I’m designing a high pass filter: 2nd order, LC high pass. I’ve noticed that I can play around with the values of the inductor and capacitor to achieve the same cut-off frequency. For example, if I use a 20 mH inductor and a 100 uF capacitor, the cutoff is about 112 Hz. You get the same result with 10 mH and 200 uF.
Are there reasons you would pick one or the other or different values for this high pass filter. Will they sound different? I’m inclined to pick a high value inductor because then I can use smaller value capacitors because the hi-end capacitors get expensive real quick as capacitance goes up, and because the cap is in the signal chain and the inductor isn’t (e.g. money spent on the cap should go further than money spent on the inductor). Furthermore, how much does the inductor type matter here? Would an expensive toroidal inductor give you anything extra compared to a cheaper iron core inductor since it isn’t in the signal chain and the current through it is relatively low?
Are there reasons you would pick one or the other or different values for this high pass filter. Will they sound different? I’m inclined to pick a high value inductor because then I can use smaller value capacitors because the hi-end capacitors get expensive real quick as capacitance goes up, and because the cap is in the signal chain and the inductor isn’t (e.g. money spent on the cap should go further than money spent on the inductor). Furthermore, how much does the inductor type matter here? Would an expensive toroidal inductor give you anything extra compared to a cheaper iron core inductor since it isn’t in the signal chain and the current through it is relatively low?
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The reason why one would pick one of the other value is impedance consideration for instance and it is not clear which combination will truly yield a better FR. The cost consideration is also important. Whatever you may be building, get your parts in the simulator first and make sure you feed the program with realistic data, otherwise it would not make much sense to bother at all. The thing with parts quality imo is that you should leave it for the very end when you will be absolutely sure about values to go with. With loudspeakers it is very much costly and too complicated for an average hobbyist to mess around with the wild impedance changes in the 100 Hz region. This is where peaks usually live and a simple CL filter does not suffice to achieve a nice high pass filter function. Your remark "because the cap is in the signal chain and the inductor isn’t..." does not really reflect reality. It it were true what you say, then taking physically inductor out of the circuit would not change the filter function one bit and we know that it does. Good luck!
LC filters have a 'Q' factor which is a measure of the damping - you normally need an RC network on the output to define the impedance (and hence the damping).
With higher L values the impedance of the filter will be higher, hence need less damping (or a higher valued damping R in other words).
If the filter's feeding a tweeter then the ESR of the inductor becomes important as it will limit the LF rejection at the tweeter terminals. The bigger the value of L, the more difficult it'll be to get low ESR.
With higher L values the impedance of the filter will be higher, hence need less damping (or a higher valued damping R in other words).
If the filter's feeding a tweeter then the ESR of the inductor becomes important as it will limit the LF rejection at the tweeter terminals. The bigger the value of L, the more difficult it'll be to get low ESR.
Thanks for your help! The high pass filter will go in between an amplifier and a 2-way crossover that was made by the manufacturer of the speakers. The reason for the high pass is to limit cone excursion and let a subwoofer handle the lower frequencies as I’ve found this to minimize distortion as high volume levels and increase clarity. I currently have a CL high pass filter there now (cap = 280 uF and inductor = 9.8 mH) but it uses cheaper parts and an electrolytic capacitor. I am upgrading the capacitor to a film capacitor since it is my understanding that film capacitors usually sound better than electrolytics in the signal chain.
As previously noted, there are different combinations of L and C that will yield the same cutoff frequency, but the Q will be different.
And as Lojzek points out, the impedance peaks and swings of a speaker near the box tuning frequency (even an octave or two) will greatly influence the effect of a 2nd order LC high pass filter.
If the speaker impedance is assumed to be a simple resistive load, the formulas on the attachment can help us understand how different combinations of L and C are related to the "Q".
As an example, let's say we have a speaker with "constant" resistance R = 8 ohms around 400 Hz. We could choose L = 4.0 mH, and C = 40 uF, so that Q = 0.80 and fo = 400 Hz. Also, the SPL at 400 Hz is 1.9 dB below the sensitivity.
Alternatively, we could use C = 30 uF, L = 5.33 mH for the same (constant) 8 ohm speaker. Again, fo = 400 Hz, but Q = 0.60. But for this combination, the SPL at 400 Hz is 4.4 dB below the sensitivity.
If you want a higher Q, raise the value for C, and lower the value of L accordingly.
Again, these formulas assume that the impedance load is constant. A 2nd order passive filter may get you in the ball park, but probably won't be ideal.
And as Lojzek points out, the impedance peaks and swings of a speaker near the box tuning frequency (even an octave or two) will greatly influence the effect of a 2nd order LC high pass filter.
If the speaker impedance is assumed to be a simple resistive load, the formulas on the attachment can help us understand how different combinations of L and C are related to the "Q".
As an example, let's say we have a speaker with "constant" resistance R = 8 ohms around 400 Hz. We could choose L = 4.0 mH, and C = 40 uF, so that Q = 0.80 and fo = 400 Hz. Also, the SPL at 400 Hz is 1.9 dB below the sensitivity.
Alternatively, we could use C = 30 uF, L = 5.33 mH for the same (constant) 8 ohm speaker. Again, fo = 400 Hz, but Q = 0.60. But for this combination, the SPL at 400 Hz is 4.4 dB below the sensitivity.
If you want a higher Q, raise the value for C, and lower the value of L accordingly.
Again, these formulas assume that the impedance load is constant. A 2nd order passive filter may get you in the ball park, but probably won't be ideal.
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