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Topic: Calculating Loudness from Efficiency/Sensitivity (Read 43362 times) previous topic - next topic
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Calculating Loudness from Efficiency/Sensitivity

Hi.

I still don't understand the whole situation regarding sensitivity/efficiency and how to calculate what is required. Please try to answer with simple words, pretend I'm an idiot. 

I noticed that there seems to be two metrics which show much power a headphone needs. The HD800's specs is "102db/1Vrms". The He-6's specs is "83.5db/mw". They seem to be different readings. I'm pretty sure Vrms is not equal to mw, lol. I found this: http://www.jensign.com/bdp95/headphones/

It's a calculator that calculates a headphone's efficiency or sensitivity. The calculator clearly states that sensitivity is db/Vrms, and efficiency is in db/mw. But other sources don't seem to agree. Audeze lists their headphones' sensitivity in db/mw. Nwavguy talked about sensitivity in db/mw. So what gives?

Here is my current understanding: A headphone requires voltage and current to function properly. Higher impedance means higher voltage needed. An example would be the DT770 600ohm headphone. Some headphones, typically planar magnetic ones, tend to have low voltage requirements but require high amounts of current, like the HE-6.

HD800 has 300ohms of "nominal impedance" and a sensitivity of 102db/1Vrms.

Here is how I would go about figuring out the demands of this headphone:
In calculator at http://www.jensign.com/bdp95/headphones/  I input 300ohms (accounting for voltage requirements of the headphone) and 102 in the sensitivity box, and I get 96.8db/mw.

10^((115-96.80)/10) = 66mw.
115 stands for 115db, plenty of dynamic range for everything. (Nwavguy considers 105-115 all that is every needed. What do YOU think?) So: At 300ohms, HD800 needs 66mw.

But of course, the amount of current and voltage required by HD800 varies depending on the frequency it is playing. The impedance definitely changes, that changes voltage requirements for sure. And higher voltage = less current required? Do we know for sure that HD800 when it's at near 600ohms during its peak, that the amount of current needed is lower, and not only lower, so low to the point that it outweighs the increase voltage requirement?

So ok, let's just assume that HD800's worst case load is 66mw. That is a current measurement, no? And "power" rating of an amp is simply max current... at a given voltage (aka, at a given impedance). So we go to the spec sheet of an amp and look at "300ohms: X amount of mw max". Yes?

Last question. An amp needs enough power and gain to make headphones play good music at a good volume. So how do the two relate? The calculations above and below show the power requirements of a headphone. So when somebody says so and so amp is able to play a music on a headphone loudly but not well a technical reason could be that an amp lacks sufficient power but has enough gain? However, wouldn't that cause clipping and all around horrible sound quality? Because I plugged my HD800s into my crappy cellphone and my Netbook that can barely run Chrome, both of which can make HD800s loud enough to make my ears hurt. Now, I know they are easy to drive, but are they really THAT easy to drive? To the point that these two horrible devices can power HD800s? (Of course, higher distortion, crappy DAC - I can hear noise floor on my cell phone, ew...)

Some more calculations if you care to doublecheck them:
He-6 is 50ohms.
Here... He-6's power requirements using Hifiman's efficiency numbers (83.5db/mw):
105db:
10^((105-83.50)/10) = 141w
110db:
446mw
115db:
1412mw

Using Innerfidelity's pessimistic measurements on He-6 instead of Hifiman's rated efficiency:
105db:
622mw
110db:
1959mw
115db:
6225mw

So... following this pattern, for LCD2: (75ohms; 93mw/db)
105db:
16mw
110db:
50mw
115db:
158mw


Thank you.

Calculating Loudness from Efficiency/Sensitivity

Reply #1 –
It is important indeed to note the difference between dB per mW ratings, and dB per Volt ratings. NwAvGuy is a good source on the subject:

http://nwavguy.blogspot.com/2011/02/headph...-impedance.html
http://nwavguy.blogspot.com/2011/02/headph...-explained.html

As headphone impedance increases, the amount of current decreases (and vice versa).

As output impedance and headphone impedance values get closer to one another, the differences in impedance on the headphones (relative to frequency) will accentuate the voltage divider effect and cause more or less audible frequency response variations. If the output impedance is low enough (smaller than 1/8th the headphone impedance), the voltage divider effect will get smaller, to the point of producing little to no frequency response variation.

An ideal setup would be a source with low output impedance, and a pair of moderately sensitive, high(ish) impedance headphones. The Sennheiser HD800's specs look pretty good, in that respect.

