My head is tangled from too many posts.
If I increase speaker efficiency by 3db, I actually double the output.
but don't I need 10 times the output to double the apparent volume?
er......... how many db would I need to increase by to double the apparent volume?
I percieve that I have lost the (simple) plot.
If I increase speaker efficiency by 3db, I actually double the output.
but don't I need 10 times the output to double the apparent volume?
er......... how many db would I need to increase by to double the apparent volume?
I percieve that I have lost the (simple) plot.
I already knew that the ear was logrithmic
I already knew that 3db equates to a doubling of power output
But that's not the answer to the question.
I already knew that 3db equates to a doubling of power output
But that's not the answer to the question.
I could live with this answer if I could get my head around the numbers, but I just can't make the numbers prove it. There is no way that I can make 2, 3, of 10 crunch into .414 using logs or natural logs. Its buggin me.
The formula for the power in watts is :
x dB = 10 log y
x = number of dB increase
y = times the power you get
Look at what the guy said, he said an increase of .414 maybe he should have said power of 1.414 times versus the original power imput.
You want it to use two times more power.
10 log 2 = 3.0102999566398119521373889472449 dB increase
The formula for the amplitude in voltage is :
z dB = 20 log w
z = number of dB increase
w = times the voltage amplitude you get
You want it to use twice the amplitude.
20 log 2 = 6.0205999132796239042747778944899 dB increase.
Understand a bit better now? Keep the questions coming!
x dB = 10 log y
x = number of dB increase
y = times the power you get
Look at what the guy said, he said an increase of .414 maybe he should have said power of 1.414 times versus the original power imput.
You want it to use two times more power.
10 log 2 = 3.0102999566398119521373889472449 dB increase
The formula for the amplitude in voltage is :
z dB = 20 log w
z = number of dB increase
w = times the voltage amplitude you get
You want it to use twice the amplitude.
20 log 2 = 6.0205999132796239042747778944899 dB increase.
Understand a bit better now? Keep the questions coming!
So for fun, since alot of people agree that 10 dB is twice as loud, you could invent a formula...
u dB = 33.219280948873623478703194294948 log v
u = number of dB increase
v = times the loudness you get
Want it twice as loud so :
33.219280948873623478703194294948 log 2 = 10.000000000000000000000000000014 dB increase
Wonderful, isn't it?
u dB = 33.219280948873623478703194294948 log v
u = number of dB increase
v = times the loudness you get
Want it twice as loud so :
33.219280948873623478703194294948 log 2 = 10.000000000000000000000000000014 dB increase
Wonderful, isn't it?
So anyway, working in 3 sig fig if we can;
Again, I get the power equation:
10 X the Log of 2/1
and the Voltage equation.;
20 x the log of the ratio 2/1 - which works for SPL
I also get the 10db required for the perception of double volume. ( said 10 times. I was wrong).
I still don't see 1.414 ( I already tried it).
I can see that;
33.2 log 2 = 10db is derived from the voltage/SPL equation, but i can't see what use it is. (un herring rouge?)
How much do I need to raise the speaker power by to increase the SPL by 10db? Or is it as simple as saying that to increase SPL by 10db I must raise the power by 10db?
So If I needed to raise the power by 10db;
20 log 3.2 = 10db ?
Again, I get the power equation:
10 X the Log of 2/1
and the Voltage equation.;
20 x the log of the ratio 2/1 - which works for SPL
I also get the 10db required for the perception of double volume. ( said 10 times. I was wrong).
I still don't see 1.414 ( I already tried it).
I can see that;
33.2 log 2 = 10db is derived from the voltage/SPL equation, but i can't see what use it is. (un herring rouge?)
How much do I need to raise the speaker power by to increase the SPL by 10db? Or is it as simple as saying that to increase SPL by 10db I must raise the power by 10db?
So If I needed to raise the power by 10db;
20 log 3.2 = 10db ?
The guy was talking about voltage amplitude... he wanted to tell you what 3 dB of increase would do to the voltage amplitude.
If we take that formula...
20 log 1.414 = 3 dB
Now, to get 10 dB more SPL, you need to increase the power by 10 dB, yes it's as simple as that.
10 dB = 10 log y
y = 10^(10/10) = 10^1 = 10
That means to get 10 dB more SPL, you need 10 times the power.
If with 1W you get 85 dB, to get 95 dB you'll need 10W, and to get 105 dB you'll need 100W, and to get 115 dB you'll need 1000W and etc...
If we take that formula...
20 log 1.414 = 3 dB
Now, to get 10 dB more SPL, you need to increase the power by 10 dB, yes it's as simple as that.
10 dB = 10 log y
y = 10^(10/10) = 10^1 = 10
That means to get 10 dB more SPL, you need 10 times the power.
If with 1W you get 85 dB, to get 95 dB you'll need 10W, and to get 105 dB you'll need 100W, and to get 115 dB you'll need 1000W and etc...
Pbassred said:
You used the voltage equation there...
