Hi,

Thank you for your time and effort to do something like that.

> The unused connector on the VCO board is appearently only a kind of
> testsocket.

I thought it was a relative midi connector which never implemented :-)

Now, about the Portamento Pot, I just checked that the lines are not cut. Without having the appropriate knowledge to read the schematics of the Monopoly, I tested it with just one thought:

While the pot is set to 0, the Portamento must be 0 to all VCOs unless it has four lines that each one goes to each VCO and in VCO4 case one line is cut so it is always on.

This was my thought for checking it.

Can you explain me a little how this pot works and what should I do to test and/or fix it Which is the line for VCO4 to shortcut and how can do this Does it contain graphite like the sliders on JX3p

Thanks

--- In korg_mono-poly@yahoogroups.com , Florian Anwander <fanwander@...> wrote:
>
> Hi
>
> >> Yes, you are right; it is exactly as you said.
> >> http://www.2shared.com/fadmin/26610218/4b21bac3/monopoly.zip.html
> > Aaaah! Things become clearer. I have to track the things in reverse. I
> > will come back later.
> ok, I had a look at that:
>
> Both groups of additional parts appearently are NOT modifications, but
> they seem to be the standard wiring of the very first series. (in my
> MonoPoly the same connections exists, but they are achieved as traces on
> the pcb!)
>
> The one on the VCF board is the the circuit around Q10, R85, R86, D7,
> and C27.
> The one on the bender board is the regular wiring of the Arpeggiator
> Off/On/Latch switch.
>
> The unused connector on the VCO board is appearently only a kind of
> testsocket.
>
>
> My most likely suspect is the portamento potentiometer itself. You wrote:
> > During the procedure, I checked all the Portamento pot lines, with
> > a polymeter and they work.
> If you measure the lines, then you won't recognize if the slider of the
> potentiometer itself is broken. Did you try to shortcut the portamento
> potentiometer for vco4
>
> Florian
>