Headphone power, impedance, SPL, ...

[bakker_be]

@amirm usually displays power values for 300 and 33 Ohm when testing headphone amps, which is understandable, as one needs to decide on standard values to enable performance comparison. It does leave me with a quandary though:

• My headphones (Focal Elear) are 80 Ohm. How do I know the power available at that impedance? I know headphone impedance isn't linear across the frequency range, so I also know that this is a "synthetic" value.
• Focal specsheet claims 104dB SPL / 1mW @ 1kHz. How do I calculate SPL for the power obtained above?

This has probably already been covered somewhere, but I haven't found it ...

 

[RayDunzl]

Here's a calculator...

http://www.digizoid.com/headphones-power.html

 

[amirm]

My headphones (Focal Elear) are 80 Ohm. How do I know the power available at that impedance?

You wouldn't from my testing. My dummy load has 50, 100 and 150 ohm and I used to report those. Out of laziness, I have cut back on that. Longer term I plan to automate the load selection so i can show all of these values. But just have not had the time to do it.

For now, a reasonable estimate is to reduce the power by a factor of 33/80 and get the power rating.

 

[garbulky]

You wouldn't from my testing. My dummy load has 50, 100 and 150 ohm and I used to report those. Out of laziness, I have cut back on that. Longer term I plan to automate the load selection so i can show all of these values. But just have not had the time to do it.

For now, a reasonable estimate is to reduce the power by a factor of 33/80 and get the power rating.

So for instance if the amp has a power of 10mwatts at 33 ohms. Then to calculate at 80 ohms you would do 10mwattsX33 divided by 80? Or is it 10 divided by 33 times 80?

 

[bakker_be]

So for instance if the amp has a power of 10mwatts at 33 ohms. Then to calculate at 80 ohms you would do 10mwattsX33 divided by 80? Or is it 10 divided by 33 times 80?

I understood it as 10 * (33/80) = 4.125 mW
This calculation would give around 50 mw for e.g. th DX3 Pro, meaning that according to the calculator mentioned by @RayDunzl it would push my Focal Elear easily beyond 120 dB SPL not good for my ears

 

[amirm]

Then to calculate at 80 ohms you would do 10mwattsX33 divided by 80?

This one as you always get less power as impedance goes up.

Power = V^2 / R. As such, all else being equal, power is inversely proportional to resistance.

 

[RayDunzl]

So for instance if the amp has a power of 10mwatts at 33 ohms. Then to calculate at 80 ohms...

Try http://www.sengpielaudio.com/calculator-ohm.htm

10mwatts at 33 ohms

Recalculate using the volts obtained above and 80 Ohms

Shows amirm's 33/80 works...

 

[bakker_be]

In a table it's very eye-opening how this all works out, with Ohm, Volt, Ampère and efficiency taken into account Thanks for enlightening me @RayDunzl & @amirm

 

[Sal1950]

Try http://www.sengpielaudio.com/calculator-ohm.htm

I bow to Ray's Scrutinizing knowledge.

 

[RayDunzl]

I bow to Ray's Scrutinizing knowledge.

I bow to Mr Sengpiel who accumulated so many formulae and made them point and click...

https://en.wikipedia.org/wiki/Eberhard_Sengpiel

The top of his audio collection:

http://www.sengpielaudio.com/Calculations03.htm

 

[Sal1950]

I bow to Mr Sengpiel who accumulated so many formulae and made them point and click...

DAMN! Just looking at the list makes my brain hurt.

 

[nhatlam96]

I would like to calculate how much dB the M500 can push for the Clear.
I know that the power is more than enough, but I would like to be able to calculate it.
1. M500: 63 milliwatt at 300ohm
2. Clear: 55 ohm, 101 dB/mW, 114 dB/V
The headphone power calculator google doc is messed up with those weird #REF!s, someone changed the values wrong.
If someone could download and fix it, that would be much appreciated.

 

[solderdude]

It sometimes isn't as simple as just calculating the power.

