Help with DB Calculations

DaCarlson

DaCarlson

GasBurning Stereo Machine
Decibels (dB)
It has been known for a very long time that human ears cannot resolve very small differences in sound pressure. Originally, it was determined that the smallest variation that is audible is 1dB - 1 decibel, or 1/10 of 1 Bel. It seems fairly commonly accepted that the actual limit is about 0.5dB, but it is not uncommon to hear that some people can (or genuinely believe they can) resolve much smaller variations. I shall not be distracted by this!

dB = 20 * log (V1 / V2)
dB = 20 * log (I1 / I2)
dB = 10 * log (P1 / P2)

As can be seen, dB calculations for voltage and current use 20 times the log (base 10) of the larger unit divided by the smaller unit. With power, a multiplication of 10 is used. Either way, a drop of 3dB represents half the power and vice versa.




I'm desperatly trying to learn the ways of audio, Could someone steer me in the right direction?



So, if I understand this correctly. Lets say we have an amp with pushing a current of 2amps. The formula would be...

Db = 20*log[(2)1/(2)2]

I came with up with an answer of 40 db. Does this sound correct? Or am I missing something?


I tried using the formula for Amperage but it doesn't work for me either.

dB = 20 * log (I1 / I2)
 
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methinks you're overlooking one of the currents, possibly I1. If 2amps is coming out there must be a lower I going in. What is it? log (I1/I2).
I'm sure all the BSEE experts will prove me wrong.
 
From what I understand 3 db is the smallest change the ear can detect and this is the stickler-if you are expecting the change. Or maybe it's if you are not expecting it.

Another point is 3 db is a doubling of amplifier power. 1 watt=3 db, 2 watts = 6 db, 4 watts = 9 db, 8 watts = 12 db etc.

So a 200 watt amp is 3 db louder than a 100 watter.
 
There is some confusion/misinformation about the smallest perceived increment. And it seems the date of the information has something to do with it. I was taught, in a perception class in 1970, that 1.5 dB was the smallest.

But most of agree that +3dB is a doubling of power/sound.

But it doesn't look like you have a scientific calculator, calculating lob base 10 properly.

BTW, I1/I2 are values, first current/second current.
 
The dB is only a measure of change. You have to have some reference to start from. When we talk about XXdB in relation to sound output that is referenced to 20uPa (20 micro Pascals of pressure).

For power you're usually looking at changes in reference to 1W output.

In your example of 2A, if you reference that to 1A then you'll get 20*log(2/1) = 6dB change. To relate that to the power equation, P=I^2*R so (factoring out the R which will remain constant) doubling the current will quadruple the power. So your equation would be 10*log(4/1) = 6dB. Same thing, just 2 different ways to get there.

I'm not sure exactly what your question was but maybe I got you a little closer? If you can be a little more specific I'll see what I can do.

Ray
 
DaCarlson said:
Decibels (dB)
It has been known for a very long time that human ears cannot resolve very small differences in sound pressure. Originally, it was determined that the smallest variation that is audible is 1dB - 1 decibel, or 1/10 of 1 Bel. It seems fairly commonly accepted that the actual limit is about 0.5dB, but it is not uncommon to hear that some people can (or genuinely believe they can) resolve much smaller variations. I shall not be distracted by this!

dB = 20 * log (V1 / V2)
dB = 20 * log (I1 / I2)
dB = 10 * log (P1 / P2)

As can be seen, dB calculations for voltage and current use 20 times the log (base 10) of the larger unit divided by the smaller unit. With power, a multiplication of 10 is used. Either way, a drop of 3dB represents half the power and vice versa.




I'm desperatly trying to learn the ways of audio, Could someone steer me in the right direction?



So, if I understand this correctly. Lets say we have an amp with pushing a current of 2amps. The formula would be...

Db = 20*log[(2)1/(2)2]

I came with up with an answer of 40 db. Does this sound correct? Or am I missing something?


I tried using the formula for Amperage but it doesn't work for me either.

dB = 20 * log (I1 / I2)
Click to expand...

Hi--
The "dB" is a mathematical ratio of two things in the same units, like two voltages, two currents or two values of watts. As a ratio then, you need two values to have any meaning.

Example: If an input to something is 1 volt and the output is 10 volts, and you want to know the dB ratio, it is: 20*Log(10 volts/1 volt)=20dB. Or, 10 volts is 20dB greater than 1 volt.

Example: If the amp was putting out 140 watts and is now putting out 7 watts, what is the difference expressed in dB? 10*Log(140 watts/7 watts) = 13dB. Or, 140 watts is 13dB greater than 7 watts, etc.

The dB then has no units of its own - it is just a mathematical convenience.

As for what people can hear as a minimal difference, the latest studies would peg that value as 0.1dB at some frequencies. Naturally, your hearing sensitivity varies with frequency and with loudness, so the 0.1dB is not an absolute in all cases, and certainly not with all people.
 
So if i undertand correctly you take the amps rating and compare it to its output at the time? For say... my amp is rated at 80 watts, im putting out 7. I would compare the two values in the equation? I'm not sure if i understand what you mean. So if a speaker is rated at 117 db what does that mean?
 
