I understand the high-level function of OTAs, that they provide an output current proportional to an input voltage. But what I don't understand, and what I can't seem to find any information on online, is exactly how they're used and what the function of the different connections is.

wikipedia's image of the schematic symbol of an OTA

The most pressing questions I have are thus:

  1. What is the function of \$I_{bias}\$? Why are there diodes between this pin and the two inputs? I've seen them called linearizing diodes; how do they linearize the response of the circuit?
  2. What is the function of \$I_{abc}\$? As I understand it, it provides a scaling factor; is this correct? How does it do so?
  3. What is the (exact or approximate) formula for \$I_{out}\$ as a function of \$V_{in+}\$, \$V_{in-}\$, \$I_{bias}\$, and \$I_{abc}\$?
  4. What are the standard application topologies for an OTA? Op amps have the standard inverting and non-inverting amplifier topologies, and slightly more complicated summing amplifier and difference amplifier, integrator and differentiator, but are there any common circuit idioms for the OTA?
  5. Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)
  • 1
    \$\begingroup\$ Some of your questions are answered in the LM13700 datasheet \$\endgroup\$ Commented Oct 15, 2018 at 13:12
  • \$\begingroup\$ @ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again. \$\endgroup\$
    – Hearth
    Commented Oct 15, 2018 at 13:15
  • \$\begingroup\$ Have you found ti.com/lit/an/sboa117a/sboa117a.pdf? \$\endgroup\$ Commented Oct 15, 2018 at 13:21
  • \$\begingroup\$ @ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do. \$\endgroup\$
    – Hearth
    Commented Oct 15, 2018 at 14:26

3 Answers 3


The most important point to note is that the input diodes and \$I_{bias}\$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.

A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.

So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal \$V_{be}\$ for \$V_{diff}\$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.

The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.

If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of \$I_{abc}\$ and \$I_{bias}\$.

You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf


Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf

Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.

For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.

Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)

Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.


The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).

Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.

There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.

Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).

There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.

A very good introductory text (Applications) can be found here:



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