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How do I convert Amir's sensitivity (mv to drive 94 SPL) to db(1vRMS)?

JanesJr1

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I cannot convert Amir's numbers for sensitivity, stated as mv required for 94 dB SPL, into db(1vRMS), for example to use in the headphone power calculator.

If I divide Amir's 94dB into the mv sensitivity rating to get db from one mv, the number gives nonsensical results (including converting mV to V). I'm missing something obvious ...
 

staticV3

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94+20*log10(1/[Amir's Vrms]) =dB SPL / 1Vrms.

Example: Amir measured the ATH-R70x at 0.318Vrms/94dB SPL (at 425Hz).
-> 94+20*log10(1/0.318)=104dB SPL/1Vrms

You can then use this calculator to convert that to dB SPL/mW.
-> 104dB/Vrms and 474Ω means 100.76 dB/mW

Btw, you can use the same method to convert innerfidelity's sensitivity measurements.
E.g.:
Screenshot_20220512-151828_Drive.jpg

90+20*log10(1/0.352)=99dB/Vrms

Edit: it was Solderdude who taught me this a while back, so all credit goes to him
 
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JanesJr1

JanesJr1

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There's a handy calculator for that here...
Thank you all for the replies! I'm going to have to go back and learn the basics so I understand the logic behind the formula. I will give it a try... if anyone has a reference for a good primary reference to learn the basics of audio electronics, let me know. I've been picking up the extremely-basic-basics on the fly, but I'm going to have to learn more in a systematic way. I never learned a single thing about electrical circuits all the way through 19 years of school. Thanks again.
 
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Thank you all for the replies! I'm going to have to go back and learn the basics so I understand the logic behind the formula. I will give it a try... if anyone has a reference for a good primary reference to learn the basics of audio electronics, let me know. I've been picking up the extremely-basic-basics on the fly, but I'm going to have to learn more in a systematic way. I never learned a single thing about electrical circuits all the way through 19 years of school. Thanks again.

From my signature ...

Audio is probably the most butt simple of all the various fields in electronics ... But is still electronics so the same basic principles apply.
 
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JanesJr1

JanesJr1

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AnalogSteph

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The formula is quite easily derived under the assumption of linearity and with some basic logarithm juggling.

If the transducer is assumed linear, these two ratios should be equal:
SPL_linear(1 V) / SPL_linear(94 dB) = V(1 V) / V(94 dB)
IOW, increasing voltage input by a factor X will increase sound output by the same factor. I hope we can agree on that. This simplification generally holds very well in the world in hi-fi, as any deviation equates to nonlinear distortion.

Now let's convert the whole shebang to dB by applying 20*log10() on both sides, while exploiting the property that log(a/b) = log(a) - log(b):

SPL(1 V) - SPL (94 dB) = 20log(V(1 V) / V(94 dB))

Some minor reshuffling gives:

SPL(1 V) = SPL (94 dB) + 20log(V(1 V) / V(94 dB)) = 94 dB SPL + 20log(1 V / V(94 dB))

That should look familiar from above. :)

If you want millivolts instead,
94 dB SPL + 20log(1 V / V(94 dB)) = 94 dB SPL + 20log((1 000 mV/mV) / (V(94 dB) [mV]/mV))
= 94 dB SPL + 20log(1 000) - 20log(V(94 dB) [mV]/mV)
= 94 dB SPL + 60 dB - 20log(V(94 dB) [mV]/mV)
= SPL(1 V)

Ta-dah! :D

The hardest part is formally keeping your units straight, as logarithms obviously work on dimensionless numbers only, so you have to expand fractions as required:
log(a[unit]/b[unit]) = log(a[unit]/unit / (b[unit]/unit)) = log(a[unit]/unit) - log(b[unit]/unit)

For those wondering why you can write "94 dB SPL + 60 dB" when it may have been drilled into them that you cannot add disparate units, let's inspect what that actually means in the linear world:
94 dB SPL = 1 Pa (Pascal)
+ 60 dB = * 1000
So 94 dB SPL + 60 dB = 1 Pa * 1000.
Plain dBs are just a ratio. Only those with a suffix have ties to actual physical units.
 

Jimbob54

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The formula is quite easily derived under the assumption of linearity and with some basic logarithm juggling.

If the transducer is assumed linear, these two ratios should be equal:
SPL_linear(1 V) / SPL_linear(94 dB) = V(1 V) / V(94 dB)
IOW, increasing voltage input by a factor X will increase sound output by the same factor. I hope we can agree on that. This simplification generally holds very well in the world in hi-fi, as any deviation equates to nonlinear distortion.

Now let's convert the whole shebang to dB by applying 20*log10() on both sides, while exploiting the property that log(a/b) = log(a) - log(b):

SPL(1 V) - SPL (94 dB) = 20log(V(1 V) / V(94 dB))

Some minor reshuffling gives:

SPL(1 V) = SPL (94 dB) + 20log(V(1 V) / V(94 dB)) = 94 dB SPL + 20log(1 V / V(94 dB))

That should look familiar from above. :)

If you want millivolts instead,
94 dB SPL + 20log(1 V / V(94 dB)) = 94 dB SPL + 20log((1 000 mV/mV) / (V(94 dB) [mV]/mV))
= 94 dB SPL + 20log(1 000) - 20log(V(94 dB) [mV]/mV)
= 94 dB SPL + 60 dB - 20log(V(94 dB) [mV]/mV)
= SPL(1 V)

Ta-dah! :D

The hardest part is formally keeping your units straight, as logarithms obviously work on dimensionless numbers only, so you have to expand fractions as required:
log(a[unit]/b[unit]) = log(a[unit]/unit / (b[unit]/unit)) = log(a[unit]/unit) - log(b[unit]/unit)

For those wondering why you can write "94 dB SPL + 60 dB" when it may have been drilled into them that you cannot add disparate units, let's inspect what that actually means in the linear world:
94 dB SPL = 1 Pa (Pascal)
+ 60 dB = * 1000
So 94 dB SPL + 60 dB = 1 Pa * 1000.
Plain dBs are just a ratio. Only those with a suffix have ties to actual physical units.
Yeah, that's why I get jealous of people who understand the science and the maths.
 
