[sdiy] OTA's, Iabc, and the magic of V to I

[Scott Bernardi]

What you are seeing when you drive the Iabc pin of the has "resistance" (i.e., a
V/I value), but is not a resistor. It is the input of a current mirror, and what
you see is the V/I ratio of a could of BE junctions of transistors (so if you
used a current mirror you would see the same thing!).  The common way of driving
an OTA Iabc input is with a pnp transistor in the feedback loop of an opamp
(which makes a current source, which doesn't care about the nonlinear resistance
seen at the OTA Iabc pin).
Take an opamp an a pnp transistor. Opamp + input goes to ground, the output goes
to the pnp base. The -input goes to the pnp emitter.  The pnp collector goes to
your OTA Iabc input. Next connect an input resistor Rin to the - input. The
value of current you feed to the OTA will be Vin/Rin.  You can only do positive
input voltages. It is also common to protect the circuit against negative
voltages by adding a diode with the anode (flat part) connected to the - input
and the cathode to the opamp output (so its junction points in the opposite
direction as the pnp's BE junction).
This would be well suited as the voltage control input for a VCA, for example.

Nils Pipenbrinck wrote:

> Hiho list!
>
> I'm on a personal quest to design and build my own sine-waveshaper (and get
> some more expericences with OTA's).
>
> Therefore I fooled around with a 3080 OTA last week. One thing I found out
> was (to my surprise) that the resistance of the amplifier bias input pin
> varies widely and nonlinear with the OTA output current.
>
> I built a simple test-circuit to plot some curves. Just some resistors and
> the OTA itself. Iabc was provided via a 100K resistor from V+. Vdiff was
> setup with a voltage divider powered by a low impendance voltage source, and
> the output was loaded with a 10K resistor (where I measured the voltage drop
> across).
>
> As soon as the output current went up (by changing the differential voltage)
> Iabc went down (in my setup by factor 3 from 300µA to 100µA). This gave a
> very nice atan shaped curves at the output. For my sine-waveshaper that's
> brilliant, initially planed to overdrive the differential input, not to load
> down the current source. I'm very satisfied with the resulting shape, but:
>
> Sometimes you don't want distortion. How to go linear and provide a voltage
> controlled current to this pin? Will a current mirror do the job for me?
>
> I'm curious, because I see the following problem:
>
> Say I have a mirror built around two trannies (T1 and T2), T1 is the
> "master", collector current is generated via a resistor from CV. T2 is the
> mirror which powers the Iabc of the OTA.
>
> When the resistance of T2's collector path changes (because the OTA does so)
> wouldn't this affect the base-emitter resistance of T1, which in turn
> changes the "master" current? If so I would have gained nothing to make the
> current to voltage converter more linear than the resistor I used in the
> first place. (it's a mirror, no current copy machine :)
>
> Gee, I'm on thin ice here, and I'm meditating over this for quite a time
> now. Trannies make me scratch head all the time because any current and
> resistance is somewhat related. The only possible way I can think of is a
> sledgehammer method: Adding an opamp, measuring the voltage drop across a
> resistor in the Iabc current path and do some kind of feedback compensation
> to get rid of the drift.
>
> There must be a simpler way - Could someone please enlighten me?
>
>         Nils 'getting headaches' Pipenbrinck
>
> Btw. I'm on a business trip for 3 days. I'm not sure if it's possible for me
> to follow the list for this time, so if you answer me, and I stay quited
> just wait a day or two.

--
Scott Bernardi
sbernardi at attbi.com