Glide circuit of the ProphetV

[jhaible]

> > And, when you use the 13700, you don't need the buffer, because you can
use
> > the one on the chip right?
>
> No... the darlington buffer can only "pull up" and has a passive
pull-down. so
> it can source lots of current, but sink is fixed... this would cause
asymmetry
> in the glide.

No. Not if you add a small enough resistor from the buffer's output to the
negative supply. An emitter follower has low output impedance regardless of
polarity,
as long as you don't steal *all* the current from the transistor.

> The diodes clamp this and the resistor limits the voltage to the clamp. as
a
> side effect this makes the ramp linear for large steps (input differential
is
> clamped at .7 volts) and sort of exponential for small steps. In practice
you
> get a linear glide.

The diodes are just for protection. They don't have anything to do with
linear
or expo glide. (when they start to conduct, the OTA's input stage is long
saturated.)

> If you remove the diodes and resistor and ground the negative input you
will get
> exponential glide

No. That's the case if you remove the diodes and use small resistors to GND
at the OTA inputs.

(and problems of offset voltage which you must trim out for DC
> accuracy)

Yes. That's why expo glide with OTAs is a bad idea.

> The feedback assures DC accuracy.

Yes. What you see at the output is the offset of the input, i.e. small.
In the expo case, the offset voltage is amplified according to the chosen
amount of negative feedback. Small feedback -> high offset,
strong feedback -> glide becomes more and more linear for larger
intervalls.

JH.