Hi folks,

anybody has some knowledge about impedance/ohm .... ?

Situation: I want to purchase an mp3 player and the last two I had were not loud enough. So my questions would be:

* Do headphones with less or more impedance/ohm sound lounder ?
* Do headphones with less or more impedance/ohm consume the batteries faster ?
* Is there any way to know how "loud" the mp3-player would be itself (independently of the headphones) before purchasing ?

Thanks a lot!

Not necessarily, although there is a tendency for headphones with significantly higher impedance, IOW 400 ohms versus 16 ohms, to play softer.

I would hope for a headphone specification like:

"xx ohms impedance
Sensitivity yyy dB for z milliwatts (mw)"

Usually z = 1

One actually sees specs like this occasionally. More dBs for less or equal milliwatts means that the earphones will play louder, all other things being equal.

The impedance spec allows converting the given sensitivity into an equivalent sensitivity for a voltage input.

Higher impedance and lower voltage per dB specifies a headphone that is likely to play more loudly with more different players.

Of course, vendors don't have to tell the absolute truth.



Usually the power usage of other parts of the MP3 player are far more significant than the headphones. Higher impedance and lower voltage per dB specifies a headphone that is likely to play more loudly with less battery drain.



If the player's characteristics are fully specified, then you have a chance of knowing this.

But sufficient specifications are not common.

For example, here are Apple's specs for the classic iPod:

http://www.apple.com/ipodclassic/specs.html

I see no way to get even the slightest clue from this information about how loudly this player plays.

I would look for a spec such as:

"Maxium output xx milliwatts (mw) into a yy ohm load"

Or even better:

"Maximum xx volts with no load and a source impedance of yy ohms"

More volts with a lower source impedance gives a player that will play more loudly with more different headphones.

But, it is all a dream as few if any players or even headphone amplifiers give relevant specs.

Usually lower impedance = more power = louder. But, some headphones have higher sensitivity than others, so you have to check the specs.

Maybe.... Lower impedance usually means more power (assuming you're listening louder), which means shorter battery live (maybe).

I'm not sure about that one...

-------------
Some Basic Electronics:
With an "ideal source", the voltage stays constant (independent of the impedance), and but the current increases when the impedance goes down. This means you get more power (Watts or milliwatts) with lower impedance. If you've read power amplifier specs, you've probably seen specs like "100W @ 8 ohms, 200W @ 4 ohms".

The catch is, that there is usually a current limit on the output (maybe a simple resistor) and there is a limit to the amount of current the battery can supply. Depending on the particular player, the power may actually drop with too-low of an impedance.

We usually don't have enough information about the player to make actual calculations, but if you're comfortable with algebra, the following may help:

Voltage, Current, and Resistance (Impedance) are related by Ohm's Law:
I = V/R
Where I=Current (Amps), V = Voltage (Volts), R=Resistance (Ohms).

Using algebra, we can re-write the other 2 versions of Ohm's law:
V=I x R
R=V/I

Then, there's the Power Formula:
P (Watts) = I x V

And, from algebra:
P = V(squared)/R
P = I(squared) x R

Thanks so much for your wise information Arnold and DVDdoug. With your guidelines, I started some further reading, and begin to understand the issue a little bit.

DVDdoug: Is V really voltage ? Because Wikipedia says, V is "voltage drop" or "the potential difference measured across the resistance" ...

Also I found one posting a bit intriguing. What do you think of it:



Because I thought low impedance consumes less battery, but here the contrary is stated ...
.

OK, looking on concrete examples:



When comparing headphone A to B,C,D, which can (in theory) be expected to be louder and why ?

It's really all the same... I think "voltage drop" is probably misleading. But, these all really mean the same thing:

"One volt applied across the headphones."
"A voltage drop of 1V across the headphones."
"A one volt potential across the headphones."

I'd use the term "voltage drop" where there are two resistors in series, or if there's an output limiting resistor on the player's output (in series with the headphones). In that case, the voltage will be divided between the resistor and the headphones (in proportion to the resistance)... If the resistance is equal to the headphone impedance, half of the voltage will be "dropped" across the resistor, and half will be "dropped" across the headphones.

Actually, I didn't know that sensitivity is referenced to voltage and and efficiency is power (milliwatts). But, if you know the impedance, you can convert between the two.

However, in order to compare a headphone that specifies efficiency to a headphone that specifies sensitivity, you'd need to know how to normalize the voltage or wattage level so that you're comparing "apples to apples"... And, we have to assume that the player maintains constant voltage with varying loads.




A difference in dB is related to a difference in voltage by the following:
dB = 20 log(V2/V1)
Or, Voltage Ratio = 10 (to the power of) ((dB2-dB1)/20)

(For example, a doubling in voltage (V2 = 2, V1 = 1) is a +6dB increase.)

