Electronic – Understanding Operational Transconductance Amplifiers

amplifierotatransconductance

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.)

Best Answer

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