This is a differential amplifier with common-mode signal as input (figure a) :
Since equal voltages vin(CM) drive both inputs simultaneously, there is almost no current through the wire between the emitters. We will then redraw figure (a) to something like figure (b). Thenafter, we can remove the connecting wire:
So the gain of Common mode signal with single-ended output will be:
$$A_{v(CM)} =\frac{R_C} {2R_E}$$
Why don't current from both v_in went through the single R_E at the same time? Why need to separate the path of two current? It's because I see no problem squeezing the two current in a single resistor if ever they can't go past each other.
Best Answer
OP has said:
and
It was interesting to me to see such a conversation between a "teacher" and a "student" (OP), which took place maybe a thousand times at school... and which has always ended in the same way - by rejecting the OP's idea and imposing the prevailing explanation.
Obviously, OP believes that there is no need to "cut" Re in two halves. And indeed this is so. Let's see why.
In this "common mode", two emitter followers are controlled by the same input voltages Vin(CM) and they drive a common load Re. So, the common current Ie = Vin/Re flows through Re. But only half of this current flows through each Rc (assuming zero differential input voltage). Thus, the (output) voltage drop created by this "half emitter current" Ic = 0.5 x Ie across Rc is VRc = 0.5 x Rc x Vin/Re.
Why then did we have to persuade OP to "cut" Re? Only because someone many years ago managed to solve the problem in this way, right?
However, I still advise OP to "cut" Re in order to keep his/her teacher satisfied... and my explanation to remain for "internal use" only. A little diplomacy is never superfluous:)