Yes, you need to increase the voltage by a factor of 3.2 and the current by the same factor... why? Because power is equal to current x voltage so 3.2 x 3.2 = 10.24.
More simple to use the power equation directly and forget about the voltage amplitude equation. You'll probably not need it.
Life is simple.
So, more efficient speakers costs less than bigger amplifiers (bank for bucks). I'm my case, its a matter of jumping from 800W to 8000W, or find drivers that are 10db more efficient. So, ok, neither is going to happen but given my cheap chinese drivers, Eminence at 98db is a worthwhile swap.
Life is simple
So, more efficient speakers costs less than bigger amplifiers (bank for bucks). I'm my case, its a matter of jumping from 800W to 8000W, or find drivers that are 10db more efficient. So, ok, neither is going to happen but given my cheap chinese drivers, Eminence at 98db is a worthwhile swap.
Life is simple
Amplifier power is quite cheap these days... 2400W cost around 600$.
Then you have horn-loading to increase the efficiency and multiple drivers increase the efficiency too. At the same power level, doubling the number of drivers increase the efficiency by 3 dB. That's why some people build line array, with 8 cheap drivers rated at 87 dB you reach 96 dB efficiency. So if that Eminence driver is more than 10 times the price of a cheap and relatively good driver, you can think about that.
Then you have horn-loading to increase the efficiency and multiple drivers increase the efficiency too. At the same power level, doubling the number of drivers increase the efficiency by 3 dB. That's why some people build line array, with 8 cheap drivers rated at 87 dB you reach 96 dB efficiency. So if that Eminence driver is more than 10 times the price of a cheap and relatively good driver, you can think about that.
Yeah. It all needs to go in one car though. 
I already run a biamped system using a single 15" and a 10" + Piezo (each side)x-over at 250Hz. The horn loaded enclosure would give me the same problem. The bucks includes the transport.
I didn't know about the 3db increase from adding drivers. I would gain something by addind a 2nd cab, since I would then be at 4ohm, but that's only worth about 10 log 425/275 = 2db
........... But if you need to drag that much gear around - its time to hire a pro!
You unwittingly told me something though. If you think that a cheap speaker is likely to be only 87db, it answers my earlier post about my anonymous, no-spec Chinese drivers. I can get 97 - 98db from an Eminence Kappa.
I already run a biamped system using a single 15" and a 10" + Piezo (each side)x-over at 250Hz. The horn loaded enclosure would give me the same problem. The bucks includes the transport. I didn't know about the 3db increase from adding drivers. I would gain something by addind a 2nd cab, since I would then be at 4ohm, but that's only worth about 10 log 425/275 = 2db
........... But if you need to drag that much gear around - its time to hire a pro!
You unwittingly told me something though. If you think that a cheap speaker is likely to be only 87db, it answers my earlier post about my anonymous, no-spec Chinese drivers. I can get 97 - 98db from an Eminence Kappa.
I was woundering the same thing and finally got the answere from Simon5....Thanks..
To continue with this line of thought....Line arrays...
How does amplification work for Line arrays? If a feed 1 driver 1watt I get 86db. I have 16 drivers in my array for an efficiency of 98db. But do I need to give each driver 1watt signal?.....So to feed my array I would need 16watts of power?
For 119db would I need 100 or 160watts amplifier?
If I biamped the array would my amplifier requirements be the same or smaller? Whats the determining factor?
Thanks for you help.
Exipnos
To continue with this line of thought....Line arrays...
How does amplification work for Line arrays? If a feed 1 driver 1watt I get 86db. I have 16 drivers in my array for an efficiency of 98db. But do I need to give each driver 1watt signal?.....So to feed my array I would need 16watts of power?
For 119db would I need 100 or 160watts amplifier?
If I biamped the array would my amplifier requirements be the same or smaller? Whats the determining factor?
Thanks for you help.
Exipnos
Pbassred said:
A 2nd cab would bring the amplifier power up by 2 dB like you calculated, but you'll gain 3 dB like I just said about the increase from adding drivers at the same power level.
2 + 3 = 5 dB gain in the end.
Note that the gain from going to 4 ohms is only available if the amplifier deliver more power at 4 ohms, you understand fast hehe!
exipnos said:
I said if you keep the power imput constant and you add drivers you gain efficiency. Your amplifier stay at 1W output.
At 1W total you'll get 98 dB output.
To get 119 dB, you will need a 128W amplifier.
100W will only get you 118 dB.
160W will get you 120 dB.
Biamping, you get the same requirements. You need the same amount of total power to get them at that loudness.
I guess I should have said 1.414. Sorry, I assumed everybody was familiar wih ohms law. Yes, these are voltage calculations, not power. Like it or not, amplifiers really put out volts, and power (watts) are just a number derived by calculating the volts out of the amplifier by the current droped over a given load. An increase in (voltage) amplitude of 1.414 gives a power increase of 3db. A doubling in amplitude gives a 6db increase (4 times the power).
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