My goodness fellas. You have a guy here confused about the way a basic low-pass filter works. Don't confuse him further!!
@goodguys The LC circuit you're describing is not uncommon. The inductor is low impedance at low frequencies, the capacitor is low impedance at high frequencies.
The series/shunt combination of the two of them in typical LC low-pass filter means the impedance seen by the amplifier will (almost) never be below the impedance of the driver itself.
Adding series resistance is a solution to a problem that doesn't exist.
Dave.
@goodguys The LC circuit you're describing is not uncommon. The inductor is low impedance at low frequencies, the capacitor is low impedance at high frequencies.
The series/shunt combination of the two of them in typical LC low-pass filter means the impedance seen by the amplifier will (almost) never be below the impedance of the driver itself.
Adding series resistance is a solution to a problem that doesn't exist.
Dave.
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That's likely to be true if the designer follows standard crossover design formulas. However if you 2X the capacitance and 1/2 the inductance you'll get the same crossover frequency, and create an impedance dip, yielding a boost just before the response falls off. This is true of both low pass and high pass filters.
When considering various L / C combinations and their combined reactance with driver,
it is worth remembering that with high value inductors, you have the ability of introducing
1 or more ohms of DCR into the winding. This can be very useful to tame amplifier stress.
If needed, general EQ can be added after this 🙂
it is worth remembering that with high value inductors, you have the ability of introducing
1 or more ohms of DCR into the winding. This can be very useful to tame amplifier stress.
If needed, general EQ can be added after this 🙂
Yeah it IS a trend. And many speakers that claim "8 ohm" are really more like "4 ohm." The thing about it is, the DC resistance is in the denominator of the efficiency equation. The lower the coil resistance, the more efficient the speaker (all other things being equal). So lower and lower impedance lets the designer kind of "cheat" to get more bass from a small enclosure etc. At normal volume levels, it doesn't matter. At higher volumes into clipping, nobody is blaming the speaker designer, it gives an excuse to "upgrade" from an AVR to a more powerful outboard amp (although then I see people talking about buying some 2x130W Emotiva or whatever, which is not really meaningful progress).
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Thanks! I don't have a scenario in mind, it's a hypothetical question. Just something i saw on you tube aroused my curiosity as i was searching around for more information on the question i posed in post #1, and i am trying to find out more this resistor dummy load thing.
How do they get away with using resistors instead of a speaker. Is it an 8 ohm resistor they are using, and what kind of impedance would an 8 ohm resistor give
PS.
I feel lucky that my Mackie M1400 doesn't 'bat an eye-lid' driving 2 ohm loads.
It's true class A/B and looking at current prices MUST BE an absolute bargain!
https://reverb.com/au/p/mackie-m1400i-fr-series-2-channel-power-amplifier
I feel lucky that my Mackie M1400 doesn't 'bat an eye-lid' driving 2 ohm loads.
It's true class A/B and looking at current prices MUST BE an absolute bargain!
https://reverb.com/au/p/mackie-m1400i-fr-series-2-channel-power-amplifier
The thing about a RESISTOR is that its value when measured with DC is called resistance.
When the same RESISTOR is measured with any AC - audio frequency, it is called impedance.
It is actually just a matter of 'terminology' that deciphers whether we are talking about AC or DC.
It is a word change used to assist an understanding.
( a resistor actually has the same impedance as it does have resistance )
When the same RESISTOR is measured with any AC - audio frequency, it is called impedance.
It is actually just a matter of 'terminology' that deciphers whether we are talking about AC or DC.
It is a word change used to assist an understanding.
( a resistor actually has the same impedance as it does have resistance )
An 8 ohm resistor is just 8 ohms. Resistance, impedance, all 8 ohms. This is because it acts like 8 ohms at all frequencies 😉
Why do you care about a test resistor, what do you want to know about it?
I wanted to know because i thought resistance and impedance were slightly different from each other. I thought impedance meant the sum of all resistive forces, IE inductive reactance and and capacitive reactance and resistance added up together equals impedance.
Maybe i got it wrong
Maybe i got it wrong
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Thanks! Speaking theoretically, if i had a drive unit with a nominal impedance of 4 ohms, it's minimum impedance should be about 2 ohms or possibly lower, maybe 1.8 or 1.5 ohms in some cases.
Some amps may be happy to drive these low impedance loads but some may not. If i were to use a high power one ohm resistor in series with the driver would that raise the impedance to appoximatly 3 ohms and make the drive unit a little more amplifier friendly.
It would but it's 100% counterproductive because it would eat up all of the response peak you were trying to buy with your LC filter. Just pick a smaller value of "C" and a bigger value of "L"
Not likely. The resistance of a 4 ohm coil is usually 3 ohms or more. The resistance typically sets the minimum.
On top of that, the crossover adds to it. It may not be doing much at the same frequency as the woofer dip, but you can at least add the resistance of the coil to it.
Beyond that, you're only really dropping the impedance if you use a high Q filter, which is not typical but is different in each case.
In an LCR circuit (R is the resistance of the speaker)
Q = R * square root of (C/L)
and Frequency = 1/(2 * pi * (square root of L * C))
So if you have an 8 ohm woofer, C = 250uF capacitor and L= 5mH inductor
The Q is 1.789 (which will create a peak of about 5dB and an impedance dip)
and the F is 142
If you add a resistor you'll eliminate most of the peak and reduce SPL across the band, which is usually a bad idea.
If you use C = 125uF capacitor and L = 10mH inductor (capacitor value divided by 2 and inductor value x 2)
the Q is half (0.9) and the F is still 142
Q = R * square root of (C/L)
and Frequency = 1/(2 * pi * (square root of L * C))
So if you have an 8 ohm woofer, C = 250uF capacitor and L= 5mH inductor
The Q is 1.789 (which will create a peak of about 5dB and an impedance dip)
and the F is 142
If you add a resistor you'll eliminate most of the peak and reduce SPL across the band, which is usually a bad idea.
If you use C = 125uF capacitor and L = 10mH inductor (capacitor value divided by 2 and inductor value x 2)
the Q is half (0.9) and the F is still 142
Where are you getting this information? Are you just making it up out of thin air????
Dave.
Because it is a part of the second-order filter.
Your math is wrong, it doesn't work like that. Read about LC second-order low-pass filter.
Of course, there are many examples of wrong loudspeaker designs (mostly at so called "high-end" speakers) with very low impedance at low (or high) frequencies, combined with very high phase angle... which, of course, demand ultra-expensive "high-end" amplifiers (weighting a ton) capable of driving 1 Ohm.
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