Tail resistor in differential gain

Question

Why doesn't the tail resistor feature in the denominator for differential gain (whereas it does for common mode gain) in the differential amplifier below? (Fig 5N.28 from Learning the Art of Electronics)

LAoE explains this by pointing out that in pure differential mode with an equal but opposite difference at each input, the voltage across the tail resistor won't change whereas it does change with a signal common to both inputs. That much is clear but I still don't understand why the tail resistor can be ignored for calcualtion of differential gain.

And what about a differential signal that isn't equal and opposite?

Answer 1

The usual explanation is one of symmetry.

^{simulate this circuit – Schematic created using CircuitLab}

Applying a differential signal means that the middle will not change its voltage because the differential pair is constructed in a symmetrical fashion. If the left pulls it up with a certain voltage, then the right will pull it down an equal amount. This means that the middle point can be considered an AC ground (0V) for differential signals.

For common mode signals, you don't have this canceling out, because both left and right pull in the same direction.

Of course, the actual circuit isn't perfectly symmetrical (it would be if the left also had a load resistor), but a BJT transistor will "shield" effects happening at its collector from the emitter (in the active region, when \$v_{CE}\$ changes, \$i_{C}\$ doesn't change much so it doesn't affect the emitter current). So the approximation is still very much valid. However, if you look really, reaaally closely, you will find a dependency on the tail resistor because of that asymmetry.

And what about a differential signal that isn't equal and opposite?

Such a signal can be broken down again in a non-zero common mode signal and a differential signal that is equal and opposite. For example:

$$\begin{align} v_+ &= 1V \\ v_- &= -2V \\ v_d &= v_+ - v_- = 3V \\ v_{cm} &= \frac{v_+ + v_-}{2} = -0.5V \end{align}$$