passerby999
WARNING: BAD TRADER
Also known as I-Love-Music
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- Apr 15, 2006
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same as title...anybody helps?
Originally Posted by creyc Any particular headphones "impedance", given as an average resistance value, determines the resistance the amp sees. Be it the internal amp in your iPod or in your Hornet headamp. Now the more the resistance, the higher the voltage will need to be from the amp to get the same amount of power. We can thank Ohms law for defining that for us. Say you have a headphone like a Sony V700, with a very low impedance. (we'll say 16 ohms) It doesn't require you turn the volume knob up very high to be loud. Does this mean this headphone is more efficient or maybe it produces sound out of thin air? Obviously not. That volume knob doesn't adjust how much power your amp puts out, nope not at all, because there are two attributes used to describe electrical energy. To keep things simple we'll just say you're merely raising and lowering the voltage when you turn the volume knob. Going back to the example, lets say the amps volume is set to put out .4 volts. The headphone load is 16 ohms. Ohms law says current = volts / ohms. current = .4v/16ohm current = .025A current = 25mA Thats 10 mWatt of power moving the diaphragm. (power = amperes x voltage) Now lets take a 600 ohm AKG. Putting out that same .4 volts (same place on the volume knob) we get: current = .4v/600ohm current = .00067A current = .667mA Thats .267 mWatt of power moving the diaphragm. Thats about 37.5 times more power output to the Sonys. Go ahead and divide 600ohms by 16ohms and you guessed it, 37.5 times more resistance on the AKGs. ... |