I'll let more knowledgeable people get into the details.

Here's my own calculator for dealing with volts, dB/mW and dB/V ratings, etc:
http://outpost.fr/audioconverter/

Calculating Loudness from Efficiency/Sensitivity

Reply #2 –
Quote
It's a calculator that calculates a headphone's efficiency or sensitivity. The calculator clearly states that sensitivity is db/Vrms, and efficiency is in db/mw.
That is correct*, but sometimes people are a little "loose" with the definitions.  The terminology isn't that critical as long as you have dB/V or dB/mW.

Since solid state amplifiers are "constant voltage" sources**, and headphones are tested with constant-voltage sources, the dB/V (sensitivity) is better for comparing the actual "loudness" of different headphones.  It doesn't really help to know that your 600 Ohm headphones are super energy-efficient....  You aren't really concerned that your 600-Ohm headphones are consuming less power (mW) than your 32-Ohm headphones, what you really want to know is if they will be loud enough.





* Technically speaking, dB/mW doesn't give you the energy efficiency, which would be percentage of acoustic mW out/electrical mW in.  But it's more related to efficiency than dB/V sensitivity.

 

** "Constant voltage" doesn't mean truly constant voltage, since normal audio signals are varying and it also varies with the volume control setting.    In this context "constant voltage" means that the output voltage doesn't change when the load impedance changes (within reasonable limits...  The voltage may not be maintained if you connect an 8-Ohm or 4-Ohm speaker.  ). 

Headphone amps are not always true constant-voltage sources.  As skamp says, if the source impedance of the headphone amp is too high (relative to the headphone)  the voltage will very as the headphone impedance varies over the frequency range, and you'll get frequency response variations.  (Usually the headphone amp manufacturer doesn't specify the source impedance, more often they specify the recommended minimum headphone impedance.)

And if the amp's voltage drops with the headphone load, the sensitivity rating could fool you.      (The efficiency rating would also mislead you, because if you aren't getting the expected voltage, you aren't getting the expected milliwatts either.)

Calculating Loudness from Efficiency/Sensitivity

Reply #3 –
Somebody told me that power ratings from amps don't mean anything because:

Quote
Can somebody fact check this please:
Power in wattage is voltage times current.

Some headphones like more current than they do voltage and vice versa.

If amp A produces 2 amps and 0.5v, then it equates to 1 watt.

If amp B produce 0.5 amps and 2 volts, then it equates to 1 watt.

So the power output might be the same, but the effects each amp has on how the headphones perform is different.

So you can't really say that 1 watt is equal to another watt.


Can somebody fact check this please?

Calculating Loudness from Efficiency/Sensitivity

Reply #4 –
If the output impedance of the amp is small relative to the headphones, then each amp will produce the same current for a given voltage.

That statement above is true but only applicable if the amp can't drive the load correctly.

Calculating Loudness from Efficiency/Sensitivity

Reply #5 –
The power that an amplifier can deliver to a load is always going to depend on the load impedance. In the case of speakers the load impedance tends to fall in a fairly narrow range, and the amplifier manufacturer can specify a given power into a given load impedance, or range of impedances.

Since headphones vary widely in impedance, it becomes a bit trickier. Luckily it is not difficult to design a headphone amplifier that delivers sufficient power over most or all of this range.

Calculating Loudness from Efficiency/Sensitivity

Reply #6 –
To add to that, because headphone impedances vary so much, usually you don't care what the output power of the amp is (and sometimes its not even specified).  Its not a directly useful number.

Instead you want to know its output impedance (which will determine if its good enough to drive a given pair of headphones), and its maximum output voltage, which will determine how loudly it can drive a pair.  From this you can also calculate the power delivered to your headphones, but that by itself is not usually an interesting number.

Calculating Loudness from Efficiency/Sensitivity

Reply #7 –
Quote
Can somebody fact check this please:
Power in wattage is voltage times current.
True.

Quote
Some headphones like more current than they do voltage and vice versa.
I'm not sure what that means...  A 600 Ohm headphone at 1mW takes more voltage and less current than a 32-Ohm headphone at 1mW.

Quote
If amp A produces 2 amps and 0.5v, then it equates to 1 watt.

If amp B produce 0.5 amps and 2 volts, then it equates to 1 watt.
That's impossible with the same pair of headphones.  The relationship between voltage, current, and impedance is determined by Ohm's Law.

