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Assuming the same ballpark dB/W, the voltage and current is irrelevant - the headphones will requires the same power - W=VI
Ideally true.
Headphone sensitivity is generally quoted in Voltage for a given output rather than watts. As the impedance increases the Current will decrease, substituting using Ohms law into the above we end up with P=V^2/R, you can see that you need to increase the Voltage by a "Squared" Factor to achieve a similar power dissipation. In reality this rarely happens as the sensitivity of the higher impedance phones only require only a little more voltage rather than the "Squared" Factor that the formular would suggest.
LPSpinner.
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