Discussion in 'Sound Science' started by proton007, Apr 25, 2012.

1. First of all I would like to say thank you proton007 for taking your time and effort to get all this information together, especially "Hydraulic Analogy" is great, there are some minor things that gets me little confused, but otherwise I think I get the point perfectly.

Second thing is that when it comes to formulas I get little lost.

Mainly here :

Use this formula to calculate the power needed:
Code:
` [color=#808080]Power = Antilog ( (Desired SPL - SPL per mW)/ 10). Note that a 3dB increase in SPL will cause a 2x increase in Power. For reference, 85 dB is considered to be the limit where long term exposure can cause hearing damage. Use 85dB to calculate the average power needed.[/color]`

Deduced that SPL is sensitivity, 85 dB is "ideal volume" but thats where it ends for me

I don't want to waste your time, but if you get some to spare, could you please show me and maybe other electrical newbies example of explained calculation ?

Got data here :

Sennheiser HD700
- Nominal Impedance : 150Ω
- Max Impedance : 370Ω @ +-120hz
- Sensitivity : 105dB (Characteristic sound pressure level (at 1 KHz) - 105 dB SPL (1 kHz, 1 Vrms))
- THD : 0.03% (1 kHz, 1 Vrms)

FiiO E12
- Drive ability : 16~300Ω (recommend)
- Output Power : >880mW@32Ω
- THD : <0.005%
- Output Impedance : <0.5Ω
- Input Sensitivity : <710 mV(Gain:high)
- Signal to Noise Ratio : >110dB
- MAX input Level : >8 Vrms
- MAX output voltage : 15.5 Vp-p
- MAX output current : 170 mA
- Channel imbalance : <0.5dB

Your opinion on these two would mean a lot to me. Thanks in advance.

Sources :

ps : Liked the avatar-bold text matched color

2. Did some research, found some data

If I understand this right than to get "maximum" out of HD700 I need amp which will provide me enough... to talk in your analogy "pressure" - 6.16Vp-p and and "amount of water" - 14.52mA right ? As for volume , 31.62mW ?

When I look at E12 specs I feel like I ordered nuclear power plant for my small garden sprinkler. (31.62mW for 120dB ... and E12 has max output 880mW ... oh guess I may be getting it while writing this  880mW at 32ohm not 150)

But still voltage and current required in table looks pretty low.

3. Flisker likes this.
4.
Oh, I feel totally dumb to be honest, because those numbers don't tell me anything.

- How did you find out that HD700 needs -1dB gain ? What does it even mean ? Isn't gain like multiplier so it is always positive ?
- What's 2Vrms at source ?

- Do you think E12 is good amp to go with HD700 ? Why ?

I guess I'll have to take some class on these things

E: "S@1V is the sensitivity with 1 V, 500 Hz sine wave input in dB SPL." HD700 has 105 but at 1000 Hz does it change things ?

5.
Sensitivity is either specified for 1 V or 1 mW. It's often not specified but V usually means Vrms = Volt root mean square.
2.83 Vpp / 2 = 1.41 Vp / sqrt(2) = 1 Vrms

It's usually specified at 1 kHz, but in my table I used 500 Hz where possible. Since the HD700 has flat mids it doesn't change, so 105 dB SPL @ 1 Vrms it is.

Your source outputs some max RMS level. CD players often output 2 V, the Apple devices about 1 V max, the Sansa Clip+ about 0.5 V max.

If we want 110 dB SPL we need to add 5 dB to the HD700's sensitivity.
+5 dB = 10^(5/20) = 1.78x * 1 V = 1.78 V

If we have a 2 V source we actually need negative gain (attenuation): 1.78/2  = 0.89x, in decibels: 20*log(0.89) = -1 dB

So if you indeed have a 2 V source you can theoretically achieve 111 dB with 0 dB gain, or 121 dB with +10 dB gain.
If your source only outputs 1 V you have to subtract 6 dB.
If it only output 0.5 V you have to subtract another 6 dB, so 12 dB in total, and so on...

edit: From what I can tell the E12 seems like a good choice, especially since it has a gain switch.

Flisker likes this.
6.
Thanks for formulas.

Is it this just about volume ? I got already enough volume from PC - ALC889 Chip - http://www.realtek.com.tw/products/productsView.aspx?Langid=1&PFid=28&Level=5&Conn=4&ProdID=173

I'am more concerned about raising audio quality, don't think that HD700 will perform at it's "best" just from ALC889 integrated chip on pc motherboard.

7. The gain values reported by jude seem to be completely off btw, because the specs say >16 dB for high gain....

edit: Yeah that's just about volume, but the E12 definitely performs better than onboard audio.

Flisker likes this.
8.
Yea, it's because early version of E12 had 0 or 10dB gain, now it's 0 or 16dB. I'am getting new so the one with 0/16.

9. Oh that's good to know.

10. Main thing I still can't figure out is, what does this

actually mean in real world audio performance

FiiO says : "Based on this formula, we have worked out the power, voltage and current for some popular headphones. In order to get the best sound effect, you can refer to the power required by 120dB sound pressure and peak voltage required by high dynamic."

Well if I look at E12 at their page, it says :

- Output Power : >880mW@32Ω
- MAX output voltage : 15.5 Vp-p
- MAX output current : 170 mA

Therefore by their information E12 should be kind of overkill to use with HD700 because it got nearly 30 times more output power than should be needed, 3 times more output voltage and 10 times more output current. So I'am thinking "what the he..". Why would someone use Schiit Lyr or some other much much stronger amp with these.

11. The table is wrong because it assumes the 105 dB SPL is based on 1 mW which it isn't. The headphone produces 105 dB SPL with 1 V.

According to E12 specs the 15.5 Vpp max output are 5.5 V.
5.5V into 150 ohms = 200 mW max.

Since we've already established that for 110 dB you need 1.78 V, into 150 ohms = 21 mW.
So the E12 produces about 10x more max power with high gain and max volume.

For 95 dB SPL the HD700 needs about 0.66 mW.

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12.
How did u get 5.5V ?

13.
15.5 / 2.8284; the 2.8284 is the square root of 2 (the ratio of the peak and RMS level for a sine wave) multiplied by 2.

By the way, according to these measurements, the E12/Mont Blanc can actually output 6.5-7 Vrms into 150 Ω. The 15.5 Vp-p is specified for 32 Ω, where the measured level was more like 5 Vrms.

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14. See #170 for conversion from Vpp to Vp to V (rms).

15.5 / 2 / 1.41 = 5.5 V

15.
Great, thanks

Yea, guess I can see it there now. Tbh I'am mentally destroyed for today and have to wake up early, will continue with thinking about all provided information tomorrow.

Thanks a lot, really appreciate your help