Doepfer

I have done some experiments using a breadboard and a Zener diode, and
by supplying a lower voltage to the diodes ?protecting? the 4053 chip,
I was able to ensure that for a 0-12V input signal no voltage greater
than the supply reached the chip. This confirmed what I expected, i.e.
that the chip then behaves the same for a 0-12V signal in as it does
for a normal +/-5V (albeit with the 12V chopped down to about 7.5V,
but with no problematic ?bleed through?).

I could suggest a modification to the lower switch (the right half of
the board) which would allow it to work with 0-12V signals as well as
the standard +/-5V (extending to cover both switches wouldn?t be such
a great step). However this will require: 2 new components ? a
resistor and a zener diode; drilling the PCB and inserting 3 pins on
which to mount the new components; cutting 3 tracks (6 for both
switches); adding 5 wires (8 for both). This is quite involved, and
wouldn?t be for the faint hearted!
The price paid in using the Zener rather than another regulator (the
Zener has a better choice of voltage over the regulator) is an
increase in current consumption of the A-150 by about 5mA ? it also
means the mods to the board are probably easier though.

Tim

[The views expressed above are entirely those of the writer and do not
represent the views, policy or understanding of any other person or
official body.]


> > I'm wondering if it's possible to mod only half of the dual switch.
> 
> Almost certainly: I don't have the full PCB diagram with me, but there
> must be a suitable point at which to cut the +8V and -8V tracks
> between both halves of the 150 circuit, and then strap the appropriate
> one (of the now isolated rails) to +12V and the other to ground. You
> would really also need to place new 10uF and 100nF capacitors between
> the new 12V rail and ground.
> 
> BUT if you are going to go to this extent, it's not really such a
> great step to sorting the problem out properly (if I'm right about
> it's cause I must add!!!), and then the whole module would work with
> +/- 8V *and* 0 - 12 V (although the switched gate/trigger would no
> longer be 0 - 12V, but I've seen nothing in any circuit so far to
> suggest that this would be a problem). The problem as I see it is
> because the diodes 'protecting' the chip won't conduct until the
> inputs are around 8.7V: this voltage is then greater than the supply
> to the chip (8V), which is what (I'm assuming) is causing it to
> misbehave. A solution would be to lower the +8V to the diodes to, say,
> +7V: then nothing above 7.7V would reach the chip, which being less
> than 8V would (hopefully!) mean it behaves properly. This could be
> achieved with another regulator of appropriate rating, but would
> require quite a few straps and cuts to tracks because of the way the
> diodes and chips hang off the +8V rail. In fairness, the FAQ solution
> is described as 'simple' - perhaps you could persuade Dieter to post a
> more 'Rolls-Royce' solution (after checking the above 'theory' out of
> course)?
> 
> Tim
> 
> [The views expressed above are entirely those of the writer and do not
> represent the views, policy or understanding of any other person or
> official body.]