Erica Synths EDU DIY EG Eurorack Module Kit - Manual (Page 16)

Why’s that important? To answer this, we’ll first have to talk about the other comparator.
Since it doesn’t have a high-pass at its input, it will simply behave like the comparator in
our previous iteration. Whenever the input voltage is above 3.8 V, we’ll get a constant
12
V at its output. The diode afterwards serves the same purpose as the one up top – it
blocks the comparator’s low state.
After this, I’ve set up another potentiometer as a variable voltage divider. This allows us to
take the 12 V during the comparator’s high phase and freely scale them to any value
between those 12 and 0 V. Whatever voltage we dial in here will be our sustain level.
Why?
Because the 100k resistor at the input bu ff er doesn’t connect straight to
ground like before, but rather to our sustain level voltage . If that sounds confusing,
let’s break it down step by step.
Our input starts out low. This means that both our
comparators’ outputs sit at –12 V. But because of the
two diodes, this doesn’t propagate, and so our bu
ff er’s
input gets pulled down to 0 V through the 100k resistor
and potentiometer. Giving us 0 V at the envelope’s
output as well. Next, let’s assume that the input goes
high. This will do two things: we’ll get a voltage spike
after our high pass, which gets converted into a short
12 V burst by the top comparator. Simultaneously, the
other comparator pushes out a constant 12 V that get
scaled down by our sustain potentiometer.
Let’s assume we’ve set it to about 50 %. This means that at the bu ff er’s input, we’ve got
our 12 V burst coming from the top, and a constant 6 V coming from the bottom
comparator. Since there’s no resistor in the top path, but a 100k in the bottom one, the
burst will „win“ and push the overall voltage at the bu
ff er’s input up to about 12 V.
Our bu ff er – being a bu ff er – will copy those 12 V and push them through the attack
potentiometer. Again assuming that we’ve dialed in a fast attack, there will be a pretty low
resistance in its path, allowing the capacitor to be charged up to about 12 V before the
burst is over. So at this point, our envelope’s output sits somewhere around 12 V – its
peak value. But because the burst is a burst, it’ll quickly die down and the top
comparator’s output will drop to –12 V.
So suddenly, with the top diode blocking, the
only voltage applied to the bu ff er’s input is our sustain level: 6 V . This means that our
bu ff er’s output will drop from around 12 to those 6 V, allowing our capacitor to partially
discharge through the release path. Because remember – the burst charged it up to
around 12 V.
Once the voltage at the capacitor has dropped to the sustain level, it will stabilize, giving
us a constant 6 V at the envelope’s output. Until the input signal swings low, our bu
ff er’s
output drops to 0 V, and the capacitor is allowed to complete its discharging process. The
result is an output curve with 4 distinct phases: attack, decay, sustain and release. Now
as you might have noticed, there are two rather big caveats here.
First, both in the
decay- and the release phase, the capacitor discharges through the same
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