Check this photo:
http://tinypic.com/r/34sfsb9/5
 I have enclosed which solder joint on Portamento pot corresponds with what. There are four pairs and I assume that there are both lines for every VCO.
 --- In
korg_mono-poly@yahoogroups.com
, "dir_marillion" <dir_marillion@...> wrote:
 >
 > Hi, I understood.
 >
 > > Btw: it does not have to be the potentiometer itself. It could be >that the corresponding solder joint is corrupted.
 >
 > I just re-opened the synth and checked all lines of the pot which correspond to the solder joints to the motherboard. No corruptions found. Will recheck.
 >
 > The Synth is disconnected from power outlet and I am checking only if the lines are ok.
 >
 > Of course I have not idea which line is for VCO4 and it seems that schematics do not show this detail...
 >
 > --- In
korg_mono-poly@yahoogroups.com
, Florian Anwander <fanwander@> wrote:
 > >
 > > Hi
 > >
 > > > Can you explain me a little how this pot works and what should I do
 > > > to test and/or fix it
 > > The working principle of the portamento circuit is very simple. There is
 > > an capacitor between ground and the control voltage line from the
 > > keyboard for the VCO pitch. If the keyboard control voltage changes,
 > > then its power will first be used to load the capacitor. This will take
 > > some time. So the voltage change will not be immediate, but will
 > > increase slowly. This causes at the VCO what we call portamento.
 > > (The same happens if the voltage drops. Then the unloading will take
 > > some time).
 > >
 > > As said, loading or unloading the capacitor always takes a little time,
 > > but if the CV comes with enough current (if the power is high enough
 > > (remember: power = voltage x current), then the loading or unloading of
 > > the capacitor is that fast, that our ear does not recognize it.
 > >
 > > The portamento potentiometer now simply acts as a variable resistor (so
 > > called rheostat) which is inserted in the CV-line before the capacitor.
 > > If the poti is turned fully counterclockwise then the resulting resistor
 > > is zero -> the capacitor will be loaded that fast, that we do not
 > > recognize it.
 > >
 > > If the poti now gets turned clockwise, the resulting resistor is
 > > increased -> the increased resistor means less current runs into the
 > > capacitor -> the capacitor will be loaded slowlier -> the pitch changes
 > > slowlier -> portamento.
 > >
 > >
 > >
 > > > Which is the line for VCO4 to shortcut and
 > > > how can do this
 > >
 > > I cannot tell you exactly which layer of this four layer potentiometer
 > > is the one for the VCO4, but I found an article in the web, which
 > > describes the principle of our problem quite good
 > >
 > >
 > > Have a look at
http://www.allaboutcircuits.com/vol_6/chpt_3/7.html
 > >
 > > in the picture
 > >
http://sub.allaboutcircuits.com/images/05151.png
 > >
http://sub.allaboutcircuits.com/images/05150.png
 > > you have to think the battery as the keyboard CV and the motor as the
 > > capacitor. That is how the portamento potentiometer is wired.
 > >
 > >
 > >
 > >
 > > The picture
 > >
http://sub.allaboutcircuits.com/images/05152.png
 > > describes, what might be broken in your portamento potentiometer.
 > >
 > > If you connect both the outer connectors of the potentiometer with each
 > > other, then it is like the potentiometers resistance would not be set to
 > > zero.
 > >
 > >
 > >
 > >
 > >
 > > > Does it contain graphite like the sliders on JX3p
 > > >
 > > Yes. All potentiometers work like that, but you can't repair it with an
 > > graphite spray!
 > >
 > > Btw: it does not have to be the potentiometer itself. It could be that
 > > the corresponding solder joint is corrupted.
 > >
 > > Florian
 > >
 >