On Thu, 23 Mar 2000, Simon Barnes wrote:
Hi,
> Francisco Jose Montilla wrote: (in reply to Jim Coon)
>
> > That doesn't happen in the amp/transducer world, AFAIK. Here
> > weren't working with DC, is variable Hz AC, and what
>
> Sorry Francisco, I'm having a lot of difficulty in understanding what you
> are trying to say. Jim's equation works with AC or DC.
>
> > actually happens if my memory serves me well is that current voltage
> > increases, and power decreases. (hence my doubt).
>
> What is "current voltage" ? Power = current * voltage.
I meant (sorry, english isn't my native language) that what excite
the headphone transducers is a variable (AC) electrical signal. I don't
have the mathematical demonstration handy, but if you go in a little
deeper, you'll notice that with higher loads (i.e. impedances) peak
voltage rises. Those equations assume a constant (i.e. DC) voltage.
> > That's why amps tend to produce better sound
> > when using higher impedances.
>
> I'd like to see some evidence for this. I suspect the impedance only really
> affects the volume (More impedance = less power)
Not exactly. Less intensity (the I in Power = Voltage * Intensity,
dunno what are the variable names used in english). Higher impedance, less
distortion (i.e. less intensity), more definition and sharper sound (more
Voltage).
My doubt is, if a 200 Ohm headphones with higher sensitivity than
a 32 Ohm headphones (I mean, so that with a given volume setting, you
perceive equal volume level pluging anyone) which will drain more battery.
I'd bet the 200 Ohm, but am not sure...
I'd try to reproduce the equation development to prove the
voltage raise.
greets,
*****---(*)---**********************************************---------->
Francisco J. Montilla System & Network administrator
[EMAIL PROTECTED] irc: pukka Seville Spain
INSFLUG (LiNUX) Coordinator: www.insflug.org - ftp.insflug.org
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