[sdiy] 3280 VCA question

[Jim Patchell]

    The circuit on my page, as you noted, the base does not go directly to
ground.  By biasing a diode, this will eliminate the dead band I mentioned that
the circuit in the P-5 schematic has.  Like I said, maybe for a VCA, the
deadband might be a good thing.

    As far as the math...if you assume that the base-emmiter junction has a zero
volt drop (this is not reality, but close enough), the outut current will be the
input voltage divided by the resistance.  I think I remember (it was only 10
minutes ago I looked, and I already forgot) that the resistor was 3.3K, so, if
the input voltage is 3.3 volts, the current will be 1mA (3.3 / 3300 = 0.001).

    -Jim

Kenneth Martinez wrote:

> Ok...looked at his pages...patchell_ca3280_5.html shows something similar,
> except that the base doesn't go straight to ground.  Still a little unsure
> of this - what's the math for its use - how do I set the resistors on either
> side of the tranny so I'll know that 5V in will produce 1mA out to the OTA's
> Iabc, for example?
>
> The Old Crow wrote:
>
> > On Fri, 30 Mar 2001, Kenneth Martinez wrote:
> >
> > > I was breadboarding a 3280 VCA, copied from the final voice VCA of the
> > > Prophet-5...
> > >
> > > I've got it working, but I don't understand the function of Q410 in
> > > controlling Iabc - what does it do, since its base is connected to
> > > ground?  How does this resistor-transistor-resistor setup differ from
> > > just connecting a single resistor?
> >
> >   That is a linear voltage to current converter.  The OTA is a
> > current-controlled device; to control it with a voltage you want to
> > convert the voltage to a current.  I think Jim Patchell has some pages
> > about using the CA3280.
> >
> > Crow
> >
> > /**/