Volume, Power, Voltage, Current, and Amplifier Output Impedance in Planar Headphones
Jan 4, 2014 at 1:06 PM Post #16 of 18
  but that is just looking at how many volt we need/how loud it gets, nothing about current.  


The main point of the first post is that the headphones take the voltage from the amplifier, and the impedance of the headphones decides how much current we get using Ohm's law, current = voltage / resistance. As long as the amplifier is within its operating limits and not clipping or distorting, it will put out the appropriate amount of current. That last part is where amplifier power comes in -- if an amp doesn't have enough power or enough current reserves, it won't put out enough current at that voltage. But any two amps that are operating within their limits at the same voltage will put the exact same current into the HE-400. 

The HE-400 has relatively high sensitivity for a planar. Something like the HE-4, HE-5, or HE-6 would need more current (as well as voltage) at the same volume. 
 
Jan 4, 2014 at 1:17 PM Post #17 of 18
  but that is just looking at how many volt we need/how loud it gets, nothing about current.  he400 might need more power than some because it combine low efficiency and low impedance. where we often only need to look strong for either V or A so real power is not mandatory.

 
Power = Voltage X Current
 
They are low impedance, low efficiency 'phones, so they will need a bit more current than a Grado for example.
 
I'm not too sure what you are trying to say.
 
Jan 4, 2014 at 3:13 PM Post #18 of 18
^ And current I is determined by voltage V and resistance R as posted before in #4.
 
As long as the amp doesn't clip you can completely ignore current.
 
 
In the water analogy: When you turn up the pump (increase volume of your amp) then the thickness (resistance) of the pipe will determine how much water (current) will flow through it.
 

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