I have an OPA548 IC which is a high-power op-amp. I am using it in non-inverting amplifier configuration as shown in the schematic. enter image description here

The input signal is a ramp signal with low voltage 0.1 V and high voltage 1 V @ 1 kHz. I have added a RC low pass filter at the output because I am trying the circuit in a breadboard and wires leading to high frequency noise. I have used a load as 50-ohm resistor for simulation and to test the output before connecting to the actual equipment. The circuit works fine.

But when I connect the actual equipment for which this circuit is designed, I get an output like something shown below: Channel 2(blue) is the voltage at opamp output. Channel 4(green) is the voltage after LPF when connected to equipment. enter image description here

If I change the signal frequency to 100 Hz the output is: enter image description here

The output for 10 Hz is normal as expected. enter image description here

At higher frequency, the output is not as expected. The load is an equipment which has some kind of piezo driver. So, I think the load is not purely resistive, but how should I modify the circuit to work as expected for the reactive load? or the filter can be modified any way to work for any load?

Please don't recommend using a buffer stage, any other suggestions would be helpful?

  • \$\begingroup\$ It seems like the signal is correct at the op-amp output. So maybe you need to take some feedback from VF1. For example, maybe a capacitor or RC in parallel with R5 could help. \$\endgroup\$
    – user57037
    Commented Dec 18, 2022 at 6:28
  • \$\begingroup\$ Is the opamp stable if you remove the output series resistor? \$\endgroup\$
    – tobalt
    Commented Dec 18, 2022 at 7:33
  • \$\begingroup\$ Kishore, nicely prepared question. +1. \$\endgroup\$
    – jonk
    Commented Dec 18, 2022 at 8:53
  • \$\begingroup\$ Kishore, just a few thoughts to start out. (I'm not engaged. Just talking out loud.) Normally, you want to have some specification for the capacitance, the resonant frequency, and supply voltage range. I think you've proven that the supply voltage range is fine. And if you are well below the resonant frequency then the piezo "thing" may likely be modeled as a capacitor. Not 'purely resistive' for sure. Just looking at the curves, you could easily work out the impedance using FFT, division, and an inverse FFT. (Which is what I'd do, perfunctorily.) But any spec info on the piezo? \$\endgroup\$
    – jonk
    Commented Dec 18, 2022 at 9:21
  • 2
    \$\begingroup\$ The down step in your sawtooth wave requires an infinite output current to discharge the capacitance of your load. That's not going to happen, so you will always have a problem. Do you require a waveform with an abrupt step in it, or can you specify a finite slew rate / bandwidth that will meet your system requirements? If so, once you have a practical waveform, you can do things like add a lowpass filter with series inductance to transform your capacitive load into a resistive one, which will obviously only work over a limited bandwidth. Either that, or add a FET pulldown directly at the load \$\endgroup\$
    – Neil_UK
    Commented Dec 18, 2022 at 10:33

1 Answer 1


I think that using such integrator circuit on opamp output with high capacitance is poor decision. It is simple cicruit, called dominant-pole compensation, used often to increase stability, but this circuit could also easilly kill good design. Output RC filter creates additional pole in Bode diagram and if cutoff frequency is pretty high (near frequency of 1 dB gain) it can add a phase shift and create instability in usually stable opamp. I saw many designs where no output capacitor is used (just resistor) and RC filter creates with input capacitance of cable (since as I understood you used long cable, it have pretty high capacitance for opamp to drive). Also you should keep in mind that amplitude of harmonis of your sawtooth signal decreasing ~1/N, where N is number of harmonic. I suggest you to use RC-snubber at the output after R5 and remove capacitor (or you can try with it while making on breadboard). You should choose values experimentally. Also you can reserve place for small (few pF) capacitor and place it between output of opamp and inverting output and try to place feedback circuit after R5. Sorry, I can't help with actual values because if you want precise circuit with opamp - it is often huge work on getting right passive element's values.


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