Johannes,
This is an extremely comprehensive answer. Many thanks for taking the
time to explain. Let me read carefully, disgest and perform some more
tests on the LFO. I'll let you know what I find!
Best wishes,
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)
> >
>