I was reading a book about amplifiers and I have read the following sentence about the following picture:

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The idea-guide is that we must access simultaneously to the Gate terminal and to the Source terminal of a transistor so that we can command it to produce a controlled current, as shown in the previous picture.

But from a practical point of view, access to those two points requires some caution in order, for example, not to alter the polarization of the transistor: in fact we could not directly connect two ideal voltage generators referred to ground to the two terminals of the transistors shown above. We should necessarily think of a coupling in AC, at least for one of the two inputs (though this is not always possible, in particular in an integrated circuit where the realization of high value capacitors is practically impossible).

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First question: can you explain me the highlighted sentences? I do not understand which is the problem: I have always seen Common - Source configurations with only a voltage source with its positive terminal connected to the Gate and its negative terminal (often GND) connected to the Source. I have never seen two voltage sources connected as it says; but also if I decide to do this, I do not see any problem).

Then it continues in this way

Note that the two inputs have different DC voltage and different impedance shown on signal. Both of these situations are uncomfortable in application practice.

Second question: why?

Then it says:

To overcome these obstacles we could think of accessing one of the two terminals of the transistor via a follower stage. The following circuit has the beneficial effect, as it is easy to verify, to bring the two inputs, Va and Vb, of the new system to the same DC value. The solution of adding a follower to the simple initial transistor is actually correct and effective. This configuration of transistors is typical of the input stage of operational amplifiers (OpAmp) and is usually called "differential stage". In fact, the input impedances of the two terminals have been matched to a high value (in the case of even infinite MOSFETs), making this stage ideal when both voltage signals are to be read. Note how in this case, thanks to symmetry, the follower is lost and who is the signal transistor.

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Third question: can you explain me how does it work in simple words?

  • \$\begingroup\$ That looks like a common gate. Vb is your bias voltage to set the Q point. Va is your input at the source. Vu is your output at the drain. \$\endgroup\$
    – user110971
    Dec 1, 2019 at 19:39
  • \$\begingroup\$ @Kinka-Byo, I wrote a short answer to your question below but there is much more to say about the philosophy of this legendary circuit solution. If you ask more specific questions, I will extend my answer with additional explanations. \$\endgroup\$ Dec 1, 2019 at 22:46

1 Answer 1


The author of the book tries to reveal the "philosophy" behind the transistor differential amplifier (aka long-tailed pair) by building the circuit... and not by giving the ultimate perfect circuit solution... which deserves our respect. Here is what this "building scenario" would look like briefly:

  1. We can drive a (FET) transistor from the side of the gate (common-source amplifier), from the side of the source (common-gate amplifier)... or from both (differential amplifier). So the bare transistor is itself a differential amplifier with two inputs - inverting (the gate) and non-inverting (the source). Then let's apply the two (grounded) input voltages VA and VB respectively to its gate and source.

  2. The problem of this "1-transistor differential amplifier" is that the input sources do not work under the same load conditions - VA is practically unloaded (it "sees" the extremely high input impedance of the gate) while VB is significantly loaded (it "sees" the very low output impedance of the source)... ie, the trouble is in the non-inverting input (the source). Then let's help it.

  3. If we connect a(nother) source follower (acting as a buffer) between the VB input voltage source and the source, we will "alleviate the suffering" of VB and put it under the same conditions as VA. Thus both they will "see" the same input impedance.


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