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.