Explaining impedance, once and for all
Aug 4, 2008 at 5:55 PM Thread Starter Post #1 of 11

JazzHands

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Hi guys,

I've read several threads on HF now concerning impedance and I'm still none the wiser, I think.

So far, this is what I'm understanding, and I've written it down to remind myself as follows:
A lower impedance means a headphone draws more power and produces a more powerful sound.

A higher impedance means a headphone draws less power and produces a less powerful sound.

The higher the impedance, the lower the loudness.

The lower the impedance, the higher the loudness.
Is this correct?

What I don't understand is how this realistically relates to headphones and earphones. If my understanding is correct, a pair of 16 ohms earphones should require more power from a portable player but will produce a more powerful sound; and a 32 ohms pair draws less power but produces a weaker sound.

The problem is, I have a pair of 32 and 16 ohms earphones from the same manufacturer (though I borrowed one pair from a friend for a test) and at the same volume with an iPod I can't tell any difference in volume between one and the other. Sound quality, I can absolutely tell the difference. Loudness, not at all.

A friend of mine has a degree in audio engineering and even he admited that impedance has always confused him.

I really want to understand it to the point that I could explain it to someone like I could explain every other aspect of headphones, and I feel I'm missing a massive piece of knowledge by not truly understanding impedance.

Please help?
 
Aug 4, 2008 at 6:10 PM Post #3 of 11
Quote:

Originally Posted by G-man /img/forum/go_quote.gif
I SWEAR there was the exact same thread like a month ago.


Yep, and I wrote a quite explicative and elaborated reply. I wish the search feature in this forum worked a bit better.

Rgrds
 
Aug 4, 2008 at 6:12 PM Post #4 of 11
Quote:

Originally Posted by Cool_Torpedo /img/forum/go_quote.gif
Yep, and I wrote a quite explicative and elaborated reply. I wish the search feature in this forum worked a bit better.

Rgrds



I remember. First time i ever felt stupid. I should have continued physics, but soo little time.
 
Aug 4, 2008 at 7:22 PM Post #6 of 11
OK, here we go, since I can't find the thread nor the reply, I'll try to explain a bit -again- how the impedance/sensitivity thing goes.

Headphones' drivers are transducers that convert AC electricity into sound. Dynamic drivers, the most common ones, use a coil -wounded wire around a hollow cylinder- into a magnetical field provided by a magnet, to convert the AC voltage into movement which is transferred to a membrane. The membrane's movement is transferred to the air particles in front of your ear. For the frequency characteristics of that vibration, and its pressure level, your brain interprets it as sound.

The voice coil of the driver has an impedance, which is the opposition it presents to the AC source (the amp or any headphone out) to the free flow of electrons thru it. The lower that impedance, the more freely the electrons travel and the closer is the scenario to a short-circuit. This means that your source of electricity needs to pump more current intensity to correctly drive the transducer. So you can take two conclussions from this:
- What makes the AC to drive any coil is its voltage. The minute variations of voltage follow the signal originally recorded.
- The current intensity is important to keep the coil excited, and you need more current intensity the lower is the impedance. There's a relation between the current voltage and the intensity which is the power measured in watts. Power is the product of the voltage and the intensity: P=V*I. This is why amps are rated for their power output and not only for their voltage capabilites.

Up to this moment there's no relation between the impedance and how loud the transducer will sound. However there's a parameter named sensitivity which tells you how loud will a transducer "sound" for a given amount of power you're feeding it. The sensitivity is rated in dB/mW for headphones, so a pair of phones delivering a SPL of 100dB/mW are more sensitive (can sound louder) than a pair rated at 90dB/mW.

So the easy or hard to drive a pair of phones is, depends on both parameters, the sensitivity and the impedance. The worst case would be a pair of phones of very low sensitivity and also a very low impedance. Why? because they'll be asking to the source more watts to sound equally loud as a more sensitive pair, and an important part of that power will be asked in the form of current intensity, which is something that most portable players, headphone outs in receivers and players, etc. aren't designed to deliver. This is the case of cans like AKG 701 or Denon D5000.
If your cans are low impedance but are very sensitive (the case of Grados and most IEMs) then despite their asking more current from the source, they still manage to sound very loud because they need very little power to do so.

Most people tend to think that low impedance equals to louder sound, but this is plainly wrong. It all depends on the sensitivity and how much power the cans need to give a high SPL. Also take into account that not all manufacturers offer their sensitivity values and not all them do in dB/mW but do in dB/mV. It's not much of a problem, you just need to convert the mV in mW knowing the phones impedance.

Rgrds
 
Aug 4, 2008 at 7:23 PM Post #7 of 11
This thread comes up every 2 weeks in a once-and-for-all fashion. we need a well-written sticky. in bold text.

All I'm gonna say is: impedance is only one variable out of at least three that determine how loud the headphone is.
 
Aug 4, 2008 at 9:43 PM Post #8 of 11
tongue.gif
Yeah a sticky would be nice

It seems to me that all it is.. is how much the headphone impedes current flow. For a given voltage/frequency, lower impedance will allow more current flow, translating to more power. Then of course there's the SPL produced by a given power/frequency.

Then the source might have it's own self impedance, which causes the voltage to drop as more current is drawn, meaning non-linear change as load (headphone) impedance changes. The source's maximum current flow (op-amps particularly) will determine the maximum power for a given impedance, provided voltage is there. Perfect voltage-sources (0 output impedance) will put out constant voltage at any load, meaning power increases with lower impedance.



I dunno.. just felt like writing stuff down because I'm sitting in front of a computer screen
bigsmile_face.gif
 
Aug 5, 2008 at 12:32 AM Post #9 of 11
There should be an impedance sticky. This comes up much too often.

In a nutshell, headphone impedance tells you very little about how hard they are to drive. This is like using a car's horsepower to figure out what the mileage would be. There is a relationship, but you need more information.

What everyone seems to miss is that amps have an output impedance. Everything that drives headphones has an output impedance, even iPods and soundcards.

The difference between an amp's output impedance and a headphone's impedance tells you how efficiently power transfers from the amp to the headphones.

Think of it as trying to connect two pipes. If you connect two one inch pipes to each other, water will flow easily from one to another. If you connect a one inch pipe to a six inch pipe, the water pressure will not be the same on the other side. The closer the two pipes match, the better the transfer of water will be.

Impedance works much the same way. Whether it is 32 ohms or 600 ohms, the power transfer will be best with an amp that has a similar output impedance. I don't have it at hand, but there is a formula where you can work out power loss from an impedance mismatch.

Now, once you know how much power actually gets to the headphones, you can use the sensitivity of the headphones (measured in dB) to determine how loud they will get given the Watts that get through.

There is more to it, but I hope you get the idea. Keep in mind that impedances are on a curve, as is power, and there are other variables. I'm still learning about those, but maybe someone here can expand on that.
 

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