[donunus]

How can I compare the sensitivity of headphones that are rated 105db per mV or per mW or Veff etc... How do I convert from one another. And how does the headphones impedance play a part in these values? This is something very valuable to all of us here on headfi and I'm sure many people here would be very happy if someone can explain this perfectly

In fact, I believe these types of things should be on a sticky here on headfi

 

[iQEM]

same here, need explanation about those too...

 

[jcx]

http://www.head-fi.org/forums/showthread.php?t=170186

has a spreadsheet that shows many headphones and calcs dB/mW and dB/V for them


some basics:
http://www.head-fi.org/forums/showth...mW#post2003921


[(1 Vrms)^2 ] / R_headphone = Power in Watts rms for 1 Vrms drive


power dB ratio (ref= 1mW): 10 * log10[ P/0.001 ]


if you have dB/mW, add power ratio dB for 1 V to get spl @ 1 V


if you have dB/V, subtract power ratio dB to get dB/mW

 

[ailevin]

As a newbie to headphone impedance/sensitivity I have found jcx's excellent explanations via search. However, sometimes there are "trivial" steps that he skips that were not apparent to me, so I wanted to comment. Also, if I get them wrong, please correct.

The "dB/mW and dB/V" were misleading to me since I expected it to be like the slope of a line where if you increase the power or increase the voltage there is a linear relationship with dB of SPL (which is sound pressure level, essentially a sound power measurement). They really should be read as SPL dB at 1mW or SPL dB at 1V rather than dB per mW or dB per V.

Impedance ("Z" often used as symbol) is resistance ("R" is common symbol) as a function of frequency and quoted impedance is usually resistance in Ohms at 1kHz. However, essentially all of the amp/power/Voltage/current calcs just use R.

There are two essential electrical relationships that are used:
V=I*R Voltage=Current times Resistance
P=V*I Power=Voltage times Current

For our headphone analysis there are a couple useful combinations of the above two equations. Solving the first one for I, I=V/R and substituting into the second one lets you calculate the power from voltage or voltage from power for a particular resistance:
P=V^2/R Power is Voltage squared divided by Resistance or
V=(P*R)^(1/2) Voltage is the square root of Power time Resistance

So if you are given dB SPL at 1mW you can calculate what Voltage that corresponds to, and if you are given dB SPL at 1V you can calculate what Power that corresponds to. Notice that Power goes like the square of Voltage so that doubling the voltage increases the power four times.

I am sure that a dB tutorial can be found elsewhere, but dB is a logarithmic measure of a ratio of two quantities with like units, so you are always implicitly referring to some voltage, power, or sound pressure. I like to think of it as so many dB up or down.

if you are comparing power to power or voltage to voltage the dB is
10*log(ratio) Ten times the logarithm to the base 10 of the ratio
If you are comparing how a power ratio varies according to a voltage ratio then
dB (Power ratio) = 20*log(Voltage ratio) Twenty times the logarithm to the base 10 of the ratio
This is because the Power ratio varies as the square of the Voltage ratio.

Hope this helps.

Alan