Quote
So the power output might be the same, but the effects each amp has on how the headphones perform is different.
Again, Ohm's law always holds...  I don't know what "perform different" means, but 1mW is 1mW.    The headphones might have different frequency response if the amp's source impedance is too high, but in that case there is voltage-loss inside the amp, less voltage applied to the headphones, and therefore less current and less power (Wattage).    And in that particular example, the power is NOT the same.

Quote
So you can't really say that 1 watt is equal to another watt.
With the same headphones, it's the same.

Calculating Loudness from Efficiency/Sensitivity

Reply #8 –
2 amps and 0.5 volts means that the load is 0.25 ohms.
0.5 amps and 2 volts means that the load is 4 ohms.

Those are not typical nominal headphone impedances.


With typical amplifiers you only control the voltage. The headphone will draw exactly the amount of current required.
For example:
2V into 30 ohms: 66.6 mA, 133 mW
2V into 600 ohms: 3.33 mA, 6.66 mW
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #9 –
Maybe what the person meant to argue was that... headphones need a proper amount of voltage and current to function optimally. P = I x V. So...  Let's say a headphone needs 4w of power and an amp can provide 4w... 4 = 4 x 1 and 4 = 1 x 4. Is that 4w being outputted from having a crapton of current and little voltage, or possibly from a ton of voltage but little current? Or does that not matter because the only thing that matters is how much power total is being outputted?



While a headphone's impedance changes according to the frequency being played, I read today that the voltage demanded by the headphone doesn't change, only the current. And higher impedance means less current required. Since P = I x V, I decreases and V stays constant, P must be lower.

According to measurements from Inner Fidelity, the impedance of HD800 varies from 350ohms to almost 650ohms.
1. How is the 300ohm figure for HD800's specs derived? At no point is the impedance lower than 350ohms.
2. While the voltage requirement doesn't increase when impedance increases from a different frequency, that only accounts for how much power the headphone needs. The amp may have difficulties powering a headphone at HD800's 650ohm peak because while the power required is lower, the output capabilities of the amp is also lower. For example, these are the max output specs for the O2:

15R load:  337mW
33R load:  613mW
150R load: 355mW
600R load: 88mW

Who is to say that HD800 during its 650ohm peak won't need more than 88mw of power?
Yeah, I know that 10^((115-96.8)/10) = 66mw... So that if I want to be able to get HD800s to play at 115db with efficiency of 96.8 db/mw, I need 66mw of power. I think this might be an explanation for the "Yeah, an amp can make headphones loud, but can it power it completely???" idea. How is this efficiency figure derived? And can I calculate the requirements of HD800 during the 650ohm peak like: 10^((115-100.1)/10) = 31mw? The 100.1 means the efficiency goes up (as expected when the impedance increases and sensitivity remains the same...) And then we compare the 31mw of power required during the 650ohm peak to the 600ohm output of 88mw for the O2?

I'm having a hard time coherently asking this question, sorry.

Calculating Loudness from Efficiency/Sensitivity

Reply #10 –
The impedance in the specs is known as "nominal impedance". (image) Manufacturers usually take the freedom to round this number.
For example, no loudspeaker has exactly 4.0 or 8.0 ohms. The actual impedance may be as much as 20% off.


Audio amplifiers are usually voltage sources. So they control the output voltage, and try to let the load draw as much current as it needs (Ohm's law).

The impedance peak of headphones you can see in the bass range is the resonant frequency. Here, the driver has the highest efficiency. Given that the frequency response of both the headphone and signal is flat, this is the point where the amplifier has the easiest job, because it needs to provide the least amount of power.


The sensitivity specs are usually measured at 1 kHz. Sensitivity can be both specified for a fixed voltage (usually 1 Vrms) or for a fixed amount of power (usually 1 mW).

1 V into the HD800 produces 102 dB SPL with a 1 kHz tone. Current draw is less than 3 mA.
At 100 Hz a 1 V signal will not only produce a higher SPL (see frequency response) but also draw less current -> less power -> easier load for the amplifier.
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #11 –
Quote from: Dark_wizzie on
Maybe what the person meant to argue was that... headphones need a proper amount of voltage and current to function optimally. P = I x V. So...  Let's say a headphone needs 4w of power and an amp can provide 4w...


P = I x V.

I= V/Z.

P=V/Z*V

P = V^2/Z.

Z is constant for a given pair of headphones.  Therefore, really you are saying that you need a given voltage.  Usually this is what people care about, and why sensitivity is often quoted in db/V. 

Quote from: Dark_wizzie on
I'm having a hard time coherently asking this question, sorry.


