Oct 29, 2012 at 10:20 PM

Quote:
I couldn't follow what you're trying to calculate, but before we even get to that, people have been using log10( ) to be shorthand for logarithm with base 10.  Somehow a 10 ended up inside the argument of your logarithms.

Senn HD 595 (0.055 Vrms for 90 dB SPL):
20*log10(1 / 0.055) = 25.2 dB,
so 1 Vrms would give 90 dB + 25.2 dB = 115.2 dB SPL

HiFiMAN HE-400 (0.129 Vrms for 90 dB SPL):
20*log10(1 / 0.129) = 17.8 dB,
so 1 Vrms would give 90 dB + 17.8 dB = 107.8 dB SPL

Thanks for correcting me.

The problem is that I suck at using my own calculator. I haven't used Log forever. The sequence for me is 20*log(1 / 0.129), for example.

I have been trying to figure out how to calculate headphone SPL based on Vrms and other specs I am unused to. A lot of headphones and amps do not report mW as I might expect.

Oct 30, 2012 at 7:25 AM

#### xnor

Quote:
Ok - the first formula makes sense and I can replicate your number. 85 dB is achieved at 0.031 Vrms for the senns, and 0.0725 for the hifimans.

Hypothetical, lets Change the Vrms to 1.

20*Log(10*(1 / 0.33)) = 30 dB. Therefore, 1 Vrms yields max SPL of 112 dB. (Hifiman)
20*Log(10*(1 / 0.055)) = 45 dB. Therefore, 1 Vrms yields max SPL of 135 dB (Sennheiser)

Though, both of those numbers are pretty close to radically high. Realistic? Or is it likely that driver limits would prevent such SPL? I am surprised that 1 V could yield that much volume considering amps like the Asgard are rated for up to 7 Vrms (but 1 Vrms at 10% THD).

As mikeaj wrote, log is the short for logarithm with base 10. So if 10^x = y then log(y) = x. Another notation is lg.

The 115 dB SPL mikeaj calculated is right and "close" to the specs (112 dB SPL @ 1Vrms, 1 kHz) provided by Sennheiser. Of course, the sound pressure level cannot scale linearly until infinity. Most drivers struggle to produce 120 dB SPL at low frequencies.

To calculate dB SPL @ 1 mW from @ 1 Vrms use this: dBSPL@1Vrms - 10*log(1000 / impedance)

For the HD595 this is 112 - 10*log(1000 / 55) = 99 dB SPL @ 1mW.

Oct 30, 2012 at 2:08 PM

Quote:

As mikeaj wrote, log is the short for logarithm with base 10. So if 10^x = y then log(y) = x. Another notation is lg.

The 115 dB SPL mikeaj calculated is right and "close" to the specs (112 dB SPL @ 1Vrms, 1 kHz) provided by Sennheiser. Of course, the sound pressure level cannot scale linearly until infinity. Most drivers struggle to produce 120 dB SPL at low frequencies.

To calculate dB SPL @ 1 mW from @ 1 Vrms use this: dBSPL@1Vrms - 10*log(1000 / impedance)

For the HD595 this is 112 - 10*log(1000 / 55) = 99 dB SPL @ 1mW.

You're the man. As noted, there is some sense that this math can under or over-estimate SPL at a given power level.

I just got my Hifimans in the mail today. I am currently assaulting them with a battery of music, and I can see that given my lower power output, the "loudness" of the recording influences whether or not I feel like I need another volt or two. The HE's are very deceptive however, as they are quite "relaxed" and smooth even at higher volumes so far. I'll be playing with them for awhile trying to decide what a realistic long-term volume need it. I am leaning toward a headstreamer atm, though I would love to get my hands on a creek OBH 11 or some other discrete honey. For the sake of simplicity however, I am leaning toward DAC combos.

Nov 2, 2012 at 11:49 AM

#### chewy4

I didn't see this being discussed in here and was wondering about something:

Are planar magnetics immune to impedence mismatches? Meaning that it is unneccesary to follow the 1/8th rule?

If you take a look at Tyll's charts on them, they have near flat FR and phase with 600Ohms impedance going in to them. But they are generally low impedance HPs.

