for the differential pair in Fig 6.2(a)... find the voltage at the emitters and at the outputs.
If, as we are given, Vbe = 0.7 at Ic = 1mA, then we must solve for Vbe at Ic=0.5mA.
0.5 = exp (dV / V_T).    Thus, dV = -0.017 volts... and our new Vbe =
thus, at Ic = 0.5mA,  Vbe = 0.683
Ve = Vcm - Vbe = -2 - 0.683 = -2.683 volts

ic1 = ic2 = alpha * 0.5 = 0.99 * 0.5 = 0.495 mA
Vc1 = Vc2 = Vcc - ic * Rc = 5 - 0.495 * 3K = 3.515 volts

"~=" means "approximately equal"
ic1 = alpha * I  ~=  1 mA
Vbe = 0.7 volts
Ve = 1 - 0.7 = 0.3 volts
Vc1 = Vcc - ic1 * Rc ~=  5 - 1mA * 3K = 2 volts
Vc2 = Vcc - 0 * Rc = 5 volts.

Ve = 0.4 volts
I = 4.6 mA
Vc1 = 5 - 4.6 * 1 = -0.4 volts
Vc2 = 5 - 0 * 1 = -5 volts

Eq. (6.4)
ie1 = I  /  (1 + exp (Vb2 - Vb1) / Vt) = 0.8 I
==>  Vb2 - Vb1 = Vt * ln (0.25) = -34.6 mV.
thus, Vb1 - Vb2 = 34.6 mV

(a) The largest common mode voltage is determined by finding the
highest base voltage such that the collector voltage = base voltage.
Vcm (max) = Vc_{1, 2} = Vcc - (I / 2) * Rc)
With this situation, however, the output signal does not have a full output swing.
so, alternatively,
if we want the largest common mode voltage and still get our full output swing,
Vcm (max) = Vc_1,2 = Vcc - I * RC

(b) if the current is steered to Q1, then
Vc1 = Vcc - I * Rc, a change of - (I*Rc) / 2
Vc2 = Vcc, a change of +(I*Rc) / 2

(c) I * Rc represents the full differential output swing.
need to pick I * Rc such that at Vcm = 3 volts, a full downward swing of
I * Rc of one of the collectors is only as low as 3 volts.
Vcm (max) = 3 = 5 - I*Rc
==> I * Rc = 2 volts

(d) base current should not exceed 2 uA.
(I / 2) / (1 + Beta)  <= 2 uA  (using alpha ~= 1)
==>    I  <=   4 (1 + Beta) uA
Thus, I = 0.404 mA
Rc =  2v / I  = 2v /  (0.4 mA)  =  5K