Without more context, I can't answer the question of why different expressions are used but do note that
$$r_e = \frac{1}{g_m}||r_{\pi}$$
so the expressions are, in fact, equivalent.
To see this, recall
$$r_{\pi}= \frac{\beta}{g_m}$$
Thus,
$$\frac{1}{g_m}||r_{\pi} = \frac{r_{\pi}}{\beta}||r_{\pi} = \frac{r_{\pi}}{1 + \beta} = r_e$$
I will try to give you some hints:
1.) Of course, you can define the output of the diff. pair at one collector node only (Vo1 or Vo2). It is just another option to use the other collector node as well and defining the amplifier output as Vo2-Vo1.
2.) Differential mode: Assuming linear operation (and this is always assumed) the current increase of the left BJT is equal to the corresponding current reduction of the right BJT. Hence, there is no current change in the common emitter resistor Re. Hence, you can treat each transistor stage as a simple common emitter stage without any emitter degeneration (the common emitter nodes behave as if they were signal grounded). The gain formula for such a simple arrangement is known.
3.) Common mode: Again, treat the BJTs as common emitter stages - however, now with emitter degeneration. Both BJT`s amplify the same signal. Again, the gain formula for a simple common emitter stage with Re feedback (degeneration) is known and can be used - however, you have to consider that the current change through Re is doubled because this resistor is common to both transistors. As a consequence, the feedback effect is doubled and the effect of resistor Re must appear as 2*Re in the corresponding formula.
4.) If you like to start with your small-signal model (and finding the gain values by yourself) the procedure is simple: For differential mode you only need to consider one of the transistors (because they are not connected (emitter signal grounded). This applies also to the common mode - however, in the gain formula you must replace Re with 2Re.
5.) Finally, an important comment: You have defined the diff. mode as V1=Vd/2 and V2=-Vd/2. This is OK. However, please note that in the first part of your question the expression for Ad does NOT apply to your definition. The given gain is for V1=Vd and V2=0. This is a non-symmetrical operation of the circuit, which is allowed but involves also a certain common mode voltage.
Best Answer
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Part (a) - REE and RC
What is the current through REE required to be?
what is the quiescent voltage at the emitters?
Correct, so now use ohms law to calculate REE (6143 ohms): -
So, for either of the two transistors, what could be the maximum and what could be the minimum current that flows when a large signal is applied?
Hint - you know that the inputs to both transistors are opposite polarity and so you can assume the "common" (or shared) emitter voltage remains at -0.7 volts hence (as previously understood), total current in REE is 700 uA.
You use Ohms law - what value of resistor produces a 1 volt p-p signal when the current is 700 uA p-p?
That's what I get!
Part (b) - common mode gain
Given that both inputs are assumed to be rising up and down together (that's what common mode means), you can calculate the change in the common emitter voltage and use that change in voltage to imply a change in emitter current. That change in emitter current is shared equally between both transistors hence, half that change of current flows through Rc.
This means that the common-mode gain is \$\dfrac{R_C}{2\cdot R_{EE}}\$ = 0.116
Part (b) - differential mode gain
Here we can ignore the shared emitter resistor REE (because the shared emitter voltage is constant) and focus on what is called the internal emitter resistor \$r_E\$. The voltage gain for a single common emitter amplifier with no emitter resistor is: -
$$\dfrac{R_C}{r_E}$$
But, because we are applying a differential signal the above voltage gain is halved because each base only receives 50% of the full differential signal hence the differential gain of a long-tailed pair (old-fashioned language) is: -
$$\dfrac{R_C}{2\cdot r_E}$$