If you're just looking to get maximum (or current) output from your system then here's the deal...The speakers will have a rating of XXdB@1W1m. That will be the sound pressure referenced to 20uPa with 1W of input and measured at 1m.

If you want to find out the maximum that the speakers will do given your amp then you take that number and add it to 10*log(max power).

Example: if you're speakers have a rated sensitivity of 90dB@1W1m and your amp can put out 80W then the max output will be 90 + 10log(80) = 90 + 19 = 109dB measured at 1m.

If your amp is idling away at 10W then you'll be at 90 + 10log(10) = 90 + 10 = 100dB at 1m.

Does that help?

Ray
 
DaCarlson said:
So if i undertand correctly you take the amps rating and compare it to its output at the time? For say... my amp is rated at 80 watts, im putting out 7. I would compare the two values in the equation? I'm not sure if i understand what you mean. So if a speaker is rated at 117 db what does that mean?
Click to expand...

Sorry if I confused you. The "dB" is nothing other than a convenient way to COMPARE two things. It can be any two things you need to compare, as long as they are both in the same unit of measure. The ones we use in electronics frequently are Volts, Amps, Watts, SPL, Sonnes, and so on.

There's no specific meaning to it, other than its convenience as a ratio. So, instead of me saying, "Wow, my output is 37,895 times bigger than my input," I can say, my output is 91dB higher than my input." Now, why do this? Because "log values" can be squeezed onto say a meter, much easier than linear values. A volt meter than could show a linear range of 1 volt to 10,000 volts would have quite an unreadable scale! But a "log meter" going from 0 to just 100 can express a range of over 100,000:1. So, likewise, if we had to draw frequency response from 20Hz to 20,000Hz on a linear scale the thing would be unreadable. Using a log scale, it compresses down to very readable scale.

Now to your example. If your amp is 80 watts maximum, and you are using 7 watts, you can say, "The output of 7 watts is 10dB below maximum output." Again, you use the dB whenever you want an easy, standard way to compare two of the same KIND of numbers.

Hope that helps!
 
JuicyMusic said:
Sorry if I confused you. The "dB" is nothing other than a convenient way to COMPARE two things. It can be any two things you need to compare, as long as they are both in the same unit of measure. The ones we use in electronics frequently are Volts, Amps, Watts, SPL, Sonnes, and so on.
Click to expand...

I remember when I first "got" dB. My buddies and I were using it for everything. "I can eat 3dB more donuts than you can!!" :D

You can use dB for any 2 similar things, just like JM said. 4 elephants is 6dB more than 1 elephant. A 6-pack is 7.8dB more than 1 beer. It's nothing but a measure of change. And a speaker with 100W input is 20dB louder than the same speaker with 1W input.

Ray
 
RayW said:
The dB is only a measure of change. You have to have some reference to start from. When we talk about XXdB in relation to sound output that is referenced to 20uPa (20 micro Pascals of pressure).

For power you're usually looking at changes in reference to 1W output.

In your example of 2A, if you reference that to 1A then you'll get 20*log(2/1) = 6dB change. To relate that to the power equation, P=I^2*R so (factoring out the R which will remain constant) doubling the current will quadruple the power. So your equation would be 10*log(4/1) = 6dB. Same thing, just 2 different ways to get there.

I'm not sure exactly what your question was but maybe I got you a little closer? If you can be a little more specific I'll see what I can do.

Ray
Click to expand...

And thank you for keeping me from pulling out one of my old text books.

Also, please all remember Ray's mention that dB is always referenced to some unit of measure.

The cable signal coming into my TV is +2dBm. That is, 2 dB above a millivolt.
 
So lets see if i get it... I have a speaker that is rated at 92.5db at 1 metre, 1 watt. So let me get this straight. I pump 35 watts into the speakers.

Then i use the equation: 92.5 + 10log(35) = 127.5db at 1 metre
 
DaCarlson said:
So lets see if i get it... I have a speaker that is rated at 92.5db at 1 metre, 1 watt. So let me get this straight. I pump 35 watts into the speakers.

Then i use the equation: 92.5 + 10log(35) = 127.5db at 1 metre
Click to expand...

Right formula, but wrong answer. I think 10Log(35) + 92.5 =107.9dB
 
Grainger49 said:
And thank you for keeping me from pulling out one of my old text books.

Also, please all remember Ray's mention that dB is always referenced to some unit of measure.

The cable signal coming into my TV is +2dBm. That is, 2 dB above a millivolt.
Click to expand...


The term dBm uses 1 milliwatt as a reference, not 1 millivolt.

Tom
 
haven't you guys noticed in the example equation that he has the same value of amps for both I1 and I2 and thefore no gain? The log of 1=0. Or am I going bonkers?
 
The original example made no sense as there was only 1 value stated and no reference (stated or implied) so we just left it alone and tried to explain the concept in general.

Dacarlson, you seem to be having some trouble with the log function in your calculations. What are you using for a calculator?

Ray
 
watts is power, the ability to do work, like power amplifier that can drive speakers. That's where the power is dissipated, not at a cd output, tape output or phono output. Tha amplfier supplies power and the speaker consumes it. Go back to your physics books.
 
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