Discussion about the Korg PolySix synthesizer

Johannes,

After a major fault finding session I would like to report back the 
following:

The first thing I did was be rather lazy and swap the transconductanc 
op-amp IC21 to see if it made any difference; no! I then checked the 
FET Q1 s-g, g-d and g-d. Similarly I checked the matched pair of bi-
polars at Q3 c-b, b-e and c-e. No problems. I then very carefully 
checked all the resistors in circuit, including VR5, which all 
checked OK. I tested C19 which 0.339uF. Not bad for an electrolytic 
which is over 25 years old! The MG was still ocillating flat out at 
3.4hz even with VR5 at it uppermost setting. I then checked the dc 
voltage present at pin 7 of IC13. Changing the MG frequency control 
from 0 (+ 4.98V) to 10 (-0.17v). But wait? Can this be right? I 
checked pin 12 of IC19 which is the mux output X3; same voltage 
reading. Working back to IC2 on the KLM-370 board I checked the mux 
input X2 at pin 15. The voltage swings accurately around the 
frequency pot wiper from +5 to -5v. However, taking a reading at the 
muxed data output on pin 3 of IC2 the voltage will only swing as 
mentioned before from +4.98 to -0.17v. Now forgive me if I am wrong 
but I understood the HA14051 worked like this. You appliy a voltage 
to one of the 8 mux inputs marked X0-7. The voltage change is 
detected and a bitmap is set on data address lines A, B and C 
according to which pin was strobed. At the other end this bitmap is 
decoded the data read on the receiving mux and the appropriate signal 
(in this case a voltage) will appear at the correct pin as determind 
by the bitmap. The fact that the voltage doesn't swing correctly 
between the two poles of the pot would suggest a fault with the 
transmitting HA14051? Am I right in assuming this? If the address 
lines were cut/missing then some of the muxed output would be never 
selected and hence always at 0 volts. This might also account for the 
other problem in that the ADS bit of the EG works (sort of!) but the 
release pot does nothing! (Again the voltage swings about the same as 
the MG frequency control)

Your expert opinion on this matter is greatly anticipated!

Best regards,
Andy

--- In PolySix@yahoogroups.com, Johannes Hausensteiner <johau@...> 
wrote:
>
> The oscillator design is of the multivibrator type. The not so 
common
> thing is that it is voltage controlled.
> The control voltage is buffered by IC13. it is converted to current
> by R16, R69, TH1 combination (temperature compensated by use of 
TH1).
> This current is mirrored by dual-transistor (for good match) Q3. The
> mirrored current is fed into control input of transconductance 
amplifier
> 1/2 of IC21. Download the data sheet of this IC to find out how it
> works. LM13700, NE5517, CA3080 are similar devices. It is an 
amplifier
> with voltage input (high impedance) and current output (high 
impedance),
> hence the name transconductance (current divided by voltage is 
called
> conductance, the reciproke of resistance). So you input a voltage 
and
> get a current on the output. That is the reason why timing capacitor
> C19 does not have any resistor associated with. The transconductance
> (= similar to gain of opamps) is controlled by the current which is
> drawn out of the control pin (pin16).
> One half of IC16 is the other vital part of the multivibrator 
circuit,
> it acts as the schmitt trigger (positive feedback).
> How does it work?
> Lets assume any voltage level on IC16/pin6 after power on. It will 
be
> more or less than the voltage on IC16/pin5. Depending whether it is
> more or less the output of IC16 will be positive or negative 
saturated,
> which is about +Vcc -1.5V or -Vcc +1.5V (+/-13.5V). This voltage is
> divided by resistor divider R53 - R52 and fed into the 
transconductance
> amp. Depending on the control current the output of IC21 delivers a
> proportional current. This current charges the timing capacitor C19
> developing a linear ramp voltage. The high impedance voltage is 
buffered
> by JFET Q1 and fed back to IC16 input. When it reaches the same 
level
> as voltage on IC16/pin5 the output switches to other saturation 
level
> and C19 is being discharged until IC16/pin6 trips pin5 and the 
output
> reverses polarity again (multivibrator principle). Since the output
> current of IC21 is controlled by the control current, which in turn 
is
> determined by the "MP SPEED" control voltage, the oscillation 
frequency
> is controlled (larger current charges/discharges the timing 
capacitor
> faster). Is it clear?
> What can go wrong? - Anything 8-/.
> I assume you did the adjustment procedure for MG speed (VR5). The
> service manual says, "If these values (50Hz) cannot be obtained 
after
> replacing IC LM13600, adjust R78 and R52 respectively."
> First check the control voltage (IC13/pin7). It should reach -5 .. 
+5V.
> Next check the output swing of IC16 (pin7) it should be at least +/-
13V.
> Next compare voltages before and after the buffer Q1; should be 
(nearly)
> equal.
> More measurements are not so easy because it is all current driven; 
you
> cannot simply measure voltages. Best thing to try is replacing 
components
> (C19, IC21, Q3, maybe TH1). If you dont find a dual transistor you 
can
> use two discretes of the same type (would be good of the same lot).
> 
> Good luck!
> 
> Johannes
> 
> 
> Andrew Jury wrote:
> > Hi!
> > 
> > I have just released with my MG (LFO) seems to be running very 
slowly 
> > with the frequency knob at 10. Before I dip in with my soldering 
iron 
> > does anyone know how the MG oscillator works? I have studied the 
> > schematic and there is nothing in the design which looks remotely 
like 
> > one! I believe the MG should run at 50Hz flat out. My 'scope says 
just 
> > over 3Hz when I measure at the VCA MOD test point. There is an RC 
> > network connected to the gate of Q1 on the KLM-367 which possibly 
> > triggers a voltage at the drain over a time period. The RC nework 
> > values (R27, C19) 10K, 0.33uF equate to a frequency of just over 
48Hz, 
> > which looks promising. Can anyone shed any light on how this bit 
of the 
> > P6 works?
> > 
> > Thanks,
> > (A)
> >
>