dB per milliWHAT? Efficiency vs. Sensitivity vs. "How loud do they really go?" - Page 2

Great work. This is a valuable resource.

sorry about that Jung, I found that link immediately after I made that post, hence the edit line on the last post.

I have a pair of the 701s. I'm looking for an amp to power them fully. What sort of specs (specific levels with reference to 701s impedance and sensitivity) from the amp should I be looking for to drive the 701s properly?

On paper, to get 110dB, K701 needs 51mW, 1782mV, and 29mA. You can start with doubling these power, e.g., with an amp capable of 100mW, 4V, and 60mA, for each channel.

Convert sensitivity in dB/V to sensitivity in dB/mW.

Another good sensitivity thread.

I just replied to a thread (see post #37) where I put "an actual conversion formula" to convert sensitivity in dB/V and impedance to sensitivity in dB/mW.

The formual gets the same results as post #18 above. I'll repeat the equation here for reference.


==== Extra Credit =============
For the geeks out there, the equation to convert dB/V to dB/mW is

dB/mW = dB/V + 10*log(R*P/V^2)

where
dB/V is the sensitivity in dB (SPL) at 1 Vrms of voltage into impedance R, 103 dB/V into 300 ohms in this case
R is the nominal impedance, 300 ohms in this case
V is the reference voltage, 1 Vrms in this case
P is the reference power, 0.001 watt in this case
dB/mW is the sensitivity in dB (SPL) at 0.001 watt of power (that is 1 milliwatt)
log is the logarithm base 10

In the above example we have:

dB/mW = 103+10*log(300*0.001/1^2) = 98.7

I don't know how u got that formula:

 

if dB/mW=10 log P/p=10 log(( V^2/R)/p)  since P=V^2/R and p=0.001Watts

 

then dB/mW=10 log(( V^2/R)/p)=10 log V^2 +10 log(1/(R*p))=20 log V +10 log(1/R*p)

 

I  know  i am wrong but i don't know where i mistook. can u help me?

 

thanks

any help please?

I will try to explain the derivation of my equation.  To avoid confusion, I’ve changed some of my nomenclature.  The V in dB/V refers the SI unit of 1 volt (in the International System of Units). The terms V1 and V3, elsewhere, refer to variables representing the potential difference (also called voltage) across the terminals of the headphones with nominal impedance R. 

======================

 

The equation to convert dB/V to dB/mW is

 

dB/mW = dB/V + 10*log(R*P2/V1^2)

 

where

dB/V is the sensitivity in dB (SPL) at 1 volt rms into impedance R, 103 dB/V into 300 ohms in this case

R is the nominal impedance, 300 ohms in this case

V1 is the reference voltage for dB/V, 1 volt rms in this case

P2 is the reference power for dB/mW, 0.001 watt in this case

dB/mW is the sensitivity in dB (SPL) at 0.001 watt of power (that is 1 milliwatt)

log is the logarithm base 10

 

For the specific case that P2 = 0.001 watt and V1 = 1 volt rms into impedance R (in ohms)

 

dB/mW = dB/V - 30 + 10*log(R) where R is in ohms. 

 

In the above example we have:

 

dB/mW = 103+10*log(300*0.001/1^2) = 97.8 or

dB/mW = 103-30+10*log(300) = 97.8. 

 

======================

To derive this formula, we first define what we mean by dB/mW and dB/V.  The term dB/mW refers to the sound pressure level (SPL) in dB(SPL) when the input power to the headphones is 1 mW = 0.001 watt.  Thus dB/mW is defined by the following equation

 

dBp(P3)= dB/mW + 10*log(P3/P2)

where

dB/mW is the SPL in dB(SPL) when the input power is 0.001 watt

P2 = 0.001 watt

P3 is some arbitrary input power level, and

dBp(P3) is the SPL in dB(SPL) at power level P3. 

 

When P3 = P2, dBp(P3) = dB/mW, in agreement with the definition of dB/mW.  As P3 increases, the SPL increases.  If P2 < P3, dB/mW < dBp(P3), as expected. 

 

Likewise, the term dB/V refers to the SPL in dB(SPL) when the input voltage is 1 volt rms into impedance R.  The definition of dB/V is

 

dBv(V3)= dB/V + 10*log([V3/V1]^2)

where

dB/V is the SPL in dB(SPL) when the input voltage is 1 volt rms into impedance R

V1 = 1 volt rms into impedance R

V3 is some arbitrary input voltage rms level into impedance R, and

dBv(V3) is the SPL in dB(SPL) at voltage level V3 into impedance R. 

 

Now if P3 = V3^2/R, then dBp(P3) and dBv(V3) correspond to the same power input level and dBp(P3) = dBv(V3).  

 

Substituting we have

 

dB/mW + 10*log(P3/P2) = dB/V + 10*log([V3/V1]^2)

 

dB/mW = dB/V + 10*log([V3/V1]^2) - 10*log(P3/P2) = dB/V + 10*log([P2/P3]*[V3/V1]^2)

dB/mW = dB/V + 10*log([P2/P3]*[{V3^2/R}/{V1^2/R}])

 

Since P3 = V3^2/R,

dB/mW = dB/V 10*log(P2/{V1^2/R})

 

dB/mW = dB/V 10*log(R*P2/V1^2)

 

Since P2 = 0.001 watt, and V1 = 1 volt rms into impedance R, we have

 

dB/mW = dB/V + 10*log(R*[0.001 watt]/[1 volt]^2). 

 

This simplifies to

 

dB/mW = dB/V - 30 + 10*log(R)

 

where

dB/mW is the sensitivity in dB (SPL) at 0.001 watt

dB/V is the sensitivity in dB (SPL) at 1 volt rms into impedance R and

R is the headphone nominal impedance in ohms. 

 

 

thank you so much for your time. I am doing an application in which I know the rms value of the voltage I am delivering at the output ( I mean the digital rms value). Then if i know the user has a headphone with a sensitivity of X dB/V, if we get 1 Volt of rms value with the voltmeter I know what SPL he has...also if I know the dB/mW and know the impedance.

 

That's why I needed that explanation.

 

Thank u again for your kind answer.

You're welcome.

 

Many thanks to your description! This is what I'm looking for, I got more deep understanding about this part, It's good for reference.