Re: volume control adapter?

["Simon Biber" <simon@basilisk2.cjb.net>]

> What the heck is a logarithmic pot?  All that I have seen are linear, or
close
> to linear.
>

The volume that we percieve is not linearly related to the sound output of
the speaker (which is related to the current driving the speaker, see
below). The ear has the ability to hear soft sounds better than you would
think. Most volume controls use a logarithmically graded potentiometer to
provide this response.

That's the difference between decibels (perceived volume difference) and
pure sound pressure (pascals).

> Also, why are you grounding one side of the pot?  This will probably just
give
> you signal loss.
>
> Try using a cheapo 1K pot from Radio Shaft and wire two leads together
toward
> one end.  Then
> hook it in series with the signal side of the speaker.  The wiring of the
two
> leads effectively cuts
> the POT in half.  All in all you will probably have to turn down the pot
> because the resistance will
> be too high toward one end.

My circuit uses the potentiometer in parallel with the speaker to make a
voltage divider. Where I said ground, I mean the other side of the signal.
This gives a minimum volume when:
5 volts / 1000 ohms = 5 mA total current (none through the speaker)
and a maximum volume when:
5 volts / 1/(1/1000 + 1/8) = 630 mA total current, of which 625 mA is
through the speaker.

Your circuit reduces it from a potentiometer to a variable resistor. You are
adding a resistor in series with the speaker, resulting in a minimum volume
of
5 volts / 1008 ohms = 4.96 mA total current through the speaker
and a maximum volume of
5 volts / 8 ohms = 625 mA total current

So your solution will provide the same maximum volume, but will never reduce
the volume to zero, as mine does. At minimum volume yours wastes 4.96 mA
versus 5 mA for mine. Not much difference I'd say!

Simon.