I am working on building a crossover filter for a set of speakers and I have started by designing a simple second-order RC low-pass filter on a breadboard.

enter image description here

Assuming the circuit is wired correctly (I have tried other configurations) I can't figure out why the cut-off frequency is not where I want it to be.

The parts I used were 2 capacitors (0.125 μF and 0.136 μF) and the resistors were both 1 kΩ with 1% tolerance. I provided 2 V to the circuit via a signal generator (Hantek) and measured the change in voltage as a function of frequency.

For the 1st order filter with the same values (+- 10%) I got a cutoff frequency of roughly 1200 Hz. Based on my math (which could be completely wrong), I expected the second-order filter to be down 3 dB at around 750 Hz, but it is down 3 dB at 400 Hz. I am unsure as to whether my issue is circuit-related, or math-related. It may be behaving as expected, I just don't know. I have never taken an engineering class.

If there is a specific formula I should be using for second-order filters since I have seen a few, please let me know. Any ideas as to why the cut-off point might be lower than expected would be greatly appreciated. I have read a few other posts related to this but wasn't sure if they were applicable here.

  • 3
    \$\begingroup\$ Are you missing a ground connection between F and C? \$\endgroup\$ Commented Feb 8, 2023 at 3:02
  • \$\begingroup\$ Have you tried simulating it? With those values in LTspice I get 3 dB down at 447 Hz. \$\endgroup\$
    – GodJihyo
    Commented Feb 8, 2023 at 6:57
  • \$\begingroup\$ I did and it behaved just like you said @GodJihyo. I clearly am observing what is expected, I just don't know the math. \$\endgroup\$ Commented Feb 8, 2023 at 22:15
  • \$\begingroup\$ Where do you intend to insert this filter? \$\endgroup\$ Commented Feb 9, 2023 at 5:40
  • \$\begingroup\$ @BruceAbbott eventually in a speaker but I need to learn how to build different filter types with various slopes before I start worrying about phase matching or finding the right cutoff points for the drivers. I bought some cheapo Dayton speakers and rewired the internals so I can just pop components in and out and see what it does. \$\endgroup\$ Commented Feb 9, 2023 at 20:53

2 Answers 2


If you take the scope away, here's the circuit schematic for what you have built:


simulate this circuit – Schematic created using CircuitLab

C2 is not actually connected on the bottom side. You need to connect it to ground.

The "ground" clip of your oscilloscope probe is not shared with the ground pin of your signal generator. It instead acts as a reference point from which voltages are measured. Even if the two grounds were referenced to mains earth, you'd the path would have to go through the oscilloscope probe, through the scope, through its power cable, into the wall, back to the earth on the signal generator, through its power cable, through it, back through the cable to your breadboard. This would introduce a ton of resistance and inductance and make things misbehave. You need to explicitly wire the bottom of C2 to ground, on your breadboard, to make the filter work properly.

Another thing to keep in mind is that these breadboards have a lot of parasitic resistance, capacitance, and inductance. Analog circuits won't work very well on them, especially ones that rely upon small/precise capacitance and resistance values.

  • \$\begingroup\$ Thank you for the reply. I wish I could send more photos because I have tried additional configurations where it was explicitly grounded the way you said and it still had the same response. I did go back and include the proper grounding wire with the generator for this schematic just to check and I still had the issue. I imagine at this point the only possibility (assuming the scope is working and the parts aren't broken) is the math. I checked the impedance of the entire circuit and it was almost exactly 2kohms which is what I expected. \$\endgroup\$ Commented Feb 8, 2023 at 6:10
  • \$\begingroup\$ From what I understand, there is a very different formula for calculating higher order filters as opposed to first order (your usual 1/2piRC). \$\endgroup\$ Commented Feb 8, 2023 at 6:11
  • \$\begingroup\$ Yeah, the 2nd order cutoff frequency is not just a simple repeat of the calculation. You need to calculate the natural frequency and the damping factor, then use that to calculate the cutoff frequency. I usually use this tool. A cutoff frequency of 1200Hz is about right. \$\endgroup\$
    – Polynomial
    Commented Feb 8, 2023 at 6:36
  • \$\begingroup\$ This is good to know. I put the values into the calculator you sent and got a center frequency of 1200 Hz. Is this the same as the -3db point? If it is, then something must be wrong with the circuit because the cutoff point that I'm measuring is eerily close to if I just multiplied the R and C values together in the first-order equation. \$\endgroup\$ Commented Feb 8, 2023 at 15:46
  • \$\begingroup\$ I also downloaded some circuit simulation software so hopefully, that will help me learn how these circuits are supposed to behave. \$\endgroup\$ Commented Feb 8, 2023 at 16:06

Looking into this a bit what I find is that while you can find formulas for cascaded RC low pass filters, they only actually work if you put a buffer between stages. The formulas for first order RC filters are based on them being driven by a zero impedance source and having an infinite impedance load. When you start hanging one stage on the output of another the normal formulas go out the window.

One thing you can do, as mentioned in a comment on the linked page, is to make the impedances in the second stage much higher than the first stage. I simulated with the second resistor value multiplied by 10 and the second capacitor value divided by 10, the results were better but still a bit off, changing the values by a factor of 100 gets the simulated results pretty close to the calculated values.

Here is the simulation result, the calculated center frequency of 1220 Hz should be at -6dB. You can see it gets closer as the second stage impedance increases. enter image description here


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