And for power:
dB = 10 log(P2/P1)
Or, Power Ratio = 10 (to the power of) ((dB2-dB1) / 10)

(Doubling the power results in a +3dB increase)

I hope got all of that correct! You know what they say... "never do math in public!"


"Efficency" seems more related to battery usage. A more efficient headphone takes less power for a given volume level.

You can think of resistance as a narrow water pipe. A skinny water pipe has higher resistance to water-current flow and it takes longer for the water tank to drain. A fat low-resistance pipe allows higher flow, and the tank empties faster. That's not a perfect analogy, because the water pipe isn't doing any "work". Back to electricity... A high wattage light bulb has lower resistance, more current flow, and consumes more power, than a low-wattage bulb.

OK... I know this was going to get complicated... I made a spreadsheet...

First I calculated the voltage for 1mW at the different impedances. SQRT(Ohms*W)
Then I calculated the dB diffeence between that voltage and one volt. 20*LOG(1/calculated Voltage) (Any reference level would work. as long as it's the same for all headphones).
Finally, I added that dB difference to the rated dB level to get the theoretical dB level at 1V for all of the headphones.

It would be slightly more complicated if some ratings were referenced to voltage and others to milliwatts.


Impedance - Power --- Voltage@1mW --- +dB@ 1V ------ Rated SPL -- SPL@ 1V
32 ----------- 0.001 ---- 0.178885438 ---- 14.94850022 --- 98 --------- 112.9485002
64 ----------- 0.001 ---- 0.252982213 ---- 11.93820026 --- 98 --------- 109.9382003
16 ----------- 0.001 ---- 0.126491106 ---- 17.95880017 --- 105 -------- 122.9588002
64 ----------- 0.001 ---- 0.252982213 ---- 11.93820026 --- 105 -------- 116.9382003


The answer is Headphone C!


That all assumes constant voltage from the player, with any headphone load.

Sorry about all of the decimal places... Too lazy to clean that up.

C by inspection because it has the lowest impedance and the highest sensitivity. When hooked to a low impedance source, it will be the loudest.

However, if the source has a very high impedance, the balance may tip to phones D.

BTW I did look at DVDougs post first, but I realized that the question had an easy answer, but also a possible other answer depending on which source is used to drive it.

In theory yes, in practice not really. They're often speced, but the specs are all but made up in the tests I've seen.

With that in mind, ignore the specs on MP3 player amp powers. Fortunately, theres not a huge range in output level (maybe 20dB between extremes with a few dB being typical in my experience).

If you want loud, just buy IEMs. They go way louder then you'll need with even the quietest MP3 players (due to design and also low impedance).

Mike, thanks for your real-world tips. And Arnold and DVDdoug, thanks for the science !
I've never been to any forum like this with people equally wise and helpful ... thanks so much !!

I'm wondering why Arnold says "However, if the source has a very high impedance, the balance may tip to phones D." as the source impedance is not part of DVDdougs calculation, how would that come into play ? I'd be happy to leave that aside if not important, because it's already quite hard to understand.

DVDdoug, I'm very happy seeing your calculations in post #7, that's exactly where I wanted to go ... I'd wish so much I could do all that maths on my own, without bothering people. The thing is, I'm reasonably good in maths, but a zero in physics (as I had an awful teacher) ... so I can solve equations and everything and follow your calculations but at the same time have no clue as to WHY they are done this way:



I can imagine that it would take A LOT of time to explain that, in which case I really don't want to ask for that.

Still there is this problem. Provided the following is correct, ...




... is corect, it should say "efficiency" in Arnold's definition, not "sensitivity".



I had not second-guessed Arnold's definition and thus used his definition to give impossible source data ("headphone A: sensitivity 98 dB for 1 mW") .... while it should have read 1 mV. I guess DVDdoug was not aware of this impossibility either during his/her calculations. So DVDdoug ended up calculating dB @ 1V, which would be the efficiency, while we should rather be looking for the phones' sensitivity, right ?

So in case the units in my source data were db for 1 mV (instead of dB for 1 mW as erroneously stated), then how would this affect the calculations .... ?

Both sensitivity and efficiency are in common use.

Here's an example of the use of sensitivity:

http://store.shure.com/store/shure/en_US/D...tID.105181400#t

"Technical Specifications:

"Sensitivity (1mW): 114 dB SPL/mW
"Impedance (1kHz): 26

I found an AKG spec sheet that used the word efficiency, and gave it in terms of dB per volt.

http://www.akg.com/mediendatenbank2/psfile...56224d61018.pdf

"SPECIFICATIONS

"Efficiency 105 dB SPL/V
"Rated impedance 62 ohms

So, theres the evidence from independent sources that use either form of specifications.

They can be converted to the other form using Ohm's law.



I had not second-guessed Arnold's definition and thus used his definition to give impossible source data ("headphone A: sensitivity 98 dB for 1 mW") .... while it should have read 1 mV.

Again efficiency is given both ways. See my examples, above.