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I am trying to understand the advantage of using a voltage-controlled current source (VCCS) to drive the bias current inputs of an LM13700 OTA, compared to a simple application of the control voltage (CV) directly through a resistor connected to the bias input.

As I understand it, if the bias current input of the LM13700 sits two diode drops above the negative supply rail, then the easiest possible thing to do is calculate a current limiting resistor that yields a bias current around the maximum 2mA for the largest expected CV. This seems to be the approach taken by all the example circuits in the LM13700 datasheet and numerous other OTA design resources, with the assumption that the CV applied to the resistor comes from a low impedance source.

My understanding is that if we want to control the bias current exponentially, then it makes perfect sense to use a linear CV an exponential VCCS, as the control current would be less susceptible to noise than trying to generate exponential voltages directly. But what is the advantage that justifies the added complexity in the case of a linear VCCS like the one proposed in Musical Applications of Microprocessors p203 for biasing the 3080 OTA

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The only one I can think of suits the need to drive multiple bias current inputs from the same CV, where we can connect more transistors in parallel and source the same amount of current from each, without dividing the current between multiple current limiting resistors (though I suppose we could also just lower the current limiting resistor values). Are there other advantages?

Moreover, I've seen another approach I don't quite understand in some Moog schematics, where the CVs are applied through a resistor following an op amp voltage buffer connected to a BJT common base current buffer. Why exactly is the current buffer necessary/desirable here?

Moog VCCS

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  • \$\begingroup\$ Isn't it all about making sure the VCO has a 1 volt (for example) per decade frequency sweep capability. This 1 volt delivers (say) 100 Hz, 2 volts delivers 200 Hz, 3 volts delivers 400 Hz, 4 volts delivers 800 Hz. If you didn't have this then one voltage that controls two VCOs set a musical fifth apart would sound rank and discordant as soon as you started raising the control voltage. \$\endgroup\$
    – Andy aka
    Commented Jul 19, 2018 at 20:58
  • \$\begingroup\$ Thanks for the response @Andy. The voltage to frequency relationship you're describing is the 1V per Octave standard, which would be obtained with an exponential VCCS. My question is about alternative methods for linear gain control, for controlling things like linear VCAs and multipliers. \$\endgroup\$ Commented Jul 19, 2018 at 21:09

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I believe the ideal OTA circuit has high open loop gain, high slew rate and and high bandwidth with low power consumption, high SNR and dynamic range and superior linearity. But this leads to tradeoffs for bias current, power , linearity and speed.

Thus current mirroring, output buffering is desireable and often used in "Super-Class AB" OTA's outputs with Flipped Voltage Followers.

ref. 60 dB range OTA www.isn.ucsd.edu/pubs/iscas04_ota.pdf

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The ABC pin acts as two diode drops up from VEE, so it has a significant tempco, and presents a nonlinearity that a plain resistor otherwise wouldn't have. Consider the impact of this upon precision circuits -- usually those used for tone control in vintage synthesizers. Without control of these errors, it would not be possible to be pitch-perfect over the full e.g. 0-10V CV range, nor as ambient and internal temperature drift throughout the day, or with the number of modules plugged in, etc.

The first figure addresses both by making the current independent of the pin voltage, using a VCCS. The second does the same, but with an exponentiator to give linear pitch rather than linear frequency. (Or maybe just to cover a wide range, if "rate" means, perhaps, slopes in an envelope generator.)

Note that the latter figure's function can also be solved by using a differential amplifier to shift the CV down, referenced to VEE instead of GND, and applying that voltage (or rather a fraction thereof) directly to the ABC pin, again with suitable current limiting. It might still be better to use a discrete transistor (or compensated (matched or monolithic) pair) as in the figure, depending on how nonideal the ABC pin is (internal resistance, leakage, other effects), and give or take resistances in common between the two amplifier sections within the IC (namely Iabc1, Iabc2, and one reflected current each) which might introduce error or coupling between them.

Using current drive simply avoids all these concerns, at design time, without having to test anything, or leave it up to future uncertainty in case the manufacturer decides to revise the chip and change unspecified characteristics that you'd be depending on. An op-amp is cheap insurance against that. No, it's not a very powerful function to use an op-amp for, but we can't use fractional op-amps just because one function isn't as fancy as another. They're cheap enough, anyway -- or, perhaps given the era, and the sheer number of them required in a full unit -- the overall product budget large enough to handle needing a few extras.

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