Amplifiers often have a current limit and always have a voltage limit.
With 32 Ohm (16 Ohm would have been even better) and 300 Ohm (which is practically no load) you can accurately calculate the power in intermediate impedances.

A way to figure out how much output power will be available is by first looking at available power at the lowest impedance.
If that is not very high you can be pretty sure the output current is limited.
When powers are specified at 16 Ohm or 32 Ohm you can make some calculations for output current limits.
When also output power is given at 300 Ohm or 600 Ohm you can easily calculate the maximum output voltage.

We can assume that manufacturers supply actual maximum output numbers but some of them simply supply incorrect numbers.

To calculate current limit: convert mW to W by dividing the mW rating by 1000.
Then divide output power (in W) by the given lowest impedance (Ohm) and then take the SQRT (square-root) of the outcome: I=√(P/R).
This will give you the max. output current the amp can deliver in Amp. When the output power is not extremely high you can assume this current is the actual limit.

Now we need the max. output voltage. As most amplifiers/sources can easily provide enough current to for headphones above 120 Ohm one can safely assume that output powers given at 300 or 600 Ohm are caused by output voltage limiting.

To calculate voltage limit: multiply output power (in W) by the given highest impedance (Ohm) and then take the SQRT (square-root) of the outcome: V=√(PxR) . This will give you the max. output voltage the amp can deliver in Volt.

With these numbers you can calculate the output power in various impedances but is not really as straight forward as it seems.
One needs to make 3 calculations based on the current and voltage and then see what the power will be for a specific impedance.

First you calculate the maximum power which is max. current x max. output voltage. The outcome is max power in mW.
This value is only valid for a specific impedance though. This differs from amp to amp and can only exist at the point where the maximum voltage is available at the point where the impedance is a certain value where it draws the maximum current AND still reaches the maximum output voltage.
This is easy to calculate: Divide the max. output voltage (V) by the current (in A) and you get the impedance in Ohms: R=(V/I).
Let's call this the optimal power impedance for now.
Note you need to convert mA to A for this. 1000mA = 1A, 100mA = 0.1A.

Once you have this optimal power impedance you can determine if you need the maximum Voltage or maximum Current to calculate output power in a certain impedance.
For impedances below the determined optimal power impedance you must use the Current limit formula below.
For impedances above the determined optimal power impedance you must use the Voltage limit formula below.

Current limit formula: Headphone impedance (Ohm) x max output current (mA) x max output current (mA) (I² x R) = power in mW in that impedance.
Voltage limit formula: first (max output Voltage (V) x max output voltage (V) ) and divide that outcome with the impedance (V²/R) = power in W in that impedance. Convert to mW by multiplying W x 1000.

Now the example above:
Q5:
max. current = 132mA (0.132A)
max. voltage = 3V
optimal power impedance = 22 Ohm
max. output power in 22 Ohm = 3V x 0.132A = 0.396W (396mW) in 22 Ohm.
Below and above 22 Ohm the output power will thus always be less than the max. output power.

This means for both 35 Ohm and 80 Ohm you need the voltage limit formula as the impedance is above 22 Ohm
35 Ohm = 0.257W (257mW)
80 Ohm = 0.113W (113mW)

For the K3 you don't know the max. voltage but given the fact that they do not give this and it is a portable device the optimal impedance most likely is between 16 and 32 Ohm.
So you can calculate the max. output voltage based on the 32 Ohm power V=√(PxR) = 1.96V
35 Ohm = 0.109W (109mW)
80 Ohm = 0.048W (48mW)

 

[nhatlam96]

to calculate output power in a certain impedance.

with certain impedance you mean headphone impedance?

 

[solderdude]

yes, impedance of a headphone.

The output power of an amp is not linear by voltage and not linear by current

Below the output power of a current limited amplifier

']www.garage1217.com/POWERSPECS/HorizonIII.png[/IMG]

Below an example of an amplifier with the same voltage limit but a higher current limit.

Below and example of an amp with a lower voltage limit and higher current limit

only look at the bluelines, the other colors show the effect of a different output resistance.

