A few SPL-related questions.
Dec 2, 2008 at 5:26 PM Thread Starter Post #1 of 3

Deiz

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I've been searching for budget headphones for a friend. Likely to be driven by an iPod in the short term, and a Fiio E5 after that.

I extrapolated off of figures given for the Fiio E3: 70 mW @ 32 ohms, 12 mW @ 300 ohms.

Formula being: Output mW = 70 - (Resistance - 32) * 0.21641.

I realize it's a gross oversimplification, as output is likely nowhere near that linear (And if you exceed 356 ohms in my model, you start generating power.) but it should work for rough estimation.

My main question is, does impedance play into SPL when you're basing things off of sensitivity?

Let's take two headphones, the PX100 (32 ohms @ 1 kHz, 114 dB @ 1 mW) and the HD650 (300 ohms @ 1 kHz, 103 dB @ 1 mW)

Given that to increase the dB by 10, you must exponentially increase the power, the output of the headphones would be 124 dB and 113 dB @ 10 mW, respectively.

Now, the Fiio E3 is capable of supplying 12 mW @ 300 ohm, so the HD650 would, if my math is correct, output 113 dB + (12-10) / 90 * 10 dB = 113.22 dB, an utterly imperceptible increase.

On the other hand, the PX100 @ 32 ohm would receive 70 mW, and 124 dB + (70-10) / 90 * 10 dB = 130.67 dB, a substantial increase.

I am unsure of whether this makes sense, but from anecdotal evidence I'd say the HD650 outputs between a quarter to a third of the PX100's SPL at the same volume setting.

Also, the headphones I'm likely pointing him to are the 66 ohm, 99 dB sensitivity ATH-AD500s - I'd assume they'd receive 62.6 mW from the Fiio E3 and thus end up with a peak SPL of 114.84 dB, which again makes sense to me because while their impedance is drastically lower than that of the HD650, they're markedly less efficient.
 
Dec 7, 2008 at 6:44 AM Post #2 of 3
Quote:

Originally Posted by Deiz /img/forum/go_quote.gif
I extrapolated off of figures given for the Fiio E3: 70 mW @ 32 ohms, 12 mW @ 300 ohms.



The simplest way is first to solve for open circuit voltage and source resistance,
assuming that the source is linear within the range of quoted values:

E = 1.96 V
Rs = 9.9 Ohm

Voltage across arbitrary load RL is now:

VL = E * RL / (RL + 9.9 Ohm)

And the power delivered to the load is:

PL = VL * VL / RL

To convert this to SPL first find the ratio of PL to 1 mW:

A = 10 * log10 (PL / 1 mW)

Now add the specified sensitivity (must be specified for 1 mW power):

SPL = sensitivity + A

This would be the SPL corresponding to full-scale sinewave at 1 kHz (sensitivity is specified at 1 kHz).
The average power when listening to music will usually be 15-20 dB less.
 
Dec 7, 2008 at 10:07 AM Post #3 of 3
i'd be very curious if you perform any tests - ive always wondered if the rated sensitivity of headphones was for the entire unit, or only the drivers
smily_headphones1.gif
 

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