High Efficiency Speakers

I am interested in building a pair of speakers that would be highly efficient. Something like the klipsch reference line. Thier floor standing loudspeakers have efficiency ratings of 98 dB/1 watt or greater. But they also have a fairly wide response range and are shielded for home theater use.

I don't seem to be able to find any drivers that would give me this capability. Tweeters are no problem, but the only way I can think of to get this kind of efficiency would be to wire together two sets of in series wired drivers and then connect the two pairs in parrales. This shoudl theoretically get me the impedance of 8 ohms and an efficiency of about 98 dB/ 1 watt.

This approuch seems liek overkill though, klipsch does it with two drivers and a horn tweeter. Could someone shed some light on this problem.

Thanks
Mark
 
I am still trying to understand the math invloved with the concept efficiency increasing with more drivers. Non of my calculations seem to support the idea. Admittedly, I am not the best at math, but it seems that if more drivers are used, the power is simply redistributed, and each speaker emits according to the amout of power it receives. If it receives less, its emits less. So 4 speakers configured as two series pairs in parallel, should distribute the power so 1/4 goes to each one, and each emits 1/4 what it would have. I used to think the efficiency went up with more drivers, though.

Now if they are all in parallel, then the power input goes up, causing an increase in total SPL, in fact 4 times more energy output since there is 4 times more input (assuming they are not driven to the point of power compression in any case, and the amp can drive the load without voltage drop). Am i still wrong about this thing?
 
DB vs SPL

You guys are close. Wiring up speakers in parallel or series deals with the resistance the output of the amp sees. The resistance the amp sees determines how much power the amp puts out. This is known as load. When you double your speakers and double your power you gain 3 db. That is going from 1 speaker to 2 to 4 to 8 to 16. Now each speaker has a rating of 90 db at 1w/1m, and if you had 16 speakers driven by a 16 watt amp you would figure 90 db + 12 more db because we doubled 4 times. 4 x 3db = 12 db. If a speaker is 90db at 1w/1m and it can handle 100 watts, then at 100 watts its db gain would be determined by watts = log x 10 = 20db . 20db + our original 90db = 110db at 1 meter. Now if we hook up 8 amps at 100 watts per channel to 16 of these speakers we gain 12 more db to give us 122 db at 1 meter. Now for every doubling of distance away from our speaker we are we lose 6 db. From 1m to 2m to 4m to 8m to 16m we just lost 24db. Seems unfair to pay so much for 3db and then lose more by stepping back a little bit. This is the reason for those monster pas at the concerts. How you would apply this to multiple drivers in a cabinet with a crossover must be answered by someone else. I don't have the design background that other members do.

As far as parallel, 4 8 ohm drivers in parallel would give you 2 ohms. This is going to make the amp attempt to output more power if it can. If it can't your shopping for another amp. Ideally, you would balance an output with a driver of the same rating. If the amp puts out 200 watts a channel into 8 ohms match it with a 200 watt, 8 ohm driver. If an amp puts out 300 watts into 4 ohms match it with two 150 watt 8 ohm drivers wired in parallel. Series and parallel are circuits that help us match drivers to amplifier outputs to more efficiently utilize our money.:D
 
Thanks for your reply PassFan.

I know understand why my idea wouldn't work as expected. I was sort of thinking that logically it didn't make sense. However some information I had acquired in the past made me think it might work. I suppose that information wasn't the whole story.

Now I would like to ask if anyone has any insight into how klipsch produces such high efficiency speakers. Thier ratings for the whole box are upwards of 98 dB at one watt. In my personal browsing of audio sites I have never seen any high efficiency designs for home audio that weren't some kind of horn loaded design.

Basically I know why my design won't work, now I need to know what will. :)


Thanks,
Mark
 
PassFan, I think you are incorrect.

speakers are linear with voltage, not power.

doubling the # of drivers and doubling the power would give you a theoretical on axis 6dB increase in output.

example, an 8 ohm speaker with a sensitivity of 92dB/2.83V/1m aka 92dB/1W/1m.

If your amplifier can output say 11.32Vrms(16W) into 8 ohms, a single driver connected would play at 104dB.

Now lets add a second identical speaker on its own identical amplifer. it too can play at 104dB by its lonesome. If the speakers play in phase, at a point, that is equidistant, and one meter away from the center of both cones the RMS air pressure at that point will double. similar to voltage, a doubling of pressure correlates to a 6dB rise in SPL. (to test this for yourself, 20 log (1/2E-5) = ~94. 20 log (2/2E-5) = 100. so double the power, and double the drivers gives an increase of 6dB. Two drivers with a total of 32 watts will give an output of 110dB.

now lets keep the same overall power as the first example.

Each amplifier now is going to output 8 watts(8Vrms) into 8 ohms. each speaker will play at 101dB alone, summing the on axis outputs at a point equidistant from both cones will yield an SPL of 107dB's. which is 3dB more efficient than a single speaker, even though the overall power is the same.

