From the circuit diagram and PCB mask in the service manual, it looks as if the input to pin 5 is the averaged output of outputs 1A and 2A (pins 13 and 14), achieved by connecting two 2.2k resistors in series between 13 and 14, with pin 5 connected to the midpoint between the two resistors. That point also then goes off to the input of the HA1452 preamps.
The BBD chip is IC4 in the following two links:
http://www.therogoffs.com/cs80/manuals/CS80_Service_Manual/57%20-%20OE2%20Circuit%20Diagram.jpghttp://www.therogoffs.com/cs80/manuals/CS80_Service_Manual/58%20-%20OE2%20Circuit%20Board%20&%20Wiring.jpg --- In yamahacs80@yahoogroups.com, "faxiomas@..." <faxiomas@...> wrote:
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> Hi folks
> I'm having a debate with a friend over the chorus schematics from the CS80's main diagram, so I need some help from anybody who remembers how the circutry in the PCB is actually made: the debate concerns the dry signal to the inputs of the MN3001 IC. I believe that the schematics are drawn well, when I see that the dry signal is sent to both IN and IN' inputs of the BBD, while keeping the phasing on through the output of the first BBD line to the input of the second. My friend thinks that the schematics have some errors: he tells that the dry signal can't be delivered to both lines at once, because if the pin next to the dry signal's cable was real, the dry signal would have bypassed the BBD, so the signal is delivered to line 1 then line 2 through the line 1's output.
> Can anybody shed some light on this? David? Christopher? Anybody else?
> Thanks for your replies...
> M
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