[sdiy] Mysterious transistor on one of the ASM-1 VCO versions

[Mike]

From: "Aaron Lanterman" <lanterma at ece.gatech.edu>
> There are two versions of the ASM-1:
>
> 1996: http://home.swipnet.se/cfmd/synths/friends/stopp/asm1vco.new.pdf
>
> 2003: http://home.swipnet.se/cfmd/synths/friends/stopp/asm1vco-1.1.pdf
>
> The main difference between the two is an NPN transistor at the output
> of the LM311, which appears in the 2003 version but not the 1996
> version.

The Elby ASM2 version has gone back to no transistor:

http://www.elby-designs.com/asm-2/vco/vco1-cco-asm2-cct.pdf

My personal view is that the transistor has been drawn in to emphasize that
the LM311 has an open collector output, and is actually meant to be part of
the IC. Note that it has no component identifier or part number. Taking a
look at the Nat Semi datasheet: http://www.national.com/ds/LM/LM311.pdf
(time for a new acronym: RTFD), shows the LM311 to have an open collector
output - so an external NPN wired as show could not actually work - the base
of the NPN would be connected to -15 Volts or  left open circuit depending
upon the state of the comparator.

The drawings don't show the connections of all the pins of the LM311 - take
a look at the Elby drawing to see what I mean.

One personal peeve about a lot of CAD circuit diagrams is the way they tend
to favour layout space over clarity. Having the Comparator in line and to
the right of the integrator and a bit of tidying would show the function and
signal flow much better.

With reagard to the stuck oscillator take note of the bit on th ASM 1
homepage about the drain/source connections of the FET.

Hope this helps

Mike




>> But this 2003 version with the PNP bewilders me. Alas, that's the one
> I happened to send the link to Greg to build, and the PNP has been
> vexing us today.
>
> He built the oscillator, but it appears the gate is stuck at around 0
> V, so the JFET is conducting, and as we change the CV, we get a DC
> change at the triangle output, but no actual oscillation. The level at
> the + input of the comparitor is less than the 5 V threshold, so the
> output of the comparator is at -15 V; my understanding of BJTs is that
> the transistor at the output is then "off", so no current is flowing
> through the collector, and the gate is more or less hooked to ground.
> And then I'm confused what would happen if the LM311 turns "on" - as
> an open collector output, it should disconnect the base, and then I'm
> not sure what the BJT would do - would it set the gate to the -15
> rail? And if so... isn't that BJT doing the opposite of what we want?
> Like it's almost acting as a logic inverter on the LM311 output.
>
> I get more confused reading the ASM-1 webpage:
>
> "When output of the CA3140 goes above Vref, the comparator LM311 will
> pull its output transistor collector (pin 7) to -15 V since its
> emitter (pin 1) is hooked to
> -15V. This will bias the JFET (Q3) into conducting range and reset the
> capacitor C2 through the Rds(ON) of the JFET."
>
> I though setting the gate to -15 V should pinch the JFET "off," not
> put it into conducting range... this is suggested by p. 190 of
> chamberlin, which says: "as ong as the integrator output is less than
> Vref, the comparitor output is negative, which keeps the FET switch
> across the integratind compacitor off, allowing it to charge. As soon
> as the integrator voltage reaches Vref... The comparator output is
> constrained to rise no further than ground but this is enough to fully
> turn on the high-thrshold-voltage FET switch."
>
> So the ASM-1 webpage seems to say that -15 V at the gate does the
> reset, and the Chamberlin book seems to say that 0 V at the gate does
> the reset.
>
> *confused Aaron is confused*