Electronic – Where does the Maximum Emitter-Base Voltage come from

The diode is in the best position, and is of an appropriate type.

It conducts when the input is negative, the same as the transistor base conducting when the input is positive. The 47K resistor is about 1/10 of a normal RS-232 load. One could also block the voltage, but then a -100V spike (ESD say) could break down the 1N4148 and break down the E-B junction, causing irreversible damage.

Also, a 1N4148 is an appropriate diode for this application. It's a "switching diode", low capacitance and fast reverse recovery. A 1N4001 would also likely work okay, at least at slow baud rates. The 200mA rating means that even if a very high voltage were to appear at the input the transistor is fully protected, at least up until the resistor arcs over,.

So as you mentioned it says that the transistor is essentially two diodes.

You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:

As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.

What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!