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Comments on What is the idea behind the transistor differential amplifier?

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What is the idea behind the transistor differential amplifier?

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It was a big challenge for me to show the basic idea behind the op-amp instrumentation amplifier by building and reinventing it step by step. But it is no less challenging to show the idea behind the simpler but legendary transistor differential amplifier (figuratively named "long-tailed pair"). I will do it here in the same inventive way by sequentially building basic transistor amplifier stages known as common emitter, common base and common collector transistor amplifier stage. I will show it conceptually on the blackboard and practically on the prototyping board. In addition, I will suggest a classification with more meaningful names - base-driven, emitter-driven, base-driven with negative feedback and base-emitter driven transistor amplifier stage.

Building the circuit

What is an amplifier?

In the literal sense of the word, there is no (energy) amplification. It is not possible to get a large power from a small one because this would violate the law of energy conservation (energy can only be converted). "Amplification" actually means "regulation". Here is a simple intuitive explanation of this trick…

Amplifier with a current output...

To make an "amplifier", we connect a regulating element (transistor T) in series with the supply voltage source Vcc and begin to control its "resistance" through the input voltage Vin. As a result, the current Ic starts to change.

Transistor amplifier with a current output

I think this is the simplest possible "functional" explanation of how a transistor works. If you do not like the word "resistance", add "nonlinear" before it. If you still do not agree, then I would ask you, "What other common property besides resistance does a transistor have to be able to regulate the current in a circuit?"

... indicated by an LED

In addition to the classic ammeter, we can monitor the current with the help of an LED (the intensity of its light gives us an idea of the current magnitude).

Transistor amplifier with a current output indicated by an LED

Amplifier with a voltage output…

But we need a voltage amplifier. For this purpose, we need to convert the output current Ic into an output voltage Vout. To do this, we include a resistor Rc in the output circuit and a "non-inverted" drop VRc = Ic.Rc appears across it ("non-inverted" in the sense that when the input voltage increases, the output also increases in absolute value).

Transistor amplifier with a voltage output

We can also use the complement of this drop to the supply voltage as an "inverted" output voltage Vce ("inverted" in the sense that when the input voltage increases, the output also increases in absolute value). This gives us the simplest transistor amplifier configuration, known as common emitter, in which the input voltage is applied to the base and the emitter is connected to ground.

The transistor input voltage as a difference of two input voltages

If we apply a second voltage Ve to the emitter, the possibilities to control the transistor increase and we can get more versions of transistor amplifier stages - common base, common collector, differential amplifier. Now the "true" input voltage Vbe of the transistor is the difference of the two voltages Vb and Ve. Vbe is a differential, "floating" (ungrounded) voltage, and Vb and Ve are single-ended voltages measured with respect to ground.

Transistor amplifier with two input voltages and a voltage output; the current is still indicated by an LED

The voltages are represented on the circuit diagram by voltage bars with proportional height (in red)... and the currents - by current loops with related thickness (in green). See also my papers What are voltages in circuits? and Where do currents flow in circuits?

Practical circuit

... diagram

The two voltages Vb and Ve are "produced" by 1 k linear potentiometers (P1 and P2) and applied to the base and emitter of the transistor (BC547C)... and a LED (VQA13) is connected in series to the collector resistor Rc. The voltages were measured by the voltmeters V1 and V2 and the collector current (transistor state) is indicated by the LED. When one of the voltages is constant, we will draw a discrete voltage divider instead of a potentiometer, but in practice we will get it through a potentiometer, on which we keep the wiper stationary.

Setup for investigating various basic transistor amplifier stages

... on the blackboard

The circuit diagram on the blackboard

Implementing the circuit on the prototyping board

The circuit on the prototyping board

Large picture of the circuit on the prototyping board

Exploring various basic transistor circuits

Base-driven (common-emitter) transistor stage

In this case, we apply some constant voltage Ve to the emitter and initially, approximately the same base voltage Vb. We then begin to vary carefully, within small limits, the base voltage as the input voltage Vin.

Investigating a base-driven (common-emitter) transistor stage

We see that when the base voltage exceeds the emitter voltage by about 600 mV, the transistor begins conducting. The collector current increases and the LED starts to glow (see the movie). The voltage drop across the collector resistor begins to grow, and its complement to Vcc, the voltage Vce, decreases. Therefore, the circuit is inverting.

