I have a Fairchild BC547 (NPN) that I'm going to use as a switch.
What parameter(s) should I look at, if I want to know how long time the transistor will take to go from cut-off to fully saturated?
Electronic – Which datasheet parameters describe the switching perfomance of a BJT
bjtdatasheetswitchingtransistors
Related Solutions
When choosing the right transistor for this job, first I'll eliminate the PNP transistors. They're a bit more complicated to use in your case. As you said, for a PNP transistor, active high becomes active low, meaning the transistor will switch on when you apply 0V from your Arduino, but it won't switch off when you apply 5V from the Arduino. You'll need to apply 12V to the base of the PNP transistor to switch off (VEB = 0).
Leaving PNP's behind, looking at the NPN's that you have availabe, only the BC547B (Ic = 100mA) couldn't handle the 480mA current that your siren needs. From the remaining 3 transistors, I'd choose the one that can handle the most current, just to be on the safe side. That would be the BC517 darlington, which can handle a maximum of 1.2A, more than enough for your siren.
Only now you'll have to worry about the gain of the BC517. But, because BC517 is a darlington transistor, it has a huge gain (hFE = 30,000), so you can easily switch on the transistor with a very small base current. If you chose to drive the base of the transistor with a 1KOhm resistor, you'll have a 3.6mA base current, which is sufficient for your purposes.
So the winner would be the BC517.
It is usually capacitance that causes problems - and it's usually the miller capacitance that is the main culprit. The miller capacitance is internally from collector to base and if the base is driven from a high impedance (10k in your diagram - really quite high to expect decent performance) the miller capacitance forms a low pass filter with the input resistor.
If miller C is 10pF and input resistance is 10kohm then these two components form a low pass filter with a cut-off at: -
\$f_C = \dfrac{1}{2\pi R C}\$ = 1.59 MHz.
Try reducing the 10k ohm input resistor to something more sensible like 330 ohms and see the difference. I simulated a BC817 as you have drawn it and it took about 2.3us to reach 11 volts when switched off. With a 1k input resistor it took less 0.33us. With 330 ohms it took about 0.25us.
If I lowered the collector load to 1k and kept the 330 ohm input resistor it took about 0.09us.
The lowering of these resistors all serve to reduce the effect of miller capacitance (amongst other things).
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Best Answer
It's the current gain bandwidth product, which your datasheet lists as 300 MHz typical.
You have your base current Ib and collector current Ic, so Ic/Ib is your required gain. If you then divide 300,000,000 by this, you will get a typical max. switching frequency. You could then take your maximum switch-on time to be half of the period of this frequency. Personally, I would derate this value considerably as it is a 'typical' and take a hundredth of it's value. If you don't get a switching time that you need, consider another transistor. (That's assuming you're designing anew and not fixing existing.)
So I would use:
$$t_{on} = \frac{I_c}{1,200,000 \times I_b}$$