Here is my task:

Calculate and sketch transfer characteristic for circuit below, for 0 < Vin < Vcc. It is known Vcc = 2.1V, VBE = 0.7V (when BJT conducts), VD = 0.7V (when diode conducts), VCES = 0.2V, Beta - > infinity, R1 = R2.

Here are my thoughts. It is not possible for BJT to be in saturation and diode to conduct because VCES = 0.2V and VD = 0.7V, so KVL is not met in that case. When BJT is in forward active mode, IC = IE and IB = 0 because of Beta - > Infinity. Here are cases I analysed:

1. BJT is in cut-off mode, diode conducts:

2. BJT in saturation, Diode is off

I can calculate now Ib (using KCL for BJT), but I'm not sure how to find range for input voltage Vin in this case. How to do this?

3. BJT in forward-active, Diode conducts

I got Vin = const in this case.

 

In what condition the diode will be if Vd < 0.7V (0.699(9)V)

 

Diode will not conduct
Related to case 2, If I prove that for some range of input voltage BJT is saturated, I also proved that in that range Diode is off, right?

 

Ok I understand. But notice that for Vce <0.7V diode is OFF but the BJT is not yet in saturation.

 

So case 2 (diode off, bjt in sat.) is not possible?

 

So case 2 (diode off, bjt in sat.) is not possible?

No, all I want to say is that when the BJT is turns ON (Vin > 1.4) and Vout = 1.4V transistor already will be in active region.
And with your strange boundary for the diode lead me to the conclusion that the diode must immediately turn off after the BJT is ON.
But this is not going to happen in real world.