Wattage, Speakers, unlimited SPL ?

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Jon8

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Hello everyone :smile:


I have a question that bugs me since few days now..

On many audio related forums, it's admitted that when you double the power (wattage) to a speaker*, you should gain 3 decibels (theorically of course, there is many other factors such thermal and mechanical limitations, etc..)

Also, it's admitted that when you double the cone area, by adding a SECOND identical speaker*, you should also gain 3db.

*speaker = transducer, raw driver, not enclosure.

BUT, most people pretend that when you connect two speakers in parallel (i.e. 2 x 8 ohms nominal in // = 4 ohms load) you get double power from the amp and double surface cone area = 6db.

Well, i disagree.

I cannot understand why the power DOUBLED in EACH voice coil :-?


Because if that's true, then we could get an unlimited sound pressure level with only 1 watt, only by using enough transducer...

By exemple, let's take a 90db 1w/1m rated driver.


90db reading w/ONE driver using 1 watt = 1 watt consumption TOTAL

93db reading w/TWO drivers using 1/2 watt ea. = 1 watt consumption TOTAL

96db reading w/FOUR drivers using 1/4 watt ea. = 1 watt consumption TOTAL

99db reading w/EIGHT drivers using 1/8 watt ea. = 1 watt consumption TOTAL


... and so on.


Something seems wrong, doesn't it ? :confused:


I think there is major difference between adding cone area on a single motor (voice coil) AND adding cone area that will end up with TWO motors to feed with power. Am i right ?

I just can find a mathematical formula or anything to verify all this...

any help will be greatly appreciated. :smile:
 
Adding a second speaker to an amplifier will double the total watts into the speakers, e.g. one speaker=10 watts, two speakers=10 watts each/20 watts total, PROVIDED that the amplifier is big enough, and that it will work correctly into the different impedance of a single speaker or two speakers in parralel.
Therefore subject to the above caveats, two speakers give twice the sound of one, not four times.

Doubling the cone area only doubles the output if each speaker cone is still driven at the same wattage. In the example you give, the total output power from the amplifier is doubled, but the power PER SPEAKER is still the same, therefore the sound output per speaker is the same, and the total sound output is doubled by the presence of two speakers instead of one. If the same power is spread among two speakers then in theory no increase in sound can result.

As an example
one speaker at 10 watts=90 db
two speakers at 5 watts each=90 db
one speaker at 20 watts=93 db
two speakers at 10 watts each=93 db

In practice the situation is more complex than this suggests since the effeciency of a speaker is not constant but varies according to the input wattage.
 
Broadgage, thank you for your answer.

That's exactly what i think.

However, it's far from being accepted in the audio forums. I've asked few people who are involved in the audio business and it seems that our calculation is incorrect, for a reason that i just don't understand.

This 6db gain thing seems to be widely accepted in our domain.

But i feel like having the possibility of unlimited SPL (which is acoustic energy) from a single watt (or even less, in fact..) it's just impossible.

A speaker efficiency (electrical energy/acoustic energy) is what ? 1% ? 3% ?

So i don't understand why it could be possible to obtain 100% and even more (!) efficiency only by adding other transducers... It just doesnt make sense to me.

Is there anyway to prove the fact that it's impossible ? A simple experience, a mathematical formula ?
 
Jon8 said:
Also, it's admitted that when you double the cone area, by adding a SECOND identical speaker*, you should also gain 3db.
Click to expand...

If yu do this, you have already doubled your output power. Either that, or you have taken some away from the first speaker.

Are you sure you aren't misunderstanding in some way. Decibels can be funny animals. If I remember correctly, if you double the voltage, thats a 6 db increase, but if you double the power, that's a 3 db increase. Or was it the other way around?

So decibels in electrical power may not equal decibels of sound pressure.
 
Jon8 you have to put it into context. Amplifiers are rated something like this 50 watts @ 16-ohms, and 100 watts @ 8-ohms. Notice I did not include 4-ohm rating.

Transistor ampliers are very sensitive to impedence, and will only be rated down to some number. For example if the lowest rated impedence is 8 ohms at 100 watts, adding another speaker to make it 4 ohms will not give you 200 watts. It means you over-loaded the amplifier.

If has to do with the DC supply voltage driving the finals. Most high-end amplifiers use class AB1 with a bi-polar power supply. The transistors have a minimum impedaence when turned fully on. This minimum impedence is the same as the minimum impedence as the speaker impedence. So lets say it is 8 ohms with a plus/minus 25 volt power supply. Do some math and see what happens when you put a 4 ohm speaker in series with 8 ohms of the transistor.

Now if you are an ole fart like me that still has a McIntosh tube amplifier that has a matching output transformer you can select your ouput impedence of 32/16/8/4 and the output is always the same. Can't do that with transistors.

What I am driving at is if you amplifier is rated down to say 200 watts @ 2 ohms, you get the most power output at 2 ohms. But if you put 4 ohms on the output you only get 100 watts, 8/50, and 16/25
 
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steve66 said:
If yu do this, you have already doubled your output power. Either that, or you have taken some away from the first speaker.

Are you sure you aren't misunderstanding in some way. Decibels can be funny animals. If I remember correctly, if you double the voltage, thats a 6 db increase, but if you double the power, that's a 3 db increase. Or was it the other way around?

So decibels in electrical power may not equal decibels of sound pressure.
Click to expand...

Double the power is indeed a 3db increase in sound, presuming constant efficiency etc.
Double the voltage would be four times the power, presuming a constant impedance. That because doubling the voltage will double the current, giving four times the power, which presuming constant efficiency is four times the sound, which is 6db.
 
