Moog 921A

[terry michaels]

Message text written by Turner Wallace
>I have puzzled over the 901A schematics for quite a while now.

Is there anyone who has a full understanding of the circuit?

Particular points of interest are:
*       The high frequency compensation
*       The optional transistor (Q10 ?) 
<

Hi Turner:

Like all sawtooth oscillators, this VCO ends up slightly lower than the
correct frequency when operated at the higher frequency end of its range. 
This is due to the finite amount of time needed to discharge the timing
capacitor and reset the sawtooth waveform.  This reset time becomes a
larger fraction of the waveform period as you go up in frequency, causing
the VCO to run slightly flat.  Also, the diodes in the exponential
converter part of the VCO have a small amount of bulk resistance, which
lowers the output current slightly at the higher end of the frequency
range.  The high frequency compensation uses the Moog-Hemsath technique to
correct for the VCO running flat at the higher part of its range, it does
this by sampling the current being delivered to the diode string in the
exponential converter (and the sawtooth oscillator core), and feeding part
of it back as a correction voltage into the front end of the exponential
converter.  If you have access to back issues fo Electronotes, the
technique is explained in detail in issue 87A, also it has been discussed
here in the past, check the synth-DIY archives.

Q10 is the output stage of a operational amplifier consisting of Q6, Q7,
Q8, and Q9, it feeds current back into the diode string to keep the top end
of the string at the same voltage as the reference end of the diode string.
 A transistor in the 901B matched to Q10 generates the exact same amount of
current for the sawtooth oscillator core.

Terry Michaels