Midbass

[HCCA 10+ year member]

I came across this a while back. I was looking into the idea of a 3-way system, up front. Most mid-bass drivers have somewhat low power handling, IMO. This one seems to be really good, by the performance numbers...though a bit pricey. Anybody use these, or have a good three way setup that would like to comment? http://criticalmassaudio.com/catalog/index.php?act=viewProd&productId=5

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[MiniVanMan 10+ year member]

That driver ran at 200 watts won't really be any louder than a 4 ohm driver at 100 watts, and it's expensive as hell.

The Peerless SLS 8's are great, and the Dayton RS225s can handle quite a bit of power. The Seas L18, or L22 will rock your world as well. All those options are well under $100.00 each.

Not to get too technical, but an 8 ohm driver running off of 100 watts will have roughly the same output as a 4 ohm driver running off of 200 watts. Given all other things being equal. It's a function of sensitivity, not power handling.

 

[ARSkemp 10+ year member]

Not to get too technical, but an 8 ohm driver running off of 100 watts will have roughly the same output as a 4 ohm driver running off of 200 watts.

no

a 100w speaker @ 4ohms will be equal in loudness to a 100w speaker @ 8ohms (all else equal)

 

[squeak9798 5,000+ posts]

That driver ran at 200 watts won't really be any louder than a 4 ohm driver at 100 watts, and it's expensive as hell.
The Peerless SLS 8's are great, and the Dayton RS225s can handle quite a bit of power. The Seas L18, or L22 will rock your world as well. All those options are well under $100.00 each.

Not to get too technical, but an 8 ohm driver running off of 100 watts will have roughly the same output as a 4 ohm driver running off of 200 watts. Given all other things being equal. It's a function of sensitivity, not power handling.

You are assuming the driver with the higher impedance automatically has a sensitivity 3db higher than that of the other driver, which isn't inherently true especially if we are comparing different drivers.

So no, your comparisons do not hold true.

That said I can't disagree with your speaker recommendations //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif And I would agree that the speaker linked by the thread starter would have to be pretty spectacular to warrant that price. I have a hard time believing that the drivers you mentioned wouldn't offer just as good if not better performance at a much lower price.

 

[MiniVanMan 10+ year member]

You are assuming the driver with the higher impedance automatically has a sensitivity 3db higher than that of the other driver, which isn't inherently true especially if we are comparing different drivers.
So no, your comparisons do not hold true.

I said all else being equal. I know there are variations.

Generally (key word again) speaking, an 8 inch driver with a higher overall sensitivity, will a lot of times have lower low end capability. A key difference to note when dealing with pro audio drivers. Just because it's 8" doesn't make it a low end beast. Pro Audio drivers are super sensitive, and loud as a result, but you're not going to get the same low end extension as a home audio, or car audio driver designed for low end duty.

So, when somebody asks for advice on a MIDBASS driver for a 3 way, I can assume a number of things. First is they want low end extension. Second is, they want output. Third is, they're not going to care much about top end extension (unless matching with domes). Now, the question comes as to how low the person wants this particular driver to go. Just 80 hz? I can recommend some 8" drivers that have great overall sensitivity, that will be louder at any impedance level than an 8" that digs an octave lower.

I know, when I point this stuff out, that I'm going on assumptions, and I should probably be clearer. I'm on a personal mission though to debunk this myth that to get louder you need lower impedances, and more power. What you need is more power, and higher sensitivity (at any impedance). We're pretty limited by physics though.

 

[MiniVanMan 10+ year member]

no
a 100w speaker @ 4ohms will be equal in loudness to a 100w speaker @ 8ohms (all else equal)

Are you saying when "giving" each speaker 100 watts, the 8 ohm and 4 ohm will be equal in output? That is not correct. All else being equal.

http://www.diymobileaudio.com/forum/showthread.php?t=31

 

[bdavies11 10+ year member]

Are you saying when "giving" each speaker 100 watts, the 8 ohm and 4 ohm will be equal in output? That is not correct. All else being equal.
http://www.diymobileaudio.com/forum/showthread.php?t=31

Um yes he is correct.

He is saying that a 4 ohm driver receiving 100 watts @ 4 ohms will have the same output as an 8 ohm driver receiving 100 watts @ 8ohms (all else equal). //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[MiniVanMan 10+ year member]

Take an example like the RS180. It comes in both a 4 ohm and an 8 ohm version.

Take an amp that does say 100 watts at 4 ohms, and 50 watts at 8 ohms.

Take both drivers and run them subsequently on the same amp. Both drivers will have relatively the same output.

