7
\$\begingroup\$

I am teaching students about the differential pair (see circuit below). I addressed many issues like the difference between differential and common mode gain, taking arbitrary vb1 and vb2 signals and expressing them with common mode + differential components, and finally using current mirrors for single-ended conversion. BJT differential amplifier with passive load. One thing I have not addressed is that the differential pair is easy to saturate. Very seldom does the differential pair operate as a linear amplifier, but rather the outputs usually are pushed to their limit (to their rails). In fact, the range of differential inputs that produce linear outputs is very small (perhaps +/- two thermal voltages as seen below). What I would like to know from you is if there are any applications of the differential pair that depend on using the linear region only? Telecom antenna preamps? Wheatstone bridges?enter image description here Thanks for your help - (this is my first stackexchange post)

\$\endgroup\$
8
  • 2
    \$\begingroup\$ I am not an IC designer, but if you replace the two resistors labelled Rc with a current mirror, I believe you will get much higher gain and extend the linear region, although it will cost you some headroom. Well worth doing much of the time. \$\endgroup\$
    – user57037
    Commented Oct 12, 2015 at 22:41
  • 2
    \$\begingroup\$ Too lazy to write a full answer, but almost any opamp with BJT input stage. Also the (hardly used nowadays) ECL. \$\endgroup\$ Commented Oct 12, 2015 at 22:42
  • \$\begingroup\$ I probably misunderstood what you meant by "linear region" in my comment. But I will leave it there anyway. \$\endgroup\$
    – user57037
    Commented Oct 12, 2015 at 22:51
  • \$\begingroup\$ Also, I don't know if this really counts as use, but the differential pair are the only transistors in the Boyle macromodel of the opamp typically used in SPICE simulations (for older opams); the rest are ideal sources, etc. The newer opapms have more sophisticated SPICE models, but the number of transistors is still usually restricted to just these two. \$\endgroup\$ Commented Oct 12, 2015 at 23:16
  • \$\begingroup\$ An alternate formulation of your question might be "Are there any applications for a differential pair running "open loop" as an amplifier (not a switch)". You'll be the judge on whether that expresses your question more clearly, but that's the interpretation I took from it :) \$\endgroup\$
    – scanny
    Commented Oct 12, 2015 at 23:18

3 Answers 3

6
\$\begingroup\$

A similar circuit is on the input of operational amplifiers. Yes-- it saturates quite easily, which is why it is usually configured in a negative feedback arrangement.

In fact, Analog Devices uses almost exactly this circuit in a document describing how to protect input stages on opamps. Real op amps are much more complicated, but its a good first order approximation.

enter image description here from http://www.analog.com/library/analogDialogue/archives/41-05/input_protection.html

Look familiar??

Notice, the output of the circuit is fed back to negative input, introducing negative feedback and keeping the overall gain in control.

\$\endgroup\$
4
  • \$\begingroup\$ Yes, of course OpAmps are THE typical application of a differential pair. They don't have to be linear at all. It's enough if the output is monotonous to the input (difference voltage). They may saturate easily. That's no problem because (as Olin mentioned) most of the time they are operated in a way that negative feedback avoids saturation (open loop amplification is not primarily relevant). \$\endgroup\$
    – Curd
    Commented Oct 13, 2015 at 1:23
  • 2
    \$\begingroup\$ @Curd, you mean monotonic. \$\endgroup\$ Commented Oct 13, 2015 at 1:35
  • 1
    \$\begingroup\$ If the output is monotonous I'd look for a more interesting input. \$\endgroup\$ Commented Oct 13, 2015 at 12:30
  • \$\begingroup\$ @Scott: yes, "monotonic". English is not my first language and in my first language we use the same adjective for both meanings :-). \$\endgroup\$
    – Curd
    Commented Oct 14, 2015 at 7:58
5
\$\begingroup\$

You'll see this circuit used in voltage controlled amplifiers. Here an audio signal is feed into one of the inputs and the current through the emitters controls the gain of the amplifier.

A google search for "discrete synth vca" will give you several example circuits. Here is a simple one:

enter image description here

\$\endgroup\$
4
\$\begingroup\$

You are right in that this circuit is not very linear. It's main feature is that it has low offset voltage between the two input. It is therefore commonly used where the gain will be large in the circuit itself, but there will be negative feedback around this circuit at a higher level. That means the differential pair is only operated at a few mV difference either way, anything more than what causes the open loop output to saturate is irrelevant.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.