 

[nhatlam96]

yes, impedance of a headphone.

The output power of an amp is not linear by voltage and not linear by current

Below the output power of a current limited amplifier

']www.garage1217.com/POWERSPECS/HorizonIII.png[/IMG]View attachment 105761

Below an example of an amplifier with the same voltage limit but a higher current limit.

View attachment 105762

Below and example of an amp with a lower voltage limit and higher current limit
View attachment 105763

only look at the bluelines, the other colors show the effect of a different output resistance.

Headphone power calculator
M500 specs:
570mW at 32Ohm -> 0,57 W
65mW at 300Ohm -> 0,065 W
Focal Clear specs:
55Ohm, 101dB/mW, 114dB/V.



Max output current I in ampere = SRT_(P/R) = SRT_(W / lowest Ohm)
SRT_(0,57 / 32) = 0,133 A

Max output voltage V in volt = SRT_(W * highest Ohm)
SRT_(0,065 * 300) = 4,416 V

Maximum power in watt = max I * max V
0,133 * 4,416 = 0,587 Watt

Optimal power Impedance R in Ohm = max V / max I
4,416 / 0,133 = 33 Ohm

Find out if max V or max I is needed for a certain headphone impedance.
headphone Impedance < optimal power impedance = current limited
headphone Impedance > optimal power impedance = voltage limited
55 Ohm > 33 Ohm = voltage limited

Current limited, power in mW: headphone impedance * (max current in mA)²
Voltage limited, power in mW: ((max voltage)² / headphone impedance) * 1000
(4,416² / 55) * 1000 = 350 mW in 55 Ohm

What should I do now?

Website: Headphone Power Calculator - Headphonesty
350 mW for Focal Clear is 126dbSPL, the value looks realistic...

 

[solderdude]

Well now you know M500 can deliver 350mW in the Clear.
This will give you the max reachable SPL (is not the same as Phon)

Clear = 104dB @ 1mW so at 350mW you can add 10log350 = 25dB to the 104 = 129dB SPL
This means it can drive this headphone effortlessly.

EDIT: you found the SPL calculator... this is indeed what you need.
Then you need to understand what these SPL levels mean.
When you find 105dB for instance one immediatly links this to average levels one is allowed to endure for a certain time.
That is NOT what the SPL is.. It is what peak levels (rms) SPL can be reached. NOT average SPL what most people assume.

When one can reach 90dB peaks one can listen to it 'normal' levels where normal is you can listen to an entire album without having to turn the volume down.
When one can reach 105dB peaks one can listen louder (without distortion) to a level that is loud. Loud enough to keep it at that level for the duration of a song.
When one can reach 120dB peaks one can only listen to this for a short while before one gets the compelling urge to turn the volume down.

Of course, the moment one EQ's the lows up by a number of dB that means less 'average' level is available.
For instance, one has a HD650 and decides they like the Harman curve (some folks actually do), certainly at normal listening levels and boosts the lowest frequencies +10dB.

Let's assume one has a device that can provide 105dB peaks then the 10dB boost means that about 95dB peaks for the mids remains.
The bass can then peak at 105dB SPL but that really isn't loud.
This means that the sound starts to distort a lot sooner (at half the perceived SPL) compared to when no EQ has been applied.

Note: to compensate for this the amp in question would have to be 10dB higher as well. This is the reason why most folks complain their amp lacks 'power' and can't driver their headphones loud. They simply lack the gain for it because digital EQ, with the +10dB bass boost, will have had a pre-amp gain of -10dB.

 

[nhatlam96]

The bass can then peak at 105dB SPL but that really isn't loud.

dB values
Bad: average 75-80, peak 90
Good: average 90-95, peak 105 (I find this quite loud tbh.)
Perfect: average 105-110, peak 120

Focal Clear + M500 = 126dBSPL, -10dB for EQ = 116dBSPL (perfect)
HD650 + M500 = 115dBSPL, -10dB for EQ = 105dBSPL (good)

The M500 is a little weak^^
For the HD650, I basically have to turn Windows Audio and the amplifier nearly all the way up.
For the Focal Clear it's like 70% Windows Audio and 100% amplifier, uff...

Anyways, I ordered the M500 now and will report back in the M500 thread...

 

[BDWoody]

It sometimes isn't as simple as just calculating the power.

Amplifiers often have a current limit and always have a voltage limit.
With 32 Ohm (16 Ohm would have been even better) and 300 Ohm (which is practically no load) you can accurately calculate the power in intermediate impedances.

A way to figure out how much output power will be available is by first looking at available power at the lowest impedance.
If that is not very high you can be pretty sure the output current is limited.
When powers are specified at 16 Ohm or 32 Ohm you can make some calculations for output current limits.
When also output power is given at 300 Ohm or 600 Ohm you can easily calculate the maximum output voltage.

We can assume that manufacturers supply actual maximum output numbers but some of them simply supply incorrect numbers.

To calculate current limit: convert mW to W by dividing the mW rating by 1000.
Then divide output power (in W) by the given lowest impedance (Ohm) and then take the SQRT (square-root) of the outcome: I=√(P/R).
This will give you the max. output current the amp can deliver in Amp. When the output power is not extremely high you can assume this current is the actual limit.

Now we need the max. output voltage. As most amplifiers/sources can easily provide enough current to for headphones above 120 Ohm one can safely assume that output powers given at 300 or 600 Ohm are caused by output voltage limiting.

To calculate voltage limit: multiply output power (in W) by the given highest impedance (Ohm) and then take the SQRT (square-root) of the outcome: V=√(PxR) . This will give you the max. output voltage the amp can deliver in Volt.

With these numbers you can calculate the output power in various impedances but is not really as straight forward as it seems.
One needs to make 3 calculations based on the current and voltage and then see what the power will be for a specific impedance.

First you calculate the maximum power which is max. current x max. output voltage. The outcome is max power in mW.
This value is only valid for a specific impedance though. This differs from amp to amp and can only exist at the point where the maximum voltage is available at the point where the impedance is a certain value where it draws the maximum current AND still reaches the maximum output voltage.
This is easy to calculate: Divide the max. output voltage (V) by the current (in A) and you get the impedance in Ohms: R=(V/I).
Let's call this the optimal power impedance for now.
Note you need to convert mA to A for this. 1000mA = 1A, 100mA = 0.1A.

Once you have this optimal power impedance you can determine if you need the maximum Voltage or maximum Current to calculate output power in a certain impedance.
For impedances below the determined optimal power impedance you must use the Current limit formula below.
For impedances above the determined optimal power impedance you must use the Voltage limit formula below.

Current limit formula: Headphone impedance (Ohm) x max output current (mA) x max output current (mA) (I² x R) = power in mW in that impedance.
Voltage limit formula: first (max output Voltage (V) x max output voltage (V) ) and divide that outcome with the impedance (V²/R) = power in W in that impedance. Convert to mW by multiplying W x 1000.

Now the example above:
Q5:
max. current = 132mA (0.132A)
max. voltage = 3V
optimal power impedance = 22 Ohm
max. output power in 22 Ohm = 3V x 0.132A = 0.396W (396mW) in 22 Ohm.
Below and above 22 Ohm the output power will thus always be less than the max. output power.

This means for both 35 Ohm and 80 Ohm you need the voltage limit formula as the impedance is above 22 Ohm
35 Ohm = 0.257W (257mW)
80 Ohm = 0.113W (113mW)

For the K3 you don't know the max. voltage but given the fact that they do not give this and it is a portable device the optimal impedance most likely is between 16 and 32 Ohm.
So you can calculate the max. output voltage based on the 32 Ohm power V=√(PxR) = 1.96V
35 Ohm = 0.109W (109mW)
80 Ohm = 0.048W (48mW)

What a great explanation.

 

[nhatlam96]

What a great explanation.

It really is, because I managed to follow it through haha