Now take the 8 ohm driver and put it on an amp that does 100 watts at 8 ohms and you will have a 3 db gain over the 4 ohm.

In order to raise the impedance you need to lengthen the wire wound around the coil. To go from 4 ohm to 8 ohm you need to double it. That causes an increase in inductance (double). Given the same magnet strength, you just doubled your BL. That will equate to an increase of 3 db sensitivity. Higher impedance, higher sensitivity. That means more output with less power. Much more efficient.

Now a real benefit, is when you go from 4 ohm to 8 ohm, the power will not be half. To say so, would be saying that you have an amp that has 100% efficiency as the load drops.

Stop thinking in terms of impedance, but rather sensitivity.

 

[bdavies11 10+ year member]

Take an example like the RS180. It comes in both a 4 ohm and an 8 ohm version.
Take an amp that does say 100 watts at 4 ohms, and 50 watts at 8 ohms.

Take both drivers and run them subsequently on the same amp. Both drivers will have relatively the same output.

Now take the 8 ohm driver and put it on an amp that does 100 watts at 8 ohms and you will have a 3 db gain over the 4 ohm.

Um, i'm not quite sure how to put this so ill just put it...

That is incorrect. //content.invisioncic.com/y282845/emoticons/uhoh.gif.c07307dd22ee7e63e22fc8e9c614d1fd.gif

 

[bdavies11 10+ year member]

Think about it bud, Your basically saying that the higher impedance drivers are more efficient. If that were true we wouldnt need 63980938 watts to get loud.

Again, If that held true, we could simply buy a 64 ohm driver and give it 6.25 watts @ 64 ohms and it would be just as loud as the 2ohm driver getting 200watts @2 ohms. //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[MiniVanMan 10+ year member]

Think about it bud, Your basically saying that the higher impedance drivers are more efficient.

That is EXACTLY what I'm saying!!!! I'm sorry you're stuck in your car audio dogma. It's an absolute FACT that when you increase the coil length within a magnetic field you get higher sensitivity. By increasing coil length, you get higher impedance.

I will go as far as to say, there are mechanical properties of every driver that need to be overcome before any real output is realized. That makes a 4 ohm driver fairly appealing when dealing with low power situations like using head units (especially stock) to power a speaker. You get more power to overcome the initial mechanical properties of the driver. That's why Mms is a factor of sensitivity. The stiffer your suspensions, and the heavier your cones (which comes with a lot of your ultra stiff cones like the RS series drivers from Dayton), the more power you'll need to overcome them before you realize output gains. High Qts drivers generally have stiffer suspensions. High Qts drivers are desireable in a car because of a more controlled low end in an infinite baffle situation. You lose a little low end capability, but your overall sound in the midbass region will be more controlled.

When you have a driver such as this, sensitivity can suffer. That's when we start to hear about things like, "this driver ***** up power, and wants more". Being limited by a 12 volt source, power at 8 ohms is harder to come by. This was especially true in the early days of car audio. It was a give and take situation. We don't have as much power to work with, and I can get decent output out of a 4 ohm driver, and since I'm stifled by the mechanical properties of these drivers. I'll use a 4 ohm driver and give it as much power as available.

Now that we have more advanced circuit topologies, and much more efficient amplifiers, we get more power for our initial 12 volts. So, in a high powered application, 100+ watts, it's actually better overall to use an 8 ohm driver. More efficiency, cooler amps, cooler speakers. Less loss = more efficiency all around.

Want to keep going? I can pull out the book.

 

[bdavies11 10+ year member]

Ok then... Im going to go buy myself an amplifier that puts out .000000000000000000000015 watts @ 837,489 ohms for 14 cents (since it only puts out .000000000000000000000015 watts), hook it up to my 4" sony 837,489 ohm speaker and go shatter pioneers 180+ spl record with it. //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[helotaxi 5,000+ posts]

Think about it bud, Your basically saying that the higher impedance drivers are more efficient. If that were true we wouldnt need 63980938 watts to get loud.
Again, If that held true, we could simply buy a 64 ohm driver and give it 6.25 watts @ 64 ohms and it would be just as loud as the 2ohm driver getting 200watts @2 ohms. //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

That's exactly what he's saying. Things to consider: the sensitivity rating that you normally see is db @ 1m @ 2.83v. 2.83V equals 1 W into an 8 ohm load. It equals 2W into a 4 ohm load and 4W into 2 ohms. If 3 speakers all have the same sensitivity rating of 93dB @ 2.83V/1m and one is 8ohm, one is 4 ohm and one is 2 ohm, which is more sensitive? The 8ohm driver is only getting 1W and is creating the same amount of sound as the other two getting 2x and 4x the power respectively. It takes power to make power and as impedance decreases and current increases, resistive losses in the amp increase. Running an 8 ohm driver would allow the amp to run cooler and draw less power from the electrical system of the car compared to the same amp trying to get the same output from a 2 ohm driver with the same rated efficiency. Factor in that the amp will probably be able to produce more voltage into the higher load and the 8ohm driver will probably get louder overall as well.

