows, with 5 columns on the left and 5 on the right. And then there are power and ground strips on both the right side and left side. I have coupled left and right power rails with jumper wire and have done the same for the left/right ground rails. \$\endgroup\$
    – acker9
    Commented Sep 1, 2021 at 23:04
  • \$\begingroup\$ Well known designer's maxim: Oscillators don't. Amplifiers do. \$\endgroup\$
    – elchambro
    Commented Sep 2, 2021 at 0:58
  • \$\begingroup\$ @acker9 As Kevin indicates, the parasitics of a wireless breadboard are horrible. Figure 5 pF between every node. Including between the opamp output and its (+) input node. Do you have any calculations handy for the various resistor and capacitor values? \$\endgroup\$
    – jonk
    Commented Sep 2, 2021 at 1:46

1 Answer 1


There is opportunity for positive feedback to Vin+ with layout issues. The ground loop to C4 must be carefully selected not to share output current with anything by capacitance or loop inductance.

The GBW product of your Op Amp must higher than a 1st order system.

In fact, my discovery was \$ GBW min= Q^2 f_o Av\$ for Q>>1 This will reduce your gain margin and transfer function if not met. Thus, in order to solve this you must cascade lower Q BPF filters to achieve your result or use the part with the required GBW.

  • This will contribute to a tendency to oscillate, but the above positive feedback is the real issue. Supply decoupling must be very close to IC.
  • Breadboard long jumpers are bad news for inductive ground loops. Although I once had success with twisted pair magnet wire soldered to resistor wire.

TI Filter designer will easily verify GBW as I figured out the sensitivity for the required Q. (Not the oscillation result though, just the required GBW)

Final Remarks

This design appears to demand a BPF performance with f0 around 820 kHz with a BW less than 160 Hz. This translates to a Q well over 5000 whereas Q > 100 is impossible to achieve any accuracy. You would need a GBW product with doesn't exist yet well over 1 GHz to achieve 0 dB gain at centre frequency and it would still likely oscillate from stray capacitance on the order of 0.5pF to the non-inverting input. At these Bandwidths all impedances must be well under 500 Ohms.

The attenuation ratio from 5k to 50 Ohms is the tell in this insane design.

If you needed a Q of > 5k you would use: Ceramic resonator or custom Xtal resonator (Q=10k to 100M $$$ OCXO) or some mechanical resonator with 10 ppm tolerances. Alternatively you could gang 1000 Op Amps each with a Q of 5 and maybe have a lot of ripples and unexpected spikes and notches from mismatched phases and 1% parts. Also the GBW of each Op Amp would need to be 25* fo for unity gain error correction to within 1 deg.

  • \$\begingroup\$ I'm accepting this answer because it seems there's unanimous consent that breadboard parasitics are the main problem. Also the pointer to the TI Filter designer tool is helpful. Your GBWmin equation is interesting but I lack enough background on this equation to make use of it. If this equation is published, or if you derived it from published material, footnoting it with a reference to a source would be great. \$\endgroup\$
    – acker9
    Commented Sep 3, 2021 at 14:07
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    \$\begingroup\$ I invented the equation based on observing many trial and calculations using TI 's filter program and backup with Falstad's afilter browser based design tool. you can do the same or take my word for it or find an error if you doubt it. It is logical to me as the phase sensitivity of an integrator GBW effect on the phase sensitivity of Q increases with Q greatly \$\endgroup\$ Commented Sep 3, 2021 at 15:04
  • \$\begingroup\$ I am not doubting you at all, I just was hoping for a little more background, which you provided. Thank you! (P.S. I meant 'unanimous consensus' in my first comment rather than 'consent', of course. Darn typos.) \$\endgroup\$
    – acker9
    Commented Sep 3, 2021 at 15:20
  • \$\begingroup\$ Just remember my formula or consider uncompensated video amps for this task \$\endgroup\$ Commented Sep 3, 2021 at 16:33

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