Analogy. An op-amp with grounded (or with fixed voltage) non-inverting input resembles the common-emitter transistor stage:

  • op-amp inverting input = transistor base

  • op-amp non-inverting input = transistor emitter

  • op-amp output = transistor collector

Emitter-driven (common-base) transistor stage

Now we swap the roles by applying a constant voltage Vb to the base and initially, approximately the same emitter voltage Ve. We then begin to vary carefully, within small limits, the emitter voltage as the input Vin.

Investigating an emitter-driven (common-base) transistor stage

For example, if the emitter voltage approaches the base voltage by less than 600 mV, the transistor starts to turn off. The collector current decreases and the LED starts to glow (see the movie). The voltage drop across the collector resistor begins to decrease and its complement to Vcc, the voltage Vce, to increase. Therefore, the circuit is non-inverting.

But here we notice an unpleasant moment - the large collector current flows through the input source. This means that the circuit has a small input resistance.

Analogy. An op-amp with grounded (or with fixed voltage) inverting input resembles the common-base transistor stage:

  • op-amp non-inverting input = transistor emitter

  • op-amp inverting input = transistor base

  • op-amp output = transistor collector

See also another movie in the laboratory on Basic Electronics (April 16, 2019).

Base-driven with feedback (common-collector) transistor stage

In this case, as the base voltage increases, the transistor begins to conduct more and more and let more and more collector current pass through the emitter resistor (the LED starts to glow more and more brightly). Figuratively speaking, the transistor "lifts" its emitter thus decreasing the input voltage. The name of this trick is "emitter degeneration", "bootstrapping" or "voltage following negative feedback".

Investigating a base-driven with feedback (common-collector) transistor stage

The voltage drop across the collector resistor grows, but this does not really matter in this case. More significantly, the emitter voltage rises and tends to equal the base voltage ("follows" it, hence the name "emitter follower").

Analogy. An op-amp follower resembles the common-collector transistor stage:

  • op-amp non-inverting input = transistor base

  • op-amp inverting input = transistor emitter

  • op-amp output = transistor emitter

Base-emitter driven transistor ("differential amplifier")

We can extend our experiments if we start varying both voltages at the same time. Thus, we imperceptibly "invent" the prototype of the differential amplifier.

Analogy. An op-amp without grounded inputs resembles the transistor differential stage:

  • op-amp inverting input = transistor base

  • op-amp non-inverting input = transistor emitter

  • op-amp output = transistor collector

We have two options to change the input voltages:

… in the same direction (common mode)

First, let's vary both voltages Vb and Ve (move both potentiometer wipers) simultaneously and in the same direction. As a result, the input voltage Vbe, the collector current Ic and the output voltage Vout do not change, i.e. the circuit does not amplify. It is interesting that, in this simple 1-transistor "differential amplifier", they don't change at all... while, in the true 2-transistor differential amplifier, they do change to some extent!

Investigating a transistor stage by applying both base and emitter inpout voltages (common mode)

… in the opposite directions (differential mode)

Let's now start varying both voltages Vb and Ve (move both wipers) simultaneously but in different directions (opposite). Now the input voltage Vbe, the collector current Ic and the output voltage Vout drastically change, ie. the circuit amplifies a lot.

Investigating a transistor stage by applying both base and emitter inpout voltages (differential mode)

True differential amplifier

This was the prelude to the true 2-transistor differential amplifier (long-tailed pair). Let's now see how we can transform the imperfect 1-transistor "differential amplifier" into true differential amplifier...

Making circuit symmetrical

Problem. Obviously, the disadvantage of the 1-transistor "differential amplifier" is its asymmetry - beta +1 times more current flows through the voltage source connected to the emitter compared to the source connected to the base.

Remedy. To make the circuit symmetrical, we can include another emitter follower between the voltage source and the emitter.

True 2-transistor differential amplifier (long-tailed pair)

Problem solved... but another problem appears. Let's see what it is..

Removing the common-mode error

Background. Our differential amplifier has a differential input but its output is not differential; it is referenced to ground, asymmetrical or, as they say, "single-ended". This is because, in most cases, we want it to be. But we want the voltage at this output to not change when both input voltages change at the same time, in the same direction and to the same extent (the so-called "common mode").

In this mode, the two emitter followers help each other to drive a common load - the emitter resistor Re. The role of this resistor is to pass the emitter currents to ground. Figuratively speaking, this resistor "pulls down" the common emitter point while the transistors "pull" it up.