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JanesJr1

JanesJr1

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The formula is quite easily derived under the assumption of linearity and with some basic logarithm juggling.

If the transducer is assumed linear, these two ratios should be equal:
SPL_linear(1 V) / SPL_linear(94 dB) = V(1 V) / V(94 dB)
IOW, increasing voltage input by a factor X will increase sound output by the same factor. I hope we can agree on that. This simplification generally holds very well in the world in hi-fi, as any deviation equates to nonlinear distortion.

Now let's convert the whole shebang to dB by applying 20*log10() on both sides, while exploiting the property that log(a/b) = log(a) - log(b):

SPL(1 V) - SPL (94 dB) = 20log(V(1 V) / V(94 dB))

Some minor reshuffling gives:

SPL(1 V) = SPL (94 dB) + 20log(V(1 V) / V(94 dB)) = 94 dB SPL + 20log(1 V / V(94 dB))

That should look familiar from above. :)

If you want millivolts instead,
94 dB SPL + 20log(1 V / V(94 dB)) = 94 dB SPL + 20log((1 000 mV/mV) / (V(94 dB) [mV]/mV))
= 94 dB SPL + 20log(1 000) - 20log(V(94 dB) [mV]/mV)
= 94 dB SPL + 60 dB - 20log(V(94 dB) [mV]/mV)
= SPL(1 V)

Ta-dah! :D

The hardest part is formally keeping your units straight, as logarithms obviously work on dimensionless numbers only, so you have to expand fractions as required:
log(a[unit]/b[unit]) = log(a[unit]/unit / (b[unit]/unit)) = log(a[unit]/unit) - log(b[unit]/unit)

For those wondering why you can write "94 dB SPL + 60 dB" when it may have been drilled into them that you cannot add disparate units, let's inspect what that actually means in the linear world:
94 dB SPL = 1 Pa (Pascal)
+ 60 dB = * 1000
So 94 dB SPL + 60 dB = 1 Pa * 1000.
Plain dBs are just a ratio. Only those with a suffix have ties to actual physical units.
AnalogSteph, thank you for a thoughtful and apparently Amir-worthy response to my question. I'm a little abashed to admit that I'm going to break open not only the basic electronics text but also my high school math text. But I promise to earn my stripes and take advantage of your advice.

Ditto to StaticV3 and Doug Blake!
 
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JanesJr1

JanesJr1

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Yeah, that's why I get jealous of people who understand the science and the maths.
JimBob, "maths"? Is ZZtop a Brit band, or is JimBob a closet chips eater?
 

Spkrdctr

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The formula is quite easily derived under the assumption of linearity and with some basic logarithm juggling.

If the transducer is assumed linear, these two ratios should be equal:
SPL_linear(1 V) / SPL_linear(94 dB) = V(1 V) / V(94 dB)
IOW, increasing voltage input by a factor X will increase sound output by the same factor. I hope we can agree on that. This simplification generally holds very well in the world in hi-fi, as any deviation equates to nonlinear distortion.

Now let's convert the whole shebang to dB by applying 20*log10() on both sides, while exploiting the property that log(a/b) = log(a) - log(b):

SPL(1 V) - SPL (94 dB) = 20log(V(1 V) / V(94 dB))

Some minor reshuffling gives:

SPL(1 V) = SPL (94 dB) + 20log(V(1 V) / V(94 dB)) = 94 dB SPL + 20log(1 V / V(94 dB))

That should look familiar from above. :)

If you want millivolts instead,
94 dB SPL + 20log(1 V / V(94 dB)) = 94 dB SPL + 20log((1 000 mV/mV) / (V(94 dB) [mV]/mV))
= 94 dB SPL + 20log(1 000) - 20log(V(94 dB) [mV]/mV)
= 94 dB SPL + 60 dB - 20log(V(94 dB) [mV]/mV)
= SPL(1 V)

Ta-dah! :D

The hardest part is formally keeping your units straight, as logarithms obviously work on dimensionless numbers only, so you have to expand fractions as required:
log(a[unit]/b[unit]) = log(a[unit]/unit / (b[unit]/unit)) = log(a[unit]/unit) - log(b[unit]/unit)

For those wondering why you can write "94 dB SPL + 60 dB" when it may have been drilled into them that you cannot add disparate units, let's inspect what that actually means in the linear world:
94 dB SPL = 1 Pa (Pascal)
+ 60 dB = * 1000
So 94 dB SPL + 60 dB = 1 Pa * 1000.
Plain dBs are just a ratio. Only those with a suffix have ties to actual physical units.
I have never seen Latin written so well. I always wanted to learn a foreign language! Oh wait, or is that a long lost language from an aboriginal people who are long forgotten? I forget which. Oh well. Good Post!
 
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