In page 92 of the Art of Electronics (or 2.3.2) the author summarized a few quantities derived from Ebers-Moll equation.
- For the small-signal impedance looking into emitter re, why Is(T) is not in the derivative?

- For the temperature dependence, the author considered Ic at constant Vbe and Vbe at constant Ic. How do we take that into account in real design? For example, if Vbe was initially 0.6V and Ic 10uA, how do they change if temperature is incresed by 10 degrees?

- How can I understand the early effect equation Ic = Ico(1 + Vce/Va)? For example, if Vce increases, Ic increases and therefore the voltage drop across the collector resistor should increase, so Vc decreases and therefore Vce decreases. It doesn't make sense to me.

Thanks very much!

 

- For the small-signal impedance looking into emitter re, why Is(T) is not in the derivative?

Have you try derived this "re" yourself ? Also keep in mind that in normal condition Is >> I_forward current, so we can omit Is.

For the temperature dependence, the author considered Ic at constant Vbe and Vbe at constant Ic. How do we take that into account in real design? For example, if Vbe was initially 0.6V and Ic 10uA, how do they change if temperature is incresed by 10 degrees?

Try read this
http://forum.allaboutcircuits.com/blog/fun-with-the-diode-equation.589/
In greener Vbe drops around 2.2mV per degree °C, beta (β) also rise with temperature around 1% per degree °C of a initial value. And Icbo doubles when temperature increase by 8°C.

How can I understand the early effect equation Ic = Ico(1 + Vce/Va)? For example, if Vce increases, Ic increases and therefore the voltage drop across the collector resistor should increase, so Vc decreases and therefore Vce decreases. It doesn't make sense to me.

I don't understand your confusion here.
Simply Early effect (base-width modulation) means that Ic current will change his value as Vce change even if Vbe (Ib) is keep constant. So we have another source of a nonlinearity.

For example if we assume that β = 100, Vcc = 10V; Rc = 1kΩ ; Ib = 50μA and Va = 50V we have.

Ico= 50μA * 100 = 5mA (without Early effect) and Vceo = 10V - 5mA*1kΩ = 5V.

But if we include Early effect Ic will change. We have

Ic = Ico*(1 + Vce/Va) (1)

Vce = Vcc - Ic*Rc (2)

And if we solve for Ic we have

Ic = (Ico* (Va + Vcc))/(Ico* Rc + Va) = Ico * (1 + Vcc/Va)/(1 + Rc/Ro) = 5mA *(1+ 10V/50V)/(1 + 1kΩ/10kΩ) = 5mA *1.2/1.1 = 5.454mA

where Ro = Va/Ico = 50V/5mA = 10kΩ this means that Early effect can be model as a resistor Ro connected from the collector to the
emitter of an “perfect” transistor.