Ignore power, convert everything to voltage, and it becomes much more simple than you're thinking.

Calculating Loudness from Efficiency/Sensitivity

Reply #12 –
Quote from: xnor on
The impedance in the specs is known as "nominal impedance". (image) Manufacturers usually take the freedom to round this number.
For example, no loudspeaker has exactly 4.0 or 8.0 ohms. The actual impedance may be as much as 20% off.


Audio amplifiers are usually voltage sources. So they control the output voltage, and try to let the load draw as much current as it needs (Ohm's law).

The impedance peak of headphones you can see in the bass range is the resonant frequency. Here, the driver has the highest efficiency. Given that the frequency response of both the headphone and signal is flat, this is the point where the amplifier has the easiest job, because it needs to provide the least amount of power.


The sensitivity specs are usually measured at 1 kHz. Sensitivity can be both specified for a fixed voltage (usually 1 Vrms) or for a fixed amount of power (usually 1 mW).

1 V into the HD800 produces 102 dB SPL with a 1 kHz tone. Current draw is less than 3 mA.
At 100 Hz a 1 V signal will not only produce a higher SPL (see frequency response) but also draw less current -> less power -> easier load for the amplifier.

Hey,

My point was that, while at the resonant frequency the headphone needs the least amount of power to work well, it's also at a high impedance. And at high impedance, the amp itself has a hard time outputting as much power. I can imagine in my head, an amp that has very good power output at 300ohms but has a very low 600ohm output, causing the amp to actually fail at an "easier load". Does that make sense?


--

Calculating Loudness from Efficiency/Sensitivity

Reply #13 –
No, sorry, it doesn't.

Given a sensitivity, let's say 100 dB SPL @ 1V, the ideal headphone would have several thousand ohms of impedance. Such a headphone does not exist, however.

High impedance means that the amp will not be loaded down.
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #14 –
Quote from: xnor on
No, sorry, it doesn't.

Given a sensitivity, let's say 100 dB SPL @ 1V, the ideal headphone would have several thousand ohms of impedance. Such a headphone does not exist, however.

High impedance means that the amp will not be loaded down.

But if we look at the power ratings of an amp, say the o2:
15R load: 337mW
33R load: 613mW
150R load: 355mW
600R load: 88mW

At 600ohms the output of O2 drops to only 88mw though. That's lower than 150 ohms or 33ohms, which I don't get. The headphones themselves will require less power to drive, absolutely, given the sensitivity is sameish and impedance is higher, but the amp itself isn't able to output as much power. (355 -> 88mw) 



Also: Is there a reason why sensitivity is measured at 1khz or is it just tradition? Wouldn't it be a good idea to measure sensitivity from a few different frequencies?

Calculating Loudness from Efficiency/Sensitivity

Reply #15 –
I like the water analogy:

Think of the amp as a water pump. The thicker the pipes (= lower impedance), the more water needs to flow (= higher current) to keep up the pressure (= voltage). We want constant pressure (voltage source). The water/power supply also needs to keep up with that water demand!

Let's say at the end of the pipe there's a water wheel. Ideally, you'd want the wheel to spin fast (high efficiency, sensitivity) with very little water coming out of the pipe.

--

1 kHz is a fairly good choice, because most headphones don't have huge dips or peaks there, it's far away from the extremes, it's below the ear/pinna resonances but above our ear's low-frequency insensitivity... Of course, a headphone with greatly boosted bass and treble will be a lot louder than one with rolled-off bass and treble, given the same sensitivity (since it's just measured at 1 kHz).

As soon as you start measuring at multiple frequencies you're basically measuring the frequency response.
You can actually plot the frequency response with an absolute y-axis (dB SPL) instead of relative dB's (sometimes "dBr"). Example for a speaker.
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #16 –
Quote from: Dark_wizzie on
But if we look at the power ratings of an amp, say the o2:
15R load: 337mW
33R load: 613mW
150R load: 355mW
600R load: 88mW


Yes, that is a power supply and usually opamp (the small integrated circuits that do the amplification) limitation.

Assume you have a power supply with +/- 10 V, so if the opamps were perfect they could output 20 V peak-to-peak, or 10 V peak, or 7 V RMS (Vpeak divided by square root of 2).
P = V^2/R

As you increase R, the power drops.
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #17 –
Let's look at those specs, converting to voltage and current:

15R, 337 mW - V = 2.2, I = 0.15
33R, 613 mW - V = 5.0, I = 0.14
150R, 355 mW - V = 7.3, I = 0.048
600R, 88 mW - V = 7.3, I = 0.012

As you can see, the amplifier is limited to either 7.3 volts or 0.15 amps, depending on the load impedance.

