Hello,
First crossover design, looking for advice. How do you get rid of 2 impedance resonance peaks on a 2 way 3rd Order Butterworth Vented Design without using an RLC circuit. Crossover point is 2Khz. The RLC circuit in my design will flatten the net impedance on both peaks but the capacitor value is 1.453 millifarads which is a 1,453 uF capacitor. So that is no good. Resonance peaks are 31.60 ohms at 26.94 Hz and 27.51 ohms at 53.61 ohms. The dip in the middle of the peaks is 11.99 ohms at 38.70. The overall flat net impedance from 5 Hz to 20Khz averages 5.80 ohms. If I leave the peaks in would this affect the sound. I will be using an 8 ohm receiver for eventual speaker hook up. see pdf. attachment.
First crossover design, looking for advice. How do you get rid of 2 impedance resonance peaks on a 2 way 3rd Order Butterworth Vented Design without using an RLC circuit. Crossover point is 2Khz. The RLC circuit in my design will flatten the net impedance on both peaks but the capacitor value is 1.453 millifarads which is a 1,453 uF capacitor. So that is no good. Resonance peaks are 31.60 ohms at 26.94 Hz and 27.51 ohms at 53.61 ohms. The dip in the middle of the peaks is 11.99 ohms at 38.70. The overall flat net impedance from 5 Hz to 20Khz averages 5.80 ohms. If I leave the peaks in would this affect the sound. I will be using an 8 ohm receiver for eventual speaker hook up. see pdf. attachment.
Those peaks in the LF do not need to be dealt with, they're there beacuse that's part of how a BR box works. There are some dumb car audio sites that will tell you otherwise, but they're completely wrong
Hi NTSB...
Pete is basically right - you do not need to flatten those 2 impedance peaks.
Also, it is impossible to with any practically realizable simple passive RLC circuit to flatten twin impedance peaks at such low frequencies.
Their presence will only affect the sound if your amplifier has a high output impedance.
The only change I would make is lengthen the port-tube, or whatever tunnel, is behind the vent into the enclosure.
That will lower the tuning frequency and give you a little extra low bass extension.
This change will also cause the magnitude of the lower-bass impedance peak to be reduced a little and the magnitude of the upper-bass impedance peak to increase a little, and that will not cause any problems, but can be audibly preferable, depending on the bass quantity and extension in the recordings you play.
Pete is basically right - you do not need to flatten those 2 impedance peaks.
Also, it is impossible to with any practically realizable simple passive RLC circuit to flatten twin impedance peaks at such low frequencies.
Their presence will only affect the sound if your amplifier has a high output impedance.
The only change I would make is lengthen the port-tube, or whatever tunnel, is behind the vent into the enclosure.
That will lower the tuning frequency and give you a little extra low bass extension.
This change will also cause the magnitude of the lower-bass impedance peak to be reduced a little and the magnitude of the upper-bass impedance peak to increase a little, and that will not cause any problems, but can be audibly preferable, depending on the bass quantity and extension in the recordings you play.
<rant on>
Unfortunately, a lot of disinformation is spread in the car audio domain. It is a shadow world. Some great builders/technologies exist there, and have made their way into mainstream hifi, but for the most part, it's all hooey where "bling" and "max SPL" are the goals rather than HQ sound. For every Patrick Bateman who's tinkering to achieve quality sound, there are a hundred teenagers thinking "Wow, shiny red metallic subwoofer!"
<rant off>
Lets not write off the need to get rid of those peaks just yet. There could be at least 2 reasons to do it.
1) If you plan to use a subwoofer where this system is passivly filtered to play the midbass.
2) If you have an amp not capable of dealing with big electrical phasechanges.
This is a way I do it when I have a midbass and low x-over: You put an inductor in series with a resistor, and they together in parallel with the (mid)bass. I have no equation for you regarding the values as I do it manually in a comuterprogram. In this case I would start maybe with a 10mH and 10Ohm. In the program you just adjust the values until you get desired result.
This will not get rid of the peaks but dampen them considerably. Just remember to use resistors capable of handling high power, at least 40-50 watts, ex 5 47 ohms in parallel.
1) If you plan to use a subwoofer where this system is passivly filtered to play the midbass.
2) If you have an amp not capable of dealing with big electrical phasechanges.
This is a way I do it when I have a midbass and low x-over: You put an inductor in series with a resistor, and they together in parallel with the (mid)bass. I have no equation for you regarding the values as I do it manually in a comuterprogram. In this case I would start maybe with a 10mH and 10Ohm. In the program you just adjust the values until you get desired result.
This will not get rid of the peaks but dampen them considerably. Just remember to use resistors capable of handling high power, at least 40-50 watts, ex 5 47 ohms in parallel.
in case 1, sealing the box is the solution I'd use; in case 2... then there's a problem with the amp. Your solution would lower efficiency considerably. If you really need an HPF there then a cap in series would be better, although I wouldn't be keen on that either...
at low frequencies these things are better dealt with at line level.
at low frequencies these things are better dealt with at line level.
A ported box can be the desired solution for many reasons. Many small speakers/sub-systems have satelites with ported satelites, and with good reason. Whatever you do with the q-values down at the bass-level vil propegate to the midrange area. In many cases you vill get a more open midrange with a ported box so a sealed box isn't necessarily a or the solution. Either way, the conductor/resistor trick will work in both cases.
A tubeamp or lower wattage transistoramp can easily have problems with large phasechanges, the speakers feedback to the amp shouldn't be underestimated.
You say a cap is the better solution. As to what? I did mention the possibility of this being a midbass in a passivly filtered solution so of course a cap would be involved. Problem is a cap is in relation to the speakers impedance so of course it will work better if you flatten the impedance closer to nominal impedance, which this curcuit will provide.
I don't agree with you that it lowers efficiency. The amp sends out its signal in voltage, then the speaker draws current as needed making up the total wattage. Even though this system draws more current, it will be more correct from a filterstandpoint and most amps will have an easier time driving the lower impedance because you have dampened the big electrical phasechanges which amps sees as a bigger challenge.
The curcuit I proposed is not uncommon and does not lower efficiency any more than a comparable RC-curcuit used to flatten impedance at higher frequencies, as opposed to this curcuit used to flatten impedance at lower frequencies.
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