This is all true for mid/high freq drivers. If you want the driver to play lower, it gets a lot more complicated because cone excursion away from the rest position where Bl is strongest changes Bl and all the other driver parameters. The larger the excursion, the larger the change. Mid/high freq drivers barely move at all and their motor strength remains nearly constant across its power and freq range.

Ok then... Im going to go buy myself an amplifier that puts out .000000000000000000000015 watts @ 837,489 ohms for 14 cents (since it only puts out .000000000000000000000015 watts), hook it up to my 4" sony 837,489 ohm speaker and go shatter pioneers 180+ spl record with it.

You are grossly over simplifying a very complex situation that you clearly don't understand. Let me know how that works out for you.

 

[bdavies11 10+ year member]

http://www.epanorama.net/documents/audio/speaker_impedance.html

Impedance and effiency

Let's look at the following situation: Take an 8 ohm speaker and wind twice the length of wire onto the voice coil. The resistance woul go up, for sure, but because there is no more wire in thegap, the electromagnetic couping coefficient, the Bl product, would also go up. And you would have, as a result, a 16 ohm speaker with essentially the same efficiency as the 8 ohm speaker, all other things being equal.

Or you could design a speaker with both a higher impedance (longer wire in the voice coil) AND a larger magnet assembly with higher flux density in the gap and get a higher impedance driver with higher electro-acoustic efficiency.

Or you could design a higher impedance driver with a stronger magnet and a lighter cone and get even more efficiency.

The point is, the rated impedance IS NOT the same as the efficiency, nor is there any direct correlation between the two. Efficiency of a given direct readiator driver is determined by the folowing relationship: l

 

[MiniVanMan 10+ year member]

http://www.epanorama.net/documents/audio/speaker_impedance.html
Impedance and effiency

Let's look at the following situation: Take an 8 ohm speaker and wind twice the length of wire onto the voice coil. The resistance woul go up, for sure, but because there is no more wire in thegap, the electromagnetic couping coefficient, the Bl product, would also go up. And you would have, as a result, a 16 ohm speaker with essentially the same efficiency as the 8 ohm speaker, all other things being equal.

Or you could design a speaker with both a higher impedance (longer wire in the voice coil) AND a larger magnet assembly with higher flux density in the gap and get a higher impedance driver with higher electro-acoustic efficiency.

Or you could design a higher impedance driver with a stronger magnet and a lighter cone and get even more efficiency.

The point is, the rated impedance IS NOT the same as the efficiency, nor is there any direct correlation between the two. Efficiency of a given direct readiator driver is determined by the folowing relationship: l

That's what I've been trying to tell you. But all other things equal, a 4 ohm driver still has to have a cone, suspension, magnet, etc. All other physical properties are still there. So, when increasing efficiency as detailed in your link, a byproduct of that is an increase in impedance. It is a byproduct.

You have a flux field in the center of your magnet. It's a fixed diameter, so you only have a fixed amount of room to work with. Take that magnet, and put a 4 ohm coil in there. Depending on the winding of the coil will dictate the inductance of the coil. The inductance will determine how it interacts with the flux field. The impedance is just a product.

I'll quote this next part.

If you take a given driver, and simply swap out voice coils, you end up with more efficiency as you increase the impedance. Take a voice coil, say 2" diameter, 1" winding length, 24AWG 4 layer, and swap it with a 2" diameter, 1" winding length, 27AWG 4 layer, and you double the impedance, but the efficiency also goes up - less mass and better packing density."

It's because the moving mass has dropped, and if desired - because of the thinner wire diameter which packs in tighter - you can put more layers in the voice coil and potentially raise the BL."

You didn't do anything to debunk anything I said, or Helotaxi said. You proved it. Now tell me, how do you increase the efficiency, or even keep it the same, of a speaker by lowering it's impedance, knowing that in order to do so, you have to unwind the coil?

What don't you get?

http://www.epanorama.net/documents/audio/speaker_impedance.html

Or you could design a higher impedance driver with a stronger magnet and a lighter cone and get even more efficiency.

Hmmmm, sounds like a 16 ohm pro-audio driver. Gee, I wonder why they would do that?