Should be good as long as the pk-pk audio voltage is less than 5V
The frequency response will depend on the input Z of the following stage -
but assuming it is high enough, the low frequency cut-off will be determined
by the input cap. value & the 2 10k resistors (in AC parallel = 5k)
The a 100nF cap would give you a (low frequency) cut-off (-6dB) of
1/(2*pi*R*C) = 318Hz.
If the following stage is of low Z, use its impedance in parallel with the
10K resistors to work out the cap required for the input side and just the
input Z to work out the output cap. Allow a fair bit of safety factor (2 -
10 times) if possible

The 10k resistors could probably be increased to 100k or even more if low
frequencies are required - but preferably don't use electrolytic caps as any
leakage would lead to distortion.

Richard

> -----Original Message-----
> From: Gennette Bruce [mailto:bruce.gennette@TAFE.NSW.EDU.AU]
> Sent: Tuesday, July 13, 1999 12:10 PM
> To: PICLIST@MITVMA.MIT.EDU
> Subject: Re: analog switch bias solution
>
>
> Dave's solution is probably the best way to go, but if you
> don't have a good
> electronics background you will need an explanation of why it
> works - so
> here goes.
>
> (use a fixed font (like Courier) at 80 chars wide to view this diagram
> properly).
>
> Audio signal, swinging       line held at     Audio signal, swinging
> -0.6V to +0.6V               half +Voltage     +1.9V to +3.1V, all DC
>
>                            --------/------- +5V
>                                    |
>                 AC jumps           -                          AC jumps
>                 capacitor         | |10K
> capacitor
>                                   | |
>                      | |          |_|                            | |
> IN o-----------------| |-----------|-------------4066
> switch-----| |---o OUT
>                      | |           -                             | |
>                                   | |
>                                   | |
>                                   |_|10K
>                                    |
>                            --------\------- 0V
>
> The incoming audio signal can jump the capacitor.  The high
> value resistors
> hold the line into the 4066 switch at a 'fixed' value above
> 0V, which the AC
> is superimposed on to. Because these resistors are high value
> they won't let
> the DC through very quickly, so the fast-changing AC is
> unaffected on the
> input side.
> After passing through the switch the AC jumps another
> capacitor onto a line
> that is not DC biased, so the original AC only gets out.
>
> Take care selecting the value of the coupling capacitors,
> otherwise some
> frequencies will be cut off (could be a good thing in some
> applications like
> intercoms, etc).
>
> Can someone else suggest some capacitor values (and what
> frequencies they
> pass/cut) ?
>
>         -----Original Message-----
>         From:   Dave VanHorn [SMTP:dvanhorn@CEDAR.NET]
>
>         Solution #2: Capacitively couple the signal in and out of the
> circuit, and
>         on the input side, use a pair of equal value
> resistors, say 10k, to
> bias the
>         4066 to 2.5V.
>