You do have to be mindful when adding more speakers changes the final load impedance, which is why I used two amplifiers instead of one in my example. When you have a final load rating that differs from 8 ohms, the power sensitivity (dB/1W/1m) will be different than the voltage sensitivity (dB/2.83V/1m).

I prefer to work with voltages since its easier for me to think of amplifiers as voltage controlled voltage sources and you can skip a step and just work with the basic ohms law, instead of converting back and forth to power all the time.

jt
 
DB

I love this forumn. I learn alot everytime I come here. Power is proportional to the square of the voltage. The db relation to voltage is therefore doubled that of the relation to power. A doubling of wattage gives a 3db increase and a doubling of voltage gives a 6db increase. It is much easier for a novice to use wattage over voltage. Now if you double your wattage from 200 to 400 you will naturally need another driver. I'm not saying anybody is wrong, I think its more a case that we're using two different methods to arrive at the same conclusion. Now given db/spl=20log(p1/p0) where p0 and p1 are the sound pressures in Newtons per square meter, Then if one spl is twice another it is 6 db greater if 10 times another it is 20 db greater. Anyway, I've gone a little deeper into this than I wanted but that is good. I stand by my earlier statements. This is the formula used by the large Concert Sound companies to size their rigs for different venues given a target db rating.;)
 
jteef

Explain to me your math on your speaker example. If 8 ohm at 92 db 1w/1m, then at 16 watts why only 101 db ( I calculate 104. In keeping with the same driver for your second example you give each speaker half the power yet you deducted 3 db. When you sum the on axis output this would have to double the single spl to gain 6 db to reach 104 db, which is what I calculated the original spl to be as opposed to your 101 db. Are you saying the spl will double with two speakers. Also I've never heard of rms applied to sound pressure before, could you explain this.
 
Audiofreak

If you double the drivers without the power then you half the power to each speaker which causes a 3db loss per speaker. Naturally if I am driving my speakers to their max eff. then if I double my power I must add more drivers. My example was not a typical case of driving 100 watt speaker with a SE tube amp. I'm assuming a perfect output match; amplifier to driver.
 
Re: Audiofreak

PassFan said:
If you double the drivers without the power then you half the power to each speaker which causes a 3db loss per speaker. Naturally if I am driving my speakers to their max eff. then if I double my power I must add more drivers. My example was not a typical case of driving 100 watt speaker with a SE tube amp. I'm assuming a perfect output match; amplifier to driver.

if you have 2 x 90db/1W/1M drivers and you connect them is paralell and give it a total of 1W the you'll get 93dB ... same power in gives an extra 3dB because the radiating area is doubled.

its the same with crossovers if both drivers are down 3dB @ the crossover frequency there is a 3dB peak on axis in the frequency response.
 
Well help me then with the math; if twice the power is a 3 db gain then half the power would be a 3db loss. Are you saying that a speaker at 90db 1w/1m will not change its output spl if you drop it to a half watt. It would then be 90 db at 1/2w/1m. You have to double your spl to gain 6db, but I don't understand were your increase is coming from. If you cut half the power don't you get half the movement of the cone. If thats the case; if I half the movement of one cone and add another to it then havn't I gone back to were I started as it relates to air being moved.

Aren't there other forces in play at the x-over point.
 
PassFan said:
Well help me then with the math;

1.if twice the power is a 3 db gain then half the power would be a 3db loss.

2.Are you saying that a speaker at 90db 1w/1m will not change its output spl if you drop it to a half watt. It would then be 90 db at 1/2w/1m.

3.You have to double your spl to gain 6db, but I don't understand were your increase is coming from. If you cut half the power don't you get half the movement of the cone. If thats the case; if I half the movement of one cone and add another to it then havn't I gone back to were I started as it relates to air being moved.

4.Aren't there other forces in play at the x-over point.

1.This is correct.

2. no the individual driver will decrease it's output according to it's efficiency which does not change.

3.its got to do with how the acoustic waves add together when on axis.

4.as far as this is concerned, no there are not.
 
Solution

Well I couldn't put this one to bed without knowing what is going on here, so I went off to read and what I read confused me even more until I thought about it. The book stated exactly as I told it here in my original post. Also it stated that two sine waves exactly in phase with each other when added will double the size of the wave. Elswhere in the book it stated when the angles of the acoustic throws converge and cover over each other that anywhere from 3db to 6db of increase will be realized. Well how can the same book give and take away at the same time. It dawned on me while getting into the shower. It has to do with the axis. Large concert sound systems are designed for maximum coverage of large areas (which we don't concern ourselves with in our listeneing rooms). They target their maximum db level in the dead zones. Their use of multiple arrays with precise off axis angles are designed for coverage not convergance. They don't worry about a 6db gain at the convergance only the db level in the dead zone. We at home are only concerned with one axis, as it relates to the listening position. That being, then our drivers are mostly all convergence as close as 1 meter away from the speaker. That is how Klipsch gets their eff.. What do you think audiofreak.
 
yep i agree with that ... as you stated the gain is upto 6dB if you are in the right spot .... well 2 x 90db/1W/1M drivers each fed 0.5W and producing a signal that is perfectly in phase with the other .... each individual driver will be down to 87dB but then you add the 6dB of gain which will bring it back to 93dB so as i there you have it .... we were both right about different things.
 
