Great, thanks! I've printed it all out. Give me a few days to order
parts and a spare A150, and I'll give it a try. I'll let you know
how it works out also.

Joe

--- In Doepfer_a100@y..., "stinchcombe_t" <tstinchcombe@q...> wrote:
> 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.]