Korg Poly800/EX800 Users

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