Other than current handling, I would look for a transistor that has a high DC current gain so that it saturates easily, with a low base current, and one that has a low \$V_{CE}\$ (sat). That is to say, when it is fully turned on, the voltage drop across it is low.
I don't know what additional current handling you are looking for, but I have a suggestion. Check out the 2SD882 medium power NPN transistor. I needed a transistor with decent DC gain, current handling and low \$V_{CE}\$ (sat). After quite a search through numerous data sheets, I settled on that one. I managed to get my hands on a bunch of Panasonic made in Japan ones, and the \$H_{FE}\$ of all of them measured over 360. (This is probably no important in this application.)
The PNP complement of this transistor is 2SB772.
Now let's look at the resistors. \$R_{SC}\$ is a current sensing resistor. This will have some tiny value, a fraction of an ohm. The value is important because it determines the trigger threshold for turning on the outboard transistor. The datasheet doesn't give a value for that in your particular circuit, but I think the value from Fig. 11 of 0.33 ohms can be reused for the Fig. 11a circuit.
Then there is the base resistor on the transistor. Its value is does not appear critical. But note that this resistor will function as the emitter feedback resistor for the switch transistor inside (Q2). The feedback which it develops oppposes the turning on of the internal transistors.
There is a reason why Fig. 11a is called "NPN switch" while 11b is "Saturated PNP Switch". The PNP topology does not develop feedback. The switch emitter is simply grounded.
The PNP looks like the superior circuit; I would go for that one.
Looks good to me. The inverse diode D1 is a good idea. If you have a minimum of 12V available you may wish to reduce R2 somewhat. This circuit has a threshold of maybe 2V, you could easily halve R2 or double R1.
In the case of momentary extreme over-voltage, the base-emitter voltage (forward biased) will not rise above a volt or so, even with 100mA. It looks like another diode in inverse parallel to D1. One of the advantages of a BJT in this application. The limitation is more likely to be the voltage rating of R1.
If you want to consider sustained overvoltage, you may have to consider the power rating of R1. If some idiot connects it to the mains (we can usually assume that about 240VAC is the most voltage idiots will have access too- idiots with access to higher voltages are sort of a self-eliminating problem) then R1 would dissipate almost 6W, so it would have to be a physically large part. You could solve that issue by increasing the value of R1 so that a smaller part could be used.
Best Answer
\$C_{ibo}\$ and \$C_{obo}\$ are composed of parasitic capacitances. A parasitic capacitance is an unavoidable property of all transistors because their terminals are made of conducting materials in very close proximity to each other. Parasitic capacitances start to affect performance when the transistor is operated at high frequencies, which is why they are minimized as much as possible.
A BJT has several parasitic capacitances between all three of its terminals.
The input and output capacitance are defined as below according to this application note from Infineon:
As for the terminology, base means that the capacitance is measured with respect to a common-base configuration, where the input is at the base of the transistor and the output is at the collector.
Input/Output denotes which terminal the capacitance is being measured from. Open/Short denotes whether or not the output - the terminal opposite to that of the terminal where the capacitance is measured from - is an open circuit or a short.
You can read the application note for more capacitance definitions.