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

CN-0271 fcutoff

zanzibar Contributor
zanzibar Jul 4, 2012 6:32 AM
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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,

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  • 1. Re: CN-0271 fcutoff
    JFitzger Analog Employee
    JFitzger Jul 5, 2012 1:10 PM (in response to zanzibar)
    Currently Being Moderated

    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

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