Problem is because they do that by current. When the input voltages change, the transistors change their collector currents so that to change the voltage drop across the emitter resistor in the same manner. As a result, the voltage drop VRc across the collector resistor Rc changes as well... and it should not...

Remedy. Maybe you expect me to suggest straight away that we replace the emitter resistor with a current source as the textbooks say? I am going to disappoint you a bit because I think it is better to naturally reach this idea ourselves... and even experiment with it on our setup.

The trick is extremely simple but clever. We want the current through the emitter resistor not to change when the voltage drop across the resistor, for example, increases... ie, the ratio VRc/Rc = Ie not to change when the numerator VRc increases. So the solution is to increase the denominator Rc to the same extent.

Modeling. We can model the trick on our setup by replacing the constant emitter resistor with a variable one. For this purpose, we can take another (third) potentiometer 1 k connected as a rheostat and start rotating it together with the other two. The result will be that Ie, Ic1 and ultimately Vout will not change... which is what we wanted to do.

Differential amplifier with dynamic emitter resistor

Implementation. Behaving in this way, the variable resistor acts as a dynamic resistor that maintains a constant current through itself (current stabilizing element). This is one way of making current sources, which is why they denote it by the current source symbol.

Differential amplifier with emitter current source

In practice, it is most often implemented through a transistor with a constant base voltage.

Differential amplifier with transistor emitter current source

Imperfections

Our experimental setup is extremely simple, but it still has some drawbacks. Let's see what they are and how we can overcome them.

  1. Potentiometers have quite high resistance and influence each other through the base-emitter junction, which seems to connect them like a piece of wire. We can reduce this negative effect by replacing them with lower resistance ones but they should be able to dissipate most of the heat.

  2. There is a danger of damage to the potentiometers or the base-emitter junction if the wipers are given in their extreme opposite positions (the source is given "short"). The solution is to insert a protecting resistor in the circuit.

Comparison

So, what is the difference between the two classifications?

As we have seen, there is always only one "true" input voltage Vbe applied between the base and emitter (across the base-emitter junction). In most applications, it is floating... but, as a rule, in circuits, input voltages should be grounded (single ended). That is why, for greater versatility and flexibility, we form Vbe as a difference between two input voltages (Vbe = Vb - Ve)... and then vary the one, other or both. We have two ways to name these configurations:

In the classic centuries old classification, we give the name of the configuration (common-emitter, common-base, etc.) according to which of the transistor terminals is with a fixed voltage (i.e., what terminal is AC grounded, passive, neutral). But it is the less important terminal, especially in the case of the so-called common collector. The bare fact that the collector is AC grounded does not mean anything because the input voltage is not applied between the base and collector. And what about the differential stage where there is no such AC grounded terminal?

In our classification here, we give the configuration name (base-driven, emitter-driven, etc.) according to which of the transistor terminals is driven by the input voltage (i.e., what terminal is active) that is more meaningful.

Web resources

Common Base Amplifier Confusion is my answer in StackExchange related to this paper

How do we investigate basic transistor amplifier stages? is the same but enlarged story in my blog Circuit Stories

What is the idea behind the op-amp instrumentation amplifier? is the same story but about the more sophisticated op-amp version of the differential amplifier.

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4 comment threads

If I understand what you said refers to my description of a differential amp (DA) with 3 NPN's , " T... (1 comment)
The simple ideas behind the differential amplifier are to ; - eliminate the Vbe diode voltage offs... (2 comments)
The common terminal is not the least important, perhaps it is often ignored how important it is. (3 comments)
The differential BJT amplifier is well-defined in simpler terms. Av=Rc/(re+Re)=Rc/(26/Ie+Re) (2 comments)
The differential BJT amplifier is well-defined in simpler terms. Av=Rc/(re+Re)=Rc/(26/Ie+Re)
TonyStewart‭ wrote almost 2 years ago

The differential BJT amplifier is well-defined in simpler terms.

Av=Rc/(re+Re)=Rc/(26/Ie+Re)

Circuit fantasist‭ wrote almost 2 years ago

@Tony, the very title of this story specifies that it is about the idea - the structure, the topology of the circuit solution. Inventing the circuit and calculating it are different things - the former is something qualitative; the second is quantitative. It is incorrect to explain qualitative things by quantitative means...