Does anyone know how much power we need for each frequency? I mean, I wanna know, how much relative power do I need for 20hz, 30hz, 50, 100 ...
Lets say; at 20hz, for each additional 3 db, I have to double the power comparing 30hz
for 30hz for each additional 3db, I have to double the power comparing 40hz.
..... etc.
I hope I could explain 🙂
Lets say; at 20hz, for each additional 3 db, I have to double the power comparing 30hz
for 30hz for each additional 3db, I have to double the power comparing 40hz.
..... etc.
I hope I could explain 🙂
Hi, depends on enclosure, drivers, all kinds of things. You can inspect this quite easily with VituixCAD for example, that shows power at any point in your system with any signal and intensity. You can look at single woofers on the enclosure tool as well.
Here an example, 18" TBW100 driver in a reflex box, I can tune the box to determine at which frequency there is more or less power, shown by the top middle graph. With constant input voltage of 2.83V this 8ohm version draws about 0.3W at 20Hz and roughly 1W at 30Hz, and so on. Might be surprising, that a lot of power is used high up in frequency, but it's also a bit boosted response there so you'd EQ it down and reduce the power. So, it kinda depends 😀
In general, where your system resonates there is no need to put much power into it to get the cone flying. Below and above, where impedance drops power is pouring in.
For a given system 10x power gives 10dB extra output assuming the system is not limited in other ways. 6db increase in output needs 4x more power, 3db double power.
Here an example, 18" TBW100 driver in a reflex box, I can tune the box to determine at which frequency there is more or less power, shown by the top middle graph. With constant input voltage of 2.83V this 8ohm version draws about 0.3W at 20Hz and roughly 1W at 30Hz, and so on. Might be surprising, that a lot of power is used high up in frequency, but it's also a bit boosted response there so you'd EQ it down and reduce the power. So, it kinda depends 😀
In general, where your system resonates there is no need to put much power into it to get the cone flying. Below and above, where impedance drops power is pouring in.
For a given system 10x power gives 10dB extra output assuming the system is not limited in other ways. 6db increase in output needs 4x more power, 3db double power.
Thank you for your reply. 🤗
I know how logarithmic relationship works between power and output ☺️
it’s impossible to move a 18” cone in Reasonable distances at 20hz by 1 watt 🙂 There is something wrong with graphs or I can’t interpret it properly ☺️
So could you tell me; 18" TBW100 driver in a reflex box in your example, how much power do we need to get 90db/1mt and 100db/1mt at 20hz?
90 and 100db at 30hz?
90 and 100db at 40hz?
You know all sensitivity measures of speakers are taken differently. What i am wondering is a general rule without counting on speaker’s specifications.
What I am trying to understand is what is the rational relationship between power and frequencies?
I know how logarithmic relationship works between power and output ☺️
it’s impossible to move a 18” cone in Reasonable distances at 20hz by 1 watt 🙂 There is something wrong with graphs or I can’t interpret it properly ☺️
So could you tell me; 18" TBW100 driver in a reflex box in your example, how much power do we need to get 90db/1mt and 100db/1mt at 20hz?
90 and 100db at 30hz?
90 and 100db at 40hz?
You know all sensitivity measures of speakers are taken differently. What i am wondering is a general rule without counting on speaker’s specifications.
What I am trying to understand is what is the rational relationship between power and frequencies?
It's the same, 6db more output at any frequency needs 4x more power at that frequency. Power at any particular frequency depends on the driver properties and in which kind of a enclosure it's in.
If at all possible, please download VituixCAD, open it, on top "Tools" menu click "enclosure tool" and off you go, you figure out this stuff in no time 🙂
Hint: big driver in a big box it needs much less power to reach any output, than on a smaller one.
If at all possible, please download VituixCAD, open it, on top "Tools" menu click "enclosure tool" and off you go, you figure out this stuff in no time 🙂
Hint: big driver in a big box it needs much less power to reach any output, than on a smaller one.
Thank you again 🤗
But I am sure that the power needed to move 18”woofer and 1” tweeter is not same. So parallelly power needed for frequencies produced by these drivers are also not same 😊
But I am sure that the power needed to move 18”woofer and 1” tweeter is not same. So parallelly power needed for frequencies produced by these drivers are also not same 😊
Mostly depends on the signals (music?) you want to reproduce.
I once made a graph showing peak signal values of different music tracks:
I will read the thread. Thank you.
If we had a full range driver and if we knew the different sensitivity values of each frequencies, (not an average in 1khz), then we could have known the answer ☺️
If we had a full range driver and if we knew the different sensitivity values of each frequencies, (not an average in 1khz), then we could have known the answer ☺️
The power distribution depends on the type of music.
But, as a generalization you need about half the power below 300 Hz, and the otherhalf above.
dave
But, as a generalization you need about half the power below 300 Hz, and the otherhalf above.
dave
Hi Dave. Thank you.
It’s not what I wanted to learn.
I saw some topics that people say that they only need 5-10watts to drive their speakers. So why do we need thousands of watts for subwoofers. ( I know we need 🙂
Is there any “general rule” of rational relationship between power and frequencies?
I know all logarithmic relationships between power and spl but it doesn’t work same for all frequencies maybe because of perceived loudness effect. I don’t know.
If yes, there should be a general formula.
Additionally, it’s clear that the power needed to move a 1” tiny tweeter and 18” big woofer is not the same.
