Using a "black box" model of an op amp, it behaves as a Class B amplifier because it provides a "push/pull" output in response to a + or - signal at the input.

However instead of a Class B configuration, the first stage of an op amp is the actually the familiar "differential configuration" with two common emitter amps connected in a "symmetric pair". The input to one of amps is grounded so the diff amp responds to the only to the difference in inputs.

However a Class B amp also responds to the difference between the input to the "complementary pair" and the ground line.

So why can't a Class B configuration be used instead of the differential? It seems more logical with the intent of a complete op amp to make its internal circuits function more like the black box model.

 

Why do you care what is in the black box?

By definition: the voltage out of an opamp is K(Vplus-Vminus), where K is a very big number like 200,000.

 

The classic class B output stage is a low-rent version of a complementary symmetry emitter follower. There are many other output stage configurations, including a single emitter follower with a constant current source load that is very similar to one type of input stage. Except for variations in crossover distortion, rail-to-rail performance, etc., all output stages are almost identical in a closed loop, non-overdriven, normal type of circuit. So Mike is right - from a black box point of view it doesn't matter what is inside. That is the definition of a black box analysis.

Separate from that, there is no electronic reason why you couldn't use a class B stage as the input of an amplifier. However, your description of a differential stage ("The input to one of amps is grounded") is not true for all, or even most applications, and a class B stage is not something that can be swapped into an input section without a significant reworking of the rest of the amplifier circuitry. And by definition the overall open loop circuit distortion will increase significantly.

ak

 

Why do you care what is in the black box?

By definition: the voltage out of an opamp is K(Vplus-Vminus), where K is a very big number like 200,000.

As an EE engineer wanna bee, it's more than knowing just that a device works, but knowing how and why it works.

In addition to providing "rote" knowledge for obtaining an academic credential, I believe knowing the theory of what's inside the black box improves the quality of the engineering profession.

 

...I believe knowing the theory of what's inside the black box improves the quality of the engineering profession.

But that is not what your post asked. You were taking issue with the way opamps are, and were proposing a different way of making them...

 

I wasn't "taking issue" with the way op amps are designed.

I was merely inquiring as to why they are designed that particular way and I was wondering just a matter of curiosity rather than being critical.

Also, being an EE technician supervisor*** for 20 years, I believe techs should be more than just "parts swappers" and skilled only in "shot gun" troubleshooting techniques. My experience is the techs with efficient troubleshooting ability are also the ones who have a good theoretical knowledge.

*** In order to get the job, I took a written examination with 100 questions about various analog and digital circuit configurations. A score of 70% and above was passing, but only 30% of the applicants did in fact pass.

 

Back to your original questions... Any amplifier topology can provide "a "push/pull" output in response to a + or - signal at the input." If that sentence is the only requirement, a class B circuit has few advantages over other circuits and several disadvantages.

A class B circuit is used for the output stage because it provides no voltage gain but lots of current gain, and the majority of opamp applications benefit from having an output that can deliver more current than whatever is driving the input, even if the closed loop gain is only unity (or less). For that same reason, a class B circuit is not desirable for in input stage because it provides no voltage gain, the primary purpose of an opamp.

ak

 

After giving further thought to my original question, it seems the differential configuration (with two input terminals) is easier to provide both an inverting input and also the non-inverting input.

However, the Class B configuration has only one input terminal (going to the base of both complementary transistors) and the only option is a non-inverting input.

 

You are confusing the input stage and the output stage of an amplifier.

Class B amplifier is designed to produce power into a low impedance load. That is not required in the input stage of an amplifier.
A Class B amplifier suffers from cross-over distortion.
Hence Class B configuration is a poor choice for the input stage.

 

So why can't a Class B configuration be used instead of the differential? It seems more logical with the intent of a complete op amp to make its internal circuits function more like the black box model.

Hi Glenn,

For amplifying DC signals a differental pair configuration has at least two good advantages:

1. Very easy to temperature compensate the input offset voltage.
2. Very easy to provide a large common mode voltage that doesn't end up at the output.

If you are only amplifying AC signals referred to ground and the gain doesn't need to be too accurate then other types of amplifier configuration should work just fine.

Ifixit