LC Low-pass filter questions

Discussion in 'General Electronics Chat' started by calansvc, Apr 17, 2016.

  1. calansvc

    Thread Starter New Member

    Apr 16, 2016
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    So I have a simple 2nd order LC low-pass filter that I'm re-purposing, and need to know what will happen if I change the cap. The problem is, I don't know what coil is in it. Here is the circuit:

    [​IMG]

    The values I know are:

    R1 = 8 ohm speaker (or 16 ohms with two speakers in series)
    C1 = 6.8 uf
    Cutoff freq = 5.5khz

    So, how do I figure out the value of the coil, given that info?

    2nd, lets say I want a new cutoff of 6.5khz. Can I just change the cap (keeping the same coil), and still be close to the same impedance? I can't go below 8 ohms, due to the amp feeding it.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you know the equation for Cut-off frequency?
     
  3. calansvc

    Thread Starter New Member

    Apr 16, 2016
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    No. I found the formula for a 1st order RL filter, but not the 2nd order one with the cap.

    If I did have the formula, I may be able to work backwards... but my algebra is about 35 years rusty :)
     
    Last edited: Apr 17, 2016
  4. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    This is a little tricky because of the placement of the resistor. This means the cutoff does not follow 1/sqrt(LC) as the resistor could have great influence on that in this circuit.

    The general procedure is to analyze the output for the amplitude, then check the amplitude in the pass band, then equate to the proper ratio of that to the square root of 2. For this problem we can use:
    Ampl=1/sqrt(2)

    So you would first calculate the amplitude Ampl, equate that to 1/sqrt(2), then solve for w. You'll get more than one w so you check to see which one(s) is/are valid and use that. The cutoff frequency in Hertz is then w/2pi.

    You can solve for L and get the required value of L to get a cutoff frequency of 5.5kHz, and it comes out to around 2e-4 Henries but i'll wait and see if you want to calculate the exact value yourself.

    Note that if R is doubled the cutoff changes by about +8 percent, but if R is halved the cutoff frequency changes by about -32 percent roughly.
     
  5. calansvc

    Thread Starter New Member

    Apr 16, 2016
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    This circuit is actually the woofer half of a crossover in an old infinity 2-way speaker. Rather than using it as a cross-over, I'm going to use it as a low pass filter on a single speaker (or two in series). But, I want to change the cutoff from what it was listed at (5.5khz), to something a little higher....or maybe make it switchable between two values.

    I'm just wanting to understand how the values relate, so I can predict what the cutoff will be with a different cap (leaving the coil the same). Or, knowing the desired cutoff, calculating a new cap value (again, leaving the original coil).
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    Where did you get the notion that you could model a speaker as a resistor? I'm just curious.
     
  7. bertus

    Administrator

    Apr 5, 2008
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  8. MrChips

    Moderator

    Oct 2, 2009
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    5.5kHz seems rather high for a 2-way speaker system.

    Use:

    <br />
<br />
C = \frac{1}{2 \pi fR\sqr{2}}<br />
<br />
<br />
L = \frac{R\sqr{2}}{2 \pi f}<br />
<br />

    where R = 10Ω for 8Ω speaker.
     
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  9. calansvc

    Thread Starter New Member

    Apr 16, 2016
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    I didn't. That was just the way someone drew the filter circuit.

    Good article. Thanks!

    That's what was listed in the specs for the Reference E-L speaker.

    Thanks!
     
  10. calansvc

    Thread Starter New Member

    Apr 16, 2016
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    For future reference, I found a really good calculator here, that also has the formulas for calculating each component listed over to the left.
     
  11. Papabravo

    Expert

    Feb 24, 2006
    10,135
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    As one of the first graphs in the ESP article shows a speaker actually looks like an inductor after the resonance peak at 25 Hz., from about 300 Hz. up to about 50K. This should hardly be surprising since the construction of the device implies it.
     
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