Impedance/Efficiency

[the_noob 10+ year member]

Well, this one's kind of long-winded, so I'll just get started. I was browsing around the internet a bit the other day and stumbled upon this website

http://www.diymobileaudio.com

I'm sure many of you have heard of it before. Anyway, in the Articles section, there is an article about 8 ohm drivers.

http://www.diymobileaudio.com/forum/viewtopic.php?t=33

This is what I am trying to understand. So, for instance, lets use the ID Chameleons, because they are the only drivers I can think of right now that come in 4 ohm and 2 ohm. In short, here is basically what the article says:

...raising the impedance actually causes efficiency to go up and lowering the impedance actually causes a loss of efficiency.

So, if I were to get the 2 ohm version and send it 150 watts @ 2 ohm, I would get the same results as if I got the 4 ohm version and sent it 75 watts @ 4 ohm??? FYI, the rms rating on the 2 ohm version is 150 watts, while the rms on the 4 ohm version is only 75 watts, so that kind of caught my eye.

This also brought up some other questions. If it works like this for midbass drivers, then I would have to assume that it works the same way for subs. So are dual 4 ohm subs more efficient than dual 2 ohms subs? Say I ordered two dual 2 ohm subs. I can wire them to either 2 ohms or 8 ohms, right? So if I wired them to 8 ohms rather than 2 ohms, would it mean that output would be the same, yet my amp would run cooler?

Click to read more...

 

[squeak9798 5,000+ posts]

Well, first thing I would point out is that I think you missed the point of his thread. His point was to say that the impedance of the driver isn't what will determine output. Just because a driver is 8ohm, it doesn't mean it will have less output than a 4ohm driver simply because it will receive less power from the amp. Go over the speaker section on here and read through some of the posts where I've recommended 8ohm speakers. What's the first thing people say? "I don't want that, it's 8ohm" because they automatically assume it'll be less loud than a 4ohm driver because of power. What cheapboy (npdang) is saying is that it isn't impedance that determines output, but efficiency. So, just because a driver is 8ohm, it doesn't mean it will be quiter than a 4ohm driver simply because it will receive less power from the amp.

2nd, I don't see where the power ratings for the ID Chameleon's are different between the two impedance options: http://www.imagedynamicsusa.com/products_page.php?id=chameleon&type=midrange

 

[the_noob 10+ year member]

Well, first thing I would point out is that I think you missed the point of his thread. His point was to say that the impedance of the driver isn't what will determine output. Just because a driver is 8ohm, it doesn't mean it will have less output than a 4ohm driver simply because it will receive less power from the amp. Go over the speaker section on here and read through some of the posts where I've recommended 8ohm speakers. What's the first thing people say? "I don't want that, it's 8ohm" because they automatically assume it'll be less loud than a 4ohm driver because of power. What cheapboy (npdang) is saying is that it isn't impedance that determines output, but efficiency. So, just because a driver is 8ohm, it doesn't mean it will be quiter than a 4ohm driver simply because it will receive less power from the amp.
2nd, I don't see where the power ratings for the ID Chameleon's are different between the two impedance options: http://www.imagedynamicsusa.com/products_page.php?id=chameleon&type=midrange

Thanks, squeak. But for the midbass drivers, it is different.

http://www.imagedynamicsusa.com/products_page.php?id=cx&type=midrange

It could be a typo, though.

Anyway, so if I see a driver with an efficiency of 90dBs @ 8 ohm, it will be twice as efficent as a driver with an efficiency of 90dB @ 4 ohm. Therefore, I can feed it half the power?

 

[squeak9798 5,000+ posts]

Anyway, so if I see a driver with an efficiency of 90dBs @ 8 ohm, it will be twice as efficent as a driver with an efficiency of 90dB @ 4 ohm. Therefore, I can feed it half the power?

No, it depends on the method that was used. That sensitivity is supposed to be measured with 1w @ 1m. But, many 4ohm speakers are rated using 2.83V input, rather than 1w (2.83V = ~2w on a 4ohm speaker. Using 2.83V is only acceptable for use on an 8ohm speaker). So, in this case, the true efficiency of the 4ohm speaker would be 87db, NOT 90db. As you can see, here the 8ohm speaker would be 3db more efficient.

But if they are both have a true efficiency of 90db, then there isn't a difference between them (as far as sensitivity is concerned).

 

[the_noob 10+ year member]

No, it depends on the method that was used. That sensitivity is supposed to be measured with 1w @ 1m. But, many 4ohm speakers are rated using 2.83V input, rather than 1w (2.83V = ~2w on a 4ohm speaker. Using 2.83V is only acceptable for use on an 8ohm speaker). So, in this case, the true efficiency of the 4ohm speaker would be 87db, NOT 90db. As you can see, here the 8ohm speaker would be 3db more efficient.
But if they are both have a true efficiency of 90db, then there isn't a difference between them (as far as sensitivity is concerned).

Well, thats kind of what I meant. 90dB's @ 8 ohm equals 87dB's @ 4 ohm equals 84dB's @ 2 ohm? But if the 4 ohm speaker is rated at 1 watt, then there would be no difference...but don't most companies measure at 2.83V?

2.83V @ 8 ohm=1 watt

2.83V @ 4 ohm=2 watts

2.83V @ 2 ohm=4 watts

Is that correct?

 

[squeak9798 5,000+ posts]

Well, thats kind of what I meant. 90dB's @ 8 ohm equals 87dB's @ 4 ohm equals 84dB's @ 2 ohm?

Huh //content.invisioncic.com/y282845/emoticons/confused.gif.e820e0216602db4765798ac39d28caa9.gif

But if the 4 ohm speaker is rated at 1 watt, then there would be no difference...

Correct

but don't most companies measure at 2.83V?

Some do, some don't. Just depends.

2.83V @ 8 ohm=1 watt2.83V @ 4 ohm=2 watts

2.83V @ 2 ohm=4 watts

Is that correct?

Yup

 

[the_noob 10+ year member]

Huh //content.invisioncic.com/y282845/emoticons/confused.gif.e820e0216602db4765798ac39d28caa9.gif

Ok, yeah, that was worded a little differently than from what I meant to say. I guess what I was trying to say was, if a company produced a driver that came in 8 ohms, 4 ohms, and 2 ohms, and the rated efficiency for all three was 90dB's @ 2.83V, then the true efficiency of them would be 90dB's for the 8 ohm driver, 87dB's for the 4 ohms driver, and 84dB's for the 2 ohm driver. True?

 

[squeak9798 5,000+ posts]

Yes.