Is a headphone amp needed?
Reply #37 – 2005-12-14 03:49:25
Well, Ohm's law says actually the opposite. P=(V^2)/R, or P=(I^2)*R. So, in fact, the greater the impedance (R ), the lower the current (I) required for same power (P). But the greater the voltage (V) needed. I'm not sure where we are in disagreement as I did state that P=(E squared/R). I would like to point out, however, if the impedance is raised in a circuit, as in this case by plugging in a different set of headphones, the power transferred to the output load (headphones) will decrease. For example, if you have a signal source coming from a headphone jack that is 6v rms and you are using phones with 24 Ohms of impedance there will be a current of 250mA which will deliver 1.5W of power to the load. If you then replaced the 24 Ohm headphones with a set of 300 Ohm headphones then your 6v signal source will be dropping accross 300 Ohms, therefore causing the current to drop to only 20mA and now only 120mW of power is transferred to the headphone load. In your example you chose to increase the voltage coming from the headphone output jack to maintain the same current level. Power=V x I so in this case, that would require your output voltage to increase from 6v to now 75v which is a huge increase and definitely isn't going to happen unless you have something like a dedicated headphone amp to drive this thing. In any event, there is a greater power demand for more expensive (higher impedance) headphones, which I believe was my point to begin with.