Quote:
Originally Posted by

**bowei006** And kep in mind that I said current. Not power. They are different things. In all actuality, very high impedence 600+Ohm headphones actually require less power BUT more current.

Do you know Ohm's law? Because it dictates the opposite of what you just stated. V (voltage) = I (current) * R (resistance)....and... P (power) = V (voltage) * I (current) = I * (I * R) using variable replacement replacing V with (I * R).

Then further the law of electon flow is voltage (potential) is *provided*, and current is *drawn* from the source based on however much is required & permitted given the resistance/impedance; with the exception that if the source is not capable of providing the current at the given voltage (i.e. source does not have the power capability), the voltage will be drawn down to the level where voltage * current = the maximum power that the source can provide. So with V held constant, the higher the impedance for the same voltage, the *less* current is drawn.

What high impedance headphones *require* categorically is higher *voltage* to drive them, and less current will be drawn, for the same milliwatt power required input to them to produce the same decibel level out. And *if* a high impedance headphone requires high current, that implies [is a derivative way of saying] it requires *high power*.

Now, apply that linear equation to headphone's audio output level, which is a function of power, not just voltage or current alone. As stated just before, for a 32 ohm version of a headphone to produce the same Db level as the 600ohm version of the same headphone, they require the *same power input* to them. Let's imagine this headphone needs 100mw to produce a certain decibel level whatever that may be (I have no idea of what the real power required to produce that sound level, but using that as an example). We know our Rs, and we know our P. Our equivalent equation in terms of current and resistance cited on the 1st line of this post becomes:

So where I1* V1 = 100mw (variable #1 representing the 600ohm headphone) and I2 * V2 = 100mw (variable #2 representing the 32 ohm headphone), using the variable replacement we have I1 * I1 * 600 = I2 * I2 * 32. And that should clearly be self evident that to make both sides equal, i1, the current necessary to drive the 600 ohm load, is *far less* than I2, the current necessary to drive the 32 ohm load.

Edited by m8o - 12/8/12 at 5:02pm