Korg Mono/Poly

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
> >
>