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HOW TO UNDERSTAND DIFFERENT IMPEDANCE RATES?!!!

Joop Lous lous at xs4all.nl_nospam
Sun Jun 8 07:52:34 EST 1997


Luc.vanHoye at ping.be wrote:

>I am trying to figger out if the impedance of an audio device has
>something to do with how loud its signal can go...
>
>for example:
>
>I have a Roland XP80 (the audio device) which allows headphones with an
>impedance between 8-150 Ohms.
>
>Now, is it so that if I should use an 20 Ohms headphone I will probably
>get a louder signal then if I should use a headphone with an 150 Ohms
>impedance?!
>
>Could anyone explain me how this works and what people mean if they say
>that a headphone is 'hard to drive'?! 
>
Let's start first by stating that there is no standard impedance of
headphones. Knowing this the engineers that design a headphone
amplifier in their audio equipment have to try and give satisfactory
output power for a wide range of headphone impedances.

Some time ago a lot of headphones were either 8 ohm or 150 ohm. Now if
you want equal power in these two type of headphones, and the output
impedance of the amplifier is the only thing one can choose, a value
of about 35 ohm would do this job.

You might want to calculate it yourself. The output power in the
headphone is:

Ph = (U * Rh )/ (Ro+Rh)^2

Where Rh = headphone impedance, Ro is the amplifiers output impedance,
and U the output voltage level (with no load). The 8 and 150 ohm
example would be:
	(U * 8)/(Ro+8)^2 = (U*150)/(Ro+150)^2

Anyway, if your Roland has a output impedance of 35 ohm, your maximum
output power can be reached if you match this impedance. Any value
further away from this 35 ohm (up or down) will decrease the output.
So 20 ohm will still give a bit more output than 8 ohm.

If you really want more power you could modify the circuits of the
Roland. Supposing the headphone amplifier might handle the additional
power load, decrease the value of the output resistor e.g. to 8 ohm. I
would not advice to remove it completely, because there might not be
any short circuit protection present.

Joop Lous.




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