Way to tell how easily driven a set of cans are ?

[Duggeh]

Quote:

Originally Posted by Zombie_X /img/forum/go_quote.gif The lower the impedance/sensitivity = easier to drive The higher the impedance/sensitivity = harder to drive This is not always entirely true though, though I don't know of any headphones that this applies to. I've just hears stories. And I may be talking out of my butt about this stuff though this is what I've been told.

We really really do need a sticky about this because its probably not only the most commonplace but most actively proliferate piece of technological misinformation on this entire forum and it continues to be so because of posts which are epitomised beautifully by the content of this one.

Recently theres a light trend to start throwing the word sensitiveity in too, which is good, I still think that a layman-language post from somebody who knows which end of a multimeter goes in which nostril would serve wonders and be linked to as a means of quick reference surprisingly often.

 

[Emooze]

Correct me if I'm wrong, but you'd take the combined output impedance, phones + amp, find out the average or rms voltage being put across them. P = V^2/R gives you watts. Divide by 1000, and use the sensitivity to calculate decibels.

Again, I'm guessing in the dark. Am I wrong?

 

[lucky]

Ask Head-fi. The group chorus will tell you which headphones you're allowed to enjoy without amps, and which you can't possibly like (nevermind your lying ears) without them. :O)

 

[Uncle Erik]

Quote:

Originally Posted by Ericular /img/forum/go_quote.gif Thanks for this info... I am intrigued by it. I read somewhere that the output impedance of the iPod Touch is 32 ohms. With this knowledge, is there a formula or calculator that would help me predict how much volume I'd get from a certain headphone if I knew the headphone's (average) impedance and sensitivity?

Found this online:

There's more discussion of impedance matching here.