[sdiy] Waveforms (was dx, chorus and Spock)

Terry Michaels 104065.2340 at compuserve.com
Fri Aug 16 20:12:34 CEST 2002


Message text written by Gene Stopp
> ...I wonder if the 921 waveforms
> are just more "ideal"? Or was it some kind of "warping"? Or was it that
in
> combination with the 904A? And then the 902's? Or the fact that the
output
> went straight to the PA? I'm sure it was a combination of many things.<

Hi Gene:

The Moog 921 and the Minimoog oscillators both use a similar design, which
has the collector of the expo transistor charging a capacitor, the
capacitor is discharged by a transistor in parallel with the cap.  As the
capacitor charges, and the sawtooth ramp rises, the collector current
decreases slightly, because the collector to base voltage is decreasing. 
This is due to the Early effect.  The result is the sawtooth ramp is not
exactly a straight line, it flattens slightly as it nears the top.  This is
plainly visible on an oscilloscope.  OTOH,  your ASM-1 VCO, and my original
VCO design as published in Electronotes, and probably some others, all  use
an integrator that is reset by a FET.  Because the collector voltage of the
expo transistor in this design doesn't change as the sawtooth ramps up, the
charge rate is constant, and the sawtooth ramp is straight as a ruler.

The Moog 901 uses a true constant current source to charge the cap, so the
ramp from that one should be pretty straight.  I don't have one on hand to
verify that.

I would guess the slightly warped sawtooth will sound a little different
than the perfectly straight sawtooth to a discriminating ear, although I'm
guessing the difference will be subtle.  The flattened sawtooth will
probably have slight differences in the amplitudes of its various harmonics
compared to a "textbook" perfect sawtooth.   It would be interesting to see
a high resolution spectral analysis of all of these well known VCO's of the
past.

Also, the triangular and sine waveforms in these VCO's are derived from the
sawtooth, so a warped sawtooth ramp will cause the triangular waveform to
be slightly warped also, and the sine will have some additional distortion.
  Again, the difference will probably be audible, but subtle.

Terry Michaels



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