[sdiy] A few questions regarding Chris Macdonald's Dual ADSR Circuit

[Chris Manders]

Hi folks

My questions relate to Chris's Dual ADSR circuit here:

http://www.evenfall.com/diy/adsrschm.html

I built this and it works . . . sort of.

I have a few questions to anyone who knows this
circuit and might be able to assist me, please.

1) First of all, and the main problem I can see with
this circuit (and for my use of the one I built) is
that the ADSR output level will never fall to zero.

This is due (I think) to the fact that the release
current passes through a diode (and then the release
pot, and op-amp). There will be a drop in potential (I
usually guess 0.6V for silicon) across this diode,
causing the timing capacitor to never fully discharge.
Does this sound right? Certainly in the circuit I
built there is a permanent 0.2 to 0.3 volts present on
the output at all times. This makes the circuit
unusuable for my purposes.

I would guess that you could 'fiddle' the circuit to
output zero by making the final op-amp (voltage
follower / buffer) into a differential amp and feed
0.6V (or so) into the inverting input to 'off-set'
this problem. Has anyone done this or knows of another
solution?

2) I built the circuit using LM358 quad low power
op-amps, and apart from the voltage problem detailed
above they work OK.

When I come to change the socketed chips in my circuit
for the correct TL084 types, I find that the circuit
no longer functions properly. The output of the op-amp
after the 'Gate In' (which provides the attack
potential, and timing capacitor charging current) is
permanent ON (15 Volts output), even when the gate
input is zero. This is just plain wrong, and I don't
know why it does this. Shorting the gate to ground
seems to re-trigger the ADSR, and the output of the
op-amp goes zero for a fraction of a second, but due
to this problem the circuit is in a permanent state of
being 'gated'. All this just from changing the Op-Amp
IC.

Any thoughts anyone?

Thanks
Chris


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