OTA vice versa

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[gez]

Signal couples via a (op-amp) buffer output's to the Iabc pins via a resistor and...?

Interesting idea! 

[gez]

It's morning here, my brain's not quite working...how do you do the modulation? 


Edit.  Wouldn't the LFO output have to be really small in order not to swamp the audio??

[gez]

Ah, I see we've got the 'ranking' back.  I'm a 'hero member' and you're a 'Sr. Member'....or are the stars Beer Tokens?? 

[stm]

anybody tried an OTA the other way `round?
usually it can`t stand more than 50mV audio,
but loves rail-to-rail CV.

so how about reversing/swapping/interchanging
"carrier" & "modulation"...?
IYKWIM

Initially it sounds like a clever idea.  Let's do some further analisys:

OTA's like a CONTROL CURRENT to command the gain of the input voltage.  This CC is obtained by placing a limiting resistor between the Gain Control Voltage signal and the Iabc pin.  Since Iabc pin voltage is essentially fixed at 1.2V above the negative supply (or GND, depending on your supply voltage), there is a reasonably good voltage-to-current conversion going on.

The problem comes now with the fact that the Iabc control current must be unipolar, so in order to apply an audio signal you have to bias your control voltage to a suitable level, most likely to be Vcc/2, so GCV = Vcc/2 + audio.

Assuming you apply your control voltage to the normal inputs of the OTA, this has to be small, let's say less than 0.6V, which is no problem.  Let's call this "new" control voltage Vc.

Thus, Opamp output will be the control voltage Vc times the (Vcc/2 + audio) signal times a conversion constant K whose value is not relevant at this point.

Output = K*Vc*(Vcc/2 + audio)

Output = K*Vc* Vcc/2   +   K*Vc*audio

As you can see, the first term is a DC level that "dances" with the control voltage Vc, which is an undesired term!
The second term is in fact proportional to the actual audio signal times the control voltage Vc, which is what you want in the first place.

The moral is that while being doable, the voltage variations of the required biasing with respect of the control voltage may limit usefulness of such a circuit arrangement.  Of course this analisys is not conclusive, but gives a clues as to why we have not seen it before, even though OTA's may be older than 20 years.

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[Dolly Parton]

She looks even better if you use the empty glasses as goggles! 

[Paul Perry (Frostwave)]

I can see how my life could have been different, if I wasn't a teetotaller... but, to get back to the OTA thing, when you see an OTA arranged as a full analog multiplier, you can see that there isn't really any difference betweent the two inputs (I mean, switch the inputs over & get the same output).
See the National Semiconductor LM13700 data sheet for an example. There is a bit of trickiness going on regarding ofsetting & cancelling bleedthru, and note that there is an assumption that the input is buffered.

[A.S.P.]

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