# Why need to separate the path of the current of these two signal source?

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.

• Notice that if we have Vb1 = Vb2 ---->Ic1 = Ic2 thus Iee = Ic1 + Ic2 = 2Ic = Iee and RE = (Vee - Vbe)/2Ic. Now we wan to "disconnect/split" the emitters but we want to keep the current at the same level as before (Ic1 = Ic2 = Iee/2) when we "split" the emitters. Thus, now Re resistor needs to be 2Re ($Re =\frac{Vee - Vbe}{\frac{Iee}{2}} = 2\frac{Vee - Vbe}{Iee}= 2Re$) to have the same Ic current as before (after we split the emitters and we add two separate RE resistors), thus now emitters are not connected together. Did you get it?
– G36
Commented Jun 26, 2020 at 17:26
• And we do this it get equivalent circuit for vin(CM) situation.
– G36
Commented Jun 26, 2020 at 17:38
• Yes, very clear Commented Jun 27, 2020 at 3:53
• @G36, Your reasoning is convincing... but I want to ask, "Why do we still have to split Re?" This is something artificial and it is better, if possible, not to do it (see my answer). Commented Jun 28, 2020 at 12:04

OP has said:

"...I see no problem squeezing the two current in a single resistor..."

and

"I thought that both current will just get past through the single resistor and the formula will just be a simple Rc/Re..."

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:)

• Why -1? It is a good answer.
– G36
Commented Jun 28, 2020 at 16:55

This splitting is only used to illustrate that you can calculate the gain easily.

If you don't do the splitting, the calculation becomes more complicated. You don't need to do that, it's just practical. Notice that this is just a model that is equivalent to the first circuit.

• I thought that both current will just get past through the single resistor and the formula will just be a simple Rc/Re, what part of it gets complicated(sorry, noob here) Commented Jun 26, 2020 at 14:30
• the current from the first transistor raises the voltage across the resistor. Then, the V_BE and V_CE of the other transistor get reduced. So, you get two coupled systems instead of two systems that you can look at individually. Commented Jun 26, 2020 at 14:31
• V_CE of the other transistor increases... Commented Jun 26, 2020 at 15:04