Discussion on impedance and difficulty of driving
Nov 25, 2008 at 2:29 AM Thread Starter Post #1 of 5

helicopter34234

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What makes one headphone harder to drive than another headphone? I would imagine that lower impedance headphones are harder to drive because they draw more current (P=I^2*R) for the same output power and the am might be current limited. Alternatively I can see that high impedance headphones could be harder to drive because (P=V^2/R) and they require higher voltage for the same output power, possibly the amp cannot provide high enough voltages and limits. And finally it could just turn out that most amps have either high or low output impedances and therefore get maximum power transfer whenever the phones are impedance matched. Which of these scenarios are the true dominating factor? Are high impedance headphones harder or easier to drive?
 
Nov 25, 2008 at 10:11 PM Post #2 of 5

deltaydeltax

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The coil reactance also has quite a bit to do with it. Since the reactance is frequency dependent, you'll see various levels of impedance with various levels of frequency. The impedance of speakers and such is measured at DC with a multi-meter. That's the number we see on headphones, it's a nominal value which isn't static across the audio spectrum. Reactance is calculated:

XL = wL

^ read, X sub L equals ohmega L

XL is reactance in ohms. w (little w looks almost like lower case ohmega) is 2*pi*f, where pi is ~3.14 and f is frequency. L is the value of our inductor in Henries.

The value of reactance you find will be at an angle of +90-degrees. In an inductor, the voltage leads the current.

Ohm's law is the most simplistic way of calculating the output power. To get the true value, you have to test the speaker/headphone at various frequencies over the audible spectrum (only a few since doing them all will surely be done after you die, it takes long). It's a bit difficult to put polar numbers into Ohm's law (numbers can't be added and subtracted in polar form, they must be converted back into rectangular).

So, to make my understanding shorter, the speakers/headphones impedance curve has a lot to do with how easily an amp can drive it. If you have an audio driver which dips into impedance levels close to DC (0Hz), your amp will experience stress. Try driving a paper clip with your amplifier, if the DC protection doesn't kick in, you'll more than likely blow your amp.

Please excuse any typing errors, I hammered this out as fast as I could and I hope I didn't bore the hell out of anyone. There is far more which can be added to my answer.
 
Nov 26, 2008 at 4:43 AM Post #3 of 5

helicopter34234

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Thanks for the explanation. I am actually pretty good with the concepts you have described, I have taken several electrical engineering courses. My question is more geared to generalization that people seem to make regarding the ease of driving a headphone based on its DC impedance.

I found some good answers Headphone Impedance - [H]ard|Forum

Some good points made

Quote:

Opamps are good at driving voltage, not so good at current. So when you load them with a low impedance it works, to a point, but you get more noise, less bass, more distortion and so on. You need to buffer the output of the system to properly drive low impedance phones.


 
Dec 13, 2008 at 10:16 PM Post #4 of 5

jcx

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