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<p>Where is the 100 nF decoupling capactiros over the power-lines of
each chip?</p>
<p>Cheers,<br>
Magnus<br>
</p>
<div class="moz-cite-prefix">On 2021-09-21 03:18, David G Dixon via
Synth-diy wrote:<br>
</div>
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<div><span class="684555200-21092021">Just for Ss and Gs, I
thought I'd show you my layout graphic for the ASR board which
has the clock circuit on it. Please see attached. Note
that red traces are the +V rail and green traces are the GND
rail (and black traces are the -V rail). The CLOCK pin of the
4024 is connected to the light blue trace. I've also attached
what I call the "Build" picture, which is what the board
actually looks like. Finally, I've attached a picture of the
actual board. Note that there is another circuit board that
this one plugs into, which has a DG333 and comparators for
controlling the "Shift/Bypass" switches on each ASR channel
electronically. That's what the 3- and 4- pin headers plug
into. The 4-pin header passes through that board and connects
to the Panel PCB. This is where the clock signal comes in and
out of this board.</span></div>
<div><span class="684555200-21092021"></span> </div>
<div><span class="684555200-21092021">The rail traces are a bit
thicker than the other traces. You will notice that the
ground traces do not connect to each other across the top of
the board, so that the ground at the 4024 actually comes up
from the bottom. I just tried jumpering the two ground traces
temporarily with alligator clips and it had no effect (and why
should it?).</span></div>
<div><span class="684555200-21092021"></span> </div>
<div><span class="684555200-21092021">Also, I should mention that
when the ASR is being clocked by an external LFO, the CD4024BE
is clocking on the positive edge only. However, when the
Manual clocking switch is used, the CD4024BE is clocking on
both edges. The clock signal at the CLOCK pin looks identical
in either case. The manual clocking switch is connected to
the +V and GND rails, and is fed to the comparator through a
1k resistor, whereas the clock input jack is connected through
a 10k resistor. Why any of this should have an effect is
beyond me. This, for me, is now more or less in the realm of
the supernatural, in terms of my ability to understand it.</span></div>
<br>
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