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I made a pi attenuator with R1 = R2 = 1.2 MΩ and R3 = 1.45 MΩ. When supplying 5.2 V peak-peak at a range of 200 kHz-1 MHz, the rate of change for the attenuation factor is very large. What can I do to change that?

The intended attenuation factor was 10x. Experimentally, I am getting a range between 25x-50x depending on the frequency supplied.

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  • \$\begingroup\$ Chadi Harmouche - Hi, You wrote an "answer" but it appears to be replying to another answer. That breaks the rules here :( || Since you asked the question, unless you are writing the full & final answer to your own question (i.e. unless you don't need further help & you solved the problem yourself using a different solution than any other answer) please don't use the box labeled "Your Answer" below. Instead, to add important technical info / clarification, please edit the question. Or comment to respond to an answer or another comment with minor / trivial / temporary updates. \$\endgroup\$
    – SamGibson ♦
    Commented Oct 4, 2023 at 2:40
  • \$\begingroup\$ (continued) (There's a difference between "answer" & "comment" here.) || Please see the advice here about: "What should I do when someone answers my question?". If that answer has worked, as you said, please consider "áccepting it" as explained in the link I gave above. || At least temporarily, despite it not strictly being the correct process, I will convert your (non-)answer into a comment, so the answer's author sees it. || The rules here differ from typical forums, please see the tour & help center. Thanks. \$\endgroup\$
    – SamGibson ♦
    Commented Oct 4, 2023 at 2:40

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More information is really needed to properly answer your question. This includes the source and load impedances and the method you used to construct the circuit. However given the frequencies you are using are in the MHz region and the values of resistance you are using are in the megohm region, the most likely source of your problem is the parasitic capacitances of the resistors and your circuit board. Note that 1 picofarad (which is low for parasitic capacitance) has an impedance of 0.159 megohms at1 MHz. Thus it will severely affect your attenuator. You need to lower the value of your resistors by at least a factor of 1000 if you want your attenuator to work at MHz frequencies.

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  • \$\begingroup\$ Thank you for your help. This worked. Sorry for lack of information. You are a lifesaver. \$\endgroup\$
    – Chadi Harmouche
    Commented Oct 4, 2023 at 2:21

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