Korg Poly800/EX800 Users

Gordon,

I just popped all of this into a PDF and it is now sitting in the Poly
800 service files section.

Welcome to posterity!

Mike.

--- In korgpolyex@yahoogroups.com, Gordon JC Pearce <gordon@...> wrote:
>
> selloutdesigns wrote:
> > I just posted a PDF of the Poly-800 Mk1 service manual in the files
> > area. I found this
> > scan at http://www.paintingwithsound.co.nz/manuals.htm and got
> > permission to post it here. Enjoy!
> 
> Great, thanks for this.  Now we can see what the deal is with the 
> joystick board ;-)
> 
> Look at page 6 of the PDF, in the bottom left of the circuit diagram. 
> There's the PCB for the joystick, drawn in a separate box.
> 
> Okay so we've got a quad opamp NJM2056D, which is probably something 
> like a TL074 or similar - not critical.
> 
> R6 and R7 provide around 2V (2.024V assuming the resistors and 5V
supply 
> are bang on) which is buffered by IC1d (following the convention on 
> datasheets and pin numbers - the actual chip section isn't written on 
> the diagram).  Note that the 2V supply goes to the non-inverting input 
> and the inverting input is tied to the output, forming a unity-gain 
> buffer.  This means that no matter how you waggle that stick, you won't 
> affect the voltage being supplied.
> 
> The centre taps of the joystick are held at 2V to provide the centre 
> "dead band" - perhaps one of the reasons for a jittery controller is 
> dirt on the track, or a noisy signal?
> 
> The MG joystick has a 2k2 resistor to +5V at one end, with the other 
> grounded.  By waving a bit of Ohm's law at it, we find that a 2k2 
> resistor in series with a 10k resistor (the total resistance of the 
> joystick track), we get 4V approximately at the top of the pot.  That 
> gives us a swing of 0V to 4V, with a dead band around 2V in the middle.
> 
> Now, what about the pitch bender?  That's a bit more complicated.  Not 
> much, but a bit.
> 
> The pitch bend pot supplies a signal going from 2V to 4V as the bend 
> range is increased, because one end is wired to the 2V supply and the 
> other is wired through a resistor to 5v.  IC2b, R3 and R4 form an 
> inverter (C2 and C3 just stop it oscillating), which has an output
going 
> from 2V to 0V as the input increases - the output is relative to the 
> non-inverting input which is wired to 2V so it's worked out as 4V - 
> Vcontrol.  This is then fed to another inverter (which is again
relative 
> to the non-inverting input) giving its output as 4V - Vpin7.  The bend 
> range pot is wired across the outputs of the two inverters.  The 
> practical upshot of all this is that as you increase the bend range
from 
> 0 to 10, the voltage from the pot goes from 2V to 4V and the voltage at 
> the ends of the pot "spreads" from 2V at both ends to 0V and 4V. 
> Confused?  Read it again and look at the diagram.  Think of it as the 
> electronic equivalent of opening a pair of scissors...
> 
> HTH
> Gordon
>