I am sorry, you guys are correct on the calculations of the math.

I had written 104dB and changed it for reasons that elude me right now.

in any case, the principle is still valid.

SPL is equal to 20 log pRMS / 2x10^-5

one equation I have relating RMS air pressure to a moving coil loudspeaker is:

pRMS = p0/(2*pi) * BL*EgRMS / (Sd*RE*MAS) *G(s) * Tu1(s)

BL is the BL product, RE is voice coil resistance, Egrms is the generator/input voltage, Sd is cone area, MAS is acoustical mass of diaphram and air load in infinite baffle, G(s) and T(s) represent the 2nd order high pass and 1st order low pass transfer functions and dont affect pRMS in the midband.

If you double EgRMS, pRMS doubles, giving rise to a 6dB increase in SPL.

however, when you double the power, EgRMS only increases by sqrt(2) = 1.41, and thus pRMS increases by 1.41 and SPL increases by 3dB. 20 * log(1.41) = ~3dB vs 20 * log(2) = 6dB.

You can also use the pressure equation above to explain why a doubling of the cone area will produce a 6dB increase in SPL.

MAS = Mass of Diaphram / (Sd^2) + 2*MA1

Since MAS is inversely related to the square of Cone Area, we end up with Sd also being directly related to Prms. MA1 is the mass of the air load impedance on the piston.

When you connect two drivers you are doubling the cone area which calls for a 6dB increase, and dividing the power between them in half, which only calls for a 3dB decrease.

Suppose you had two 92dB/1w/1m 8 ohm speakers with the amp I referred to above ( capable of 11.32V(16W) into 8 ohms). If you wire the coils in series, you'll be left with a 16 ohm load for the amp. 5.66V will appear across the coil of each speaker which is one doubling of 2.83V, a 6dB increase.

92+6dB = 98dB per speaker. Adding the 6dB for the increase in cone area brings us to 104dB. Note that the amplifier is only supplying 8 watts total or 4 watts to each speaker.

I will redo the math for the next one since i wrote the wrong numbers down in my other post. 2 speakers each on their own 8W amplifier(8Vrms output)

The voltage is doubled from 2.83 to 5.66 and then 8/5.66 = 1.41

2*1.41 = 2.83...20 * Log(2.83) = 9dB

so you get 92+9+6 = 107dB

a 3dB increase in efficiency compared to one speaker with 16 watts input at 104 dB.


Now consider two 8 ohm speakers with their coils wired in parallel. Here each speaker will be receiving 16 watts each.

The voltage doubles twice from 2.83V to 11.32V. 20*log(4) = 12dB + 6 more because there are two of them.

92+12+6 = 110dB assuming your amplifier is capable of delivering the extra current required.

Sorry if i made this a lot more confusing than it really is, I hope I'm never a teacher...

jt
 
Wow, I think I understand this right now. You guys said a mouthful, I am just trying to obsorb it all.

These are the speakers I was talking about in my original post:

http://www.klipsch.com/products/comparison.asp?line=Home&subcategory=floorstanding

I don't mean to be rude and restate my original question, because I really do appreciate all the information that I am getting from your debate. However what amazes me about these speakers is thier efficiency with only two mid/bass drivers and a tweeter. It doesn't seem that from your talks I could achieve this with only two drivers, especially not any that I have seen. I beleive about 95 db/1w/1m woudl be the best I could do because I have been unable to find any drivers with spl above 92 db/1w/1m. So now that you have worked out the theory I am wondering if there is a way to bring it to a real enclosure.

I do realize that the increase in efficiency from 95 db which I am saying I could probobly get to 98 or 100 db which the klipch speakers achieve will really only seem like a small increase in volume. However I am just interested in the concept of creating the most efficient speaker possible. Any other advice would be appreciated.

Mark
 
Ok, well odds are that klipsch measures the efficiency @ just the right point in order to get such high numbers.... either that or they just simulate it on computer. with 2 drivers covering the same passband the efficiency /1W/1M that you wish to achieve must be 3dB greater than the efficiency of a single driver.... so if you want 100dB/1W/1M you'll need to find drivers that have an efficiency of 97dB/1W/1M .... now here comes the killer......... baffle step contributes a 6dB loss in the lower octaves and is dependant on the width of the enclosure so now the 2 mid/bass drivers must be about 103dB/1W/1M and then you must equalize the response in the crossover