I wanted to find the corner frequency of this circuit : enter image description here

I run the simulation but couldn't find the frequency using .MEAS AC f FIND frequency WHEN V(o)=-3dB command. I am getting this error in SPICE Error Log :

Measurement "f" FAIL'ed

Date: Wed Jan 27 11:59:28 2021
Total elapsed time: 0.067 seconds.

tnom = 27
temp = 27
method = trap
totiter = 0
traniter = 0
tranpoints = 0
accept = 0
rejected = 0
matrix size = 3
fillins = 0
solver = Normal

I am able to find approx corner frequency at -3dB using AC analysis using this command .ac dec 1000 1 1Meg

  • 1
    \$\begingroup\$ "3 dB" is meaningless. 3 dB of what? 100 kV? 13 nA? 1.1 mW? Also you'll need mag(V(o)), and maybe even .opt meascplxfmt=cartesian added. Look in the help under LTspice > Dot Commands > .MEASURE for an example of measuring the bandwidth of a bandpass filter and see if you can't adapt that to your case. \$\endgroup\$ Jan 27, 2021 at 8:56
  • \$\begingroup\$ @aconcernedcitizen Thank you for your comment. Actually, from -3dB, I meant -3dB of V(i) by assuming that .MEASURE works like how we perform AC analysis and taking approx value. So I changed V(o) -> mag(Vo) and -3dB -> 10^(-3/20)= 0.707945. And it worked successfully and I got frequency value that is quite close to approx value from plot \$\endgroup\$ Jan 27, 2021 at 10:33
  • 1
    \$\begingroup\$ @xyz please post your result as an answer to help further readers find the solution. Comments are not the correct format for that, as they might get removed. It is perfectly fine to answer your own question here. \$\endgroup\$
    – Arsenal
    Jan 27, 2021 at 12:00

2 Answers 2


I modified the .MEASURE command. As -3dB doesn't make sense so I specified the absolute value of voltage V(o) i.e. $$-3 = 20\log_{10}(\frac{V_o}{V_i})\\ \therefore V_o = 0.707945 \space \space[\because V_{i}=1]\\ $$

and Final command would be .MEASURE AC f FIND frequency WHEN mag(V(o))=0.7079457843841379 which gives f=158.778 printed in View > Spice Error Log

Final Circuit : enter image description here


Here is what I did to find the corner frequency at -3dB of a low-pass filter :

.meas gmax max(mag(V(out)))
.meas AC fc FROM 0 TARG mag(V(out))=gmax/sqrt(2) FALL=1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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