Nov 2, 2012 at 12:00 PM

#### Steve Eddy

##### Member of the Trade: The Audio GuildAka: TempAccount555
Pretty much, yup. Planars are a virtually purely resistive load. They don't have that same low frequency resonance that dynamic drivers do. That's why my LCD-2 Rev. 2's sound just fine when driven by the 390 ohm output impedance of my TEAC A-H500 integrated amp's headphone jack.

se

Nov 2, 2012 at 3:45 PM

#### NA Blur

I figured I would supply a comment here about one of your equations.

"
A few notes for those who want to dig deeper:
Electrical:
1. Impedance is a complex (number) load, Z = R + jX, where R is 'resistance' and L is 'reactance', both of them represented in Ohms. This can also be written as Magnitude,|Z| and Phase θ."

You talk about L as the reactance, but your equation uses X.

Nov 2, 2012 at 4:28 PM

#### mikeaj

Quote:
I figured I would supply a comment here about one of your equations.

"
A few notes for those who want to dig deeper:
Electrical:
1. Impedance is a complex (number) load, Z = R + jX, where R is 'resistance' and L is 'reactance', both of them represented in Ohms. This can also be written as Magnitude,|Z| and Phase θ."

You talk about L as the reactance, but your equation uses X.

Somebody skipped a step.  L is the common symbol for inductance.  X is generically used for the magnitude of the imaginary component, the reactance (measured in ohms).  For an inductor, it's omega * L, with omega being the frequency (angular, in radians).  For a capacitor, it's - 1 / (omega * C), with C being the capacitance.  See here:
http://en.wikipedia.org/wiki/Electrical_reactance

Nov 2, 2012 at 4:55 PM

#### Steve Eddy

##### Member of the Trade: The Audio GuildAka: TempAccount555
Looks like a simple typo to me. He wrote L instead of X.

se

Nov 2, 2012 at 5:42 PM

#### chewy4

Quote:
Pretty much, yup. Planars are a virtually purely resistive load. They don't have that same low frequency resonance that dynamic drivers do. That's why my LCD-2 Rev. 2's sound just fine when driven by the 390 ohm output impedance of my TEAC A-H500 integrated amp's headphone jack.

se

Awesome, thanks.

Nov 4, 2012 at 9:31 PM

#### proton007

Quote:
Looks like a simple typo to me. He wrote L instead of X.

se

Fixed it. Thanks to SE and NA Blur

Dec 9, 2012 at 1:32 AM

#### oscar704

Code:
`` Power = Antilog ( (Required SPL - SPL per mW)/ 10).``
does anyone know what is the "Required SPL" is? im guessing its Dynamic range but i want to confirm, if not, what is it and why?

Dec 9, 2012 at 5:20 AM

#### mikeaj

Code:
``Power = Antilog ( (Required SPL - SPL per mW)/ 10).``

does anyone know what is the "Required SPL" is? im guessing its Dynamic range but i want to confirm, if not, what is it and why?

Read that "required" as "desired". SPL is the sound pressure level, which is the sound pressure in dB referenced to 0 dB being the lower limit of audibility (a chart may help). So hopefully as you might expect, the louder you listen, the more power needs to be delivered to the headphones.

If you are listening to a recording with a wide dynamic range, the loud parts are much louder than the average or soft parts, and the peaks may be briefly much louder than the rest. In that case, then you could require more power (higher peak volume) just to keep the average level about the same compared to a different recording with a smaller dynamic range.

Dec 9, 2012 at 7:09 AM

#### stv014

The best way to find out the desired SPL is to actually measure it. Find the highest volume setting you would want to use while listening to music, then, without changing anything else, replace the music with a sound file containing 1 kHz sine wave at 0 dBFS level (lower level is OK, too, but you will need to compensate the measured/calculated SPL accordingly). You can then either measure the voltage on the headphones using a splitter and a DMM (make sure that the DMM is accurate at the frequency of the test tone) and calculate the SPL from the voltage and the sensitivity of the headphones, or use an SPL meter to measure the actual loudness. The accuracy will be limited by not knowing the sensitivity of the headphones exactly in the first case, and in the second case the measured value will be affected by factors like the seal and distance from the driver, but it is possible to get a reasonable estimate. Add a few dB of headroom for amplifier power requirement calculations. You may end up being surprised how little power is really needed to drive headphones at a non-ear damaging level.

Dec 9, 2012 at 12:11 PM