proton007
Headphoneus Supremus
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- Feb 9, 2012
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Quote:
proton: wut?
Higher impedance = higher volts; lower current.
Lower impedance = lower volts; higher current.
There's a reason 'stats are "current-less," and that orthos are "current hungry."
You cannot arbitrarily have "lots of" or even "extra" current, it's dictated by R and V (I = V/R). You get to pick V, the headphones pick R, and I is dependent. And then power = I*V. If you only need X watts based on sensitivity (which is dictated by the headphones), you will only use as much current as Ohm's Law dictates you will. If the amplifier can deliver >X then it will not go into current-limiting, clipping, etc. If it can deliver <X it will have a bad day. For many headphones, X is a very small value relative to what many amplifiers can do.
I feel like I'm missing your point though...
Thanks for the question...it got me thinking, see if this makes sense:
The impedance is fixed by the headphone, i.e. R. (although that also varies acc. to frequency, but for the sake of explanation, assume it to be fixed).
Power and loudness are related in a subjective manner, but the generally accepted rule is that 2x increase in loudness requires 10x increase in Power, and causes 10dB worth of change.
Hence, by P = V^2 / R, for the same R, we need ~3x the Voltage.
However there is also the efficiency of the speaker, so this might be >3.
Now P = VI. If P of an amp is fixed, to increase V, it needs to increase I.
Hmmm... So I guess both the V and I can be a limiting factor for an amp. Increasing V beyond the limit will cause the signal to clip, and 'I' will cause the amp to hit the current limit.
Going by this logic, a speaker with high R will need higher I to have this corresponding increase in V.
The circuitry's performance is limited by the V and I, whichever one is a limiting factor.
For most modern designs, I guess its the former, and is reached earlier than the latter.