1 Reply Latest reply: Jul 5, 2012 1:10 PM by JFitzger RSS

    CN-0271 fcutoff

    zanzibar

      Hi,

       

      "The two 100 Ω resistors and the 1 μF capacitor form a differential filter with a cutoff frequency of 800 Hz. The two 0.01 μF capacitors form common-mode filters with a cutoff frequency of 160 kHz."

       

      Please can you provide the equations to get both cutoff frequencies ?

       

      Thanks,

        • 1. Re: CN-0271 fcutoff
          JFitzger

          The equation to determine the cutoff frequency of the different filter created by the two 100Ω resistors and the 1μF capacitor is

           

          1 / (2 * pi * R * C)  or 1 / (2 * 3.14159 * 100 * 1e-6) = 1592 Hz

           

          This is then divided by 2 since it is shared between the two inputs resulting in 796 Hz

           

          The equation for the common-mode filter is exactly the same, except the value is not divided by two since the filter is not shared between the two inputs.

           

          1 / (2 * 3.14159 * 100 * 0.01e-6) = 159,155 Hz