The power would be a maximum of 7.3 * 0.15 = 1.1W at a load impedance of 7.3 / 0.15 = 49 ohms, and less at any other impedance.

Calculating Loudness from Efficiency/Sensitivity

Reply #18 –
Quote from: xnor on
Quote from: Dark_wizzie on
But if we look at the power ratings of an amp, say the o2:
15R load: 337mW
33R load: 613mW
150R load: 355mW
600R load: 88mW


Yes, that is a power supply and usually opamp (the small integrated circuits that do the amplification) limitation.

Assume you have a power supply with +/- 10 V, so if the opamps were perfect they could output 20 V peak-to-peak, or 10 V peak, or 7 V RMS (Vpeak divided by square root of 2).
P = V^2/R

As you increase R, the power drops.

Yes, so... Let's imagine:

Headphone A:
Nominal impedance: 300ohms. Requires 200mw.
Resonant Frequency: 600ohms. Requires only 100mw.

Amp B's Max Output:
300ohms: 600mw
600ohms: 80mw

While the headphone itself requires less power, the amp ends up having trouble with the headphone at resonant frequency despite it requiring the least amount of power at its resonant frequency.

This is not the case with the HD800s and the O2, but hypothetically wouldn't this situation be possible?

Sorry if I seem to be on repeat, I'm just trying to understand. 

Calculating Loudness from Efficiency/Sensitivity

Reply #19 –
"Requires 200 mW" for what? Let's say to reach 110 dB SPL at 1 kHz, where the impedance is 300 ohms.

P = V^2/R
so
V = sqrt(P*R) = sqrt(0.2 * 350) = 7.75 V

If the frequency response of the headphone is flat, then at the resonant frequency, where the impedance is 600 ohms, it will require:
7.75^2 / 600 = 0.1 W = 100 mW

--

Now let's look at the amp.

600 mW into 300 ohms = 13.4 V max
80 mW into 600 ohms = 7 V max ?

This would be a very weird amp. You could say your amp is limited to also 7 V into 300 ohms = 163 mW into 300 ohms.
With the headphone above, you won't be able to achieve 110 dB SPL at either frequency, but 109 dB SPL.    Again assuming a flat frequency response.


If anything, it is the other way around. At too low impedances the amplifier simply will not be able to supply the current needed anymore. The result is clipping, lots of nasty distortion.
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #20 –
Quote from: xnor on
"Requires 200 mW" for what? Let's say to reach 110 dB SPL at 1 kHz, where the impedance is 300 ohms.

P = V^2/R
so
V = sqrt(P*R) = sqrt(0.2 * 350) = 7.75 V

If the frequency response of the headphone is flat, then at the resonant frequency, where the impedance is 600 ohms, it will require:
7.75^2 / 600 = 0.1 W = 100 mW

--

Now let's look at the amps.

600 mW into 300 ohms = 13.4 V
80 mW into 600 ohms = 7 V

So this doesn't work. You could say your amp is limited to also 7 V into 300 ohms = 163 mW into 300 ohms.

Ok, so you're saying that mathematically it's a situation that just won't happen. I see now. 

Calculating Loudness from Efficiency/Sensitivity

Reply #21 –
Indeed. It's not impossible, but I have never seen anyone build such a headphone amp. (I'd rather not confuse you with even more technical details.)
"I hear it when I see it."

Calculating Loudness from Efficiency/Sensitivity

Reply #22 –
Note that the O2 has what NwAvGuy calls a "current limiter", to protect headphones from getting too loud. I guess it is reflected in the power ratings.

Calculating Loudness from Efficiency/Sensitivity

Reply #23 –
Quote from: skamp on
Note that the O2 has what NwAvGuy calls a "current limiter", to protect headphones from getting too loud. I guess it is reflected in the power ratings.
Current limiters are pretty common, they limit the current delivered and protect your circuits from frying due to short-circuits.
It's only audiophile if it's inconvenient.

Calculating Loudness from Efficiency/Sensitivity

Reply #24 –
As for the max SPL you want to achieve: it depends greatly on what kind of music you listen to.

Dynamic classical with low average level ... maybe 110 dB.
Highly compressed pop, rock ... probably less than 100 dB is enough for sustained SPLs that will cause hearing loss.


With most headphones a few milliwatts are enough to reach dangerous levels.
"I hear it when I see it."