Using the Analog Devices Filter Design Tool, I have created a band pass filter for a 1MHz input with 20dB of gain. I have implemented the circuit in LTSpice with the expected behaviour, however when I implement it on a breadboard I am not getting the correct amount of gain (but the filtering seems to work when I vary the input frequency).

For my voltage reference I am using a MAX6001 1.25V reference. I have tested this with different bypass capacitors, finding little difference.

My thoughts are that perhaps I have wired the circuit wrong?

Simulated output and circuit: enter image description here enter image description here


Output from Picoscope (Red input signal, blue output signal): enter image description here

  • 2
    \$\begingroup\$ Do you have any idea how much capacitance sits on that wireless protoboard, between each row? Try wiring all that as dead-bug or manhattan style and see, for starters. Or choose a design that isn't depending upon 47 pF caps. Maybe something at least 10 times larger, maybe more? \$\endgroup\$ Commented Jul 17, 2023 at 9:34
  • \$\begingroup\$ @periblepsis, Interesting, thank you for your reply. Due to the nature of the design, I wish to keep the capacitance low. From what you are saying, the lower than expected gain is from the use of the breadboard, rather than the circuit itself? \$\endgroup\$
    – LK26
    Commented Jul 17, 2023 at 9:38
  • \$\begingroup\$ @LK26 what's that requirement that keeps the capacitances low? Note how that also makes the resistors low-ohmic (i.e., "strong"), meaning you're significantly loading the outputs of the opamps, which probably is a bad idea, as well. \$\endgroup\$ Commented Jul 17, 2023 at 9:39
  • \$\begingroup\$ @MarcusMüller I need my design (whole design, not just this circuit) to be intrinsically safe, so wish to keep capacitance low \$\endgroup\$
    – LK26
    Commented Jul 17, 2023 at 9:41
  • 1
    \$\begingroup\$ 180 MHz op-amps "on breadboard" likely explains it. You need a proper PCB which is designed very carefully to avoid parasitic capacitances and inductances from affecting the performance. \$\endgroup\$
    – Justme
    Commented Jul 17, 2023 at 9:42

1 Answer 1


The midband-gain of the 1st stage for the shown circuit for two equal capacitors is (assuming ideal opamp properties): -

Gmax,1=R5/2R4=4.16 (12.4 dB).

Similarly, for the 2nd stage we get

Gmax,2=22k/5.4k=4.075 (12.2 dB)

Question 1: Did you use the filter tool correctly (because your design goal was 20dB gain) ?

Question 2: What did you measure for each stage (separately)?

Question 3: Did you check the slew rate of the opamp used?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.