Doepfer

So, here is my modification for the A-150 to get rid of the
bleed-through problem experienced when trying to switch 0-12V
gate/trigger type signals. The mod will allow both normal +/-5V
audio-type signals and also 0-12V gate-type signals to be switched
properly, but note the 0-12V signals connected through the switch will
be reduced to about 0-7.5V. I don't think this will unduly load the
source of the gate/trigger signal, and judging from most of the
circuits I've looked at, it is unlikely to prevent the gate/trigger
signal from doing what it should at its intended destination. (The mod
doesn't alter the negative supply, therefore the bleed-through problem
is likely to remain should you try and switch signals less than -8V.)
Note also that it will increase the power consumption of the module by
about 5mA. 

From these instructions it looks like a lot of work, but it actually
took a lot less time to do the mod than it did to write this lot out!!

It actually turned out easier to modify the top switch (the left
circuit on the board) and not the bottom one as I thought in a
previous post. Extending to modify both switches just means cutting 3
more tracks and adding 3 more wires, i.e. it does not double the work
needed. The basic idea is quite simple: put a resistor and zener diode
across the 8V supply to give a new, lower voltage of 6.8V; this is
then applied to the diodes D1, D4 & D5 'protecting' the chip, which
will then stop voltages greater than the supply voltage of 8V reaching
the chip when large signals (such as gate/triggers) are applied to the
switch (see FAQ page on Doepfer site to get a copy of the schematic).

New components needed (I'm lucky in that I live near to a Maplin's
store, so in brackets I've given their order codes for the parts,
which may be of use to others in the UK, and also possibly those
outside since you can check these parts on their website,
www.maplin.co.uk. They also do a good mail order service.):
 - a 240 ohm resistor (nominally 0.25 watt, Maplin ones are 0.6W) (M240R)
 - a 6.8V zener diode/voltage regulator diode BZX55C 6V8 (QH10L)
 - 3 PCB pins - Maplin code JW59P pin strip is the single version of
the paired pin strips used by Doepfer for the power connector, and can
be cut as required (Maplin do other pins, but these are sold in bags
of 100 (!), e.g. FL20W, FL23A)
 - small amount of thin, insulated copper wire (BL09K)

Identify the following on the board (hold so that 'DOEPFER A-150' text
on the component side is the right way up):
 1. +8V supply rail - this is the main track running along the top
edge of the board: its marked with '+8V' on the solder side where the
capacitor C1 is soldered to it; on top you'll see '+8V' at left-most
pin of IC2/78L08 - underneath you can follow this along and see it's
the same track.
 2. 10K resistor R2 for the left switch is the left-most component on
the board, colour coded: brown, black, orange, gold (note the Doepfer
schematic erroneously shows this as connected to +12V - it's not, it's
connected to the +8V rail).
 3. Diodes D1, D4 & D5 for the left switch: D4 and D5 are immediately
on either side of the left 4053 chip, the black stripes on the diodes
are at the ends connected to the +8V rail; D1 is the left-most of the
4 diodes to the left of the chip, again the black stripe end is
connected to +8V.
 4. Pins on connector (to power supply) which are grounded. From the
right hand end, working towards the left: first pair of pins is -12V;
next 3 pairs are ground pins we want (connected together by a track on
the PCB); next is +12V; last 3 pairs are not connected to anything
(see connector on Doepfer schematic to clarify if any doubt).

Drill 3 holes in a straight line, suitable for insertion of the pins,
about 3 to 4mm in from the top edge and about 15mm apart (but not too
close to the +8V rail on underside). (With the 0.6mm bit that comes
with Maplin's drill NF96H this was a simple task.) Call the pins
'left', 'middle' and 'right': I placed the left one through the 'R' of
'DOEPFER'; middle one between the '0' of '150' and the rectangle
around 'A-100 modular...'; and the right one between the rectangle and
'C2 100n'.

Before inserting the pins, cut the tracks (scrape away the copper
using a sharp pointed object, like a penknife: hold the board up to
the light to check that the track is actually cut):
 1. cut the track where diode D1 connects to the +8V rail
 2. cut the track where D5 connects to the +8V rail
 3. the +8V rail connects to the end pin of the 4053 chip, and then
connects to the next pin to it on the one side, and to diode D4 on the
other: cut the track between the diode and the end pin of the chip (be
careful not to damage the very narrow track close by!)

Insert the pins so that the longer end projects on the solder side of
the board (using the JW59P pins, I put the little plastic lug on the
component side). Solder wires on the underside of the board, from the
projecting end of the appropriate component to the appropriate pin:
 1. solder a piece of wire from each end of diodes D1, D4 & D5 (the
end next to the cut = same end with black stripe on diode body) to the
bottom of the middle pin (be careful not to get the diodes too hot, it
could destroy them). Attach all 3 wires to the middle pin, and then
solder in one go (or all 6 in one go if doing both switches). Also be
careful not to solder from D4 on to the nearby narrow track.
 2. solder a wire from the end of R2 where it joins the +8V rail to
the bottom of the left pin.
 3. solder  a wire from one of the ground pins on the power connector
(one of the middle pair probably easiest) to the bottom of the right pin.

On top of the board, solder the 240 ohm resistor between the left pin
and the middle pin. Solder the zener diode between the middle pin and
the right pin, with the black stripe nearest the middle pin (again,
probably best to solder the resistor and zener to the middle pin in
one go, and again take care not to overheat the zener!).

That's it for modifying the left switch. To do the right switch as
well, cut the tracks where the diodes D1, D4 & D5 of the right circuit
connect to the 8V rail, and attach wires from the diodes to the middle
pin.

To summarize, you should have the following: +8V supply from R2 to 240
ohm resistor at left pin; 6.8V at middle pin (junction of resistor and
black stripe end of zener) to black stripe ends of diodes D1, D4 & D5,
which are disconnected from the +8V rail; non-stripe end of zener at
right pin to ground at power connector.

Check: all cuts really are cut; no tracks unintentionally cut; no
stray blobs of solder bridging any tracks.

To test I suggest something like the following (if modifying one
switch, you can check against the unmodified one; if doing both, a
'before' and 'after' test would probably be a good idea!). Patch a
suitable CV source to the switch CV (e.g. A-129/3, A-174, A-176); an
LFO square output to A-160 'trig in'; 160 'divide by 2' output (so
this is now 0-12V) to I/O2 on A-150; O/I to CV2 on A-110, listen to an
output of the VCO. With  I/O1 unconnected, regardless of the switch
position, on an unmodified switch you should be able to hear the
alternating pitch due to the 0-12V pulses (cut down to about 8.5V)
always coming through the switch (adjust LFO frequency to a suitable
slow rate). If you ground I/O1 i.e. attach a CV source at 0V, the
affect is much less when the switch is 'open', but you can still hear
some variation in pitch caused by the bleed-through. After the mod,
you should only hear the pitch change when the switch is 'closed',
i.e. when the clock signal is switched through; in the 'open' position
the pitch should now be steady.

If anyone knows if a gate/trigger signal of only 0-7.5V *is* likely to
cause problems (especially Dieter if he is listening in!), then please
put me straight ASAP! I also suppose I should add some kind of
disclaimer: if you do this mod, you'll certainly invalidate any
warranty on the module, and I accept no responsibility if you blow up
your module/your A-100/your pet budgie/yourself/your house etc. etc.
(!!!).

That being said, if any other clarification is needed, let me know,
and finally, good luck!

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.]