I am not capable of calculating with measurements tools. So a general formula is what I am looking for.
Thank you all for your time.
It’s not what I wanted to learn.
I saw some topics that people say that they only need 5-10watts to drive their speakers. So why do we need thousands of watts for subwoofers. ( I know we need 🙂
Is there any “general rule” of rational relationship between power and frequencies?
I know all logarithmic relationships between power and spl but it doesn’t work same for all frequencies maybe because of perceived loudness effect. I don’t know.
If yes, there should be a general formula.
Additionally, it’s clear that the power needed to move a 1” tiny tweeter and 18” big woofer is not the same.
I am not capable of calculating with measurements tools. So a general formula is what I am looking for.
Thank you all for your time.
Typical modern subwoofers are usually not very sensitive, bass indeed can suck up a whole lot of power. And even more if you EQ. You need an amplifier with sufficient power, but even then thye could clip, how they recover from clipping really affects things. Most power i have used with mysubs was a 4 x 100w amp.
Remember that a just noticable increase in volume, 3dB, takes 2 x the power, twice as loud, 10dB, takes 10x the power. So how loud you play, or if you turnip thetas, can quickly burn watts.
You will find a wholeniche where higher sensitivity professional woofers are used.
dave
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If 300Hz is the half power point perhaps 3000Hz could be considered the tenth power point so the tweeter would then only "need" 10% of the half power.
So if using 100W for bass and 100W for tops then a tweeter might only need 10W. Purely guess work on my part tho as I never power a tweeter on it's own
So if using 100W for bass and 100W for tops then a tweeter might only need 10W. Purely guess work on my part tho as I never power a tweeter on it's own
Every driver has a impedance curve, and voltage is coming in.
Last time I checked ohms law didnt change.
because of full space losses and the way our ears perceive sound
anywhere from 3 to 12 dB is added to the bass.
depends on listening levels.
Subwoofers dont need thousands of watts at home.
tweeters dont move as much because the wavelengths are smaller.
regardless the impedance is what it is.
Most the energy going in turns into heat. As with any common direct radiator
Last time I checked ohms law didnt change.
because of full space losses and the way our ears perceive sound
anywhere from 3 to 12 dB is added to the bass.
depends on listening levels.
Subwoofers dont need thousands of watts at home.
tweeters dont move as much because the wavelengths are smaller.
regardless the impedance is what it is.
Most the energy going in turns into heat. As with any common direct radiator
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Here are some music spectra that I have measured.
I believe that speakers are generally designed to handle the spectra of pink noise above about 100Hz.
Ed
I believe that speakers are generally designed to handle the spectra of pink noise above about 100Hz.
Ed
Yeah, SPL is ultimately about volume displacement, simplified it means cone area * excursion. To maintain same SPL octave lower in frequency, the volume displacement needs to quadruble. So, since any woofer or tweeter has static size, as it stays ~the same size no matter what frequency it tries to reproduce, it means higher frequencies need way less excursion than lows to maintain flat SPL. This means highs can get away with way less power to move smaller cone less, than at same SPL at some lower frequency bigger cone needs to move way more.
Two octaves down volume displacement is 16x already, so either the cone area must have gone up or the excursion, which would need more power in general. Sensitivity is often the voltage sensitivity, how much SPL is measured at 1m at standard 2.83V or 1V or what ever input. For highs this is very low, compared to lows. For example, if you have 90db at 100Hz, and count 7 octaves up it's 12800Hz which needs 4^7 = 16384 times less volume displacement to reach 90db. Typical 15" driver has about 850cm^2 cone area, while a 1" tweeter dome is roughly 5cm^2, this is difference of 170. 16384 / 170 is roughly 100, so the woofer still needs to have about 100x excursion of the tweeter even though it's much larger, to product same SPL at 100Hz as the tweeter does at 12800Hz.
Consequently 12800Hz wavelength is roughly 2,7cm, so almost equals diameter of the tweeter. Wavelength of 100Hz is 3.4m, so if you wanted a woofer to be have as little excursion at 100Hz as tweeter has at 12800Hz, the woofer would need to be roughly 3.2m in diameter!🙂 15" is way less, like 10x, hense much more excursion required.
Well, these numbers seem so outrageous I might have error, so please welcome anyone to double check 🙂 But it's along the lines, lows are very tough to reproduce due to very very long wavelenght.
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Thanks for checking 😊 I use some rounding, so some of the numbers are approximate, while some more accurate which is kinda bad practice. Magnitude should be the same though, with rounding or not, so in this sense there is room to be bit sloppy to save time.
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the numbers are very exact considering a 25 mm dome:
frequency: 12800 Hz / 2^7 = 100 Hz
diameter: 2.5 cm * 2^7 = 320 cm
as the surface is diameter squared the numbers include the quadrupling of displacement for half frequency.
Very simple table is represented here: Power Distribution and SPL
It's not 100% accurate, but acceptable for fast calculations!
It is for 2-way speakers, but can re-calculate and for 3-way.
It's not 100% accurate, but acceptable for fast calculations!
It is for 2-way speakers, but can re-calculate and for 3-way.
The amp is voltage controlled. So voltage in results in a amplified voltage out.
But impedance of speaker is not constant over frequency. The system in example has a resonance at 20hz. Then the real part of the impedance is huge. And then the current from the amp is very low. And that means the power needed from amp at 20hz is low