Moog 904A lowpass filter Question

[Terry Michaels]

Message text written by "Dana.Scott"
>I really should look at the schematic before I answer the question, But
I'll
answer from memory at the risk of sounding stupid.  :-)
One of the pins of the 3086, pin 13 I think, is connected to the substrate
of the IC.
For the transistors in the 3086 to work correctly, none of the other pins
of
the same 3086 should ever be more negative than the substrate pin. The
extra
transistor should be connected  to pass current to the substrate pin in an
attempt to keep it as negative as possible if there is a circuit fault that
severely unbalances the ladder.

-Dana Scott<

Hi Dana:

I don't think that explains what ARP did on the 4034 and 4035.  The diode
connected transistors are the ones connected to pins 1, 2 and 3.  Pin 13 is
the emitter of one of the 5 transistors in the CA3046/CA3086, and is the
substrate connection, but that is not the  transistor that is connected as
a diode. Even if it was a matter of making the substrate negative, there
would be no reason to connect up the base and collector also, only the
emitter (substrate connection).

The diode connected transistors each share an emitter with another
transistor in the ladder (pin 3), thereby requiring at least the emitters
to be connected to the ladder circuitry.  So, as Joachim said, maybe there
is a reason based on the circuit board layout for connecting pins 1 and 2
also (the diode connected transistor).  It might have been convenient or
necessary for routing connections around the CA3086.

BTW, does anyone know for sure it is OK to leave the substrate connected
transistor (pins 13,12, and 14) floating rather than hard wiring pin 13 to
v- ?  In that case it would seem to me that diode action with the other
transistors would bring the substrate at least as negative as any other
transistor.

Terry Michaels