dylan@cs.washington.edu (Dylan McNamee) (11/07/89)
I'm interested in designing a high quality digitally controlled
attenuator for audio use. I read about a method used by Berning Audio
that uses no potentiometers in the signal path... Of course the
schematics weren't published in the review. I think I may have a
scheme, which is based on a voltage divider, that may do the trick.
in-----|
/
\ 20k
/
\
|
+------+------+------+------+----------out
| | | | |
\ \ \ \ \
40k / ? / ? / ? / ? /
\ \ \ \ \
/ / / / /
| | | | |
gnd A B C D
Points A, B, C and D are connected to transistors, which either
leave the points floating, or connect the resistor to ground (which
changes the voltage division). If digital control wasn't needed,
a plain old potentiometer could be used; the audio signal only
travels through the single fixed resistor, which can be really
high quality.
My question is: Would this work? Could someone send me some
equations for such a voltage divider, so I can figure out what the
resistances ought to be. Finally, does anyone have any ideas or
changes? (I think this could be the basis for some really nice
attenuators, especially for remote control.)
Thanks for any advice--(I'll post a summary to the net, if there
is any interest.) This newsgroup has been really educational.
dylan mcnamee
dylan@cs.washington.eduyahoo@unix.cis.pitt.edu (Kenneth L Moore) (11/09/89)
In article <9718@june.cs.washington.edu> dylan@cs.washington.edu (Dylan McNamee) writes: >My question is: Would this work? Could someone send me some >equations for such a voltage divider, so I can figure out what the >resistances ought to be. >dylan mcnamee >dylan@cs.washington.edu Resistors in series just add. R (total) = R1 + R2 .... Rn Resistors in parallel add as the reciprocal of the resistances. 1/R (total) = 1/R1 + 1/R2 + ... 1/Rn Or using mhos, the conductors in parallel just add so: G (total) = G1 + G2 + ... Gn -- yahoo@unix.cis.pitt.edu (Kenneth L Moore) @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ Ken @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@