It's like this:

It's like this:

Pdb = 10log(P2/P1)

Pdb = 20log(V2/V1) constant Z

Increasing voltage by a factor of 1.414 doubles the power, 3dB

Increasing voltage by a factor of 2 quadruples the power, 6dB.
 
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so that would be correct ?


90db 1w/1m (1 driver) = 1 watt total consumption
90db ?w/1m (2 drivers) = 1 watt total consumption
93db 2w/1m (1 driver) = 2 watts total consumption
93db 1w/1m (2 drivers) = 2 watts total consumption
96db 2w/1m (2 drivers) = 4 watts total consumption


Some speakers manufacturers use the 2.83vrms as a reference to express the sensitivity of a speaker instead of Watt value, even if it can be really misleading if not 8 ohms speakers because if 2.83v @ 8 ohms = 1 watt, it's not true with 4 ohms speaker... It's 2 watts.

All that probably started for the marketing department.. :roll:

It makes things so confusing that now people believes that you could get an unlimited sound pressure level with 1 watt (or 1 milliwatt...).

Do you have guys any idea of how to actually prove all this ? I mean, with an experience on field ?

Since the impedance on a speaker is variable and rarely purely resistive, it's difficult to control the exact power (watt) it takes from the amplifier. But what i may need it's a simple and efficient way to actually control the WATTAGE from the amplifier; not less not more than X value.

In other word i need to control the voltage AND the current, no matter what the impedance is.

How can i do that ?
 
Broadgage & Rattus: That's what I was trying to remember. Thanks for the refresher.

It seems pretty obvious that if you add a second speaker in parallel to an existing one, you are NOT going to get more than twice the output power. The amplifier's output voltage won't increase, (it will actually decrease as Dereck pointed out). But assume the best case and the output voltage stays constant: Two speakers in parallel would draw twice the current (that's easy to show with ohms law). Twice the current with the same voltage is twice the output power.

So if volume is directly proportional to wattage, that's twice the volume.

Steve
 
The whole situation gets complicated quickly!
a) Almost all loudspeakers impedance varies with frequency.
b) Almost all loudspeakers output SPL varies with frequency.
c) Almost all audio amplifiers have close to 0 Ohms output or source impedance. (well some some tweeky vacuum tube amps don't) That means that the amplifier will output the same voltage into any reasonable speaker impedance.
d) Speaker impedance may be up to 45degrees capacitive or inductive.
So building some sort of circuit that will limit Wattage or VA is tricky. But if you manage to build this circuit it will sound terrible when it goes into clipping. Early solid state amplifiers had V & A limiters to keep the output transistors in their Safe Operating Area and they sounded bad at high volume.

On another note: The combined loudspeaker SPL will very depending on:
a) Distance between speakers.
b) Audio frequency.
c) Position of SPL meter.
 
i agree with you speedskater.

It's not an easy thing to isolate the watt value with all this.

But i think it's possible somehow.


Otherwise, how the famous XXdb 1w/1m would be possible ? :-?
 
Speedskater said:
c) Almost all audio amplifiers have close to 0 Ohms output or source impedance. (well some some tweeky vacuum tube amps don't) That means that the amplifier will output the same voltage into any reasonable speaker impedance.
Click to expand...
This is an inacurate statement IMO. I do not know of any transistors that approach 0 ohms when forward biased fully on. Some FET's reach the 2 ohm level (operated im parallel) which would be the maximum forward transfer ratio and consequently the lowest speaker impedence for maximum power.

This directly relates to the amplifiers specifications. If the amp specs 4 -ohms for the maximum power output, means the source impedence is 4 ohms.

Most of the 2-ohm amplifiers I see are the ones made for automobiles. and 4 and 8 ohm's seems to be the norm for home applications.
 
FWIW from this ole audiophile is to buy speakers with the highest effeciency. I have KlipchHorns the most effecient speaker ever made rated at 105 db spl @ 1-watt. They will rattle the neighbors windows with a 10-watt amplifier. :D
 
dereckbc said:
This is an inacurate statement IMO. I do not know of any transistors that approach 0 ohms when forward biased fully on.
Click to expand...
Actually, modern transistor amplifiers do have a sub-1-ohm output impedance; many in the hundredths of an ohm. Typically, the greater the amplifier's power rating, the lower the output impedance.

The lower the output impedance, the higher the damping factor, which is the amplifier's ability to not only start moving the speaker in response to an input signal, but its ability to halt the speaker's motion when the input signal ceases.

This directly relates to the amplifiers specifications. If the amp specs 4 -ohms for the maximum power output, means the source impedence is 4 ohms.
Click to expand...
Theoretically, maximum power transfer is acheived when the source output stage, the transmission line, and the destination input stage impedances all match. An amplifer is a constant-voltage source, like our power systems; the source impedance is way lower than the load's.

In the real world, a low output impedance driving a high input impedance is better for a higher signal-to-noise ratio (which matters more for line-level voltages than speaker-level voltages) and more voltage delivered to the load.

Normal speakers are a compromise in design. A given power level requires an amplifier with both current and voltage capabilities, and the levels we use today are based on the electronics we have at our disposal.

Most of the 2-ohm amplifiers I see are the ones made for automobiles. and 4 and 8 ohm's seems to be the norm for home applications.
Click to expand...
That's because a car electrical system, while low on voltage, has an abundance of current available. A lower speaker impedance, which requires an increase in current to deliver a given power level, thrives in a current-rich environment.

Tube amplifiers, on the other hand, are high-voltage/low-current sources, and many are (or, I should say, were) rated for best power into 16 ohms. Remember that a higher voltage is required to drive a given current level into a higher impedance.
 
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