Two Amplifier problem 1
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Part (a) - REE and RC

What is the current through REE required to be?


  @Andyaka 2 times IE – Saahil2000


what is the quiescent voltage at the emitters?


  @Andyaka -0.7V? – Saahil2000


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.


  So the max is 700uA and the min is 0? How does this relate to finding
  the maximum output voltage I am confused – Saahil2000


You use Ohms law - what value of resistor produces a 1 volt p-p signal when the current is 700 uA p-p?


  Rc = 1.428 k ohms – Saahil2000


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: -

$$\dfrac{R_C}{2\cdot r_E}$$