Doepfer

> > Roland Mayer informed me about a simple solution for
> > the described A-117
> > problem.
> > 
> > He found the reason for the problem described: the
> > A-117 clock input is DC
> > coupled (simply a resistor that is connected to the
> > base of a standard npn
> > transistor). This leads to problems if AC coupled
> > signals are used as the
> > capacitors in the AC coupling circuit cause charging
> > the transistor base.
> > The combination A-111/A-117 is such an example as
> > the waveform outputs of
> > the A-111 are AC coupled. One solution is to bypass
> > the capacitors in the
> > A-111 output section (i.e. short circuit the
> > capacitors). But the more
> > general an more simple solution is to add a 100k
> > resistor that connects the
> > A-117 clock input to GND (e.g. simply across the
> > jack socket). We will
> > modify all A-117 in the future and add this resistor
> > to all A-117 clock
> > inputs.
> > 
> > We will add this item to the FAQ and DIY section in
> > the near future.

Dieter,
Thanks for this - yes, a much simpler solution: indeed I tried it 
out, and for my A-111 it does seem to solve the problem (using the 
square output only). However with my A-110's it is still marginal (at 
many frequencies you can hear the 'noise' breaking into a clearly 
periodic tone). Anyhow I have added a comment to my instructions 
pointing this out to people, so that they know a solution may be 
possible with an easier fix.

(I shall have to read up on the mechanism as to why this helps: with 
the resistor, SPICE simulation shows a significant improvement on the 
rise times at the collector, but little difference on the fall times, 
which ties in with my current understanding that base-collector 
capacitance has a greater effect as the transistor comes out of 
saturation. Since the 4006 is clocked on the high to low transition, 
I thus don't see why it should help at all!)

Regards,
Tim