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I'm having trouble getting this CA3080 OTA to work properly. :-(

I watched this video ( https://www.youtube.com/watch?v=BbaJ6TOmyVo ) which is quite helpful, but I'm missing a few cues here...

In the enclosed picture you find the circuit as I think he uses, on the right my adaptation (basically changing a potmeter for an input coming from an 0-5v CV)

I built everything on a breadboard. The only difference with my built is I changed the transistor for a BC327 (for now, tommorow I'll receive the 2n3906).

On my scope the sine shows up nice and clean but when lowering the volume with the R6-pot it shifts upwards and distorts the wave (or is that supposed to happen?). In the included video it doesn't seem to do that (but it's only audible).

I did leave out the resistor at the output of the ca3080 cause its directly connected to a tl072. Should I put in a resistor of 100k between the output of ca3080 and the input of the tl072 orso?

I could do with some extra eyes helping me out. Would someone mind checking my circuit?

OTA CA3080 design

user3810130
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1 Answers1

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It appears that a voltage source is providing an input signal to this voltage-controlled-amplifier, and a voltage source is desired as an output signal.

The CA3080 provides an output current at pin 6. A resistive load (to ground) is often used to convert this current into a voltage. This load resistor is missing. The voltage follower connected to this point then provides a near-ideal voltage source as output.
The value of this load resistor affects the overall gain - a large value gives a high gain. The OP's circuit has nearly infinite resistance at this point, giving extremely high gain. An amplifier with infinite gain is more likely a limiter.

R9 (10k) is somewhat redundant. It is a good precaution while experimenting since it helps protect CA3080 from mishaps, by limiting IABC into pin 5 to a reasonable value less than a few milliamps.

Be aware that the control voltage source CV has a control range from about +0.5V up to +5V. A dead span from zero up to about +0.5V provides no amplifier bias current (IABC) into CA3080's pin 5 IABC other than transistor leakage current.

glen_geek
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  • True, forgot about the current output of the OTA, I'll add a 47k resistor connected to ground to it. I'm not really happy with that transistor being there, doesn't that just switch on/off? Could you explain to me what that effectivly does in this circuit? I've seen a OTA circuit somewhere without the transistor but a 27k resistor connected to pin 5 of the OTA as input for the CV. – user3810130 Mar 22 '20 at 18:38
  • The **control current** into pin 5 *can be used as a switch*. For example, as the sampling signal in a **sample&hold**. It can also be used as a gain-control element. Or if you wish, as one input of a two-quadrant multiplier. No current into pin 5 yields no output signal. 500uA into pin 5 yields maximum transconductance (nearly 0.01 Siemens). The problem with pin 5 is that it is referenced to negative DC supply rail. So current generated by a simple resistor has no noise immunity from -ve supply. Your PNP *does provide noise immunity* for this current (it is ground-referenced). – glen_geek Mar 22 '20 at 20:00