Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

3db frequency for decoupling caps (application note)

  1. Aug 10, 2012 #1
    I was reading a app note on bypass caps - http://www.ti.com/lit/an/scaa048/scaa048.pdf
    I'm a bit confused as to how the 3db cut off formula is Z/2∏L, isn't it just 1/2∏L?
    Please see page 4 of app note.
  2. jcsd
  3. Aug 10, 2012 #2


    User Avatar

    Staff: Mentor

    Re: 3db frequency

    The units wouldn't be right for an equation f = 1/2∏L, for one thing.

    I get a slightly different answer from the app note, but I may not be understanding how they are setting this up. It would be nice if they would have shown a diagram of their Z effective load resistance and the L of the wiring supplying the current, but whatever.

    This is what I did:

    Assume the source power supply is connected to the load circuit through the wiring inductance L. Assume that the effective load impedance (resistance) is the allowed ΔV divided by the anticipated worst-case ΔI.

    Then you have a lowpass filter formed by that LR (LZ) circuit, with the low passband frequencies being supplied by the power supply, and ΔI frequencies above the corner frequency requiring bypass capacitors in parallel with the load Z in order to supply those high-frequency currents.

    Then the corner -3dB (1/√2) is calculated:

    [tex]\frac{V_o}{V_i} = \frac{Z}{Z + sL} = \frac{1}{√2}[/tex]

    [tex]Z√2 = Z + 2πfL[/tex]

    [tex]f = \frac{Z(√2-1)}{2πL}[/tex]

    So my -3dB frequency is slightly lower than their Z/2∏L, but they may have just been rounding it off a bit (or I may have made an error).

    EDIT -- Actually, to do it right, you need to take the complex nature of the equations into account. See the Gain equation here, for example:


    Last edited: Aug 10, 2012
  4. Aug 21, 2012 #3

    I get Z/2∏L. I'm new to the forum so i'm not familiar with how to make the derivation look nice, but in short in finding the 3dB frequency you must consider the imaginary portion in the denominator (s = jω) .

    Z+sL must be replaced with its magnitude √Z2+(Lω)2
  5. Aug 21, 2012 #4
    I don't see it in p4, but in general, for inductor

    [tex] X_L=2\pi f L\;\Rightarrow \;f=\frac{X_L}{2\pi L} [/tex]

    At break frequency,

    [tex]Z=X_L\;\Rightarrow \; f_{-3db}=\frac{Z}{2\pi L}[/tex]

    Is that what you are referring to?
  6. Aug 23, 2012 #5
    Why is Z=X_L at break frequency?

    Break freq = corner freq, right?

    The equation is on page 2 of the pdf.
  7. Aug 23, 2012 #6
    Yes, break frequency is just corner frequency.....in general. It's just general saying, break frequency is usually the frequency where the reactance equal to magnitude of the resistance. eg. if you have R and L in series to form a high pass filter where you take the output at the junction between the R and L. Then the +3DB frequency is as I described. Or in parallel R and L high pass filter where at low frequency the signal is shorted out by the L. But at break frequency where ZL = R then that's the break frequency. For specific, you really have to look at the circuit as the R might be a combination of different resistors or even other reactance.

    As for why it has to be R=XL, good question. I think it's more definition than anything. This is again "in general" where the amplitude decrease to 0.707 of the original amplitude for low pass. OR increase to 1.414 of the original amplitude for high pass. I don't want to get too specific for all cases as I am not expert in definitions, just working on the application side of it.
    Last edited: Aug 23, 2012
  8. Aug 23, 2012 #7
    Well the LP filter forms a voltage (or power) divider and the cutoff is defined at -3dB (i.e. the power is halved), so the divider arms have to be equal.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook