How To Test Ouput of MP3 Players

[oppizzippo]

Originally Posted by saratoga

...Of course if you're asking this online I suggest that you ask someone experienced with electronics to supervise you before you try it. AC voltage can be quite dangerous.

Totally, Please don't go poking around inside electronics with you multimeter unless you know what your poking at. I am a trained electrician, if you have questions feel free to ask, but be careful when playing with electronics.

Cory

[CrowdAds]

[NewGuy]

Originally Posted by saratoga

No he is correct. Well closer to correct. Halving voltage subtracts, rather then divides. However since we're talking about amplitude, it subtracts 6dB rather then 3dB (3dB is for power).

So 121-6 = 115dB.

Damn, so I can't listen in full volume without going deaf by the time Im 30... And I hoped :P

And as to whether my phone's sensitivity is actualy 121 dB - I would like to believe that since we're talking about AKG and not 0.99$ Coby "professional" earphones.

Anyway I'll try to perform the output power test on different volume scales (the difference in output between 10/40 and 40/40 for example) and I'll post the results. the only problem is to find 16 ohm resistors...

[saratoga]

If you just want to know how much the volume knob changes per step, then you don't need a resistor. Open circuit (unloaded) measurements will be fine for that.

[NewGuy]

All right. So this is what I did:
I couldn't find any 16 ohm resistors so I just used the earphones as load. I took a 3.5 mm connector with naked wires on the other end and plugged it to my Meizu's phone jack. Then I connected the wires to my earphone's jack (a half - assed connection, using tape). And then I simply connected the multimeter to those wires to check the voltage on the phones.

I created 5 different frequency mp3 files - 50Hz,200,1K,5K and 12K (with maximum amplitude) and let the player play them. The interesting thing is that the 1K file produced the lowest output voltage, and the furter the frequency was from 1K, the more voltage was produced, but the difference wasn't that drastic - maybe up to 20%, except the lowest basses - they mroduced a lot more voltage.

I checked with different volume levels and got the following:
10/40 - 10-12mV (60 for bass)
20/40 - 40-50mV (230 for bass)
30/40 - 90-120 mV (300 for bass)
40/40 - 300-325mV (325 for bass)
you can see that output voltage roughly triples itself every 10 notches...

However when playing songs (and not using frequency files) the output voltage was much lower - less then half the voltage i wrote above.

With this data, we can calculate the player's output (let's take 325mV as max output voltage):
P=V^2/r =6.6mW
A lot less than the official specs of 18mW output. Weird.

I also did the same with 32 ohm phones and got voltage values roughly 50% higher. I guess that it makes sense cause at a constant output power, when r is twice bigger, the voltage is bigger by sqrt(2) which is about 1.5.

That's how I spent the last 2 hours of my life...

[saratoga]

So its a little less then 1 dB per step, same as most rockbox'ed devices. (1 or 1.5dB per step is pretty common on DACs).

[ninjackn]

Originally Posted by saratoga

dB are always unitless, so hes ok.

Yes "dB" is unit-less but the provided spec is in dB_SPL/V units.

Originally Posted by saratoga

No he is correct. Well closer to correct. Halving voltage subtracts, rather then divides. However since we're talking about amplitude, it subtracts 6dB rather then 3dB (3dB is for power).

Going back to what we said about how dBs are unit less, the provided spec by the manufacturer is relative to SPL or sound pressure level. So his MP3 player is outputting ~0.5V which with his headphones outputs at 60dB_SPL. This is all assuming the measurements of the headphones were measured in accordiance with IEC 60268-7 specification. Infact he should have been using a 500Hz tone when measuring the output power of the DAP but I figured given the accuracy of the multimeter and such that the values he would get for our intended purposes is closed enough. But you're right about me being off by a factor of 2 for converting 0.5Vrms to dBV. I'm just too used to doing things in dBm and I was careless.

Originally Posted by saratoga

A multimeter will work fine with an appliance. It doesn't change the working of the system, since in voltage mode the impedance is nearly infinite (draws no current) and in current mode the impedance is nearly zero (drops no voltage).

Yes a multimeter changes the circuit. IF the multimeter was magically ideal then in voltmeter mode it would have infinite resistance and in ammeter mode it would had zero resistance. This is far from the case, especially with a cheaper multimeter. With a good multimeter you can expect somewhere in the average of 10Mohm resistance in voltmeter mod and maybe 3 ohms in ammeter mode.

[saratoga]

Originally Posted by ninjackn

Yes "dB" is unit-less but the provided spec is in dB_SPL/V units.

You're a little mixed up here. dB is unitless, so theres really no units. SPL refers to the reference level which is needed to convert to absolute amplitude/power, but its not a unit, which is why the math you took issue with is actually correct.

Originally Posted by ninjackn

Going back to what we said about how dBs are unit less, the provided spec by the manufacturer is relative to SPL or sound pressure level.

This actually doesn't matter. Could be relative to anything at all and the math is the same. Thats why dB is so useful for these things.

Originally Posted by ninjackn

So his MP3 player is outputting ~0.5V which with his headphones outputs at 60dB_SPL.

You're just a bit off

121-6 = 115dB at 0.5v input. Again, log scales don't work the same as linear.

Originally Posted by ninjackn

But you're right about me being off by a factor of 2 for converting 0.5Vrms to dBV. I'm just too used to doing things in dBm and I was careless.

I'm less concerned about you being off by a factor of 2 then you being off by a factor 1000 in that post!

Originally Posted by ninjackn

This is far from the case, especially with a cheaper multimeter. With a good multimeter you can expect somewhere in the average of 10Mohm resistance in voltmeter mod

You're complaining about 10Mohm resistance? You'll have more current flowing through the air between the leads then through the DMM. Thats basically infinite.

Originally Posted by ninjackn

and maybe 3 ohms in ammeter mode.

3ohm !!!

Think about what your'e saying. A 3 ohm resistance would mean that when using the 20 amp pin on a typical DMM you could dissipate 20^2*3 = 1200 watts ! You'd vaporize your unit. The actual resistance is varied (its what changing the scale does) so that its always small compared the circuit resistance over the spec'ed voltage range. 3 ohm is what you might see on the micro to milliamp scales. By the time you go up to amp scales you're at milliohms of resistance.

[ninjackn]

Originally Posted by saratoga

You're a little mixed up here. dB is unitless, so theres really no units. SPL refers to the reference level which is needed to convert to absolute amplitude/power, but its not a unit, which is why the math you took issue with is actually correct.

I looked into it and I stand by my original statement that the units are wrong. I don't see how"dB/V" or "dB/mW" turns into dB_{relative to 1 V/Pa}. It's a stupid unit.

Originally Posted by saratoga

You're complaining about 10Mohm resistance? You'll have more current flowing through the air between the leads then through the DMM. Thats basically infinite.

I said 10Mohm for a good multimeter. I've seen "cheap" ones with less than 1Mohm.

Originally Posted by saratoga

Think about what your'e saying. A 3 ohm resistance would mean that when using the 20 amp pin on a typical DMM you could dissipate 20^2*3 = 1200 watts ! You'd vaporize your unit. The actual resistance is varied (its what changing the scale does) so that its always small compared the circuit resistance over the spec'ed voltage range.

Yes, 20A on a typical DMM would vaporize it (hopefully the fuse would kick in before.) I know the workings and failures of dmms all too well which is why i'm saying you can't just say it's resistance is infinite or 0 especially when you take the test leads into account. I took the liberty of assuming we're "measuring" audio in the reals of milliamps which is why I said MAYBE 3ohms.

[saratoga]

Originally Posted by ninjackn

I looked into it and I stand by my original statement that the units are wrong. I don't see how"dB/V" or "dB/mW" turns into dB_{relative to 1 V/Pa}. It's a stupid unit.

Here, I'll do it for you taking into account units (assuming the sensitivity is actually 121dB/W not per volt since that would give sensitivity odd units):

121dB(SPL)/W = 1.2589/W
(0.5V)^2*1.2589/W = 0.3147
10*log10(0.3147 /10^-12) = 115 dB SPL


And again without units:

121dB +20log10(0.5V/1V) = 115dB

Same result.

Basically, the "units" as you call them aren't really units. They're just the reference level. So saying 121dB SPL just means its referenced to a power of 1 picowatt. Since all dB are then relative, it doesn't really matter what the absolute reference level is until you try to convert back from dB to watts.

And so this post actually works (except maybe the 3dB part):

Originally Posted by NewGuy

now, according to phone's specs, their sensitivity is 121dB/V, so at max volume 118 dB of sound are delivered (-3dB is half the SPL value)

Interestingly, his original 3dB/118 part makes sense if the sensitivity is really in dB/V, but I think thats a typo since it would mean the response of his headphones went with the square root of power, which would be a bit odd.

Originally Posted by ninjackn

I said 10Mohm for a good multimeter. I've seen "cheap" ones with less than 1Mohm.

Looking online the cheapest DMM radioshack sells is 10MOhm. I don't think it gets any cheaper/crapier then that. And why would it? You're acting like its somehow hard to make transistors with big resistances, but in fact its easy . . .

Originally Posted by ninjackn

I took the liberty of assuming we're "measuring" audio in the reals of milliamps which is why I said MAYBE 3ohms.

The quote you took issue with:

"A multimeter will work fine with an appliance. "

[ninjackn]

Damn it, my post was eaten.

Originally Posted by saratoga

Basically, the "units" as you call them aren't really units. They're just the reference level. So saying 121dB SPL just means its referenced to a power of 1 picowatt. Since all dB are then relative, it doesn't really matter what the absolute reference level is until you try to convert back from dB to watts.

Interestingly, his original 3dB/118 part makes sense if the sensitivity is really in dB/V, but I think thats a typo since it would mean the response of his headphones went with the square root of power, which would be a bit odd.

Feel free to swap W, mW and V interchangeably for my purposes. The "units" are "X dB/V." I miss-interpret this as "(X dB_SPL)/V". It would make more sense if it was "X dB_(SPL/V)" or "X dB (SPL/V)" or "X dB @ 1V" because that gives me and idea what all the ratios and references are referring to. Take for example if you have a programmable amplifier that increases the output 1 dB (SPL) per code. The unit might be short handed to "X dB/LSB". How are you suppose to understand if the output pressure increases logarithmically or linearly with each code step? So the units are wrong.


Take a look for your self, if you happen to randomly look at one single headphone specification:
http://www.akg.com/personal/K_272_HD...how,specs.html
You see it's in this magical "dB/V" or "dB/mW" and there's nothing preventing you from making the same assumptions I made.


But if you look at other headphone specs (which I did before my previous post):
Sennheiser
Beyerdynamic
Etymotic
You see they don't use dB/V or dB/mW or dB/W. Thus is why I said the units are wrong.




As to the issue of the DMM, here what i should have said to prevent our pissing match:

Originally Posted by TomCat39

I remember he said the impedance meter is needed because just using a multimeter to read the impedance with the appliance in operation, changes the whole circuit and gives you an incorrect reading. I guess, with headphones, the change is negligible?

You shouldn't use the multimeter to measure impedance of an appliance or headphone in operation because indeed does change the circuit. The multimeter would likely add in it's own small test current to the system and get an incorrect reading because it's trying to do measure the test current mixed in with the current being outputted by the system.

If you want to measure the resistance of an appliance in operation then you would need two multimeters, one to measure current and another to measure voltage and calculate the resistance from that. If you want to measure the impedance of the system then you need the system to output signals at different frequencies and calculate it's equivalent resistance, capacitance and inductance.

You should also take the resistance introduced by the multimeter and test leads into account when doing calculations.

[saratoga]

Originally Posted by ninjackn

Take for example if you have a programmable amplifier that increases the output 1 dB (SPL) per code.

"1 dB (SPL) per code"

I don't think you've understood what you've read. It just increases it 1dB. Saying "(SPL)" doesn't make sense sense increases are relative and don't have a reference level.

Originally Posted by ninjackn

The unit might be short handed to "X dB/LSB". How are you suppose to understand if the output pressure increases logarithmically or linearly with each code step?

Err dB are logarithmic. Always. Why did you just post a half dozen times about them if you didn't know what a dB was? Hell, you even tried to explain them to someone

Originally Posted by ninjackn

You see it's in this magical "dB/V" or "dB/mW" and there's nothing preventing you from making the same assumptions I made.

Well knowing what a dB was would definitely tend to stop one from making those assumptions . . .

Originally Posted by ninjackn

As to the issue of the DMM, here what i should have said to prevent our pissing match:

Don't take this so personally. Its not a pissing match. Its not a contest at all. Its just me explaining to you how DMMs work.

Originally Posted by ninjackn

You shouldn't use the multimeter to measure impedance of an appliance or headphone in operation because indeed does change the circuit. The multimeter would likely add in it's own small test current to the system and get an incorrect reading because it's trying to do measure the test current mixed in with the current being outputted by the system.

Well sure if you don't know how to use a DMM

But you can easily make that measurement without injecting any current at all. Just use the DMM as an ammeter and apply Ohms law. Then it works fine. Incidentally, this is why I said "current mode" in the post you quoted above and not "impedance mode"

[ninjackn]

Originally Posted by saratoga

"1 dB (SPL) per code"[]
I don't think you've understood what you've read. It just increases it 1dB. Saying "(SPL)" doesn't make sense sense increases are relative and don't have a reference level.

Err dB are logarithmic. Always. Why did you just post a half dozen times about them if you didn't know what a dB was? Hell, you even tried to explain them to someone

I know what a dB is, I didn't know what a dB/V is.

Expanding on the 1dB per LSB example, saying "1dB per LSB" makes sense to me, I understand that the output sound pressure level (in pascals) increases logarithmically in relationship to the increase in code.

If you write "1dB per LSB" as "1dB/LSB" I continue to think the same way.
That is to calculate the output:
20*log10(P_out/P_ref)/LSB * code = output dB

I thought it would be the same with voltage:

20*log10(P_out/P_ref)/V_ref * V_out = output dB

Which is not the case because it should be:

20*log10((P_out/P_ref) / (V_out/V_ref)) = output dB

So I read "X dB/V" as X dB per volt where as if they used "X dB @ 1V" or anything more informative than dB/V then I (and a few other members of this forum) wouldn't have made the assumption of X dB per volt. In the end I have an issue with using "dB/V" to describe headphone sensitivity and I think it's wrong.

[saratoga]

Originally Posted by ninjackn

I know what a dB is, I didn't know what a dB/V is.

"decibels per volt"

Originally Posted by ninjackn

Expanding on the 1dB per LSB example, saying "1dB per LSB" makes sense to me, I understand that the output sound pressure level (in pascals) increases logarithmically in relationship to the increase in code.

If you write "1dB per LSB" as "1dB/LSB" I continue to think the same way.
That is to calculate the output:
20*log10(P_out/P_ref)/LSB * code = output dB

That works because the voltage steps are logarithmically spaced . . . i.e. they are already logs.

Originally Posted by ninjackn

I thought it would be the same with voltage:

20*log10(P_out/P_ref)/V_ref * V_out = output dB

No thats not right. And if P_out is in units of watts, then its actually 10*log10().

Originally Posted by ninjackn

Which is not the case because it should be:

20*log10((P_out/P_ref) / (V_out/V_ref)) = output dB

Thats not right either. You need to convert V_out to dB and subtract, or else not use dB at all.

Let me put it this way, if you divide a logarithm by a number then you haven't done your math right. They add or subtract. Division isn't something thats very meaningful on them.

Originally Posted by ninjackn

So I read "X dB/V" as X dB per volt where as if they used "X dB @ 1V" or anything more informative than dB/V then I (and a few other members of this forum) wouldn't have made the assumption of X dB per volt.

No thats exactly how you should have read it. You just didn't do the math right once you read it. See my example above.

Originally Posted by ninjackn

In the end I have an issue with using "dB/V" to describe headphone sensitivity and I think it's wrong.

Well as I said above it gives nonlinear scaling of intensity with power, but if thats how the headphones actually work then I don't see how its better or worse then any other way of saying it.

[ninjackn]

Originally Posted by saratoga

No thats not right. And if P_out is in units of watts, then its actually 10*log10().

No, P is units of pressure.

Originally Posted by saratoga

Thats not right either. You need to convert V_out to dB and subtract, or else not use dB at all.

Let me put it this way, if you divide a logarithm by a number then you haven't done your math right. They add or subtract. Division isn't something thats very meaningful on them.

Oh, but it IS correct:

20*log10((P_out/P_ref) / (V_out/V_ref)) = output dB

First off we agree that if you have a sensitivity of X dB @ 1V (or "X dB/V" using AKG spec. notation) and if you have a voltage of Y dB applied then the output is = X dB - Y dB.

According properties of logarithm, division is equivalent to subtraction. We see this because dividing two exponential of the same base is equal to subtracting the exponents, that is (C^A)/(C^B)=C^(A-B).

Thus my math IS correct:
20*log10((P_out/P_ref) / (V_out/V_ref)) =
20*log10(P_out/P_ref) - 20*log10(V_out/V_ref) = X dB - Y dB = output dB.

This is also why I said i was off by a "factor of 2" because i'm used to working with power as milliwatts and converting it to dBm. To convert X milliwatts to dBm it's 10*log10(mW_power/1mW).

Often when dealing with voltages it's assumed that the reference has the same impedance and the measurement that was taken so it's converted to power:

P = V^2/Z (As it's been said on this thread may times)

10*log((V_out^2/Z) / (V_ref^2/Z))
we see that the impedance cancels out
10*log((V_out^2) / (V_ref^2))
the squares can be pushed out:
10*log((V_out/V_ref)^2)
exponential in logs becomes multiplications and ta-dum:
20*log(V_out/V_ref)


So to clarify my point of view on "dB/V" further, I did not know if dB/V was a relationship of sound pressure in dB to volts or if it was a relationship of sound pressure in pascals to volts. I assumed that it was sound pressure in dB to volts which because I read the units "dB/V" as "dB per Volt" which is incorrect because it's actually describing dB of Pascals per 1V.